research article a two-stage model queueing with no waiting...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 679369 5 pageshttpdxdoiorg1011552013679369

Research ArticleA Two-Stage Model Queueing with No WaitingLine between Channels

Vedat SaLlam1 and Muumljgan Zobu2

1 Department of Statistics Ondokuz Mayıs University 55139 Samsun Turkey2Department of Statistics Amasya University 05100 Amasya Turkey

Correspondence should be addressed to Mujgan Zobu mujganzobuhotmailcom

Received 5 December 2012 Revised 22 April 2013 Accepted 20 May 2013

Academic Editor Suiyang Khoo

Copyright copy 2013 V Saglam and M Zobu 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 consider a new queuing model with sequential two stations (stages) single server at each station where no queue is allowed atstation 2 andwith no restriction at station 1There is a FCFS service discipline inwhich the input stream is Poisson having rate 120582Theservice time of any customer at server i (119894 = 1 2) is exponential with parameter 120583

119894The state probabilities and loss probability of this

model are givenThe performance measures are obtained and optimized and additionally the model is simulatedThe simulationresults exact results and optimal results of the performance measures are numerically computed for different parameters

1 Introduction

New models of queuing theory have been needed latelyconcerning the developments in areas such as productionline communication and computer systems One of thesenecessary models is for a tandem queuing system Manyimportant studies have been done in this area The meanwaiting time and the mean customer number at two tandemchannels (servers) which are Poisson arrival and exponentialservice time were given in [1] The mean customer numberdistribution of waiting time and the probabilities of variousnumbers of the tandem queuing system at every stage ofthe Poisson arrival and the exponential service time of thetandem queuing system were found in [2] In [3] it wasproved that if arrivals to the system are Poisson process withthe parameter then the output of this system is also Poissonprocess with a parameter 120582 A more complicated examplewith network analysis was studied in [4] In queuing theoryit is usually assumed that service channels are homogeneousHowever it is seen that in real queuing systems servicechannels sometimes are heterogeneous Understanding suchsystems is important for finding solutions to both theoreticaland technical problems Under the condition that the sum

of service rates is fixed homogeneous systems have beencompared with the heterogeneous systems for performancemeasures in [5ndash9] The measures of effectiveness for tandemqueueswith blockingwere calculated according to an approx-imation method and simulated in [10] The tandem queueswith one server in the first queue and 119899 ge 1 servers in thesecond queue where the arrivals to the system with Poissonprocess having parameter 120582 and there is no waiting roombetween the two stages were analyzed and some probabilitiesfor the number of customers were found in [11] In theliterature usually for some similar models to ours blockedtandemqueueing systems have been studied and probabilitiesof number of customers have been obtained Approximateand simulation results of performance measures have beenobtained In our proposed model there is no restriction forthe first stage there is no waiting room between both stagesand no blocking (ie customers leave the system after hav-ing service in the first stage if the second stage is busy)Probabilities of being-nonbeing of customers in the first andsecond stages performancemeasures and the optimal valuesof these measures are theoretically obtained by analyzing ourproposed method Also we have compared some results byobtaining simulation results

2 Mathematical Problems in Engineering

In real life while there is waiting case in the service sys-tems because of obligations and urgency and unavailabilityof desired features the loss may occur in the beginning ofsecond stage as in our proposed model Therefore we havedecided to construct such a model and analyze itThe follow-ing two examples can be proposed as potential applicationsof our model under some conditions

The following examples can be proposed as the applica-tions of our model on some topics

(a) Fruit-Vegetable Packing Line There is a company whichexports fresh fruits and vegetables This company has aproduct packing linewhich consists of two stages for a specialexport fruit The product comes to first stage for qualitycontrol Later if the product is not in the desirable size orquality then it is taken away from the system (the loss occurs)The product is sent to the second stage to be packed with nowaiting time if it qualifies the desirable size and quality

