emgt 501 fall 2005 midterm exam solutions
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EMGT 501
Fall 2005
Midterm Exam
SOLUTIONS
1.a
M1 = units of component 1 manufactured
M2 = units of component 2 manufactured
M3 = units of component 3 manufactured
P1 = units of component 1 purchasedP2 = units of component 2 purchasedP3 = units of component 3 purchased
Min 321321 00.780.850.675.200.550.4 PPPMMM
3500,3
2000,4
1000,6
/000,18525.1
000,1535.1
Pr600,21432..
33
22
11
321
321
321
ComponentPM
ComponentPM
ComponentPM
PackagingTestingMMM
AssemblyMMM
oductionMMMts
M1, M2, M3, P1, P2, P3 0
b
SourceManufacturePurchase
Component 120004000
Component 24000
0
Component 314002100
Total Cost: $73,550
c
Max 654321 u3500u4000u6000u18000u15000u21600
0.7
8.8
5.6
75.2534
525.13
5.45.12..
6
5
4
6321
5321
4321
u
u
u
uuuu
uuuu
uuuuts
all 0,, 321 uuu URSuuu :,, 654
d.
u1* = 0.9063, u2* = 0, u3* = 0.1250, u4* = 6.5u5* = 7.9688 and u6* = 7.000
If a dual variable is positive on optimality, then its corresponding constraint in primal formulation becomes binding (=). Similarly, if a primal variable is positive on optimality, then its corresponding constraint in dual formulation becomes binding (=).
2.a
,9,3,0,1,0 **4
*3
*2
*1 zxxxx
b
93
123
3
1
4
5
21
11
21
11
1
1
1
bBC
bB
B
B
NOTE:
04
1
14
15
21
11
c ,11z,4x,0x,1x,0x **4
*3
*2
*1
The objective is changed from 9 to 11
3.a
Min 321 200700550 uuu
0,,
1213
3124
6312
4245.1..
321
321
321
321
321
uuu
uuu
uuu
uuu
uuuts
c. Since u1 = 3/10, u2 = 0, and u3 = 54/30, machines A and C (uj > 0) are operating at capacity. Machine C is the priority machine since each hour is worth 54/30.
b. Optimal solution: u1 = 3/10, u2 = 0, u3 = 54/30The (z-c)j values for the four surplus variables of the dual show x1 = 0, x2 = 25, x3 = 125, and x4 = 0.
A
B
FinishStart
F
G
C
E
D
I
H
4.a
b
Activity Expected Time Variance
A 4.83 0.25
B 4.00 0.44
C 6.00 0.11
D 8.83 0.25
E 4.00 0.44
F 2.00 0.11
G 7.83 0.69
H 8.00 0.44
I 4.00 0.11
Activity EarliestStart
LatestStart
EarliestFinish
LatestFinish
Slack CriticalActivity
A 0.00 0.00 4.83 4.83 0.00 Yes
B 0.00 0.83 4.00 4.83 0.83
C 4.83 5.67 10.83 11.67 0.83
D 4.83 4.83 13.67 13.67 0.00 Yes
E 4.00 17.67 8.00 21.67 13.67
F 10.83 11.67 12.83 13.67 0.83
G 13.67 13.83 21.50 21.67 0.17
H 13.67 13.67 21.67 21.67 0.00 Yes
I 21.67 21.67 25.67 25.67 0.00 Yes
c Critical Path: A-D-H-I
d E(T)= tA + tD + tH + tI
= 4.83 + 8.83 + 8 + 4 = 25.66 days
e 2 = A2 + D
2 + H2 + I
2
= 0.25 + 0.25 + 0.44 + 0.11 = 1.05
Using the normal distribution,
zE T
25 25 2566
1050 65
( ) .
..
