Queuing Theory
Jackson Networks
Network Model
Consider a simple 3 stage model where the output of
2 queues becomes the input process for a third
Station A
Station C
Station B
Network Model
Station A
Station C
Station B
Proposition 1: rate in = rate out, that is, if lA and lB
are the input rates for station A and B respectively,
then the input rate for station C is lA+lB.
Proposition 2: exponential inter-arrivals at A and B
provide exponential inter-arrivals at station C.
Network Model
Station A
Station C
Station B
Instructor Derivation of the minimum of 2
exponentials.
P{min(X1, X 2 )  t}  el1t el2t  e(l1l2 )t
provided, X1, X 2 exponentially distributed
Jackson Network
Station 1
(1  1 j )l1
1
l1
2
s1
Station v
l
1 j l1 
j
Station j
lj
.
.
1
2
1
sj
j l
2
sv
(1  j )l
Jackson Network
• All external arrivals to each station must follow a
Poisson process (exponential inter-arrivals)
• All service times must be exponentially distributed
• All queues must have unlimited capacity
• When a job leaves a station, the probability that it
will go to another station is independent of its past
history and of the location of any other job
1. Calculate the input arrival rate to each station
2. Treat each station independently as an M/M/s
queue
Calculate Arrival Rates
m
li   i   kilk
k 1
where
li  calculate arrival rate at station i
 i  external arrival rate at station i
ki  probability a job from station k goes to station i
Example
Jobs submitted to a computer center must first
pass through an input processor before arriving
at the central processor. 80% of jobs get passed
on the central processor and 20% of jobs are
rejected. Of the jobs that pass through the
central processor, 60% are returned to the
customer and 40% are passed to the printer.
Jobs arrive at the station at a rate of 10 per
minute. Calculate the arrival rate to each
station.
Example
.6
.2
10
Input
.8
Central
l1   1  11l1  21l2  31l3
l2   2  12 l1  22 l2  32 l3
l3   3  13l1  23l2  33l3
l1 10  0  0  0 10
l2  0  .8l1  0  0  8
l3  0  0  .4l2  0  3.2
.4
Printer
Example (cont.)
• For the computer center the processing times
are 10 seconds for the input processor, 5
seconds for the central processor, and 70
seconds for the printer. Our task is to
determine the number of parallel stations
(multiple servers) to have at each station to
balance workload.
Example
10
Input
8
Central
3.2
Printer
M / M / s1
M / M / s2
M / M / s3
l 10
 6
l 8
 12
l  3.2
 6/ 7
For our initial try, we will solve for s1 = 2, s2 = 1, s3 = 4
10
Input
M /M /2
l 10
 6
Metrics
Model
l

r
po
Wq
Lq
8
Central
3.2
Printer
M / M /1
l 8
M /M /4
l  3.2
  12
 6/ 7
Input
M/M/2
10
6
0.833
0.091
3.78
0.378
Central
M/M/1
8
12
0.667
0.333
1.33
0.167
Printer
M/M/4
3.2
6/7
0.933
0.009
8.68
2.71
Metrics
Model
l

r
po
Wq
Lq
Input
M/M/2
10
6
0.833
0.091
3.78
0.378
Central
M/M/1
8
12
0.667
0.333
1.33
0.167
Printer
M/M/4
3.2
6/7
0.933
0.009
8.68
2.71
If these numbers are correct, clearly Lq and Wq
indicate the bottleneck is at the printer station. We
may wish to add a printer if speedy return of
printouts is required. Secondarily, overall processing
may be increased by adding another input processor.
Job Shop Example
• An electronics firm has 3 different products in
a job shop environment. The job shop has six
different machines with multiple machines at
5 of the 6 stations.
Product Order rate
Flow
1
30/month
ABDF
2
10/month
ABEF
3
20/month
ACEF
Summary information is on the network below.
Job Shop Example
 A  60
Machine B
s=2
 = 22
Machine A
s=3
 = 25
Machine D
s=3
 = 11
Machine E
s=2
 = 23
Machine C
s=1
 = 29
Machine F
s=4
 = 20
Job Shop Example
 A  60
Machine B
s=2
 = 22
Machine A
s=3
 = 25
Machine D
s=3
 = 11
Machine E
s=2
 = 23
Machine F
s=4
 = 20
Machine C
s=1
 = 29
l A   A   AAl A  BAlB  CA lC  DA lD  EAlE  FAlF
lB   B   ABl A  BBlB  CB lC  DB lD  EB lE  FBlF
lC   C   AC l A  BC lB  CC lC  DC lD  EC lE  FC lF
lD   D   ADl A  BDlB  CD lC  DD lD  EDlE  FDlF
lE   E   AE l A  BE lB  CE lC  DE lD  EE lE  FE lF
lF   F   AF l A  BF lB  CF lC   DF lD  EF lE  FF lF
Job Shop Example
 A  60
Machine B
s=2
 = 22
Machine A
s=3
 = 25
Machine D
s=3
 = 11
Machine E
s=2
 = 23
Machine F
s=4
 = 20
Machine C
s=1
 = 29
l A  60  0 l A  0 lB  0lC  0lD  0lE  0lF  60
lB  0  (40 / 60)l A  0lB  0lC  0lD  0lE  0lF  40
lC  0  (20 / 60)l A  0lB  0lC  0lD  0lE  0lF  20
lD  0  0l A  (30 / 40)lB  0lC  0lD  0lE  0lF  30
lE  0  0l A  (10 / 40)lB  (20 / 20)lC  0lD  0lE  0lF  30
lF  0  0l A  0lB  0lC  (30 / 30)lD  (30 / 30)lE  0lF  60
Job Shop Example
 A  60
Machine B
s=2
 = 22
Machine A
s=3
 = 25
Machine D
s=3
 = 11
Machine E
s=2
 = 23
Machine F
s=4
 = 20
Machine C
s=1
 = 29
Lets calculate metrics for Machine B
M/M/2 model with lB  40
  22
l
40
r