(b) VoIP (Voice Over Internet Protocol) Internet telephonyrefers to communications servicesmdashvoice fax SMS andorvoice-messaging applicationsmdashthat are transported via an IPnetwork rather than the public switched telephone network(PSTN) The steps involved in originating a VoIP telephonecall are signalling andmedia channel setup digitization of theanalogy voice signal encoding packetization and transmis-sion as Internet Protocol (IP) packets over a packet-switchednetwork On the receiving side similar steps (usually in thereverse order) such as reception of the IP packets decodingof the packets and digital-to-analogy conversion reproducethe original voice stream

More generally the server transfers voice messaging tothe recipient In this system the server is first service therecipient is second service If the recipient is busy and thenthe server destroys voice messaging at that moment

2 The Model

Let 1198991and 119899

2be the number of customers in the first and

second stages respectively at any time of 119905 including thosebeing served where 119899

1= 0 1 2 119899

2= 0 1

For stage 119894 let 120585119894(119905) be defined as follows

120585119894 (119905) =

1198991 119894 = 1

1198992 119894 = 2

(1)

We can denote

11987511989911198992(119905) = Prob 120585

1 (119905) = 1198991 1205852 (119905) = 1198992 (2)

The random process

120585119894 (119905) 119894 = 1 2 119905 ge 0 (3)

is a continuous-time two-dimensionalMarkov chain For 119899 ge1 the state space of this chain becomes

119864 = (0 0) (0 1) (119899 0) (119899 0) (4)

We wish to find the steady-state probability 11990111989911198992

11990111989911198992

= lim119905rarrinfin

11987511989911198992(119905) (5)

The usual procedure leads to the steady-state equations forthis Markov chain

0 = minus12058211990100+ 120583211990110 (6)

0 = minus (120582 + 1205832) 11990101+ 120583111990110+ 120583111990111 (7)

0 = minus (120582 + 1205831) 11990111989910+ 1205821199011198991minus10

+ 120583211990111989911 1198991ge 1 (8)

0 = minus (120582 + 1205831+ 1205832) 11990111989911+ 1205821199011198991minus11

+ 1205831(1199011198991+10

+ 1199011198991+10) 119899

1ge 1

(9)

We define 120588119894= 120582120583

119894for 119894 = 1 2 and 120588 = 120582(120583

1+ 1205832)

21 State Probabilities The probability 1199011198991

denotes the prob-ability of finding 119899

1customers in the first stage at an arbitrary

point in time (see [12]) We can write these as

1199011198991

= 11990111989910+ 11990111989911= 1205881198991

1(1 minus 120588

1) 119899

1ge 0 (10)

11990100+ 11990101= 1 minus 120588

1 (11)

Using this equation and (6) we obtain the following

11990101=120582 (1 minus 120588

1)

120582 + 1205832

=(1 minus 120588

1) 1205882

1 + 1205882

11990100=1205832

12058211990101=1 minus 1205881

1 + 1205882

(12)

By substituting the expression 1199011198991+1

in (9) we get

(120582 + 1205831+ 1205832) 11990111989911= 1205821199011198991minus11

+ 12058311199011198991+1 1198991ge 1 (13)

Let us take 119886 = 120588(1 + 120588) and choose 11990111989911a place to put 119910

1198991

in (13) for simplicity In this case the following equation canbe obtained

1199101198991

= 1198861199101198991minus1+ 1198861199101198991

1198991ge 1 (14)

Both sides of (14) are divided into 1198861198991 and later the index 1198991

is changed to 119896 Then we sum this obtained value

1199101198991

1198861198991minus 1199100=

1198991

sum

119896=1

119901119896

119886119896minus1=1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

(1 minus (1205881119886)) 1198861198991

(15)

11990111989911minus 119886119899111990101= 119886

1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

119886 minus 1205881

(16)

11990111989911= 119886

1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

119886 minus 1205881

+(1 minus 120588

1) 1205882

1 + 1205882

1198861198991 (17)

Mathematical Problems in Engineering 3

When the value of 119886 in (17) is substituted the following equa-tion is obtained

11990111989911=(1 minus 120588

1) 1205882

1 + 1205882

1205881198991

1 (18)