From Appendix, area for z = -0.65 is 0.2422.Probability of 25 days or less = 0.5000 - 0.2422 = 0.2578
5.a 2 2(7200)(150)
* 1078.127200(1 / )
1 (0.18)(14.50)25000
o
h
DCQ
D P C
b Number of production runs = D / Q* = 7200 / 1078.12 = 6.68
250 250(1078.12)37.43
7200
QT
D [days] C
Maximum Inventory
72001 1 (1078.12) 767.62
25000
DQ
P
e
d
Production run length = 1078.12
10.78/ 250 25000 / 250
Q
P
days
7200(15) 432
250 250
Dr dm m
g
f Holding Cost
1 1 72001 1 (1078.12)(0.18)(14.50) $1001.74
2 2 25000h
DQC
P
Ordering cost 7200
(150) $1001.741078.12o
DC
Q
Total Cost = $2,003.48
6.a
2'
(1 / ) 'oDC
QD P IC
* 02 25000
(1 / ) (1 / )o
h
DC DCQ
D P C D P IC
C = current cost per unitC ' = 1.23 C new cost per unit
Let Q' = new optimal production lot size
2 1(1 / ) '' 1' 0.9017
* ' 1.23 1.232 1
(1 / )
o
o
DC
D P ICQ C CCQ C CDC
CD P IC
Q' = 0.9017(Q*) = 0.9017(5000) = 4509
Queueing Theory
The Basic Structure of Queueing Models
The Basic Queueing Process:Customers are generated over time by an input source.
The customers enter a queueing system.
A required service is performed in the service mechanism.
Customers
Inputsource
QueueService
mechanismServed
customers
Queueing system
Input Source (Calling Population):
The size of Input Source (Calling Population) is assumed infinite because the calculations are far easier.
The pattern by which customers are generated is assumed to be a Poisson process.
The probability distribution of the time
between consecutive arrivals is an
exponential distribution.
The time between consecutive arrivals is
referred to as the interarrival time.
Queue:
The queue is where customers wait before being served.
A queue is characterized by the maximum permissible number of customers that it can contain.
The assumption of an infinite queue is the standard one for most queueing models.
Queue Discipline:
The queue discipline refers to the order in which members of the queue are selected for service.
For example,
(a) First-come-first-served
(b) Random
Service Mechanism:
The service mechanism consists of one or more service facilities, each of which contains one or more parallel service channels, called servers.
The time at a service facility is referred to as the service time.
The service-time is assumed to be the exponential distribution.
Elementary Queueing Process
C C C C C C C
CCCC
SS ServiceS facilityS
Customers
Queueing system
Queue
Served customers
Served customers
Where
M = exponential distribution (Markovian)
D = degenerate distribution (constant times)
= Erlang distribution (shape parameter = k)
G = general distribution(any arbitrary
distribution allowed)
kE
//
Distribution of service times
Number of servers
Distribution of interarrival times
sMM //
Both interarrival and service times have an exponential distribution. The number of servers is s .
Interarrival time is an exponential distribution. No restriction on service time. The number of servers is exactly 1.
1// GM
Terminology and Notation
State of system = # of customers in queueing system.
Queue length = # of customers waiting for
service to begin.
N(t) = # of customers in queueing
system at time t (t 0)
= probability of exactly n customers
in queueing system at time t.
)(tPn
# of servers in queueing system.
A mean arrival rate (the expected number of arrivals per unit time) of new customers when n customers are in system.
A mean service rate for overall system (the expected number of customers completing service per unit time) when n customers are in system.
Note: represents a combined rate at which all busy servers (those serving customers) achieve service completions.
s
n
n
n
When is a constant for all n, it is expressed by
When the mean service rate per busy server is a constant for all n 1, this constant is denoted by .
Under these circumstances, and are the expected interarrival time and the expected service time.
is the utilization factor for the service facility.
)/( s
/1 /1
n
The state of the system will be greatly affected by the initial state and by the time that has since elapsed.
The system is said to be in a transient condition.
After sufficient time has elapsed, the state of the system becomes essentially independent of the initial state and the elapsed time.
The system has reached a steady-state condition, where the probability distribution of the state of the system remains the same over time.
The probability of exactly n customers in
queueing system.
The expected number of customers in
queueing system
The expected queue length (excludes
customers being served)
nP
L.
0
n
nnP
qL.)(
sn
nPsn
A waiting time in system (includes service time) for each individual customer.
A waiting time in queue (excludes service time) for each individual customer.
w
).(wEW
qw
).( qq wEW
Relationships between and,,, qLWL .qW
.WL
Assume that is a constant for all n.
