Solutions Manual
Discrete-Event System Simulation
Third Edition
Jerry Banks
John S. Carson II
Barry L. Nelson
David M. Nicol
August 31, 2000
Contents
1 Introduction to Simulation 1
2 Simulation Examples 5
3 General Principles 16
4 Simulation Software 17
5 Statistical Models in Simulation 18
6 Queueing Models 32
7 Random-Number Generation 39
8 Random-Variate Generation 46
9 Input Modeling 51
10 Verification and Validation of Simulation Models 55
11 Output Analysis for a Single Model 57
12 Comparison and Evaluation of Alternative System Designs 60
13 Simulation of Manufacturing and Material Handling Systems 65
14 Simulation of Computer Systems 66
1
Foreword
There are approximately three hundred exercises for solution in the text. These exercises emphasize principles
of discrete-event simulation and provide practice in utilizing concepts found in the text.
Answers provided here are selective, in that not every problem in every chapter is solved. Answers in
some instances are suggestive rather than complete. These two caveats hold particularly in chapters where
building of computer simulation models is required. The solutions manual will give the instructor a basis
for assisting the student and judging the student’s progress. Some instructors may interpret an exercise
differently than we do, or utilize an alternate solution method; they are at liberty to do so. We have
provided solutions that our students have found to be understandable.
When computer solutions are provided they will be found on the text web site, www.bcnn.net,rather
than here. We have invited simulation software vendors to submit solutions to a number of modeling and
analysis problems; these solutions will also be found on the web site. Instructors are encouraged to submit
solutions to the web site as well.
Jerry Banks
John S. Carson II
Barry L. Nelson
David M. Nicol
Chapter 1
Introduction to Simulation
For additional solutions check the course web site at www.bcnn.net.
1. Solution to Exercise 1:
SYSTEM ENTITIES ATTRIBUTES ACTIVITIES EVENTS STATE VARIABLES
a. Small appliance Appliances Type of appliance Repairing Arrival of Number of appliances
repair shop
the appliance ajob waiting to be repaired
Age of appliance
Completion Status of repair person
Nature of problem of a job busy or idle
b. Cafeteria Diners Size of appetite Selecting food Arrival at Number of diners
service line in waiting line
Entree preference Paying for food Departures Number of servers
from service working
line
c. Grocery store Shoppers Length of grocery Checking out Arrival at Number of shoppers
list checkout in line
counters Number of checkout
lanes in operation
Departure from
checkout counter
d. Laundromat Wa shing Breakdown rate Repairing Occurrence of Number of machines
machine a machine breakdowns running
Number of machines in
Completion repair
of service Number of Machines
waiting for repair
1
CHAPTER 1. INTRODUCTION TO SIMULATION 2
SYSTEM ENTITIES ATTRIBUTES ACTIVITIES EVENTS STATE VARIABLES
e. Fast food Customers Size of order Placing the Arrival at Number of customers
restaurant desired order the counter waiting
Paying for Completion Number of positions
the order of purchase operating
f. Hospital Patients Attention level Providing Arrival of Number of patients
emergency room required service the patient waiting
required
Departure of Number of physicians
the patient working
g. Taxicab company Fares Origination Traveling Pick-up Number of busy taxi cabs
of fare
Destination Number of fares
Drop-off waiting to be picked up
of fare
h. Automobile Robot Speed Spot welding Breaking Availability of
assembly line welders down machines
Breakdown rate
3. Abbreviated solution to Exercise 3:
Iteration Problem Formulation Setting of Objectives
and Overall Project Plan
1 Cars arriving at the in-
tersection are controlled
by a traffic light. The
cars may go straight,
turn left, or turn right.
How should the traffic light be se-
quenced? Criterion for evaluating
effectiveness: average delay time of
cars. Resources required: 2 people
for 5 days for data collection, 1 per-
son for 2 days for data analysis, 1
person for 3 days for model build-
ing, 1 person for 2 days for running
the model, 1 person for 3 days for
implementation.
2 Same as 1 above plus the
following: Right on red
is allowed after full stop
provided no pedestrians
are crossing and no vehi-
cle is approaching the in-
tersection.
How should the traffic light be se-
quenced? Criterion for evaluating
effectiveness: average delay time of
cars. Resources required: 2 people
for 8 days for data collection, 1 per-
son for 3 days for data analysis, 1
person for 4 days for model build-
ing, 1 person for 2 days for running
the model, 1 person for 3 days for
implementation.
