research article analysis of a multiserver queueing-inventory...
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Research ArticleAnalysis of a Multiserver Queueing-Inventory System
A Krishnamoorthy1 R Manikandan2 and Dhanya Shajin1
1Department of Mathematics Cochin University of Science and Technology Kochi 682 022 India2Department of Mathematics Indian Institute of Science Bangalore 560 012 India
Correspondence should be addressed to A Krishnamoorthy achyuthacusatgmailcom
Received 30 May 2014 Revised 24 September 2014 Accepted 28 October 2014
Academic Editor Ahmed Ghoniem
Copyright copy 2015 A Krishnamoorthy et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We attempt to derive the steady-state distribution of the119872119872119888 queueing-inventory system with positive service time First weanalyze the case of 119888 = 2 servers which are assumed to be homogeneous and that the service time follows exponential distributionThe inventory replenishment follows the (119904 119876) policy We obtain a product form solution of the steady-state distribution under theassumption that customers do not join the system when the inventory level is zero An average system cost function is constructedand the optimal pair (119904 119876) and the corresponding expected minimum cost are computed As in the case of119872119872119888 retrial queuewith 119888 ge 3 we conjecture that119872119872119888 for 119888 ge 3 queueing-inventory problems do not have analytical solution So we proceed toanalyze such cases using algorithmic approachWederive an explicit expression for the stability condition of the systemConditionaldistribution of the inventory level conditioned on the number of customers in the system and conditional distribution of thenumber of customers conditioned on the inventory level are derived The distribution of two consecutive 119904 to 119904 transitions of theinventory level (ie the first return time to 119904) is computed We also obtain several system performance measures
1 Introduction
The notion of inventory with positive service time was firstintroduced by Sigman and Simchi-Levi [1] They assumedarbitrarily distributed service time exponentially distributedreplenishment lead time with customer arrival forming aPoisson process Under the condition of stability of thesystem they investigate several performance characteristicsIn the context of arbitrarily distributed lead time the readersattention is invited to a very recent paper by Saffari et al [2]where the authors provide a product form solution for systemstate probability distribution under the assumption that nocustomer joins the system when inventory level is zero
Reference [1] by Sigman and Simchi-Levi was followed by[3] of Berman et al with deterministic service time whereinthey formulated the model as a dynamic programming prob-lem A review paper by Krishnamoorthy et al [4] providesthe details of the research developments on queueing theorywith positive service time Schwarz et al [5] were the first toproduce product form solutions for single server queueing-inventory problem with exponentially distributed servicetime as well as lead time and Poisson input of customers
They arrived at product form solution for the system state dis-tribution Nevertheless this is achieved under the assumptionthat customers do not joinwhen the inventory level is zero (ofcourse [2] of Saffari et al is the extension of this to arbitrarydistributed lead time) This is despite the strong correlationbetween the lead time and the number of customers joiningthe system during that time Subsequently several authorsmade the above assumption in their investigations to come upwith product form solution the details of which could be seenbelow Krishnamoorthy and Viswanath [6] subsume Schwarzet al [5] by extending the latter to production inventorywith positive service time References [7] of Sivakumarand Arivarignan [8] of Krishnamoorthy and Narayanan[9] of Deepak et al [10] of Schwarz and Daduna [11] ofSchwarz et al and [12] of Krenzler and Daduna are a fewother significant contributions to inventory with positiveservice time Protection of production and service stages ina queueing-inventory model with Erlang distributed serviceand interproduction time is analyzed by Krishnamoorthyet al [13]
Classical queue with inventoried items for service is alsostudied by Saffari et al [14] where the control policy followed
Hindawi Publishing CorporationAdvances in Operations ResearchVolume 2015 Article ID 747328 16 pageshttpdxdoiorg1011552015747328
2 Advances in Operations Research
is (119904 119876) and lead time is mixed exponential distributionCustomers arriving during zero inventory are lost foreverThis leads to a product form solution for the system sate prob-ability Schwarz et al [11] consider queueing networks withattached inventory They consider rerouting of customersserved out from a particular station when the immediatelyfollowing station has zero inventoryThus no customer is lostto the systemThe authors derive joint stationary distributionof queue length and on-hand inventory at various stationsin explicit product form A recent contribution of interestto inventory with positive service time involving a randomenvironment is by Krenzler and Daduna [15] wherein also astochastic decomposition of the system is established Theyprove a necessary and sufficient condition for a product formsteady-state distribution of the joint queueing-environmentprocess to exist A still more recent paper by Krenzler andDaduna [12] investigates inventory with positive service timein a random environment embedded in a Markov chainThey provide a counter example to show that the steady-state distribution of an 1198721198661infin system with (119904 119878) policyand lost sales need not have a product form Neverthelessin general loss systems in a random environment have aproduct form steady-state distribution They also introducea blocking set where all activities other than replenishmentstay suspended whenever theMarkov chain is in that setThisresulted in arriving at a product form solution to the systemstate distribution
The work on multiserver queueing-inventory systems isscarce Nair et al [16] consider an inventory system withnumber of servers varying from 119904 + 1 to 119878 depending on theinventory position Another contribution is by Yadavalli et al[17] wherein the authors consider a finite customer sourcesystem (this paper contains a few additional references tomultiserver inventory system)
In all work quoted above customers are provided an itemfrom the inventory on completion of service Neverthelessthere are several situations where a customer may not beservedmay not purchase the item with probability one atthe end of his service For example customers who maybuy an item arrive at a retail shop where there are one ormore (finite number) servers (sales executives) The serversexplain to each customer the features of product The timerequired for this may be regarded as the service time Afterlistening to the server each customer independently of theothers decides whether to buy the item (probability 120574) orleaves the systemwithout purchasing the item A less realisticexample is as follows a candidate appears for an interviewagainst a position At the end of the interview the candidatedecides to accept the offer of job with probability 120574 and withcomplementary probability rejects it In this case the job istaken as an inventory In this connection one may refer toKrishnamoorthy et al [18] for some recent developments
We arrange the presentation of this paper as indicatedbelow in Section 2 the1198721198722 queueing-inventory problemis mathematically formulated The product form solutionof the steady-state probability distribution including someimportant performance measures is obtained in Section 3Further we numerically investigate the optimal (119904 119876) pairvalues and theminimal cost for different values of 120574 Section 5
discusses the 119872119872119888 with 119888 (greater than or equal to3 but less than 119904) queueing-inventory problems by usingalgorithmic approach Section 6 gives some conditional prob-ability distributions and a few performance measures forthe 119888(ge3) server case Section 7 analyzes the distribution ofthe inventory cycle time In Section 8 the optimal 119888 andthe corresponding minimal cost for different values of 120574 areinvestigated Furtherwe look for the optimal (119904 119876) pair valuesthat would result in cost minimization for different pairs ofvalues of 120574 and 119888
2 Mathematical Modelling of the1198721198722Queueing-Inventory Problem
First we consider an1198721198722 queueing-inventory systemwithpositive service time Customer arrival process is assumed tobe Poisson with rate 120582 Each customer requires a single itemhaving randomduration of service which follows exponentialdistributionwith parameter120583 However it is not essential thatinventory is provided to the customer at the end of his serviceMore precisely the item is servedwith probability 120574 at the endof a service or else it is not provided with probability 1 minus 120574 Acrucial assumption of thismodel is that customers do not jointhe systemwhen the inventory level is zeroWhen the numberof customers is at least two and not less than two items arein inventory the service rate is 2120583 When the inventory levelreaches a prespecified value 119904 gt 0 a replenishment orderis placed for 119876 units with 119876 gt 119904 We fix 119878 = 119876 + 119904 as themaximumnumber of items that could be held in the system atany given timeThe lead time follows exponential distributionwith parameter 120573 Then X(119905) | 119905 ge 0 = (N(119905)I(119905)) | 119905 ge0 is a CTMC with state spaceΩ
1= ⋃infin
119894=0L(119894) whereL(119894) is
called the 119894th level In each of the levels the number of itemsin the inventory can be anything from 0 to 119878 Accordingly wewriteL(119894) = (119894 0) (119894 119876+119904)The infinitesimal generatorW1 of this CTMC X(119905) | 119905 ge 0 is
W1 =
[[[[[
[
119861001198600
11986120119861101198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]
]
(1)
where 11986100
contains transition rates within L(0) 11986120
repre-sents the transitions from level 1 to level 0 119861
10contains the
transitions within level 1 1198600represents the transition from
level 119894 to level 119894+1 119894 ge 11198601represents the transitions within
L(119894) for 119894 ge 2 and 1198602represents transitions from L(119894) to
L(119894 minus 1) 119894 ge 2 The transition rates are
[11986100]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[11986120]119896119897=
120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119878(1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 1198780 otherwise
Advances in Operations Research 3
[11986110]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 119896 = 1 2 119904minus (120582 + 120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 = 1minus (120582 + 120573 + 2120583) for 119897 = 119896 119896 = 2 3 119904minus (120582 + 2120583) for 119897 = 119896
119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897=
120574120583 for 119897 = 119896 minus 1 119896 = 1(1 minus 120574) 120583 for 119897 = 119896 = 12120574120583 for 119897 = 119896 minus 1 119896 = 2 3 1198782 (1 minus 120574) 120583 for 119897 = 119896 119896 = 2 3 1198780 otherwise
(2)
Note that all entries (block matrices) in W1 are of the sameorder namely 119878 + 1 and these matrices contain transitionrates within level (in the case of diagonal entries) and betweenlevels (in the case of off-diagonal entries)
21 Analysis of the System In this section we perform thesteady-state analysis of the queueing-inventory model understudy by first establishing the stability condition of thequeueing-inventory system Define 119860 = 119860
0+ 1198601+ 1198602
This is the infinitesimal generator of the finite state CTMCcorresponding to the inventory level 0 1 2 119878 for anylevel 119894 (ge1) Let 120577 denote the steady-state probability vectorof 119860 That is
120577119860 = 0 120577e = 1 (3)
Write
120577 = (1205770 1205771 120577
119904 120577
119876 120577
119878) (4)
We have
119860 =
[[[[[[[[
[
minus120573 120573
120574120583 minus (120573 + 120574120583)
2120574120583 minus (120573 + 2120574120583) dd d
2120574120583 minus (120573 + 2120574120583) 120573
2120574120583 minus2120574120583
d d2120574120583 minus2120574120583
]]]]]]]]
]
(5)
Then using (3) we get the components of the vector 120577 explic-itly as
1205770= 1 +
120573
120574120583[1 + (
120573 + 120574120583
120574120583)
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
+ (119876 minus 119904 minus 2) (120573 + 2120574120583
2120574120583)
119904minus1
]
+120573
2120574120583(120573 + 120574120583
120574120583)
times [(120573 + 2120574120583
2120574120583)
119904minus1