We get the probability 11990111989910from (10) and (18)

11990111989910=(1 minus 120588

1)

1 + 1205882

1205881198991

1 (19)

22 Loss Probability The customerrsquos loss probability is givenas

119901119871=

infin

sum

1198991=0

11990111989911=

1205882

1 + 1205882

(20)

In other way the formula (20) can be obtained from ldquoErlangrsquosloss formulardquo or ldquoErlangrsquos B formulardquo for the119872119872119888119888 queueIn [13] it is denoted as 119861(119888 120588

2) and formulated as the follows

119861 (119888 1205882) =

1205882119888

sum119888

119899=01205881198992119899 (21)

where 1205882is the utilization factor Substituting 119888 = 1 in (21)

we have (20)

3 The Measures of Performance andOptimization of Performance Measures

31 TheMean Sojourn Time Let 119879 be a random variable thatdescribes the sojourn time of customers in the system Usingthe law of total expectation we can write is as follows

119864 (119879) = 119864 (119879 | 119860) 119875 (119860) + 119864 (119879 | 119860)119875 (119860) (22)

where 119875(119860) is the probability of the loss of a customer Nowit is clear that

119864 (119879 | 119860) =1

1205831minus 120582

119864 (119879 | 119860) =1

1205831minus 120582

+1

1205832

(23)

Thus

119864 (119879) =1205831+ 1205832

(1205831minus 120582) (120583

2+ 120582)

(24)

Our main results about the problem of minimizing the meansojourn time can be explained by the following theorem

Theorem 1 If sum of two service rates 1205831+ 1205832= 120583 is fixed

then the mean sojourn time of this tandem system attains itsminimum value for 120583

1= 1205832 + 120582 and 120583

2= 1205832 minus 120582

Proof We will prove the theorem by using the followinginequality

(

119898

prod

I=1

119886119894)

1119898

le1

119898

119898

sum

119894=1

119886119894 forall119886

119894gt 0 forall119898 isin Z

+ (25)

From inequality (25) we have

1

(1205831minus 120582) (120583

2+ 120582)

ge4

1205832 (26)

If we replace the expressions 1205831+1205832= 120583 and 41205832 in equality

(24) we obtain the minimum value of 119864(119879) as follows

min119864 (119879) = 4

120583 (27)

where the equality (27) is provided with 1205831= 1205832 + 120582 and

1205832= 1205832 minus 120582

32 The Mean Number of Customers Let 119873 be the randomvariable that describes the number of customers in the sys-tem

119864 (119873) =

infin

sum

1198991=0

1

sum

1198992=0

(1198991+ 1198992) 11990111989911198992

=120582 (1205831+ 1205832)

(1205831minus 120582) (120583

2+ 120582)

(28a)

or

119864 (119873) = 120582119864 (119879) (28b)

The mean number of customers in the system is opti-mized fromTheorem 1 and the equality (28b) as below

min119864 (119873) = 120582min119864 (119879) = 4120582120583 (29)

The independence of the number of customers can beexpressed by the following theorem

Theorem 2 If the random variables 1198731and 119873

2are taken as

the number of customers in the first and second stages respec-tively then119873

1and119873

2are independent random variables

Proof The joint probability mass functions of 1198731and 119873

2

random variables is

11990111989911198992

= 119875 (1198731= 1198991 1198732= 1198992) 119899

1= 0 1 2 119899

2= 0 1

(30)

If 1198992= 0 and (119899

2= 1) are substituted in (30) the equations

(19) and (18) are obtained respectivelyThe marginal probability mass function of119873

1is given as

(10)


Table 1 For 120582 = 030 1205831= 080 120583

2= 100

Iterationnumber

Simulation results Exact results Optimal results119864(119879) 119864(119873) 119864(119879) 119864(119873) 119864(119879) 119864(119873)

100 28163 08452 27692 08308 22222 066671000 28111 08433 27692 08308 22222 066675000 28124 08436 27692 08308 22222 06667