3 Same as 2 above plus the
following: Trucks arrive
at the intersection. Ve-
hicles break down in the
intersection making one
lane impassable. Acci-
dents occur blocking traf-
fic for varying amounts of
time.
How should the traffic light be
sequenced? Should the road be
widened to 4 lanes? Method of eval-
uating effectiveness: average delay
time of all vehicles. Resources re-
quired: 2 people for 10 days for data
collection, 1 person for 5 days for
data analysis, 1 person for 5 days for
model building, 1 person for 3 days
for running the model, 1 person for
4 days for implementation.
CHAPTER 1. INTRODUCTION TO SIMULATION 3
4. Solution to Exercise 4:
Data Collection (step 4) - Storage of raw data in a file would allow rapid accessibility and a large memory
at a very low cost. The data could be easily augmented as it is being collected. Analysis of the data
could also be performed using currently available software.
Model Translation (step 5) - Many simulation languages are now available (see Chapter 4).
Validation (step 7) - Validation is partially a statistical exercise. Statistical packages are available for this
purpose.
Experimental Design (step 3) - Same response as for step 7.
Production Runs (step 9) - See discussion of step 5 above.
Documentation and Reporting (step 11) - Software is available for documentation assistance and for report
preparation.
5. Data Needed
Number of guests attending
Time required for boiling water
Time required to cook pasta
Time required to dice onions, bell peppers, mushrooms
Time required to saute onions, bell peppers, mushrooms, ground beef
Time required to add necessary condiments and spices
Time required to add tomato sauce, tomatoes, tomato paste
Time required to simmer sauce
Time required to set the table
Time required to drain pasta
Time required to dish out the pasta and sauce
Events
Begin cooking
Complete pasta cooking
Complete sauce cooking
Simultaneous
Arrival of dinner guests
Begin eating
Activities
Boiling the water
Cooking the pasta
Cooking sauce
Serving the guests
State variables
Number of dinner guests
Status of the water (boiling or not boiling)
Status of the pasta (done or not done)
Status of the sauce (done or not done)
7. Event
Deposit
Withdrawal
CHAPTER 1. INTRODUCTION TO SIMULATION 4
Activities
Writing a check
Cashing a check
Making a deposit
Verifying the account balance
Reconciling the checkbook with the bank statement
Chapter 2
Simulation Examples
For additional solutions check the course web site at www.bcnn.net.
4. Solution to Exercise 4:
L =
∞
i=0
iT
i
/T where
L = time weighted average number of customers in the system
T
i
= total time during [0,T] in which the system contains exactly i customers
L =
4
i=0
iT
i
/86 = [0(18) + 1(32) + 2(20) + 3(14) + 4(2)]/86 = 1.419 customers
L
Q
=
∞
i=0
iT
Q
i
/T where
L
Q
= time weighted average number of customers waiting during [0,T]
T
Q
i
= Total time during [0,T]inwhichexactlyi customers are waiting in the queue
L
Q
= [0(50) + 1(20) + 2(14) + 3(2)]/86 = .628 customers
6. Solution to Exercise 6:
New Service Distribution for Able
Service Probability Cumulative RD
Time
Probability Assignment
3 .30 .30 01-30
4
.30 .60 31-60
5
.25 .85 61-85
6
.15 1.00 86-00
6a.
Able Baker
Inter- Arrival RD for Time Time Time Time
Number RD for Arrival Clock Service Service Service Service Service Service Service Time in
Arrival Time Time Begins Time Ends Begins Time Ends Queue
1 - - 0 95 0 6 6 0
2
26 2 2 25 2 3 5 0
3
98 4 6 51 6 4 10 0
4
90 4 10 92 10 6 16 0
5
26 2 12 89 12 6 18 0
6
42 2 14 38 16 4 20 2
7
74 3 17 13 18 3 21 1
8
80 3 20 61 20 5 25 0
·
·
·
25 16 1 55 87 6 63 2
5
CHAPTER 2. SIMULATION EXAMPLES 6
Typical results of a simulation:
Able serves only 12 cars rather than 16 as in the previous simulation.
Average time in queue = 1.5 minutes.
6b. Simulation for Able, Baker and Charlie using some random digits.
Able Baker Charlie
Inter- Arrival RD for Time Time Time Time Time Time
Number RD for Arrival Clock Service Service Service Service Service Service Service Service Service Service Time in
Arrival Time Time Begins Time Ends Begins Time Ends Begins Time Ends Queue
1 - - 0 95 0 6 6 0
2 26 2 2 25 2 3 5 0
3 98 4 6 51 6 4 10 0
4
90 4 10 92 10 6 16 0
5
26 2 12 89 12 6 18 14 4 18 0
6
42 2 14 38 20 2
7
74 3 17 13 17 3 0
8
80 3 20 61 20 5 25 0
·
·
·
25 16 1 25 55 55 6 61 0
26
74 4 59 47 59 4 63
Typical results of a simulation:
Baker still has first shot at cars and thus has the most, or 12.