minus (120574120583
120573 + 120574120583)
+
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
((120573 + 2120574120583
2120574120583)
119904minus119894+1
minus 1)]
minus1
120577119894=
120573
1205741205831205770 for 119894 = 1
120573
120574120583(120573 + 120574120583
2120574120583)
times(120573 + 2120574120583
2120574120583)
119894minus2
1205770 for 119894 = 2 3 119904 + 1
120577119894+1 for 119894 = 119904 + 1
119904 + 2 119876 minus 1
120573
2120574120583[(120573 + 120574120583
120574120583)
times(120573 + 2120574120583
2120574120583)
119904minus1
minus 1]1205770 for 119894 = 119876 + 1
120577119876+119894=120573
2120574120583(120573 + 120574120583
120574120583)(120573 + 2120574120583
2120574120583)
119894minus2
times [(120573 + 2120574120583
2120574120583)
119904minus(119894minus1)
minus 1]1205770 119894 = 2 3 119904
(6)
Since theMarkov chain X(119905) | 119905 ge 0 is an LIQBD it is stableif and only if the left drift rate exceeds the right drift rateThatis
1205771198600e lt 120577119860
2e (7)
Thus we have the following lemma for stability of the systemunder study
Lemma 1 The stability condition of the 1198721198722 queueing-inventory system under consideration is given by 120582 lt 120583[2 minus1205731205770120574120583(1 minus 120577
0)]
Proof From the well-known result by Neuts [19] on thepositive recurrence of the Markov chain associated with 119860we have 120577119860
0e lt 120577119860
2e for theMarkov chain to be stableWith
a bit of algebra this simplifies to120582 lt 120583[2minus1205731205770120574120583(1minus120577
0)]
For future reference we define 1205881as
1205881=
120582
120583[2 minus 1205731205770120574120583 (1 minus 120577
0)] (8)
4 Advances in Operations Research
3 Computation of the Steady-State Probability
For computing the steady-state probability vector of theprocess X(119905) | 119905 ge 0 we first consider a queueing-inventorysystemwith unlimited supply of inventory items (ie classical1198721198722 queueing system) The rest of the assumptions suchas those on the arrival process and lead time are the sameas given earlier Designate the Markov chain so obtained asN(119905) | 119905 ge 0 whereN(119905) is the number of customers in thesystem at time 119905 Its infinitesimal generatorG1 is given by
G1 =
[[[[[
[
minus120582 120582
120583 minus (120582 + 120583) 120582
2120583 minus (120582 + 2120583) 120582
2120583 minus (120582 + 2120583) 120582
d d d
]]]]]
]
(9)Let120587 be the steady-state probability vector ofG1 Partitioning120587 by levels we write 120587 as
120587 = (120587012058711205872 ) (10)
Then the steady-state vector must satisfy
120587G1 = 0 120587e = 1 (11)
From the relation (11) we get the vector120587 explicitly as follows
120587119894=
[1 +120582
120583(1 minus
120582
2120583)
minus1
]
minus1
for 119894 = 0
120582
1205831205870
for 119894 = 1
1
2119894minus1(120582
120583)
119894
1205870
for 119894 ge 2
(12)
Further we consider an inventory system with negligibleservice time and no backlog of demands The assumptionssuch as those on the arrival process and lead time are the sameas given in the description of the model Denote this Markovchain by I(119905) | 119905 ge 0 Here I(119905) is the inventory level attime 119905 Its infinitesimal generatorG2 is given by
G2 =
0
1
119904
119876
119878
0 1 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((
(
minus120573 120573
120574120582 minus (120574120582 + 120573)
d d d120574120582 minus (120574120582 + 120573) 120573
120574120582 minus120574120582
d d120574120582 minus120574120582
120574120582 minus120574120582
))))))))
)
(13)
Let120595 = (1205950 1205951 120595
119878) be the steady-state probability vector
of the process I(119905) | 119905 ge 0 Then 120595 satisfies the relations
120595G2 = 0 120595e = 1 (14)
That is at arbitrary epochs the inventory level distribution120595119895
is given by
120595119895=
[1 + 119876120573
120574120582(120573 + 120574120582
120574120582)
119904
]
minus1
119895 = 0
120573
120574120582(120573 + 120574120582
120574120582)
119895minus1
1205950 119895 = 1 2 119904
120573
120574120582(120573 + 120574120582
120574120582)
119904
1205950 119895 = 119904 + 1
119904 + 2 119876
120573
120574120582(120573 + 120574120582
120574120582)
119895minus119876minus1
times((120573 + 120574120582
120574120582)
119904minus(119895minus119876minus1)
minus 1)1205950 119895 = 119876 + 1
119876 + 2 119878
(15)
Using the components of the probability vector120595 wewill findthe steady-state probability vector of the original system Letx be the steady-state probability vector of the original systemThen the steady-state vector must satisfy the set of equations
xW1 = 0 xe = 1 (16)
Partition x by levels as
x = (x0 x1 x2 ) (17)
where the subvectors of x are further partitioned as
xi = (119909119894 (0) 119909119894 (1) 119909119894 (2) 119909119894 (3) 119909119894 (119878)) 119894 ge 0 (18)
Then by using the relation xW1 = 0 we get
minus120573119909119894(0) + 120574120583119909
119894+1(1) = 0 119894 ge 0
120582119909119894(119895) minus (120582 + 2120583 + 120573)119909
119894+1(119895) + 2(1 minus 120574)120583119909
119894+2(119895)
+ 2120574120583119909119894+2(119895 + 1) = 0 119894 ge 1 2 le 119895 le 119876 minus 1
Advances in Operations Research 5
120582119909119894(119895) + 120573119909
119894+1(119895 minus 119876) minus (120582 + 2120583)119909
119894+1(119895)
+ 2(1 minus 120574)120583119909119894+2(119895) + 2120574120583119909
119894+2(119895 + 1) = 0
119894 ge 1 119876 le 119895 le 119878 minus 1
120582119909119894(119878) + 120573119909
119894+1(119904) minus (120582 + 2120583)119909
119894+1(119878)
+ 2(1 minus 120574)120583119909119894+2(119878) = 0 119894 ge 1
minus (120582 + 120573)1199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
1 le 119895 le 119904
minus 1205821199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119904 + 1 le 119895 le 119876 minus 1
1205731199090(119895 minus 119876) minus 120582119909
0(119895) + (1 minus 120574) 120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119876 le 119895 le 119878 minus 1
1205731199090(119904) minus 120582119909
0(119878) + (1 minus 120574) 120583119909
1(119878) = 0
1205821199090(119895) minus (120582 + 120573 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 2 le 119895 le 119904
1205821199090(119895) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119904 + 1 le 119895 le 119876 minus 1
1205821199090(119895) + 120573119909
1(119895 minus 119876) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119876 le 119895 le 119878 minus 1
1205821199090(119878) + 120573119909
1(119904) minus (120582 + 120583)119909
1(119878) + 2 (1 minus 120574) 120583119909
2(119878) = 0
(19)
We assume a solution of the form
119909119894(119895) = 120572
minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 (20)
for constantsΘ119894119895 and then verify that the system of equations
given in (16) is satisfiedThe constants Θ119894
119895rsquos are given by
Θ119894
0= 1 119894 ge 0
Θ119894
1=
1
120574 119894 = 1
2
120574 119894 ge 2
Θ0
119895= (1
120574)
119895
1 le 119895 le 119878 minus 1
Θ119894
2=
(120573 + 120574120582
120573 + 120582)1
1205742 119894 = 1 2
(2120573 + (1 + 120574) 120582
120573 + 120582)1
1205742 119894 ge 3
Θ119894
119895=
(1
120574 (120573 + 120582))120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
3 le 119895 le 119904 + 1
((120573 + 120574120582)
120574 (120573 + 120582))Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
3 le 119895 le 119904 + 1
(1
120574120582)120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
119904 + 2 le 119895 le 119876
(120573 + 120574120582
120574120582)Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
119904 + 2 le 119895 le 119876
(21)
where 120575119894119895minus1= (120582 + 2120583 + 120573)Θ
119894minus1
119895minus1minus 2120583Θ
119894minus2
119895minus1minus (1 minus 120574)120582Θ
119894
119895minus1
Consider
Θ119894
119876+119896
=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894
119876+119896minus1(120573 + 120582
120582)
119904
minus 120582] 3 le 119894 le 2119876
119896 = 1
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894minus2
119876+119896minus1120574120582
times(120573 + 120582
120582)
119904
minus 120582] 119894 ge 2119876 + 1
119896 = 1
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120585119894
119876+119896minus1
times [(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ119894minus1
119896minus1] 3 le 119894 le 2(119876 + 119896 minus 1)
2 le 119896 le 119904
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120574120582[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
timesΘ119894minus2
119876+119896minus1minus 120573Θ119894minus1
119896minus1] 119894 ge 2(119876 + 119896) minus 1
2 le 119896 le 119904
(22)
where 120585119894119876+119896minus1
= (120582+2120583)Θ119894minus1
119876+119896minus1minus2120583Θ
119894minus2
119876+119896minus1minus (1minus120574)120582Θ
119894
119876+119896minus1
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Advances in Operations Research
is (119904 119876) and lead time is mixed exponential distributionCustomers arriving during zero inventory are lost foreverThis leads to a product form solution for the system sate prob-ability Schwarz et al [11] consider queueing networks withattached inventory They consider rerouting of customersserved out from a particular station when the immediatelyfollowing station has zero inventoryThus no customer is lostto the systemThe authors derive joint stationary distributionof queue length and on-hand inventory at various stationsin explicit product form A recent contribution of interestto inventory with positive service time involving a randomenvironment is by Krenzler and Daduna [15] wherein also astochastic decomposition of the system is established Theyprove a necessary and sufficient condition for a product formsteady-state distribution of the joint queueing-environmentprocess to exist A still more recent paper by Krenzler andDaduna [12] investigates inventory with positive service timein a random environment embedded in a Markov chainThey provide a counter example to show that the steady-state distribution of an 1198721198661infin system with (119904 119878) policyand lost sales need not have a product form Neverthelessin general loss systems in a random environment have aproduct form steady-state distribution They also introducea blocking set where all activities other than replenishmentstay suspended whenever theMarkov chain is in that setThisresulted in arriving at a product form solution to the systemstate distribution
The work on multiserver queueing-inventory systems isscarce Nair et al [16] consider an inventory system withnumber of servers varying from 119904 + 1 to 119878 depending on theinventory position Another contribution is by Yadavalli et al[17] wherein the authors consider a finite customer sourcesystem (this paper contains a few additional references tomultiserver inventory system)
In all work quoted above customers are provided an itemfrom the inventory on completion of service Neverthelessthere are several situations where a customer may not beservedmay not purchase the item with probability one atthe end of his service For example customers who maybuy an item arrive at a retail shop where there are one ormore (finite number) servers (sales executives) The serversexplain to each customer the features of product The timerequired for this may be regarded as the service time Afterlistening to the server each customer independently of theothers decides whether to buy the item (probability 120574) orleaves the systemwithout purchasing the item A less realisticexample is as follows a candidate appears for an interviewagainst a position At the end of the interview the candidatedecides to accept the offer of job with probability 120574 and withcomplementary probability rejects it In this case the job istaken as an inventory In this connection one may refer toKrishnamoorthy et al [18] for some recent developments
We arrange the presentation of this paper as indicatedbelow in Section 2 the1198721198722 queueing-inventory problemis mathematically formulated The product form solutionof the steady-state probability distribution including someimportant performance measures is obtained in Section 3Further we numerically investigate the optimal (119904 119876) pairvalues and theminimal cost for different values of 120574 Section 5
discusses the 119872119872119888 with 119888 (greater than or equal to3 but less than 119904) queueing-inventory problems by usingalgorithmic approach Section 6 gives some conditional prob-ability distributions and a few performance measures forthe 119888(ge3) server case Section 7 analyzes the distribution ofthe inventory cycle time In Section 8 the optimal 119888 andthe corresponding minimal cost for different values of 120574 areinvestigated Furtherwe look for the optimal (119904 119876) pair valuesthat would result in cost minimization for different pairs ofvalues of 120574 and 119888
