Problem 13.10
Average interarriv al time  7.5 minutes
1
1
 0.13 3 part/minut e

average interarriv al time 7.5 minutes
60 minutes
1
 8 parts/hour

or 
1 hour
7.5 minutes
  Service rate  10 parts/hour
  Arrival rate 
Average service time 
1


1
 0.1 hour  6 minutes
10 parts/hour
82
2
 3.2 parts

Average number of parts waiting to be worked on  Lq 
 (    ) 10(10  8)
 8
 80%
Percentage of time the operator is working  utilizatio n factor  
 10

Percentage of time the machine is idle  1 - U  1 -  0.2  20%

STORM DATA SET LISTING
PROBLEM DESCRIPTION PARAMETERS
Title : bubba - taylor problem 13-10
Number of independent queueing problems : 2
DETAILED PROBLEM DATA LISTING FOR
bubba - taylor problem 13-10
Unit of Time
# SERVERS
SOURCE POP
ARR RATE
SERV DIST
SERV TIME
SERV STD
WAIT CAP
# CUSTMERS
WAIT COST
COST/SERV
LOSTCUST C
Minute
QUEUE 1
1
INF
0.133333
EXP
6
.
.
.
.
.
.
Hour
QUEUE 2
1
INF
8
EXP
0.1
.
.
.
.
.
.
Unit of Time: Minute
bubba - taylor problem 13-10
QUEUE 1 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
0.1333
0.1667
79.9998
3.2000
3.9999
0.8000
23.9997
29.9997
Unit of Time: Hour
bubba - taylor problem 13-10
QUEUE 2 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
8.0000
10.0000
80.0000
3.2000
4.0000
0.8000
0.4000
0.5000
Problem 13.22
  Arrival rate  10 customers/ hour
1 hour
 0.16 6 customer/m inute
60 minutes
Constant service time  4.5 minutes  0.075 hour
or  10 customers/ hour 
1
1
 0.2 2 customer/m inute

average service time 4.5 minutes
60 minutes
1
 13. 3 customer/h our

or 
1 hour
4.5 minutes
  Standard deviation of service time  0
  Service rate 
Average length of the waiting line  Lq 
2 2     

21   


2
 
10 2
2

2  (    ) 2(13. 3 )(13. 3  10)
 1.125 drivers
Average wiating time  Wq 
Lq


1.125
 0.1125 hour  6.75 minutes
10
STORM DATA SET LISTING
PROBLEM DESCRIPTION PARAMETERS
Title : bubba - taylor problem 13-22
Number of independent queueing problems : 2
DETAILED PROBLEM DATA LISTING FOR
bubba - taylor problem 13-22
Unit of Time
# SERVERS
SOURCE POP
ARR RATE
SERV DIST
SERV TIME
SERV STD
WAIT CAP
# CUSTMERS
WAIT COST
COST/SERV
LOSTCUST C
Hour
QUEUE 1
1
INF
10
DET
0.075
.
.
.
.
.
.
Minute
QUEUE 2
1
INF
0.166667
DET
4.5
.
.
.
.
.
.
Unit of Time: Hour
bubba - taylor problem 13-22
QUEUE 1 : M / D / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
10.0000
13.3333
75.0000
1.1250
1.8750
0.7500
0.1125
0.1875
Unit of Time: Minute
bubba - taylor problem 13-22
QUEUE 2 : M / D / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
0.1667
0.2222
75.0001
1.1250
1.8750
0.7500
6.7501
11.2501
Problem 13.24
Original arrival rate : original  149 trucks/da y
New arrival rate :  new  179 trucks/da y
1 hour
1 day

 0.005 5 day per truck
60 minutes 24 hours
1
60 minutes 24 hours
  service rate 


 180 trucks/da y
8 minutes
1 hour
1 day
Constant service time  8 minutes 
STORM DATA SET LISTING
PROBLEM DESCRIPTION PARAMETERS
Title : bubba - problem 13.24
Number of independent queueing problems : 2
DETAILED PROBLEM DATA LISTING FOR
bubba - problem 13.24
# SERVERS
SOURCE POP
ARR RATE
SERV DIST
SERV TIME
SERV STD
WAIT CAP
# CUSTMERS
WAIT COST
COST/SERV
LOSTCUST C
Original
QUEUE 1
1
INF
149
DET
5.5556E-03
.
.
.
.
.
.
New Arrival Rate
QUEUE 2
1
INF
179
DET
5.5556E-03
.
.
.
.
.
.
bubba - problem 13.24
QUEUE 1 : M / D / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
149.0000
179.9986
82.7784
1.9894
2.8172
0.8278
0.0134
0.0189
bubba - problem 13.24
QUEUE 2 : M / D / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
1
179.0000
179.9986
99.4452
89.1316
90.1261
0.9945
0.4979
0.5035
Average time a truck will have to wait to be unloaded
= expected time in the queue
24 hours
 0.4979 day 
 11.9496 hours
1 day
Problem 13.26
c = 4 servers (cabs)
Average interarriv al time  8 minutes
1
1

 0.125 part/minut e
average interarriv al time 8 minutes
1
60 minutes
or 

 7.5 passengers /hour
8 minutes
1 hour
Average service time (exponenti ally distribute d)  20 minutes
  Arrival rate 
  service rate per server 
1
1