Table 2 For 120582 = 01 1205831= 09 120583

2= 09

Iterationnumber


100 22661 02263 22500 02250 22220 022221000 22599 02261 22500 02250 22220 022225000 22500 02260 22500 02250 22220 02222

Table 3 For 120582 = 001 1205831= 06 120583

2= 07

Iterationnumber


100 31042 00311 31034 00310 30769 003081000 31034 00310 31034 00310 30769 003085000 31043 00310 31034 00310 30769 00308

Themarginal probability mass function of1198732is obtained

from Erlangrsquos B formula or (20)

1199011015840

0= 119875 (119873

2= 0) =

1

1 + 1205882

(31)

1199011015840

1= 119875 (119873

2= 1) =

1205882

1 + 1205882

(32)

11990111989911198992

= 1199011198991

1199011015840

1198992

1198991= 0 1 2 119899

2= 0 1 (33)

If (10) and (31) are substituted in (30) the equation (19) isobtained and if (10) and (32) are substituted in (30) theequation (18) is obtained Thus the independence of 119873

1and

1198732has been demonstrated

4 Numerical Results

The random arrivals and service times were generated fromexponential distribution as seconds by using MATLAB 7100(R2010a) programming for this proposedmodelThe numberof customers taken was 10000 and was performed in threeiterations Performance measures are calculated for differentvalues of 120588

119894(119894 = 1 2) and three different iterations steps that

is 100 1000 and 5000 These results were shown in Tables 12 and 3

5 Conclusions

A new queuing discipline is given for a Markov model whichconsists of two consecutive channels and no waiting line

between channels In this model steady-state equations themean sojourn time the mean number of customers andloss probability are obtained Additionally two theorems aregivenwhich are about optimization of performancemeasuresand the independent of the number of customers respec-tively Performance measures are calculated for different val-ues of 120588

119894(119894 = 1 2) and for three different iterations steps that

is 100 1000 and 5000 Moreover results of these measuresare compared in the tables above It has been seen thatthe simulation results approximated the theoretical resultsAlthough the iteration number is increased the simulationresults of performance measures have not changed Howeveras 120588119894(119894 = 1 2) converges to zero both the simulation results

and exact results approximately are equal to optimal resultsThus it is said that our proposed queueing model operateswell

For further research a model in which a customer whocompleted his service in channel 1 blocks channel 1 withprobability 120587 or leaves the systemwith probability 1minus120587 whilechannel 2 is busy can be studied

References

[1] G G OrsquoBrien ldquoThe solution of some queuing problemsrdquo Jour-nal of the Society For Industrial and Applied Mathematics vol 2pp 133ndash142 1954

[2] R R P Jackson ldquoQueuing system with phase-type servicerdquoOperational Research Quarterly vol 5 pp 109ndash120 1954

[3] P J Burke ldquoThe output of a queuing systemrdquo OperationsResearch vol 4 pp 699ndash704 1956


[4] J R Jackson ldquoNetwork of waiting linesrdquo Operation Researchvol 5 pp 518ndash524 1957

[5] R B Cooper Introduction to Queuing Theory The MacmillanCompany New York NY USA 1972

[6] D Gross and C M Harris Fundamentals of Queueing TheoryJohn Wiley amp Sons New York NY USA 3rd edition 1998

[7] V Saglam and H Torun ldquoOn optimization of stochastic servicesystemwith two heterogeneous channelsrdquo International Journalof Applied Mathematics vol 17 no 1 pp 1ndash6 2005

[8] V Saglam and A Shahbazov ldquoMinimizing loss probability inqueuing systems with heterogeneous serversrdquo Iranian Journalof Science and Technology vol 31 no 2 pp 199ndash206 2007

[9] A H Taha Operating Research Macmillian Publishing NewYork NY USA 1982

[10] A Brandwajn and Y-L L Jow ldquoAn approximation method fortandem queues with blockingrdquoOperations Research vol 36 no1 pp 73ndash83 1988

[11] A A Akinsete ldquoBlocked network of tandem queues with with-drawalrdquo Kragujevac Journal of Mathematics vol 23 pp 63ndash732001