2 Mathematical Modelling of the1198721198722Queueing-Inventory Problem
First we consider an1198721198722 queueing-inventory systemwithpositive service time Customer arrival process is assumed tobe Poisson with rate 120582 Each customer requires a single itemhaving randomduration of service which follows exponentialdistributionwith parameter120583 However it is not essential thatinventory is provided to the customer at the end of his serviceMore precisely the item is servedwith probability 120574 at the endof a service or else it is not provided with probability 1 minus 120574 Acrucial assumption of thismodel is that customers do not jointhe systemwhen the inventory level is zeroWhen the numberof customers is at least two and not less than two items arein inventory the service rate is 2120583 When the inventory levelreaches a prespecified value 119904 gt 0 a replenishment orderis placed for 119876 units with 119876 gt 119904 We fix 119878 = 119876 + 119904 as themaximumnumber of items that could be held in the system atany given timeThe lead time follows exponential distributionwith parameter 120573 Then X(119905) | 119905 ge 0 = (N(119905)I(119905)) | 119905 ge0 is a CTMC with state spaceΩ
1= ⋃infin
119894=0L(119894) whereL(119894) is
called the 119894th level In each of the levels the number of itemsin the inventory can be anything from 0 to 119878 Accordingly wewriteL(119894) = (119894 0) (119894 119876+119904)The infinitesimal generatorW1 of this CTMC X(119905) | 119905 ge 0 is
W1 =
[[[[[
[
119861001198600
11986120119861101198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]
]
(1)
where 11986100
contains transition rates within L(0) 11986120
repre-sents the transitions from level 1 to level 0 119861
10contains the
transitions within level 1 1198600represents the transition from
level 119894 to level 119894+1 119894 ge 11198601represents the transitions within
L(119894) for 119894 ge 2 and 1198602represents transitions from L(119894) to
L(119894 minus 1) 119894 ge 2 The transition rates are
[11986100]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[11986120]119896119897=
120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119878(1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 1198780 otherwise
Advances in Operations Research 3
[11986110]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 119896 = 1 2 119904minus (120582 + 120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 = 1minus (120582 + 120573 + 2120583) for 119897 = 119896 119896 = 2 3 119904minus (120582 + 2120583) for 119897 = 119896
119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897=
120574120583 for 119897 = 119896 minus 1 119896 = 1(1 minus 120574) 120583 for 119897 = 119896 = 12120574120583 for 119897 = 119896 minus 1 119896 = 2 3 1198782 (1 minus 120574) 120583 for 119897 = 119896 119896 = 2 3 1198780 otherwise
(2)
Note that all entries (block matrices) in W1 are of the sameorder namely 119878 + 1 and these matrices contain transitionrates within level (in the case of diagonal entries) and betweenlevels (in the case of off-diagonal entries)
21 Analysis of the System In this section we perform thesteady-state analysis of the queueing-inventory model understudy by first establishing the stability condition of thequeueing-inventory system Define 119860 = 119860
0+ 1198601+ 1198602
This is the infinitesimal generator of the finite state CTMCcorresponding to the inventory level 0 1 2 119878 for anylevel 119894 (ge1) Let 120577 denote the steady-state probability vectorof 119860 That is
120577119860 = 0 120577e = 1 (3)
Write
120577 = (1205770 1205771 120577
119904 120577
119876 120577
119878) (4)
We have
119860 =
[[[[[[[[
[
minus120573 120573
120574120583 minus (120573 + 120574120583)
2120574120583 minus (120573 + 2120574120583) dd d
2120574120583 minus (120573 + 2120574120583) 120573
2120574120583 minus2120574120583
d d2120574120583 minus2120574120583
]]]]]]]]
]
(5)
Then using (3) we get the components of the vector 120577 explic-itly as
1205770= 1 +
120573
120574120583[1 + (
120573 + 120574120583
120574120583)
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
+ (119876 minus 119904 minus 2) (120573 + 2120574120583
2120574120583)
119904minus1
]
+120573
2120574120583(120573 + 120574120583
120574120583)
times [(120573 + 2120574120583
2120574120583)
119904minus1
minus (120574120583
120573 + 120574120583)
+
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
((120573 + 2120574120583
2120574120583)
119904minus119894+1
minus 1)]
minus1
120577119894=
120573
1205741205831205770 for 119894 = 1
120573
120574120583(120573 + 120574120583
2120574120583)
times(120573 + 2120574120583
2120574120583)
119894minus2
1205770 for 119894 = 2 3 119904 + 1
120577119894+1 for 119894 = 119904 + 1
119904 + 2 119876 minus 1
120573
2120574120583[(120573 + 120574120583
120574120583)
times(120573 + 2120574120583
2120574120583)
119904minus1
minus 1]1205770 for 119894 = 119876 + 1
120577119876+119894=120573
2120574120583(120573 + 120574120583
120574120583)(120573 + 2120574120583
2120574120583)
119894minus2
times [(120573 + 2120574120583
2120574120583)
119904minus(119894minus1)
minus 1]1205770 119894 = 2 3 119904
(6)
Since theMarkov chain X(119905) | 119905 ge 0 is an LIQBD it is stableif and only if the left drift rate exceeds the right drift rateThatis
1205771198600e lt 120577119860
2e (7)
Thus we have the following lemma for stability of the systemunder study
Lemma 1 The stability condition of the 1198721198722 queueing-inventory system under consideration is given by 120582 lt 120583[2 minus1205731205770120574120583(1 minus 120577
0)]
Proof From the well-known result by Neuts [19] on thepositive recurrence of the Markov chain associated with 119860we have 120577119860
0e lt 120577119860
2e for theMarkov chain to be stableWith
a bit of algebra this simplifies to120582 lt 120583[2minus1205731205770120574120583(1minus120577
0)]
For future reference we define 1205881as
1205881=
120582
120583[2 minus 1205731205770120574120583 (1 minus 120577
0)] (8)
4 Advances in Operations Research
3 Computation of the Steady-State Probability
For computing the steady-state probability vector of theprocess X(119905) | 119905 ge 0 we first consider a queueing-inventorysystemwith unlimited supply of inventory items (ie classical1198721198722 queueing system) The rest of the assumptions suchas those on the arrival process and lead time are the sameas given earlier Designate the Markov chain so obtained asN(119905) | 119905 ge 0 whereN(119905) is the number of customers in thesystem at time 119905 Its infinitesimal generatorG1 is given by
G1 =
[[[[[
[
minus120582 120582
120583 minus (120582 + 120583) 120582
2120583 minus (120582 + 2120583) 120582
2120583 minus (120582 + 2120583) 120582
d d d
]]]]]
]
(9)Let120587 be the steady-state probability vector ofG1 Partitioning120587 by levels we write 120587 as
120587 = (120587012058711205872 ) (10)
Then the steady-state vector must satisfy
120587G1 = 0 120587e = 1 (11)
From the relation (11) we get the vector120587 explicitly as follows
120587119894=
[1 +120582
120583(1 minus
120582
2120583)
minus1
]
minus1
for 119894 = 0
120582
1205831205870
for 119894 = 1
1
2119894minus1(120582
120583)
119894
1205870
for 119894 ge 2
(12)
Further we consider an inventory system with negligibleservice time and no backlog of demands The assumptionssuch as those on the arrival process and lead time are the sameas given in the description of the model Denote this Markovchain by I(119905) | 119905 ge 0 Here I(119905) is the inventory level attime 119905 Its infinitesimal generatorG2 is given by
G2 =
0
1
119904
119876
119878
0 1 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((
(
minus120573 120573
120574120582 minus (120574120582 + 120573)
d d d120574120582 minus (120574120582 + 120573) 120573
120574120582 minus120574120582
d d120574120582 minus120574120582
120574120582 minus120574120582
))))))))
)
(13)
Let120595 = (1205950 1205951 120595
119878) be the steady-state probability vector
of the process I(119905) | 119905 ge 0 Then 120595 satisfies the relations
120595G2 = 0 120595e = 1 (14)
That is at arbitrary epochs the inventory level distribution120595119895
is given by
120595119895=
[1 + 119876120573
120574120582(120573 + 120574120582
120574120582)
119904
]
minus1
119895 = 0
120573
120574120582(120573 + 120574120582
120574120582)
119895minus1
1205950 119895 = 1 2 119904
120573
120574120582(120573 + 120574120582
120574120582)
119904
1205950 119895 = 119904 + 1
119904 + 2 119876
120573
120574120582(120573 + 120574120582
120574120582)
119895minus119876minus1
times((120573 + 120574120582
120574120582)
119904minus(119895minus119876minus1)
minus 1)1205950 119895 = 119876 + 1
119876 + 2 119878
(15)
Using the components of the probability vector120595 wewill findthe steady-state probability vector of the original system Letx be the steady-state probability vector of the original systemThen the steady-state vector must satisfy the set of equations
xW1 = 0 xe = 1 (16)
Partition x by levels as
x = (x0 x1 x2 ) (17)
where the subvectors of x are further partitioned as
xi = (119909119894 (0) 119909119894 (1) 119909119894 (2) 119909119894 (3) 119909119894 (119878)) 119894 ge 0 (18)
Then by using the relation xW1 = 0 we get
minus120573119909119894(0) + 120574120583119909
119894+1(1) = 0 119894 ge 0
120582119909119894(119895) minus (120582 + 2120583 + 120573)119909
119894+1(119895) + 2(1 minus 120574)120583119909
119894+2(119895)
+ 2120574120583119909119894+2(119895 + 1) = 0 119894 ge 1 2 le 119895 le 119876 minus 1
Advances in Operations Research 5
120582119909119894(119895) + 120573119909
119894+1(119895 minus 119876) minus (120582 + 2120583)119909
119894+1(119895)
+ 2(1 minus 120574)120583119909119894+2(119895) + 2120574120583119909
119894+2(119895 + 1) = 0
119894 ge 1 119876 le 119895 le 119878 minus 1
120582119909119894(119878) + 120573119909
119894+1(119904) minus (120582 + 2120583)119909
119894+1(119878)
+ 2(1 minus 120574)120583119909119894+2(119878) = 0 119894 ge 1
minus (120582 + 120573)1199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
1 le 119895 le 119904
minus 1205821199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119904 + 1 le 119895 le 119876 minus 1
1205731199090(119895 minus 119876) minus 120582119909
0(119895) + (1 minus 120574) 120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119876 le 119895 le 119878 minus 1
1205731199090(119904) minus 120582119909
0(119878) + (1 minus 120574) 120583119909
1(119878) = 0
1205821199090(119895) minus (120582 + 120573 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 2 le 119895 le 119904
1205821199090(119895) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119904 + 1 le 119895 le 119876 minus 1
1205821199090(119895) + 120573119909
1(119895 minus 119876) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119876 le 119895 le 119878 minus 1
1205821199090(119878) + 120573119909
1(119904) minus (120582 + 120583)119909
1(119878) + 2 (1 minus 120574) 120583119909
2(119878) = 0
(19)
We assume a solution of the form
119909119894(119895) = 120572
minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 (20)
for constantsΘ119894119895 and then verify that the system of equations
given in (16) is satisfiedThe constants Θ119894
119895rsquos are given by
Θ119894
0= 1 119894 ge 0
Θ119894
1=
1
120574 119894 = 1
2
120574 119894 ge 2
Θ0
119895= (1
120574)
119895
1 le 119895 le 119878 minus 1
Θ119894
2=
(120573 + 120574120582
120573 + 120582)1
1205742 119894 = 1 2
(2120573 + (1 + 120574) 120582
120573 + 120582)1
1205742 119894 ge 3
Θ119894
119895=
(1
120574 (120573 + 120582))120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
3 le 119895 le 119904 + 1
((120573 + 120574120582)
120574 (120573 + 120582))Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
3 le 119895 le 119904 + 1
(1
120574120582)120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
119904 + 2 le 119895 le 119876
(120573 + 120574120582
120574120582)Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
119904 + 2 le 119895 le 119876
(21)
where 120575119894119895minus1= (120582 + 2120583 + 120573)Θ
119894minus1
119895minus1minus 2120583Θ
119894minus2
119895minus1minus (1 minus 120574)120582Θ
119894
119895minus1
Consider
Θ119894
119876+119896
=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894
119876+119896minus1(120573 + 120582
120582)
119904
minus 120582] 3 le 119894 le 2119876
119896 = 1
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894minus2
119876+119896minus1120574120582
times(120573 + 120582
120582)
119904
minus 120582] 119894 ge 2119876 + 1
119896 = 1
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120585119894
119876+119896minus1
times [(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ119894minus1