 0.05 part/minut e
average service time 20 minutes
1
60 minutes
or 

 3 passengers /hour
20 minutes
1 hour
STORM DATA SET LISTING
PROBLEM DESCRIPTION PARAMETERS
Title : bubba - taylor problem 10.26
Number of independent queueing problems : 2
DETAILED PROBLEM DATA LISTING FOR
bubba - taylor problem 10.26
Unit of time
# SERVERS
SOURCE POP
ARR RATE
SERV DIST
SERV TIME
SERV STD
WAIT CAP
# CUSTMERS
WAIT COST
COST/SERV
LOSTCUST C
Hour
QUEUE 1
4
INF
7.5
EXP
0.333333
.
.
.
.
.
.
Minute
QUEUE 2
4
INF
0.125
EXP
20
.
.
.
.
.
.
bubba - taylor problem 10.26
QUEUE 1 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
4
Mean arrival rate . . . . . . . . . . . . . .
7.5000
Mean service rate per server . . . . . . . .
3.0000
Mean server utilization (%) . . . . . . . . . 62.4999
Expected number of customers in queue . . . .
0.5331
Expected number of customers in system . . .
3.0331
Probability that a customer must wait . . . .
0.3199
Expected time in the queue . . . . . . . . .
0.0711
Expected time in the system . . . . . . . . .
0.4044
Average number of customers waiting for service
= expected number of customers in queue = 0.533 customer
Average time a customer must wait for a cab
= expected time in the queue
= 0.0711 hour  4.266 minutes
bubba - taylor problem 10.26
QUEUE 2 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
4
Mean arrival rate . . . . . . . . . . . . . .
0.1250
Mean service rate per server . . . . . . . .
0.0500
Mean server utilization (%) . . . . . . . . . 62.5000
Expected number of customers in queue . . . .
0.5331
Expected number of customers in system . . .
3.0331
Probability that a customer must wait . . . .
0.3199
Expected time in the queue . . . . . . . . .
4.2648
Expected time in the system . . . . . . . . . 24.2648
Average number of customers waiting for service
= expected number of customers in queue = 0.533 customer
Average time a customer must wait for a cab
= expected time in the queue
= 4.2648 minutes
Problem 13.27
c = 4 servers (stations)
  Arrival rate  40 customers/ hour
Average service time (exponenti ally distribute d)  4 minutes
1
1
  service rate per server 

 0.25 part/minut e
average service time 4 minutes
1
60 minutes
or 

 15 customers/ hour
4 minutes
1 hour
STORM DATA SET LISTING
PROBLEM DESCRIPTION PARAMETERS
Title : bubba - problem 13.27
Number of independent queueing problems : 2
DETAILED PROBLEM DATA LISTING FOR
bubba - problem 13.27
Unit of time
# SERVERS
SOURCE POP
ARR RATE
SERV DIST
SERV TIME
SERV STD
WAIT CAP
# CUSTMERS
WAIT COST
COST/SERV
LOSTCUST C
Hour
QUEUE 1
4
INF
40
EXP
0.066667
.
.
.
.
.
.
Minute
QUEUE 2
4
INF
0.666667
EXP
4
.
.
.
.
.
.
bubba - problem 13.27
QUEUE 1 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
4
40.0000
14.9999
66.6670
0.7569
3.4235
0.3784
0.0189
0.0856
bubba - problem 13.27
QUEUE 1 : M / M / c
PROBABILITY DISTRIBUTION OF NUMBER IN SYSTEM
Number Prob 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
+----+----+----+----+----+----+----+----+----+----+
0 0.0599|***
|
1 0.1596|********--|
2 0.2129|***********----------|
3 0.1892|*********+--------------------|
4 0.1261|******+-----------------------------|
5 0.0841|****+------------------------------------|
6 0.0561|***----------------------------------------- |
7 0.0374|**-------------------------------------------- |
8 0.0249|*+---------------------------------------------- |
9 0.0166|*----------------------------------------------- |
10 0.0111|*------------------------------------------------|
11 0.0074|+------------------------------------------------|
OVER 0.0147|*------------------------------------------------|
+----+----+----+----+----+----+----+----+----+----+
(a) Postal workers’ idle time = 100% - mean server utilization = (100-66.667)%
 33.3%
(b) Mean customer waiting time = expected time in the queue
= 0.0189 hour  1.134 minutes
(c) Percentage of time a customer can walk in without waiting at all
P0  0.0599  6%
bubba - problem 13.27
QUEUE 2 : M / M / c
QUEUE STATISTICS
Number of identical servers . . . . . . . . .
Mean arrival rate . . . . . . . . . . . . . .
Mean service rate per server . . . . . . . .
Mean server utilization (%) . . . . . . . . .
Expected number of customers in queue . . . .
Expected number of customers in system . . .
Probability that a customer must wait . . . .
Expected time in the queue . . . . . . . . .
Expected time in the system . . . . . . . . .
4
0.6667
0.2500
66.6667
0.7568
3.4235
0.3784
1.1353
5.1353
bubba - problem 13.27
QUEUE 2 : M / M / c
PROBABILITY DISTRIBUTION OF NUMBER IN SYSTEM
Number Prob 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
+----+----+----+----+----+----+----+----+----+----+
0 0.0599|***
|
1 0.1596|********--|
2 0.2129|***********----------|
3 0.1892|*********+--------------------|
4 0.1261|******+-----------------------------|
5 0.0841|****+------------------------------------|
6 0.0561|***----------------------------------------- |
7 0.0374|**-------------------------------------------- |
8 0.0249|*+---------------------------------------------- |
9 0.0166|*----------------------------------------------- |
10 0.0111|*------------------------------------------------|
11 0.0074|+------------------------------------------------|
OVER 0.0147|*------------------------------------------------|
+----+----+----+----+----+----+----+----+----+----+
(a) Postal workers’ idle time = 100% - mean server utilization = (100-66.6667)%
 33.3%
(b) Mean customer waiting time = expected time in the queue
= 1.1353 minutes
(c) Percentage of time a customer can walk in without waiting at all
P0  0.0599  6%