[12] U N Bhat An Introduction to Queueing Theory Boston MassUSA 2008

[13] W J Stewart Probability Markov Chains Queues and Simula-tion Princeton University Press Princeton NJ USA 2009

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Stochastic AnalysisInternational Journal of


In real life while there is waiting case in the service sys-tems because of obligations and urgency and unavailabilityof desired features the loss may occur in the beginning ofsecond stage as in our proposed model Therefore we havedecided to construct such a model and analyze itThe follow-ing two examples can be proposed as potential applicationsof our model under some conditions

The following examples can be proposed as the applica-tions of our model on some topics

(a) Fruit-Vegetable Packing Line There is a company whichexports fresh fruits and vegetables This company has aproduct packing linewhich consists of two stages for a specialexport fruit The product comes to first stage for qualitycontrol Later if the product is not in the desirable size orquality then it is taken away from the system (the loss occurs)The product is sent to the second stage to be packed with nowaiting time if it qualifies the desirable size and quality

(b) VoIP (Voice Over Internet Protocol) Internet telephonyrefers to communications servicesmdashvoice fax SMS andorvoice-messaging applicationsmdashthat are transported via an IPnetwork rather than the public switched telephone network(PSTN) The steps involved in originating a VoIP telephonecall are signalling andmedia channel setup digitization of theanalogy voice signal encoding packetization and transmis-sion as Internet Protocol (IP) packets over a packet-switchednetwork On the receiving side similar steps (usually in thereverse order) such as reception of the IP packets decodingof the packets and digital-to-analogy conversion reproducethe original voice stream

More generally the server transfers voice messaging tothe recipient In this system the server is first service therecipient is second service If the recipient is busy and thenthe server destroys voice messaging at that moment

2 The Model

Let 1198991and 119899

2be the number of customers in the first and

second stages respectively at any time of 119905 including thosebeing served where 119899

1= 0 1 2 119899

2= 0 1

For stage 119894 let 120585119894(119905) be defined as follows

120585119894 (119905) =

1198991 119894 = 1

1198992 119894 = 2

(1)

We can denote

11987511989911198992(119905) = Prob 120585

1 (119905) = 1198991 1205852 (119905) = 1198992 (2)

The random process

120585119894 (119905) 119894 = 1 2 119905 ge 0 (3)

is a continuous-time two-dimensionalMarkov chain For 119899 ge1 the state space of this chain becomes

119864 = (0 0) (0 1) (119899 0) (119899 0) (4)

We wish to find the steady-state probability 11990111989911198992

11990111989911198992

= lim119905rarrinfin

11987511989911198992(119905) (5)

The usual procedure leads to the steady-state equations forthis Markov chain

0 = minus12058211990100+ 120583211990110 (6)

0 = minus (120582 + 1205832) 11990101+ 120583111990110+ 120583111990111 (7)

0 = minus (120582 + 1205831) 11990111989910+ 1205821199011198991minus10

+ 120583211990111989911 1198991ge 1 (8)

0 = minus (120582 + 1205831+ 1205832) 11990111989911+ 1205821199011198991minus11

+ 1205831(1199011198991+10

+ 1199011198991+10) 119899

1ge 1

(9)

We define 120588119894= 120582120583

119894for 119894 = 1 2 and 120588 = 120582(120583

1+ 1205832)

21 State Probabilities The probability 1199011198991

denotes the prob-ability of finding 119899

1customers in the first stage at an arbitrary

point in time (see [12]) We can write these as

1199011198991

= 11990111989910+ 11990111989911= 1205881198991

1(1 minus 120588

1) 119899

1ge 0 (10)

11990100+ 11990101= 1 minus 120588

1 (11)

Using this equation and (6) we obtain the following

11990101=120582 (1 minus 120588

1)

120582 + 1205832

=(1 minus 120588

1) 1205882

1 + 1205882

11990100=1205832

12058211990101=1 minus 1205881

1 + 1205882

(12)