119896minus1] 3 le 119894 le 2(119876 + 119896 minus 1)
2 le 119896 le 119904
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120574120582[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
timesΘ119894minus2
119876+119896minus1minus 120573Θ119894minus1
119896minus1] 119894 ge 2(119876 + 119896) minus 1
2 le 119896 le 119904
(22)
where 120585119894119876+119896minus1
= (120582+2120583)Θ119894minus1
119876+119896minus1minus2120583Θ
119894minus2
119876+119896minus1minus (1minus120574)120582Θ
119894
119876+119896minus1
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 3
[11986110]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 119896 = 1 2 119904minus (120582 + 120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897=
minus120573 for 119897 = 119896 = 0minus (120582 + 120573 + 120583) for 119897 = 119896 = 1minus (120582 + 120573 + 2120583) for 119897 = 119896 119896 = 2 3 119904minus (120582 + 2120583) for 119897 = 119896
119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897=
120574120583 for 119897 = 119896 minus 1 119896 = 1(1 minus 120574) 120583 for 119897 = 119896 = 12120574120583 for 119897 = 119896 minus 1 119896 = 2 3 1198782 (1 minus 120574) 120583 for 119897 = 119896 119896 = 2 3 1198780 otherwise
(2)
Note that all entries (block matrices) in W1 are of the sameorder namely 119878 + 1 and these matrices contain transitionrates within level (in the case of diagonal entries) and betweenlevels (in the case of off-diagonal entries)
21 Analysis of the System In this section we perform thesteady-state analysis of the queueing-inventory model understudy by first establishing the stability condition of thequeueing-inventory system Define 119860 = 119860
0+ 1198601+ 1198602
This is the infinitesimal generator of the finite state CTMCcorresponding to the inventory level 0 1 2 119878 for anylevel 119894 (ge1) Let 120577 denote the steady-state probability vectorof 119860 That is
120577119860 = 0 120577e = 1 (3)
Write
120577 = (1205770 1205771 120577
119904 120577
119876 120577
119878) (4)
We have
119860 =
[[[[[[[[
[
minus120573 120573
120574120583 minus (120573 + 120574120583)
2120574120583 minus (120573 + 2120574120583) dd d
2120574120583 minus (120573 + 2120574120583) 120573
2120574120583 minus2120574120583
d d2120574120583 minus2120574120583
]]]]]]]]
]
(5)
Then using (3) we get the components of the vector 120577 explic-itly as
1205770= 1 +
120573
120574120583[1 + (
120573 + 120574120583
120574120583)
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
+ (119876 minus 119904 minus 2) (120573 + 2120574120583
2120574120583)
119904minus1
]
+120573
2120574120583(120573 + 120574120583
120574120583)
times [(120573 + 2120574120583
2120574120583)
119904minus1
minus (120574120583
120573 + 120574120583)
+
119904
sum
119894=0
(120573 + 2120574120583
2120574120583)
119894minus2
((120573 + 2120574120583
2120574120583)
119904minus119894+1
minus 1)]
minus1
120577119894=
120573
1205741205831205770 for 119894 = 1
120573
120574120583(120573 + 120574120583
2120574120583)
times(120573 + 2120574120583
2120574120583)
119894minus2
1205770 for 119894 = 2 3 119904 + 1
120577119894+1 for 119894 = 119904 + 1
119904 + 2 119876 minus 1
120573
2120574120583[(120573 + 120574120583
120574120583)
times(120573 + 2120574120583
2120574120583)
119904minus1
minus 1]1205770 for 119894 = 119876 + 1
120577119876+119894=120573
2120574120583(120573 + 120574120583
120574120583)(120573 + 2120574120583
2120574120583)
119894minus2
times [(120573 + 2120574120583
2120574120583)
119904minus(119894minus1)
minus 1]1205770 119894 = 2 3 119904
(6)
Since theMarkov chain X(119905) | 119905 ge 0 is an LIQBD it is stableif and only if the left drift rate exceeds the right drift rateThatis
1205771198600e lt 120577119860
2e (7)
Thus we have the following lemma for stability of the systemunder study
Lemma 1 The stability condition of the 1198721198722 queueing-inventory system under consideration is given by 120582 lt 120583[2 minus1205731205770120574120583(1 minus 120577
0)]
Proof From the well-known result by Neuts [19] on thepositive recurrence of the Markov chain associated with 119860we have 120577119860
0e lt 120577119860
2e for theMarkov chain to be stableWith
a bit of algebra this simplifies to120582 lt 120583[2minus1205731205770120574120583(1minus120577
0)]
For future reference we define 1205881as
1205881=
120582
120583[2 minus 1205731205770120574120583 (1 minus 120577
0)] (8)
4 Advances in Operations Research
3 Computation of the Steady-State Probability
For computing the steady-state probability vector of theprocess X(119905) | 119905 ge 0 we first consider a queueing-inventorysystemwith unlimited supply of inventory items (ie classical1198721198722 queueing system) The rest of the assumptions suchas those on the arrival process and lead time are the sameas given earlier Designate the Markov chain so obtained asN(119905) | 119905 ge 0 whereN(119905) is the number of customers in thesystem at time 119905 Its infinitesimal generatorG1 is given by
G1 =
[[[[[
[
minus120582 120582
120583 minus (120582 + 120583) 120582
2120583 minus (120582 + 2120583) 120582
2120583 minus (120582 + 2120583) 120582
d d d
]]]]]
]
(9)Let120587 be the steady-state probability vector ofG1 Partitioning120587 by levels we write 120587 as
120587 = (120587012058711205872 ) (10)
Then the steady-state vector must satisfy
120587G1 = 0 120587e = 1 (11)
From the relation (11) we get the vector120587 explicitly as follows
120587119894=
[1 +120582
120583(1 minus
120582
2120583)
minus1
]
minus1
for 119894 = 0
120582
1205831205870
for 119894 = 1
1
2119894minus1(120582
120583)
119894
1205870
for 119894 ge 2
(12)
Further we consider an inventory system with negligibleservice time and no backlog of demands The assumptionssuch as those on the arrival process and lead time are the sameas given in the description of the model Denote this Markovchain by I(119905) | 119905 ge 0 Here I(119905) is the inventory level attime 119905 Its infinitesimal generatorG2 is given by
G2 =
0
1
119904
119876
119878
0 1 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((
(
minus120573 120573
120574120582 minus (120574120582 + 120573)
d d d120574120582 minus (120574120582 + 120573) 120573
120574120582 minus120574120582
d d120574120582 minus120574120582
120574120582 minus120574120582
))))))))
)
(13)
Let120595 = (1205950 1205951 120595
119878) be the steady-state probability vector
of the process I(119905) | 119905 ge 0 Then 120595 satisfies the relations
120595G2 = 0 120595e = 1 (14)
That is at arbitrary epochs the inventory level distribution120595119895
is given by
120595119895=
[1 + 119876120573
120574120582(120573 + 120574120582
120574120582)
119904
]
minus1
119895 = 0
120573
120574120582(120573 + 120574120582
120574120582)
119895minus1
1205950 119895 = 1 2 119904
120573
120574120582(120573 + 120574120582
120574120582)
119904
1205950 119895 = 119904 + 1
119904 + 2 119876
120573
120574120582(120573 + 120574120582
120574120582)
119895minus119876minus1
times((120573 + 120574120582
120574120582)
119904minus(119895minus119876minus1)
minus 1)1205950 119895 = 119876 + 1
119876 + 2 119878
(15)
Using the components of the probability vector120595 wewill findthe steady-state probability vector of the original system Letx be the steady-state probability vector of the original systemThen the steady-state vector must satisfy the set of equations
xW1 = 0 xe = 1 (16)
Partition x by levels as
x = (x0 x1 x2 ) (17)
where the subvectors of x are further partitioned as
xi = (119909119894 (0) 119909119894 (1) 119909119894 (2) 119909119894 (3) 119909119894 (119878)) 119894 ge 0 (18)
Then by using the relation xW1 = 0 we get
minus120573119909119894(0) + 120574120583119909
119894+1(1) = 0 119894 ge 0
120582119909119894(119895) minus (120582 + 2120583 + 120573)119909
119894+1(119895) + 2(1 minus 120574)120583119909
119894+2(119895)
+ 2120574120583119909119894+2(119895 + 1) = 0 119894 ge 1 2 le 119895 le 119876 minus 1
Advances in Operations Research 5
120582119909119894(119895) + 120573119909
119894+1(119895 minus 119876) minus (120582 + 2120583)119909
119894+1(119895)
+ 2(1 minus 120574)120583119909119894+2(119895) + 2120574120583119909
119894+2(119895 + 1) = 0
119894 ge 1 119876 le 119895 le 119878 minus 1
120582119909119894(119878) + 120573119909
119894+1(119904) minus (120582 + 2120583)119909
119894+1(119878)
+ 2(1 minus 120574)120583119909119894+2(119878) = 0 119894 ge 1
minus (120582 + 120573)1199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
1 le 119895 le 119904
minus 1205821199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119904 + 1 le 119895 le 119876 minus 1
1205731199090(119895 minus 119876) minus 120582119909
0(119895) + (1 minus 120574) 120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119876 le 119895 le 119878 minus 1
1205731199090(119904) minus 120582119909
0(119878) + (1 minus 120574) 120583119909
1(119878) = 0
1205821199090(119895) minus (120582 + 120573 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 2 le 119895 le 119904
1205821199090(119895) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119904 + 1 le 119895 le 119876 minus 1
1205821199090(119895) + 120573119909
1(119895 minus 119876) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119876 le 119895 le 119878 minus 1
1205821199090(119878) + 120573119909
1(119904) minus (120582 + 120583)119909
1(119878) + 2 (1 minus 120574) 120583119909
2(119878) = 0
(19)
We assume a solution of the form
119909119894(119895) = 120572
minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 (20)
for constantsΘ119894119895 and then verify that the system of equations
given in (16) is satisfiedThe constants Θ119894
119895rsquos are given by
Θ119894
0= 1 119894 ge 0
Θ119894
1=
1
120574 119894 = 1
2
120574 119894 ge 2
Θ0
119895= (1
120574)
119895
1 le 119895 le 119878 minus 1
Θ119894
2=
(120573 + 120574120582
120573 + 120582)1
1205742 119894 = 1 2
(2120573 + (1 + 120574) 120582
120573 + 120582)1
1205742 119894 ge 3
Θ119894
119895=
(1
120574 (120573 + 120582))120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
3 le 119895 le 119904 + 1
((120573 + 120574120582)
120574 (120573 + 120582))Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
3 le 119895 le 119904 + 1
(1
120574120582)120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
119904 + 2 le 119895 le 119876
(120573 + 120574120582
120574120582)Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
119904 + 2 le 119895 le 119876
(21)
where 120575119894119895minus1= (120582 + 2120583 + 120573)Θ
119894minus1
119895minus1minus 2120583Θ
119894minus2
119895minus1minus (1 minus 120574)120582Θ
119894
119895minus1
Consider
Θ119894
119876+119896
=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894
119876+119896minus1(120573 + 120582
120582)
119904
minus 120582] 3 le 119894 le 2119876
119896 = 1
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894minus2
119876+119896minus1120574120582
times(120573 + 120582
120582)
119904
minus 120582] 119894 ge 2119876 + 1
119896 = 1
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120585119894
119876+119896minus1
times [(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ119894minus1
119896minus1] 3 le 119894 le 2(119876 + 119896 minus 1)
2 le 119896 le 119904
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120574120582[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
timesΘ119894minus2
119876+119896minus1minus 120573Θ119894minus1
119896minus1] 119894 ge 2(119876 + 119896) minus 1
2 le 119896 le 119904
(22)
where 120585119894119876+119896minus1
= (120582+2120583)Θ119894minus1
119876+119896minus1minus2120583Θ
119894minus2
119876+119896minus1minus (1minus120574)120582Θ
119894
119876+119896minus1
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Advances in Operations Research
3 Computation of the Steady-State Probability
For computing the steady-state probability vector of theprocess X(119905) | 119905 ge 0 we first consider a queueing-inventorysystemwith unlimited supply of inventory items (ie classical1198721198722 queueing system) The rest of the assumptions suchas those on the arrival process and lead time are the sameas given earlier Designate the Markov chain so obtained asN(119905) | 119905 ge 0 whereN(119905) is the number of customers in thesystem at time 119905 Its infinitesimal generatorG1 is given by
G1 =
[[[[[
[
minus120582 120582
120583 minus (120582 + 120583) 120582
2120583 minus (120582 + 2120583) 120582
2120583 minus (120582 + 2120583) 120582
d d d
]]]]]
]
(9)Let120587 be the steady-state probability vector ofG1 Partitioning120587 by levels we write 120587 as
120587 = (120587012058711205872 ) (10)
Then the steady-state vector must satisfy
120587G1 = 0 120587e = 1 (11)
From the relation (11) we get the vector120587 explicitly as follows
120587119894=
[1 +120582
120583(1 minus
120582
2120583)
minus1
]
minus1
for 119894 = 0
120582
1205831205870
for 119894 = 1
1
2119894minus1(120582