In a steady-state queueing process,
n
.qq WL
Assume that the mean service time is a constant, for all It follows that,
.1
qWW
.1n1
The Role of the Exponential Distribution
01
)( TE
)(tfT
t
,0tfor0
0tfore)t(ft
T
An exponential distribution has the following probability density function:
Relationship to the Poisson distribution
Suppose that the time between consecutive arrivals has an exponential distribution with parameter .
Let X(t) be the number of occurrences by time t (t 0)
The number of arrivals follows
,!
)()(
n
etntXP
tn
for n = 0, 1, 2, …;
The Birth-and-Death Process
Most elementary queueing models assume that the inputs and outputs of the queueing system occur according to the birth-and-death process.
In the context of queueing theory, the term birth refers to the arrival of a new customer into the queueing system, and death refers to the departure of a served customer.
The assumptions of the birth-and-death process are the following:
Assumption 1.
Given N(t) = n, the current probability distribution of the remaining time until the next birth is exponential.
Assumption 2.
Given N(t) = n, the current probability distribution of the remaining time until the next death is exponential
,...).2,1,0( nn
,...).2,1,0( nn
Assumption 3.The random variable of assumption 1 (the remaining time until the next birth) and the random variable variable of assumption 2 (the remaining time until the next death) are mutually independent.
The next transition in the state of the process is either
(a single birth)
(a single death),
depending on whether the former or latter random variable is smaller.
1 nn
1 nn
The birth-and-death process is a special type of continuous time Markov chain.
State: 0 1 2 3 n-2 n-1 n n+1
2n 1n n210
1 2 3 1n n 1n
n n and are mean rates.
Starting at time 0, suppose that a count is made of the number of the times that the process enters this state and the number of times it leaves this state, as demoted below:
the number of times that
process enters state n by time t.
the number of times that
process leaves state n by time t.
)(tEn
)(tLn
Rate In = Rate Out Principle.
For any state of the system n (n = 0,1,2,…),
average entering rate = average leaving rate.
The equation expressing this principle is called the balance equation for state n.
State
0
1
2
n – 1
n
0011 PP
Rate In = Rate Out
1112200 )( PPP
2223311 )( PPP
11122 )( nnnnnnn PPP
nnnnnnn PPP )(1111
)(1
)(1
11223
23
23
00112
12
12
01
01
PPPP
PPPP
PP
0123
0122
3
2
012
011
2
1
PP
PP
State:
0:
1:
2:
To simplify notation, let
,11
021
nn
nnnC for n = 1,2,
…
and then define for n = 0.
Thus, the steady-state probabilities are
1nC
,0PCP nn for n = 0,1,2,…
The requirement that
10
nnP
implies that
,100
PCn
n
so that
.1
00
nnCP
The definitions of L and specify thatqL
.)(,0
sn
nqn
n PsnLnPL
,
LW
LW
.0
n
nnP
is the average arrival rate. is the mean arrival rate while the system is in state n. is the proportion of time for state n,
nnP
The M/M/s Model
A M/M/s model assumes that all interarrival times
are independently and identically distributed
according to an exponential distribution, that all
service times are independent and identically
distributed according to another exponential
distribution, and that the number of service is s
(any positive integer).
In this model the queueing system’s mean arrival rate and mean service rate per busy server are constant ( and ) regardless of the state of the system.
State: 0 1 2 3 n-2 n-1 n n+1
(a) Single-server case (s=1)
n n
State: 0 1 2 3 s-2 s-1 s s+1
2 3 )1( s s s
(b) Multiple-server case (s > 1)
for n = 0,1,2,…
for n = 1,2,…s
for n = s, s+1,...
,
,
s
nn
n
When the maximum mean service rate exceeds
the mean arrival rate, that is, when
a queueing system fitting this model will
eventually reach a steady-state condition.
s
1s
Results for the Single-Server Case (M/M/1).
For s = 1, the factors for the birth-and-death process reduce to
nC
,n
n
nC
Therefor,
,0PP nn .1
1
11
1
00
n
nP
Thus,,)1( n
nP
1
1
1
d
d)1(
d
d)1(
)(d
d)1(
)1(nL
0n
n
0n
n
0n
n
.)(
)1(1
)1(
2
0
1
PL
PnLn
nq
Similarly,
When , the mean arrival rate exceeds the mean service rate, the preceding solution “blows up” and grow without bound.