By substituting the expression 1199011198991+1

in (9) we get

(120582 + 1205831+ 1205832) 11990111989911= 1205821199011198991minus11

+ 12058311199011198991+1 1198991ge 1 (13)

Let us take 119886 = 120588(1 + 120588) and choose 11990111989911a place to put 119910

1198991

in (13) for simplicity In this case the following equation canbe obtained

1199101198991

= 1198861199101198991minus1+ 1198861199101198991

1198991ge 1 (14)

Both sides of (14) are divided into 1198861198991 and later the index 1198991

is changed to 119896 Then we sum this obtained value

1199101198991

1198861198991minus 1199100=

1198991

sum

119896=1

119901119896

119886119896minus1=1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

(1 minus (1205881119886)) 1198861198991

(15)

11990111989911minus 119886119899111990101= 119886

1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

119886 minus 1205881

(16)

11990111989911= 119886

1205881(1 minus 120588

1) (1198861198991 minus 1205881198991

1)

119886 minus 1205881

+(1 minus 120588

1) 1205882

1 + 1205882

1198861198991 (17)



11990111989911=(1 minus 120588

1) 1205882

1 + 1205882

1205881198991

1 (18)


11990111989910=(1 minus 120588

1)

1 + 1205882

1205881198991

1 (19)


119901119871=

infin

sum

1198991=0

11990111989911=

1205882

1 + 1205882

(20)



119861 (119888 1205882) =

1205882119888

sum119888

119899=01205881198992119899 (21)


we have (20)



119864 (119879) = 119864 (119879 | 119860) 119875 (119860) + 119864 (119879 | 119860)119875 (119860) (22)


119864 (119879 | 119860) =1

1205831minus 120582

119864 (119879 | 119860) =1

1205831minus 120582

+1

1205832

(23)

Thus

119864 (119879) =1205831+ 1205832

(1205831minus 120582) (120583

2+ 120582)

(24)




1= 1205832 + 120582 and 120583

2= 1205832 minus 120582


(

119898

prod

I=1

119886119894)

1119898

le1

119898

119898

sum

119894=1

119886119894 forall119886


+ (25)


1

(1205831minus 120582) (120583

2+ 120582)

ge4

1205832 (26)



min119864 (119879) = 4

120583 (27)


1205832= 1205832 minus 120582


119864 (119873) =

infin

sum

1198991=0

1

sum

1198992=0

(1198991+ 1198992) 11990111989911198992

=120582 (1205831+ 1205832)

(1205831minus 120582) (120583

2+ 120582)

(28a)

or

119864 (119873) = 120582119864 (119879) (28b)


min119864 (119873) = 120582min119864 (119879) = 4120582120583 (29)



2are taken as


1and119873



2

random variables is

11990111989911198992

= 119875 (1198731= 1198991 1198732= 1198992) 119899

1= 0 1 2 119899

2= 0 1

(30)

If 1198992= 0 and (119899



1is given as

(10)


Table 1 For 120582 = 030 1205831= 080 120583

2= 100

Iterationnumber


100 28163 08452 27692 08308 22222 066671000 28111 08433 27692 08308 22222 066675000 28124 08436 27692 08308 22222 06667

Table 2 For 120582 = 01 1205831= 09 120583

2= 09

Iterationnumber


100 22661 02263 22500 02250 22220 022221000 22599 02261 22500 02250 22220 022225000 22500 02260 22500 02250 22220 02222

Table 3 For 120582 = 001 1205831= 06 120583

2= 07

Iterationnumber


100 31042 00311 31034 00310 30769 003081000 31034 00310 31034 00310 30769 003085000 31043 00310 31034 00310 30769 00308



1199011015840

0= 119875 (119873

2= 0) =

1

1 + 1205882

(31)

1199011015840

1= 119875 (119873

2= 1) =

1205882

1 + 1205882

(32)

11990111989911198992

= 1199011198991

1199011015840

1198992

1198991= 0 1 2 119899

2= 0 1 (33)