120583)
119894
1205870
for 119894 ge 2
(12)
Further we consider an inventory system with negligibleservice time and no backlog of demands The assumptionssuch as those on the arrival process and lead time are the sameas given in the description of the model Denote this Markovchain by I(119905) | 119905 ge 0 Here I(119905) is the inventory level attime 119905 Its infinitesimal generatorG2 is given by
G2 =
0
1
119904
119876
119878
0 1 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((
(
minus120573 120573
120574120582 minus (120574120582 + 120573)
d d d120574120582 minus (120574120582 + 120573) 120573
120574120582 minus120574120582
d d120574120582 minus120574120582
120574120582 minus120574120582
))))))))
)
(13)
Let120595 = (1205950 1205951 120595
119878) be the steady-state probability vector
of the process I(119905) | 119905 ge 0 Then 120595 satisfies the relations
120595G2 = 0 120595e = 1 (14)
That is at arbitrary epochs the inventory level distribution120595119895
is given by
120595119895=
[1 + 119876120573
120574120582(120573 + 120574120582
120574120582)
119904
]
minus1
119895 = 0
120573
120574120582(120573 + 120574120582
120574120582)
119895minus1
1205950 119895 = 1 2 119904
120573
120574120582(120573 + 120574120582
120574120582)
119904
1205950 119895 = 119904 + 1
119904 + 2 119876
120573
120574120582(120573 + 120574120582
120574120582)
119895minus119876minus1
times((120573 + 120574120582
120574120582)
119904minus(119895minus119876minus1)
minus 1)1205950 119895 = 119876 + 1
119876 + 2 119878
(15)
Using the components of the probability vector120595 wewill findthe steady-state probability vector of the original system Letx be the steady-state probability vector of the original systemThen the steady-state vector must satisfy the set of equations
xW1 = 0 xe = 1 (16)
Partition x by levels as
x = (x0 x1 x2 ) (17)
where the subvectors of x are further partitioned as
xi = (119909119894 (0) 119909119894 (1) 119909119894 (2) 119909119894 (3) 119909119894 (119878)) 119894 ge 0 (18)
Then by using the relation xW1 = 0 we get
minus120573119909119894(0) + 120574120583119909
119894+1(1) = 0 119894 ge 0
120582119909119894(119895) minus (120582 + 2120583 + 120573)119909
119894+1(119895) + 2(1 minus 120574)120583119909
119894+2(119895)
+ 2120574120583119909119894+2(119895 + 1) = 0 119894 ge 1 2 le 119895 le 119876 minus 1
Advances in Operations Research 5
120582119909119894(119895) + 120573119909
119894+1(119895 minus 119876) minus (120582 + 2120583)119909
119894+1(119895)
+ 2(1 minus 120574)120583119909119894+2(119895) + 2120574120583119909
119894+2(119895 + 1) = 0
119894 ge 1 119876 le 119895 le 119878 minus 1
120582119909119894(119878) + 120573119909
119894+1(119904) minus (120582 + 2120583)119909
119894+1(119878)
+ 2(1 minus 120574)120583119909119894+2(119878) = 0 119894 ge 1
minus (120582 + 120573)1199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
1 le 119895 le 119904
minus 1205821199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119904 + 1 le 119895 le 119876 minus 1
1205731199090(119895 minus 119876) minus 120582119909
0(119895) + (1 minus 120574) 120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119876 le 119895 le 119878 minus 1
1205731199090(119904) minus 120582119909
0(119878) + (1 minus 120574) 120583119909
1(119878) = 0
1205821199090(119895) minus (120582 + 120573 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 2 le 119895 le 119904
1205821199090(119895) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119904 + 1 le 119895 le 119876 minus 1
1205821199090(119895) + 120573119909
1(119895 minus 119876) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119876 le 119895 le 119878 minus 1
1205821199090(119878) + 120573119909
1(119904) minus (120582 + 120583)119909
1(119878) + 2 (1 minus 120574) 120583119909
2(119878) = 0
(19)
We assume a solution of the form
119909119894(119895) = 120572
minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 (20)
for constantsΘ119894119895 and then verify that the system of equations
given in (16) is satisfiedThe constants Θ119894
119895rsquos are given by
Θ119894
0= 1 119894 ge 0
Θ119894
1=
1
120574 119894 = 1
2
120574 119894 ge 2
Θ0
119895= (1
120574)
119895
1 le 119895 le 119878 minus 1
Θ119894
2=
(120573 + 120574120582
120573 + 120582)1
1205742 119894 = 1 2
(2120573 + (1 + 120574) 120582
120573 + 120582)1
1205742 119894 ge 3
Θ119894
119895=
(1
120574 (120573 + 120582))120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
3 le 119895 le 119904 + 1
((120573 + 120574120582)
120574 (120573 + 120582))Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
3 le 119895 le 119904 + 1
(1
120574120582)120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
119904 + 2 le 119895 le 119876
(120573 + 120574120582
120574120582)Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
119904 + 2 le 119895 le 119876
(21)
where 120575119894119895minus1= (120582 + 2120583 + 120573)Θ
119894minus1
119895minus1minus 2120583Θ
119894minus2
119895minus1minus (1 minus 120574)120582Θ
119894
119895minus1
Consider
Θ119894
119876+119896
=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894
119876+119896minus1(120573 + 120582
120582)
119904
minus 120582] 3 le 119894 le 2119876
119896 = 1
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894minus2
119876+119896minus1120574120582
times(120573 + 120582
120582)
119904
minus 120582] 119894 ge 2119876 + 1
119896 = 1
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120585119894
119876+119896minus1
times [(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ119894minus1
119896minus1] 3 le 119894 le 2(119876 + 119896 minus 1)
2 le 119896 le 119904
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120574120582[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
timesΘ119894minus2
119876+119896minus1minus 120573Θ119894minus1
119896minus1] 119894 ge 2(119876 + 119896) minus 1
2 le 119896 le 119904
(22)
where 120585119894119876+119896minus1
= (120582+2120583)Θ119894minus1
119876+119896minus1minus2120583Θ
119894minus2
119876+119896minus1minus (1minus120574)120582Θ
119894
119876+119896minus1
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Stochastic AnalysisInternational Journal of
Advances in Operations Research 5
120582119909119894(119895) + 120573119909
119894+1(119895 minus 119876) minus (120582 + 2120583)119909
119894+1(119895)
+ 2(1 minus 120574)120583119909119894+2(119895) + 2120574120583119909
119894+2(119895 + 1) = 0
119894 ge 1 119876 le 119895 le 119878 minus 1
120582119909119894(119878) + 120573119909
119894+1(119904) minus (120582 + 2120583)119909
119894+1(119878)
+ 2(1 minus 120574)120583119909119894+2(119878) = 0 119894 ge 1
minus (120582 + 120573)1199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
1 le 119895 le 119904
minus 1205821199090(119895) + (1 minus 120574)120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119904 + 1 le 119895 le 119876 minus 1
1205731199090(119895 minus 119876) minus 120582119909
0(119895) + (1 minus 120574) 120583119909
1(119895) + 120574120583119909
1(119895 + 1) = 0
119876 le 119895 le 119878 minus 1
1205731199090(119904) minus 120582119909
0(119878) + (1 minus 120574) 120583119909
1(119878) = 0
1205821199090(119895) minus (120582 + 120573 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 2 le 119895 le 119904
1205821199090(119895) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119904 + 1 le 119895 le 119876 minus 1
1205821199090(119895) + 120573119909
1(119895 minus 119876) minus (120582 + 120583) 119909
1(119895) + 2 (1 minus 120574) 120583119909
2(119895)
+ 21205741205831199092(119895 + 1) = 0 119876 le 119895 le 119878 minus 1
1205821199090(119878) + 120573119909
1(119904) minus (120582 + 120583)119909
1(119878) + 2 (1 minus 120574) 120583119909
2(119878) = 0
(19)
We assume a solution of the form
119909119894(119895) = 120572
minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 (20)
for constantsΘ119894119895 and then verify that the system of equations
given in (16) is satisfiedThe constants Θ119894
119895rsquos are given by
Θ119894
0= 1 119894 ge 0
Θ119894
1=
1
120574 119894 = 1
2
120574 119894 ge 2
Θ0
119895= (1
120574)
119895
1 le 119895 le 119878 minus 1
Θ119894
2=
(120573 + 120574120582
120573 + 120582)1
1205742 119894 = 1 2
(2120573 + (1 + 120574) 120582
120573 + 120582)1
1205742 119894 ge 3
Θ119894
119895=
(1
120574 (120573 + 120582))120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
3 le 119895 le 119904 + 1
((120573 + 120574120582)
120574 (120573 + 120582))Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
3 le 119895 le 119904 + 1
(1
120574120582)120575119894
119895minus1 3 le 119894 le 2 (119895 minus 1)
119904 + 2 le 119895 le 119876
(120573 + 120574120582
120574120582)Θ119894minus2
119895minus1 119894 ge 2119895 minus 1
119904 + 2 le 119895 le 119876
(21)
where 120575119894119895minus1= (120582 + 2120583 + 120573)Θ
119894minus1
119895minus1minus 2120583Θ
119894minus2
119895minus1minus (1 minus 120574)120582Θ
119894
119895minus1
Consider
Θ119894
119876+119896
=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894
119876+119896minus1(120573 + 120582
120582)
119904
minus 120582] 3 le 119894 le 2119876
119896 = 1
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120585119894minus2
119876+119896minus1120574120582
times(120573 + 120582
120582)
119904
minus 120582] 119894 ge 2119876 + 1
119896 = 1
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120585119894
119876+119896minus1
times [(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ119894minus1
119896minus1] 3 le 119894 le 2(119876 + 119896 minus 1)
2 le 119896 le 119904
1
120574120582 (120573 + 120582)
times[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120574120582[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
timesΘ119894minus2
119876+119896minus1minus 120573Θ119894minus1
119896minus1] 119894 ge 2(119876 + 119896) minus 1
2 le 119896 le 119904
(22)
where 120585119894119876+119896minus1
= (120582+2120583)Θ119894minus1
119876+119896minus1minus2120583Θ
119894minus2
119876+119896minus1minus (1minus120574)120582Θ
119894
119876+119896minus1
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Advances in Operations Research
Consider
Θ1
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573)Θ0
119895minus1
minus (1 minus 120574) 120582Θ1
119895minus1] 3 le 119895 le 119904 + 1
1
120574[Θ0
119895minus1minus (1 minus 120574)Θ
1
119895minus1] 119904 + 2 le 119895 le 119876
Θ0
119878= [Θ0
119904minus (1 minus 120574)Θ
1
119878]
Θ2
119895=
1
120574 (120573 + 120582)
times[(120582 + 120573 + 120583)Θ1
119895minus1
minus120583Θ0
119895minus1minus (1 minus 120574) 120582Θ
2
119895minus1] 3 le 119895 le 119904 + 1
1
120574120582120599119895minus1 119904 + 2 le 119895 le 119876
Θ2
119876+119896=
1
120574120582[(120573 + 120582
120582)
119904
minus 1]
minus1
times[120599119876(120573 + 120583
120583)
119904
minus 120573] 119896 = 1
1
120574 (120573 + 120582)[(120573 + 120582
120582)
119904minus(119896minus1)
minus 1]
minus1
times[120599119895minus1[(120573 + 120582
120582)
119904minus(119896minus2)
minus 1]
minus120573Θ1
119896minus1] 2 le 119896 le 119904
(23)
where 120599119895= (120582 + 120583)Θ
1
119895minus 120583Θ0
119895minus (1 minus 120574)120582Θ
2
119895 119904 minus 1 le 119895 le 119878
Thus we haveinfin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 + 119876
120573
120574120582(120573 + 120574120582
120574120582)
119904
(24)
If we note xe = 1 and (20) we have
120572minus1
infin
sum
119894=0
119876+119904
sum
119895=0
Θ119894
119895120587119894120595119895= 1 (25)
Write 120572 = 1 + 119876(120573120574120582)((120573 + 120574120582)120574120582)119904 Then dividing eachΘ119894
119895120587119894120595119895by 120572 we get the steady-state probability vector of the
original systemThus we arrive at our main theorem
Theorem 2 Suppose that the condition 1205881lt 1 holds Then
the components of the steady-state probability vector of theprocess X(119905) | 119905 ge 0 with generator matrix W1 are 119909119894(119895) =120572minus1Θ119894
119895120587119894120595119895 119894 ge 0 0 le 119895 le 119878 the probabilities 120587
119894correspond to
the distribution of number of customers in the system as givenin (12) and the probabilities 120595
119895are obtained in (15)
The consequence of Theorem 2 is that the two-dimen-sional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number
of customers in an 1198721198722 queue and the other to thenumber of items in the inventory
31 Performance Measures
(i) Mean number of customers in the system is as follows
119871119904= 120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=0
119894Θ119894
119895120587119894120595119895) (26)
(ii) Mean number of customers in the queue is as follows
119871119902= 120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
(119894 minus 2)Θ119894
119895120587119894120595119895) (27)
(iii) Mean inventory level in the system is as follows