Assuming , we can derive the probability distribution of the waiting time in the system (w) for a random arrival when the queue discipline is first-come-first-served.
If this arrival finds n customers in the system, then the arrival will have to wait through n + 1 exponential service time, including his or her own.
,)1( tuetwP
Which reduces after considerable manipulation to
The surprising conclusion is that w has an exponential distribution with parameter .
Therefore,
1
)1(
1)(wEW
These results include service time in the waiting time.
Consider the waiting time in the queue (so excluding service time) for a random arrival when the queue discipline is first-come-first-served.
If the arrival finds no customers already in the system, then the arrival is served immediately, so that
qw
.10 0 PwP q
Results for the Multiple-Server Case (s > 1).
When s > 1, the factors become
sn
nsns
n
n
ssss
nC
!
)/(
!
)/(!
)/(
nC
for n = 0,1,2,…,s
for n = s, s+1,…
.)s(1
1!s
)/(!n)/(
1
s!s)/(
!n)/(
11P
1s
0n
sn
1s
1n sn
snsn
0
Consequently, if [so that ], then
s 1)( s
Where the n = 0 term in the last summation yields the correct value of 1 because of the convention that n! = 1 when n = 0. These factors also give
0
0
!
/!
/
Pss
PnP
sn
n
n
n
sn 0
.sn
if
if
Furthermore,
;)1(!s
)/(P
1
1
d
d
!s
)(P
d
d
!s
)(P
d
d
!s
)(PP
!s
)/(j
jPP)sn(L
2
s0
s
00j
js
0
0j
js
00j
0j
s
0jjs
snnq
.1
;1
;
q
LWL
WW
Lw
Example
A management engineer in the Country Hospital has made a proposal that a second doctor should be assigned to take care of increasing number of patients.
She has concluded that the emergency cases arrive pretty much at random (a Poisson input process), so that interarrival times have an exponential distribution and the time spent by a doctor treating the cases approximately follows an exponential distribution.
Therefore, she has chosen the M/M/s model.
By projecting the available data for the early evening shift into next year, she estimates that patients will arrive at an average rate of 1 every hour.
A doctor requires an average of 20 minutes to treat each patient.
Thus, with one hour as the unit of time,
2
1
2
11
hour per customer
3
11
hour per customer
3 customer per hour
so that
2 customer per hour
The two alternatives being considered are to continue having just one doctor during this shift ( s = 1) or to add a second doctor ( s = 1).
In both cases,
,1s
so that the system should approach a steady-state condition.
W
W
L
L
nP
P
q
q
n 21
0
3
2
31
92
n32
31
34
2
32
1 hour
for
s = 1 s = 2
31
21
31
n31
121
43
241
hour
hourhour
83
The Finite Queue Variation of the M/M/s Model]
(Called the M/M/s/K Model)
Queueing systems sometimes have a finite queue; i.e., the number of customers in the system is not permitted to exceed some specified number (denoted K) so the queue capacity is K - s.
Any customer that arrives while the queue is “full” is refused entry into the system and so leaves forever.
From the viewpoint of the birth-and-death process, the mean input rate into the system becomes zero at these times.
The one modification is needed
0
n
for n = 0, 1, 2,…, K-1
for n K.
Because for some values of n, a queueing system that fits this model always will eventually reach a steady-state condition, even when
0n
.1 s
The Finite Calling Population Variation of the M/M/s Model
The only deviation from the M/M/s model is that the input source is limited; i.e., the size of the calling population is finite.
For this case, let N denote the size of the calling population.
When the number of customers in the queueing system is n (n = 0, 1, 2,…, N), there are only N - n potential customers remaining in the input source.
State: 0 1 2 n-2 n-1 n N-1 N
)2( nN)1( nN)1( N
N
(a) Single-server case ( s = 1)
,
,0
,)(
n
n
nN for n = 0, 1, 2, …, N
for n N
for n = 1, 2, ...
State: 0 1 2 s-2 s-1 s N-1 N
)2( sN)1( sN)1( N
N
2 )1( s s s
(a) Multiple-server case ( s > 1)
,
,
,0
,)(
s
n
nN
n
n
for n = 0, 1, 2, …, N
for n N
for n = 1, 2, …, s
for n = s, s + 1, ...