1and


4 Numerical Results




5 Conclusions







References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











11990111989911=(1 minus 120588

1) 1205882

1 + 1205882

1205881198991

1 (18)


11990111989910=(1 minus 120588

1)

1 + 1205882

1205881198991

1 (19)


119901119871=

infin

sum

1198991=0

11990111989911=

1205882

1 + 1205882

(20)



119861 (119888 1205882) =

1205882119888

sum119888

119899=01205881198992119899 (21)


we have (20)



119864 (119879) = 119864 (119879 | 119860) 119875 (119860) + 119864 (119879 | 119860)119875 (119860) (22)


119864 (119879 | 119860) =1

1205831minus 120582

119864 (119879 | 119860) =1

1205831minus 120582

+1

1205832

(23)

Thus

119864 (119879) =1205831+ 1205832

(1205831minus 120582) (120583

2+ 120582)

(24)




1= 1205832 + 120582 and 120583

2= 1205832 minus 120582


(

119898

prod

I=1

119886119894)

1119898

le1

119898

119898

sum

119894=1

119886119894 forall119886


+ (25)


1

(1205831minus 120582) (120583

2+ 120582)

ge4

1205832 (26)



min119864 (119879) = 4

120583 (27)


1205832= 1205832 minus 120582


119864 (119873) =

infin

sum

1198991=0

1

sum

1198992=0

(1198991+ 1198992) 11990111989911198992

=120582 (1205831+ 1205832)

(1205831minus 120582) (120583

2+ 120582)

(28a)

or

119864 (119873) = 120582119864 (119879) (28b)


min119864 (119873) = 120582min119864 (119879) = 4120582120583 (29)



2are taken as


1and119873



2

random variables is

11990111989911198992

= 119875 (1198731= 1198991 1198732= 1198992) 119899

1= 0 1 2 119899

2= 0 1

(30)

If 1198992= 0 and (119899



1is given as

(10)


Table 1 For 120582 = 030 1205831= 080 120583

2= 100

Iterationnumber


100 28163 08452 27692 08308 22222 066671000 28111 08433 27692 08308 22222 066675000 28124 08436 27692 08308 22222 06667

Table 2 For 120582 = 01 1205831= 09 120583

2= 09

Iterationnumber


100 22661 02263 22500 02250 22220 022221000 22599 02261 22500 02250 22220 022225000 22500 02260 22500 02250 22220 02222

Table 3 For 120582 = 001 1205831= 06 120583

2= 07

Iterationnumber


100 31042 00311 31034 00310 30769 003081000 31034 00310 31034 00310 30769 003085000 31043 00310 31034 00310 30769 00308



1199011015840

0= 119875 (119873

2= 0) =

1

1 + 1205882

(31)

1199011015840

1= 119875 (119873

2= 1) =

1205882

1 + 1205882

(32)

11990111989911198992

= 1199011198991

1199011015840

1198992

1198991= 0 1 2 119899

2= 0 1 (33)


1and


4 Numerical Results




5 Conclusions







References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Table 1 For 120582 = 030 1205831= 080 120583

2= 100

Iterationnumber


100 28163 08452 27692 08308 22222 066671000 28111 08433 27692 08308 22222 066675000 28124 08436 27692 08308 22222 06667

Table 2 For 120582 = 01 1205831= 09 120583

2= 09

Iterationnumber


100 22661 02263 22500 02250 22220 022221000 22599 02261 22500 02250 22220 022225000 22500 02260 22500 02250 22220 02222

Table 3 For 120582 = 001 1205831= 06 120583

2= 07

Iterationnumber


100 31042 00311 31034 00310 30769 003081000 31034 00310 31034 00310 30769 003085000 31043 00310 31034 00310 30769 00308



1199011015840

0= 119875 (119873

2= 0) =

1

1 + 1205882

(31)

1199011015840

1= 119875 (119873

2= 1) =

1205882

1 + 1205882

(32)

11990111989911198992

= 1199011198991

1199011015840

1198992

1198991= 0 1 2 119899

2= 0 1 (33)


1and


4 Numerical Results




5 Conclusions







References






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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