119868119898= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119895Θ119894
119895120587119894120595119895) (28)
(iv) Mean number of busy servers is as follows
119875BS = 120572minus1([
[
infin
sum
119894=2
Θ119894
11205871198941205951+
119876+119904
sum
119895=2
Θ1
1198951205871120595119895+ Θ1
112058711205951]
]
+2[
[
infin
sum
119894=3
Θ119894
21205871198941205952+
119876+119904
sum
119895=3
Θ2
1198951205872120595119895+ Θ2
212058721205952]
]
)
(29)
(v) Depletion rate of inventory is as follows
119863inv = 120574120583120572minus1(
infin
sum
119894=1
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (30)
(vi) Mean number of replenishments per time unit is asfollows
119877119903= 120573120572minus1(
infin
sum
119894=0
119904
sum
119895=0
Θ119894
119895120587119894120595119895) (31)
(vii) Mean number of departures per unit time is as fol-lows
119863119898= 120583120572minus1(
infin
sum
119894=1
Θ119894
11205871198941205951+
119876+119904
sum
119895=1
Θ1
1198951205871120595119895)
+ 2120583120572minus1(
infin
sum
119894=2
119876+119904
sum
119895=2
Θ119894
119895120587119894120595119895)
(32)
(viii) Expected loss rate of customers is as follows
119864loss = 120582120572minus1(
infin
sum
119894=0
Θ119894
01205871198941205950) (33)
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 7
(ix) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(x) Effective arrival rate is as follows
120582119860= 120582120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
Θ119894
119895120587119894120595119895) (34)
(xi) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiii) Mean number of customers waiting in the systemwhen inventory is available is as follows
= 120572minus1(
infin
sum
119894=0
119876+119904
sum
119895=1
119894Θ119894
119895120587119894120595119895) (35)
(xiv) Mean number of customers waiting in the systemduring the stock out period is as follows
119882 = 120572
minus1(
infin
sum
119894=0
119894Θ119894
01205871198941205950) (36)
4 Optimization Problem I
In this section we provide the optimal values of the inventorylevel 119904 and the fixed order quantity 119876 Now for computingthe minimal costs of1198721198722 queueing-inventory model weintroduce the cost functionF(2 119904 119876) defined by
F(2 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882
+ (119870 + 119876 sdot 1198883) sdot 119877119903+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(37)
where 119870 is fixed cost for placing an order 1198881is the cost
incurred due to loss per customer 1198882is waiting cost per unit
time per customer during the stock out period 1198883is variable
procurement cost per item 1198884is the cost incurred per busy
server 1198885is the cost incurred per idle server and ℎ is unit
holding cost of inventory per unit per unit timeWe assign thefollowing values to the parameters 120582 = 5 120583 = 3 120573 = 1 119870 =$500 119888
1= $100 119888
2= $50 119888
3= $25 119888
4= $10 119888
5= $20 and
ℎ = $2 Using MATLAB program we computed the optimalpairs (119904 119876) and also the corresponding minimum cost (inDollars) Here 120574 is varied from 01 to 1 each time increasingit by 01 unit The optimal pair (119904 119876) and the correspondingcost (minimum) are given in Table 1
5119872119872119888 (119888 ge 3) Queueing-Inventory System
Next we consider 119872119872119888 queueing-inventory system withpositive service time for 3 le 119888 le 119904 We keep the modelassumptions the same as in Section 2 Hence the service rateis 119894120583 for 119894 varying from 0 to 119888 depending on the availability ofthe inventory and customersWhen the number of customers
is at least 119888 and not less than 119888 items are in the inventory theservice rate is 119888120583 Write Y(119905) | 119905 ge 0 = (N(119905)I(119905)) |119905 ge 0 Then Y(119905) | 119905 ge 0 is a CTMC with state spaceΩ2= ⋃infin
119894=0L(119894) whereL(119894) is the collection of statesL(119894) =
(119894 0) (119894 119876+ 119904) as defined in Section 2The infinitesimalgeneratorW2 of the CTMC Y(119905) | 119905 ge 0 is
W2 =
[[[[[[[[[[[[[[[[[[[[[
[
119861 1198600
1198601
21198601
11198600
1198602
21198602
11198600
d d d
119860119888minus2
2119860119888minus2
11198600
119860119888minus1
2119860119888minus1
11198600
119860211986011198600
119860211986011198600sdot sdot sdot
d d d
]]]]]]]]]]]]]]]]]]]]]
]
(38)
and the transition rates are
[119861]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573) for 119897 = 119896 119896 = 1 2 119904
minus120582 for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
[1198600]119896119897= 120582 for 119897 = 119896 119896 = 1 2 1198780 otherwise
[1198601]119896119897
=
minus120573 for 119897 = 119896 = 0minus(120582 + 120573 + 119894120583) for 119897 = 119896 119896 = 1 2 119888minus (120582 + 120573 + 119888120583) for 119897 = 119896 119896 = 119888 + 1 119888 + 2 119904minus (120582 + 119888120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878120573 for 119897 = 119896 + 119876 119896 = 0 1 1199040 otherwise
[1198602]119896119897
=
119894120574120583 for 119897 = 119896 minus 1 119896 = 119888 119888 + 1 119888 + 2 119878119894120574120583 for 119897 = 119896 minus 1 119896 = 1 2 119888 minus 1119888 (1 minus 120574) 120583 for 119897 = 119896 119896 = 119888 119888 + 1 119888 + 2 119878119894 (1 minus 120574) 120583 for 119897 = 119896 119896 = 1 2 119888 minus 10 otherwise
(39)
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Advances in Operations Research
Table 1 Optimal (119904 119876) pair and minimum cost
120574 01 02 03 04 05 06 07 08 09 1Optimal (119904 119876) pairand minimum cost
(3 15) (3 21) (3 27) (3 33) (3 39) (3 43) (5 46) (5 53) (6 53) (6 58)82684 10687 13057 15376 17629 19810 21904 23903 25826 27718
For119898 = 1 2 119888 minus 1
[119860119898
2]119896119897
=
119898120574120583 for 119897 = 119896 minus 1 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 minus 1 119898 gt 119896 119896 = 1 2 119878
119898 (1 minus 120574) 120583 for 119897 = 119896 119898 le 119896 119896 = 1 2 119878
119896120574120583 for 119897 = 119896 119898 gt 119896 119896 = 1 2 119878
0 otherwise
[119860119898
1]119896119897
=
minus120573 for 119897 = 119896 = 0
minus (120582 + 120573 + 119898120583) for 119897 = 119896 119898 le 119896 119896 = 1 2 119904
minus (120582 + 120573 + 119896120583) for 119897 = 119896 119898 gt 119896 ge 1
minus (120582 + 119898120583) for 119897 = 119896 119896 = 119904 + 1 119904 + 2 119878
120573 for 119897 = 119896 + 119876 119896 = 0 1 119904
0 otherwise
(40)
51 System Stability and Computation of Steady-State Prob-ability Vector The Markov chain under consideration is aLIQBD process For this chain to be stable it is necessary andsufficient that
1205851198600e lt 120585119860
2e (41)
where 120585 is the unique nonnegative vector satisfying
120585119860 = 0 120585e = 1 (42)
and 119860 = 1198600+ 1198601+ 1198602is the infinitesimal generator of
the finite state CTMC on the set 0 1 119878 Write 120585 as(1205850 1205851 120585
119878) Then we get from (42) the components of the
probability vector 120585 explicitly as
1205850= 1 +
119888minus1
sum
119894=1
119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583[1 +
119904minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]
+1205732
119888120574120583[1 +
119904minus2
sum
119894=1
119894
prod
119896=1
120573 + 119896120574120583
119896120574120583]
minus1
120585119894=
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 1 le 119894 le 119888
(120573 + 119888120574120583
119888120574120583)
119894minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205850 for 119888 + 1 le 119894 le 119904 + 1
120585119894+1 for 119904 + 1 le 119894 le 119876 minus 1
120585119876+119894=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205850 for 119894 = 1
[
[
(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[
[
1 +
119894minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
]
]
1205850 for 2 le 119894 le 119904
(43)
From the relation (41) we have the following
Lemma 3 The stability condition of the queueing-inventorysystem under study is given by 120588
2lt 1 where 120588
2= 120582(1 minus
1205850)120583[sum
119888minus1
119895=1119895120585119895+ 119888sum119876+119904
119895=119888120585119895]
Proof The proof is on the same lines as that of Lemma 1
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 9
Next we compute the steady-state probability vector ofW2 under the stability condition Let y denote the steady-state probability vector of the generatorW2 So ymust satisfythe relations
yW2 = 0 ye = 1 (44)
Let us partition y by levels as
y = (y0 y1 y2 ) (45)
where the subvectors of y are further partitioned as
yi = (119910119894(0) 119910119894(1) 119910119894(2) 119910119894(119878)) 119894 ge 0 (46)
The steady-state probability vector y is obtained as
yi+cminus1 = ycminus1119877119894 119894 ge 1 (47)
where 119877 is the minimal nonnegative solution to the matrixquadratic equation
11987721198602+ 1198771198601+ 1198600= 0 (48)
and the vectors y0 y1 ycminus1 can be obtained by solving thefollowing equations
y0119861 + y11198601
2= 0
yiminus11198600 + yi119860119894
1+ yi+1119860
119894+1
2= 0 1 le 119894 le 119888 minus 1
ycminus21198600 + ycminus1(119860119888minus1
1+ 1198771198602) = 0
(49)
Now from (49) we get
y0 = y11198601
2(minus119861)minus1= y1119860
1
2(minus1198601015840
0)
minus1
y1 = minusy21198602
2[1198601
1+ 1198601
2(minus1198601015840
0)
minus1
1198600]
minus1
= y21198602
2(minus1198601015840
0)
minus1
yi = yi+1119860119894+1
2(minus1198601015840
0)
minus1
0 le 119894 le 119888 minus 1
(50)
where
1198601015840
119894=
119861 119894 = 0
119860119894
1+ 119860119894
2(minus1198601015840
119894minus1)
minus1
1198600 1 le 119894 le 119888
(51)
subject to normalizing condition
119888minus2
sum
119894=1
yi + ycminus1(119868 minus 119877)minus1e = 1 (52)
Since 119877 cannot be computed explicitly we explore thepossibility of algorithmic computation Thus one can uselogarithmic reduction algorithm as given by Latouche andRamaswami [20] for computing 119877 We list here only themain steps involved in logarithmic reduction algorithm forcomputation of 119877
Logarithmic Reduction Algorithm for 119877
Step 0 119867 larr (minus1198601)minus11198600 119871 larr (minus119860
1)minus11198602119866 = 119871 and 119879 = 119867
Step 1 Consider
119880 = 119867119871 + 119871119867
119872 = 1198672
119867 larr997888 (119868 minus 119880)minus1119872
119872 larr997888 1198712
119871 larr997888 (119868 minus 119880)minus1119872
119866 larr997888 119866 + 119879119871
119879 larr997888 119879119867
(53)
Continue Step 1 until e minus 119866einfinlt 120598
Step 2 119877 = minus1198600(1198601+ 1198600119866)minus1
6 Conditional Probability Distributions
We could arrive at an analytical expression for system stateprobabilities of 1198721198722 queueing-inventory system How-ever for the119872119872119888 queueing-inventory system with 119888 ge 3the system state distribution does not seem to have closedform owing to the strong dependence between the inventorylevel number of customers and the number of servers in thesystem In this section we provide conditional probabilitiesof the number of items in the inventory given the numberof customers in the system and also that of the number ofcustomers in the system conditioned on the number of itemsin the inventory
61 Conditional Probability Distribution of the Inventory LevelConditioned on the Number of Customers in the System Let120578 = (120578
0 1205781 120578
119878) be the probability distribution of the
inventory level conditioned on the number of customers inthe system Then we get explicit form for the conditionalprobability distribution of the inventory level conditioned onthe number of customers in the system We formulate theresult in the following lemma
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Advances in Operations Research
Lemma 4 Assume that 119894 is the number of customers in thesystem at some point of time Conditional on this we computethe inventory level distribution 120578
119895where there are 119895 items in the
inventory We consider two cases as follows
(i) When 119894 lt 119888 the inventory level probability distributionis given by
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 119891119900119903 119894 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(54)
(ii) When 119894 ge 119888 the inventory level probability distributionis derived by
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1)120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1)120574120583)1205780 119891119900119903 119888 + 1 le 119895 le 119904 + 1
120578119895+1 119891119900119903 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 119891119900119903 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 119891119900119903 2 le 119895 le 119904
(55)