[1] Single-Server case ( s = 1), , nn
Birth-Death Process
0 1 2 n-1 n n+1
State012
n
01 PP Rate In = Rate Out
120 )( PPP 231 )( PPP
nnn PPP )(11
nnn
nn
nn
C
PPcPP
PP
PP
PP
PPP
PP
)( where
)(
)()(
)(
)(
)(
000
02
021
001
001
011
2
01
2
2
(3) 1
L
)1(L
(2) )1(
(1) 11
1)(
00
0
0
0
00
00
0
n
n
nn
nnn
n
n
n
n
n
n
nn
nnP
PP
P
PPP
(6) )(
1
(5) 11
)4( )(
)1(2
1
WL
W
LW
PnL
nnq
Example
# 6
10
60#
10
1
)510(10
5
)(
# 125
60#
5
1
2
1
10
5
)510(10
5
)(
1510
5
5.02
1
10
5
? , , ,
# 10 # 5
22
MHW
MinHourLW
L
L
WWLLHH
q
q
[2] Multiple-Server case ( s > 1)
)(
)(0 ,
nss
snnnn
Birth-Death Process
0 1 2 3 s-2 s-1 s s+1
2 3 )1( s s s
s
)2( s
State
0
1
2
s - 1
s
s + 1
01 PP Rate In = Rate Out
120 )(2 PPP
231 )2(3 PPP
12 )1( sss PsPsP
sss PsPsP )(11
12 )( sss PsPsP
02
21
02
0021
0021
0121
2
01
)(
)(
)(
)(
2
2
2
PP
PP
PP
PPP
PP
03
02
223
0
2
2
2
123
)(!3
1
2
22
3
1
2)2(
3
1
)2(3
1
P
P
P
PPP
1!
)(
!
)(
)( !
)(
)10( !
)(
)( !
)(
)10( !
)(
0
1
00
0
0
Pssn
P
nsss
snnC
nsPss
P
snPn
P
snsn
ns
n
n
nn
sn
n
n
n
sn
n
n
n
n
)( !
)(
)10( !
)(
11
!
)(
!
)(
1
)(!
)(
!
)(1
1
0
0
1
0
1
1
0
nsPss
snPnP
sn
sn
P
sn
n
n
n
s
ss
n
n
sn
sns
ss
n
n
)
1(
1
)1(!
)(
)(
2
0
0
q
q
s
jjs
snnq
WWL
WW
LW
ss
P
jPPsnL
Question 1
Mom-and-Pop’s Grocery Store has a small adjacent parking lot with three parking spaces reserved for the store’s customers. During store hours, cars enter the lot and use one of the spaces at a mean rate of 2 per hour. For n = 0, 1, 2, 3, the probability Pn that exactly n spaces currently are being used is P0 = 0.2, P1 = 0.3, P2 = 0.3, P3 = 0.2.
(a) Describe how this parking lot can be interpreted as being a queueing system. In particular, identify the customers and the servers. What is the service being provided? What constitutes a service time? What is the queue capacity?
(b) Determine the basic measures of performance - L, Lq, W, and Wq - for this queueing system.
(c) Use the results from part (b) to determine the average length of time that a car remains in a parking space.
Question 2
Consider the birth-and-death process with all
and for n = 3, 4, …
(a) Display the rate diagram.
(b) Calculate P0, P1, P2, P3, and Pn for n = 4, 5, ...
(c) Calculate L, Lq, W, and Wq.
,1 ,2 ,3 210 ), ,2 ,1( 2 nn
0n
Question 3
A certain small grocery store has a single checkout stand with a full-time cashier. Customers arrive at the stand “randomly” (i.e., a Poisson input process) at a mean rate of 30 per hour. When there is only one customer at the stand, she is processed by the cashier alone, with an expected service time of 1.5 minutes. However, the stock boy has been given standard instructions that whenever there is more than one customer at the stand, he is to help the cashier by bagging the groceries. This help reduces the expected time required to process a customer to 1 minute. In both cases, the service-time distribution is exponential.
(a) Construct the rate diagram for this queueing system.
(b) What is the steady-state probability distribution of the number of customers at the checkout stand?
(c) Derive L for this system. (Hint: Refer to the derivation of L for the M/M/1 model at the beginning of Sec. 17.6.) Use this information to determine Lq, W, and Wq.
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