 0.91
s 2(20)
Use formulas in Fig. 16.5 p. 564 or use charts
p o  .04
r  0.91
L  10 .4
r  0.91
Repeat for Machines A, C, D, E, F
Machine D
s=3
 = 11
Machine B
s=2
 = 22
 A  60
Machine A
s=3
 = 25
Machine E
s=2
 = 23
Machine F
s=4
 = 20
Machine C
s=1
 = 29
Metrics
Model
l

r
po
Wq
Lq
W
L
A
M/M/3
60
25
0.800
0.06
0.043
2.589
0.083
4.989
B
M/M/2
40
22
0.909
0.04
0.216
8.658
0.262
10.476
C
M/M/1
20
29
0.690
0.31
0.077
1.533
0.111
2.222
D
M/M/3
30
11
0.909
0.02
0.278
8.332
0.369
11.059
E
M/M/2
30
23
0.652
0.21
0.032
0.965
0.076
2.27
F
M/M/4
60
20
0.750
0.042
0.025
1.528
0.075
4.528
Interesting Application to Manufacturing
Metrics
Model
l

r
po
Wq
Lq
W
L
A
M/M/3
60
25
0.800
0.06
0.043
2.589
0.083
4.989
B
M/M/2
40
22
0.909
0.04
0.216
8.658
0.262
10.476
C
M/M/1
20
29
0.690
0.31
0.077
1.533
0.111
2.222
D
M/M/3
30
11
0.909
0.02
0.278
8.332
0.369
11.059
E
M/M/2
30
23
0.652
0.21
0.032
0.965
0.076
2.27
F
M/M/4
60
20
0.750
0.042
0.025
1.528
0.075
4.528
Note that lead time is just the time in the system which for product 1
which has sequence ABDF is
W = total time in system = lead time
= WA + WB + WD + WF = .789
WIP = work in process = parts per month x lead time
= 30(.789) = 23.67
Repeat for Machines A, C, D, E, F
 A  60
Machine B
s=2
 = 22
Machine A
s=3
 = 25
Machine D
s=3
 = 11
Machine E
s=2
 = 23
Machine F
s=4
 = 20
Machine C
s=1
 = 29
Monthly
Product Orders Sequence
1
30
ABDF
2
10
ABEF
3
20
ACEF
Lead
Time
0.789
0.496
0.345
Queue
Time
0.562
0.316
0.177
WIP
23.67
4.96
6.9
Some Recommendations
Metrics
Model
l

r
po
Wq
Lq
W
L
A
M/M/3
60
25
0.800
0.06
0.043
2.589
0.083
4.989
B
M/M/2
40
22
0.909
0.04
0.216
8.658
0.262
10.476
C
M/M/1
20
29
0.690
0.31
0.077
1.533
0.111
2.222
D
M/M/3
30
11
0.909
0.02
0.278
8.332
0.369
11.059
E
M/M/2
30
23
0.652
0.21
0.032
0.965
0.076
2.27
F
M/M/4
60
20
0.750
0.042
0.025
1.528
0.075
4.528
If lead time is too slow (WIP too high), one possibility is to add
additional machining to the bottleneck area. This occurs at stations D
and B. Note that this assumes the cost of the machines is not a critical
part of the decision. This may or may not need to be included in a
final recommendation.