Proof Let Γ1 be the infinitesimal generator of the corre-sponding Markov chain
(i) Case of 119894 lt 119888
The infinitesimal generator Γ1 is given by
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 11
Γ1 =
0
1
2
119894
119888
119904
119876
119878
0 1 sdot sdot sdot 119894 sdot sdot sdot 119888 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573)
d d119894120574120583 minus (119894120574120583 + 120573) 120573
119894120574120583 minus119894120574120583
d d119894120574120583 minus119894120574120583
119894120574120583 minus119894120574120583
))))))))))))))))
)
(56)
and the inventory level distribution 120578 can be obtained fromthe equations 120578Γ1 = 0 and 120578e = 1 and we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119894
(120573 + 119894120574120583
119894120574120583)
119895minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119894 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583]1205780 for 119895 = 1
[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
times
119894minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119894120574120583
times[1 +
119895minus2
sum
119894=0
119894
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 for 2 le 119895 le 119904
(57)
where
1205780=
1 +
119894minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119894minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119894120574120583
119894120574120583)
119904+1minus119894
minus 1]
+ 119876(120573 + 119894120574120583
119894120574120583)
119904+1minus119894 119894minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119894120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119894120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(58)
(ii) Case of 119894 ge 119888
The infinitesimal generator Γ2 is given by
Γ2 =
0
1
2
119888
119894
119904
119876
119878
0 1 sdot sdot sdot 119888 sdot sdot sdot 119894 sdot sdot sdot 119904 sdot sdot sdot 119876 sdot sdot sdot 119878
((((((((((((((((
(
minus120573 120573
120574120583 minus (120574120583 + 120573)
2120574120583 minus (2120574120583 + 120573)
d d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573)
d d119888120574120583 minus (119888120574120583 + 120573) 120573
119888120574120583 minus119888120574120583
d d119888120574120583 minus119888120574120583
119888120574120583 minus119888120574120583
))))))))))))))))
)
(59)
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Stochastic AnalysisInternational Journal of
12 Advances in Operations Research
By solving the equations 120578Γ2 = 0 and 120578e = 1 we get
120578119895=
119895minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 1 le 119895 le 119888
(120573 + 119888120574120583
119888120574120583)
119895minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583)1205780 for 119888 + 1 le 119895 le 119904 + 1
120578119895+1 for 119904 + 1 le 119895 le 119876 minus 1
120578119876+119895=
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583]1205780 for 119895 = 1
[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
times
119888minus1
prod
119896=0
(120573 + 119896120574120583
(119896 + 1) 120574120583) minus
120573
119888120574120583
times[1 +
119895minus2
sum
119888=0
119888
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583]]1205780 2 le 119895 le 119904
(60)
where
1205780=
1 +
119888minus1
sum
119895=1
119895minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
+
119888minus1
prod
119896=1
120573 + 119896120574120583
119896120574120583[(120573 + 119888120574120583
119888120574120583)
119904+1minus119888
minus 1]
+ 119876(120573 + 119888120574120583
119888120574120583)
119904+1minus119888 119888minus1
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
minus119904120573
119888120574120583
[
[
1 +
119904minus2
sum
119895=0
119895
prod
119896=0
120573 + 119896120574120583
(119896 + 1) 120574120583
]
]
+1205732
119888120574120583
[
[
1 +
119904minus2
sum
119895=1
119895
prod
119896=1
120573 + 119896120574120583
119896120574120583
]
]
minus1
(61)
62 Conditional Probability Distribution of the Number ofCustomers Given the Number of Items in the Inventory Let119901119894 119894 ge 0 denote the probability that there are 119894 customers in
the system conditioned on the inventory level at 119895 We havethree different cases
(i) When 119895 = 0
119901119894=
120583120574
120582 + 120583 + 120573
times Prob[1 item in inventory 1 customer
and the inventory is served]
+ Prob[No item in inventory
119894 customers in the system]
=120583120574
120582 + 120583 + 120573119910119894+1(1) + 119910
119894(0) 119894 ge 0
(62)
(ii) When 0 lt 119895 lt 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895) 1 le 119894 lt 119895
120582
120582 + 119895120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119895 + 1) 120583120574
120582 + (119895 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119895120583 (1 minus 120574)
120582 + 119895120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119895120583
120582 + 119895120583 + 1205731[119895le119904]
]119910119894(119895) 119894 ge 119895
(63)
The first term on the right hand side of the case of 1 le 119894 le119895 in (63) has two factors the former represent probability ofan arrival before service completion as well as replenishmentwhen therewere 119894minus1 customers and 119895 inventory in the systemSimilar explanations stand for the remaining terms and alsofor other expressions for 119901
119894
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 13
(iii) When 119895 ge 119888
119901119894=
120583120574
120582 + 120583 + 1205731[119895+1le119904]
1199101(119895 + 1)
+120583 (1 minus 120574)
120582 + 120583 + 1205731[119895le119904]
1199101(119895)
+[1 minus120582
120582 + 1205731[119895le119904]
]1199100(119895)
+
1205731[119895gt119876]
120582 + 1205731[119895gt119876]
1199100(119895 minus 119876) 119894 = 0
120582
120582 + (119894 minus 1) 120583 + 1205731[119895le119904]
119910119894minus1(119895)
+(119894 + 1) 120583120574
120582 + (119894 + 1) 120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+(119894 + 1) 120583 (1 minus 120574)
120582 + (119894 + 1) 120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119894120583
120582 + 119894120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119894120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 1 le 119894 lt 119888
120582
120582 + 119888120583 + 1205731[119895le119904]
119910119894minus1(119895)
+119888120583120574
120582 + 119888120583 + 1205731[119895+1le119904]
119910119894+1(119895 + 1)
+119888120583 (1 minus 120574)
120582 + 119888120583 + 1205731[119895le119904]
119910119894+1(119895)
+[1 minus120582 + 119888120583
120582 + 119888120583 + 1205731[119895le119904]
]119910119894(119895)
+
1205731[119895gt119876]
120582 + 119888120583 + 1205731[119895gt119876]
119910119894(119895 minus 119876) 119894 ge 119888
(64)
where 1[119895le119904]
and 1[119895gt119876]
indicate whether the replenishmentprocess is on
63 Performance Measures
(i) Mean number of customers in the system is 119871119904=
suminfin
119894=1sum119876+119904
119895=0119894119910119894(119895)
(ii) Mean number of customers in the queue is 119871119902=
suminfin
119894=119888+1sum119876+119904
119895=0(119894 minus 119888)119910
119894(119895)
(iii) Mean inventory level in the system is 119868119898
=
suminfin
119894=0sum119876+119904
119895=1119895119910119894(119895)
(iv) Mean number of busy servers is as follows
119875BS =119888
sum
119896=1
119896[
[
infin
sum
119894=119896+1
119910119894(119896) +
119876+119904
sum
119895=119896+1
119910119896(119895) + 119910
119896(119896)]
]
(65)
(v) Mean number of idle servers is 119875IS = (119888 minus suminfin
119894=0119910119894(0))
(vi) Depletion rate of inventory is 119863inv =
120574120583(suminfin
119894=1sum119876+119904
119895=1119910119894(119895))
(vii) Mean number of replenishments per time unit is119877119903=
120573(suminfin
119894=0sum119904
119895=0119910119894(119895))
(viii) Mean number of departures per unit time is as fol-lows
119863119898=
119888minus1
sum
119896=1
[
[
119896120583(
infin
sum
119894=119896
119910119894(119896) +
119876+119904
sum
119895=119896
119910119896(119895))]
]
+ 119888120583[
[
infin
sum
119894=119888
119876+119904
sum
119895=119888
119910119894(119895)]
]
(66)
(ix) Expected loss rate of customers is 119864loss =
120582(suminfin
119894=0119910119894(0))
(x) Expected loss rate of customers when the inventorylevel is zero per cycle is 119864119888loss = 119864loss119877119903
(xi) Mean number of customers arriving per unit time isas follows
120582119860= 120582(
infin
sum
119894=0
119876+119904
sum
119895=1
119910119894(119895)) (67)
(xii) Mean sojourn time of the customers in the system is119882119904= 119871119904120582119860
(xiii) Mean waiting time of a customer in the queue is119882119902=
119871119902120582119860
(xiv) Mean number of customers waiting in the systemwhen inventory is available is = sum
infin
119894=1sum119876+119904
119895=1119894119910119894(119895)
(xv) Mean number of customers waiting in the systemduring the stock out period is 119882 = sum
infin
119894=1119894119910119894(0)
7 Analysis of Inventory Cycle Time
We define the inventory cycle time random variable Γcycle asthe time interval between two consecutive instants at whichthe inventory level drops to 119904 Thus Γcycle is a random variablewhose distribution depends on the number of customers atthe time when inventory level dropped to 119904 at the beginningof the cycle and the inventory level process prior to replen-ishment We proceed with the assumption that 120574 = 1 If thenumber of customers present in the system is at least 119876 + 119888when the order for replenishment is placed then we neednot have to look at future arrivals to get a nice form for thecycle time distribution In fact it is sufficient that there areat least 119876 customers at that epoch However in this case theservice rate during lead time may drop below 119888120583 even whenthere are at least 119888 items in the inventory This is so sincenumber of customersmay go below 119888Thuswe look at variouspossibilities below
71 When the Number of Customers ℓ ge 119876+ 119888 When thenumber of customers is at least 119876 + 119888 future arrivals neednot be considered The service rate of the119872119872119888 queueing-inventory system depends on the number of customersnumber of servers and number of items in the inventoryThuswe consider the following cases
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Advances in Operations Research
Case 1 (replenishment occurs before inventory level hits 119888minus1)We consider the state (ℓ 119904) as the starting state thus theinventory level decreases from 119904 to a particular level 119904 minus 119896 for119896 varying from 0 to 119904 minus 119888 due to service completion at rate 119888120583during the lead time At level 119904 minus 119896 the replenishment occursand it is absorbed to Δ
1 where the absorbing state is defined
as Δ1 = (ℓminus119896 119876+119904minus119896) | 0 le 119896 le 119904minus119888Therefore the time
until absorption to Δ1 follows Erlang distribution of order
119896 with parameter 119888120583 and it is denoted by 119864(119888120583 119896) Now thenumber of customers in the system is ℓ minus 119896 or larger with thecorresponding inventory level119876+119904minus119896 for 119896 varying from 0 to119904minus119888 Similarly the inventory level reaches 119904 from119876+119904minus119896with119876minus119896 service completions all of which have rate 119888120583 This timeduration also follows Erlang distribution of order119876minus119896Writethis as 119864(119888120583 119876 minus 119896) Thus under the condition that there areat least119876+ 119888 customers at the beginning of the cycle and thatthe inventory level does not fall below 119888 the inventory cycletime Γcycle has Erlang distribution of order119876with parameter119888120583 That is
Γcycle sim 119864 (119888120583 119896) lowast 119864 (119888120583 119876 minus 119896)
sim 119864(119888120583 119876)
(68)
where the symbol ldquosimrdquo stands for ldquohaving distributionrdquo Theprobability of replenishment taking place before inventorylevel that drops to 119888 minus 1 is given by intinfin
0sum119904minus119888
119896=0(119890minus120583V(120583V)119896120573119890minus120573V
119896)119889V
Case 2 (replenishment after hitting 119888 minus 1 but not zero) Theinventory level decreases from 119904 to 119896 when 119896 varies from 1 to119888minus1Thefirst 119904minus119888+1 services are at the same rate 119888120583Thereafterit slows down to (119888minus1)120583 andfinally to 119896120583 when replenishmentoccurs Consequently the inventory level rises to 119876 + 119896Now here onwards the service rate stays at 119888120583 Thus in thecycle the distribution of the time until replenishment takesplace is the convolution of generalized Erlang distributionand that of an Erlang distribution 119864(119876 + 119896 minus 119904 119888120583) Theconditional distribution of replenishment realization after119904minus119896minus1 service is completed but before (119904minus119896)th is completedcan be computed as in Case 1 At the same level 119904 minus 119896 thereplenishment will occur and it is absorbed to Δ
2 where
the absorbing state is defined as Δ2 = (ℓ minus (119904 minus 119896) 119876 +
119896) | 1 le 119896 le 119888 minus 1 Thus the time until absorption toΔ2 follows generalized Erlang distribution with parameters
119888120583 (119888minus1)120583 (119896+1)120583 of order 119904minus119896 and 119896 vary from 1 to 119888minus1It is denoted byG119864(119888120583 (119888minus1)120583 (119896+1)120583 119904minus119896)Then fromΔ2 the inventory level reaches 119904 due to service completion
with parameter 119888120583 Thus the time duration follows Erlangdistribution with parameter 119888120583 of order 119876 + 119896 minus 119904 with 119896varying from 1 to 119888 minus 1 That is 119864(119888120583 119876 + 119896 minus 119904) Hencethe inventory cycle time Γcycle follows generalized Erlangdistribution of order 119876 Therefore Γcycle is defined as
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1)120583 (119896 + 1)120583 119904 minus 119896)
lowast 119864(119888120583 119876 + 119896 minus 119904)
(69)
whereG119864(sdot) stands for generalized Erlang distribution
Case 3 (replenishment after inventory level reaching zero)Then the inventory level reaches 0 from the level 119904 due toservice completion with parameters 119888120583 (repeated 119904 minus 119888 + 1times) Thus the time until absorption to Δ
3 = (ℓ minus
119904 119876) follows generalized Erlang distribution of order 119904 andparameters 119888120583 (119888minus1)120583 120583When the inventory level hits 0the system becomes idle for a random duration of time whichfollows exponential distribution with parameter 120573 Afterreplenishment the system starts service and consequently theinventory level reaches 119904 from 119876 due to service completionwith parameter 119888120583 This part has Erlang distribution withparameter 119888120583 and order119876minus 119904 Thus Γcycle follows generalizedErlang distribution of order 119876 That is
Γcycle sim G119864(119888120583 119888120583⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
119904minus119888+1 times (119888 minus 1) 120583 120583 119904)
lowast exp(120573) lowast 119864(119888120583 119876 minus 119904)
(70)
The cases we are going to consider hereafter result in cycletime distribution that are phase type with not necessarilyunique representation However one can sort out the problemof minimal representation Obviously this is the one whichconsiders that many arrivals are needed to have exactly 119876services in this cycle
72 When the Number of Customers ℓ lt 119876+ 119888 In this case wemay have to consider future arrivals as well since numberof customers available at the start of the cycle may be suchthat the service rate falls below 119888120583 Thus the cycle time willhavemore general distribution namely the phase typeWe goabout doing this Our procedure is such that the moment wehave enough customers to serve during the remaining part ofthe cycle we stop considering future arrivals Thus considera Markov chain on the state space
(119904 ℓ) (119904 minus 1 ℓ minus 1) (0 ℓ minus 119904) (119904 ℓ + 1)
(119904 minus 1 ℓ) (0 ℓ minus 119904 + 1) (119904 + 119876 ℓ)
(119904 + 119876 minus ℓ 0) (119904 + 119876 ℓ + 1) (119904 ℓ)
(119904 + 119876 119904 + 119876 minus ℓ minus 1) (119904 + 119876 minus 1 ℓ minus 1)
(119904 + 119876 119904 + 119876 minus ℓ) (119904 119904 minus ℓ minus 1)
(71)
The initial state (119904 ℓ) thus the initial probability vector willhave one at the position corresponding to (119904 ℓ) and the restof the elements zero The absorption state in this Markovchain is (119904 lowast) where lowast belongs to 0 1 2 119876 + ℓ minus 119904 andis a departure epoch Let T be the block with transitionsamong transient states and letTlowast be the column vector withtransition rates to the absorbing states as elements Thenthe cycle time has distribution 1 minus 120572119890T119905e where 120572 is theinitial probability vector with 1 at the position indicatingthe inventory level 119904 as first coordinate and the numberof customers (=ℓ) at the beginning of the cycle as secondcoordinate Note that the phase type representation obtained
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 15
Table 2 Optimal server 119888 and minimum cost
120574 01 02 03 04 05 06 07 08 09 1
Optimal 119888 andminimum cost
6 6 6 6 6 6 6 6 6 614878 16636 18393 20151 21909 23666 25424 27182 28940 30698
Table 3 Optimal (119904 119876) values and minimum cost
119888120574
01 02 03 04 05 06 07 08 09 1
3 (4 12) (4 15) (4 19) (4 23) (4 27) (4 31) (4 37) (4 39) (4 42) (4 45)95857 11357 13186 15149 17155 19159 21150 23082 24987 26855
4 (5 15) (5 19) (5 23) (5 27) (5 31) (5 34) (5 38) (5 41) (5 45) (5 48)12643 14725 16670 18606 20542 22469 24374 26255 28105 29925
5 (6 16) (6 22) (6 26) (6 31) (6 34) (6 38) (6 42) (6 45) (6 48) (6 52)15411 17753 19843 21840 23791 25707 27593 29449 31275 33072
6 (7 16) (7 23) (7 28) (7 32) (7 36) (7 40) (7 43) (7 46) (7 49) (7 53)17790 20228 22379 24406 26365 28279 30156 32000 33813 35597
7 (8 16) (8 23) (8 28) (8 32) (8 36) (8 40) (8 43) (8 46) (8 49) (8 53)20030 22490 24658 26691 28647 30552 32418 34249 36051 37824
8 (9 16) (9 23) (9 28) (9 32) (9 36) (9 40) (9 43) (9 46) (9 49) (9 53)22230 24698 26868 28897 30845 32901 34592 36412 38202 39965
9 (10 16) (10 23) (10 28) (10 32) (10 36) (10 40) (10 43) (10 46) (10 49) (10 53)24429 26897 29065 31087 33025 34908 36749 38557 40336 42089
10 (11 16) (11 23) (11 28) (11 32) (11 36) (11 40) (11 43) (11 46) (11 49) (11 53)26627 29094 31260 33276 35206 37078 38908 40705 42473 44217
is not unique since the service rate strongly depends onboth inventory level and number of customers in the systemThe case of ℓ lt 119904 here again the procedure is similar tothat corresponding to ℓ ge 119904 but less than 119876 + 119888 Theinitial state is (119904 ℓ) After exactly 119876 service completionswith a replenishment within this cycle and with arrivalstruncated at that epoch which ensure rate 119888120583 for as manyservices as possible the absorption state of the Markov chaingenerated corresponds to a departure epoch with 119904 items inthe inventory Here again the cycle time has a PH distributionwith representation which is not unique because the servicerates may change depending on the number of customers inthe system and the number of items in the inventory
8 Optimization Problem II
We look for the optimal pair of control variables in the modeldiscussed above Now for computing the minimal cost of(119904 119876)model we introduce the cost functionF(119888 119904 119876) whichis defined by
F(119888 119904 119876) = ℎ sdot 119868119898+ 1198881sdot 119864loss + 1198882 sdot
119882 + (119870 + 119876 sdot 119888
3) sdot 119877119903
+ 1198884sdot 119875BS + 1198885 sdot (119888 minus 119875BS)
(72)
where 119904 = 40 119878 = 81 and 119870 1198881 1198882 1198883 1198884 1198885 and ℎ are the
same input parameters as described in Section 4 We provideoptimal 119888 and corresponding minimum cost for various 120574values From Table 2 we notice that the optimal value of 119888 is 6for various 120574 values presumably because of the high holdingcost
In Table 3 we examine the optimal pair (119904 119876) and thecorresponding minimum cost for various 120574 and 119888 keepingother parameters fixed (as in Section 4)
9 Conclusions
In this paper we studied multiserver queueing-inventorysystem with positive service time First we considered twoserver queueing-inventory systems where the steady-statedistributions are obtained in product form Further we anal-yse queueing-inventory system with more than two serversAs observed in [21] by Falin and Templeton we conjecturethat119872119872119888 for 119888 ge 3 queueing-inventory system does nothave analytical solution So such cases are analyzed by algo-rithmic approach Conditional distribution of the inventorylevel conditioned on the number of customers in the systemand conditional distribution of the number of customersconditioned on the inventory level are derived We alsoprovided the cycle time distribution of two consecutive 119904 to119904 transitions of the inventory level (ie the first return time
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Advances in Operations Research
to 119904) We have computed the optimal number of servers to beemployed and also computed optimal (119904 119876) pair values andthe corresponding minimum cost
Notations
The following notations and abbreviations are used in thesequel
N(119905) Number of customers in the systemat time 119905
I(119905) Inventory level in the system at time 119905e (1 1 1)
1015840 a column vector of 1rsquos ofappropriate order
CTMC Continuous time Markov chainLIQBD Level independent quasi-birth-death
process
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors thank the anonymous reviewers and the editorfor helpful comments that improved the quality of thepaper This research is supported by Kerala State Council forScience Technology amp Environment to A Krishnamoorthyand Dhanya Shajin (no 001KESS2013CSTE) and by theUniversity Grants Commission Government of India underDr D S Kothari Postdoctoral Fellowship Programme to RManikandan
References
[1] K Sigman and D Simchi-Levi ldquoLight traffic heuristic for anMG1 queue with limited inventoryrdquo Annals of OperationsResearch vol 40 no 1 pp 371ndash380 1992
[2] M Saffari S Asmussen and R Haji ldquoThe MM1 queue withinventory lost sale and general lead timesrdquo Queueing SystemsTheory and Applications vol 75 no 1 pp 65ndash77 2013
[3] O Berman E H Kaplan and D G Shimshak ldquoDeterministicapproximations for inventory management at service facilitiesrdquoIIE Transactions vol 25 no 5 pp 98ndash104 1993
[4] A Krishnamoorthy B Lakshmy and RManikandan ldquoA surveyon inventory models with positive service timerdquo OPSEARCHvol 48 no 2 pp 153ndash169 2011
[5] M Schwarz C Sauer H Daduna R Kulik and R SzeklildquoMM1 queueing systems with inventoryrdquo Queueing SystemsTheory and Applications vol 54 no 1 pp 55ndash78 2006
[6] A Krishnamoorthy and N C Viswanath ldquoStochastic decom-position in production inventory with service timerdquo EuropeanJournal of Operational Research vol 228 no 2 pp 358ndash3662013
[7] B Sivakumar and G Arivarignan ldquoA stochastic inventorysystem with postponed demandsrdquo Performance Evaluation vol66 no 1 pp 47ndash58 2009
[8] A Krishnamoorthy and V C Narayanan ldquoProduction inven-tory with service time and vacation to the serverrdquo IMA Journalof Management Mathematics vol 22 no 1 pp 33ndash45 2011
[9] T G Deepak A Krishnamoorthy V C Narayanan and KVineetha ldquoInventory with service time and transfer of cus-tomers andinventoryrdquo Annals of Operations Research vol 160pp 191ndash213 2008
[10] M Schwarz and H Daduna ldquoQueueing systems with inventorymanagement with random lead times and with backorderingrdquoMathematical Methods of Operations Research vol 64 no 3 pp383ndash414 2006
[11] M Schwarz C Wichelhaus and H Daduna ldquoProduct formmodels for queueing networks with an inventoryrdquo StochasticModels vol 23 no 4 pp 627ndash663 2007
[12] R Krenzler and H Daduna Loss Systems in a RandomEnvironment-Embedded Markov Chains Analysis UniversitatHamburg Hamburg Germany 2013
[13] A Krishnamoorthy S S Nair and V C Narayanan ldquoProduc-tion inventory with service time and interruptionsrdquo Interna-tional Journal of Systems Science 2013
[14] M Saffari R Haji and F Hassanzadeh ldquoA queueing systemwith inventory andmixed exponentially distributed lead timesrdquoInternational Journal of Advanced Manufacturing Technologyvol 53 pp 1231ndash1237 2011
[15] R Krenzler and H Daduna ldquoLoss systems in a randomenvironment steady state analysisrdquo Queueing Systems Theoryand Applications 2014
[16] A N Nair M J Jacob and A Krishnamoorthy ldquoThe multiserver MM(sS) queueing inventory systemrdquo Annals of Oper-ations Research 2013
[17] V S S Yadavalli B SivakumarGArivarignan andOAdetunjildquoA finite source multi-server inventory system with servicefacilityrdquo Computers Industrial Engineering vol 63 pp 739ndash7532012
[18] A Krishnamoorthy R Manikandan and B Lakshmy ldquoArevisit to queueing-inventory systemwith positive service timerdquoAnnals of Operations Research 2013
[19] M F Neuts Matrix-Geometric Solutions in Stochastic ModelsAn Algorithmic Approach Dover New York NY USA 2ndedition 1994
[20] G Latouche and V Ramaswami ldquoA logarithmic reduction algo-rithm for quasi-birth-death processesrdquo Journal of Applied Prob-ability vol 30 no 3 pp 650ndash674 1993
[21] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of