Able serves 8 cars, and Charlie gets the leftovers, or 6 cars.
There is no waiting time in the queue.
10. Profit = Revenue from retail sales - Cost of bagels made + Revenue from grocery store sales - Lost
profit.
Let Q = number of dozens baked/day
S =
i
0
i
,where0
i
= Order quantity in dozens for the ith customer
Q − S = grocery store sales in dozens, Q>S
S − Q = dozens of excess demand, S>Q
Profit = $5.40 min(S, Q) − $3.80Q +$2.70(Q − S) − $1.60(S − Q)
Number of
Probability Cumulative RD
Customers Probability Assignment
8 .35 .35 01-35
10 .30 .65 36-65
12 .25 .90 66-90
14 .10 1.00 91-100
Dozens Probability Cumulative RD
Ordered Probability Assignment
1 .4 .4 1-4
2 .3 .7 5-7
3 .2 .9 8-9
4 .1 1.0 0
CHAPTER 2. SIMULATION EXAMPLES 7
Pre-analysis
E(Number of Customers) = .35(8) + .30(10) + .25(12) + .10(14)
=10.20
E(Dozens ordered) = .4(1) + .3(2) + .2(3) + .1(4) = 2
E(Dozens sold) =
¯
S =(10.20)(2) = 20.4
E(Profit) = $5.40Min(
¯
S,Q) − $3.80Q +$2.70(Q −
¯
S) − $1.60(
¯
S − Q)
=$5.40Min(20.4,Q) − $3.80Q +$2.70(Q − 20.4)
−$0.67(20.4 − Q)
E(Profit|Q =0) = 0− 0+$1.60(20.4) = −$32.64
E(Profit|Q = 10) = $5.40(10) − $3.80(10) + 0 − $1.60(20.4 − 10)
= −$0.64
E(Profit|Q = 20) = $5.40(20) − $3.80(20) + 0 − $1.60(20.4 − 20)
= $15.36
E(Profit|Q = 30) = $5.40(20.4) − $3.80(30) + $2.70(30 − 20.4) − 0
= $22.08
E(Profit|Q = 40) = $5.40(20.4) − $3.80(40) + $2.70(40 − 20.4) − 0
= $11.08
The pre-analysis, based on expectation only, indicates that simulation of the policies Q =20, 30, and 40
should be sufficient to determine the policy. The simulation should begin with Q = 30, then proceed to
Q = 40, then, most likely to Q = 20.
Initially, conduct a simulation for Q =20, 30 and 40. If the profit is maximized when Q = 30, it will
become the policy recommendation.
The problem requests that the simulation for each policy should run for 5 days. This is a very short run
length to make a policy decision.
Q =30
CHAPTER 2. SIMULATION EXAMPLES 8
Day RD for Number of RD for Dozens Revenue Lost
Customer Customers Demand Ordered from Profit $
Retail $
1 44 10 8 3 16.20 0
2 1 5.40 0
4 1 5.40 0
8 3 16.20 0
1 1 5.40 0
6 2 10.80 0
3 1 5.40 0
0 4 21.60 0
2 1 5.40 0
0 4 21.60 0
21 113.40 0
For Day 1,
Profit = $113.40 − $152.00 + $24.30 − 0 = $14.30
Days 2, 3, 4 and 5 are now analyzed and the five day total profit is determined.
11. Solution to Exercise 11:
Daily
Probability Cumulative RD
Demand Probability Assignment
0 .33 .33 01-33
1 .25 .58 34-58
2 .20 .78 59-78
3 .12 .90 79-90
4 .10 1.00 91-00
Lead Probability Cumulative RD
Time Probability Assignment
1 .3 .3 1-3
2 .5 .8 4-8
3 .2 1.0 9-0
CHAPTER 2. SIMULATION EXAMPLES 9
Cycle Day Beginning RD for Demand Ending Shortage Order RD for Days Until
Inventory Demand Inventory Quantity Quantity Lead Time Order Arrives
1 1 12 56 1 11 0
2 11 30 0 11 0
3 11 79 3 8 0
4 8 84 3 5 0
5 5 20 0 5 0
6 5 10 0 5 0
7 5 83 3 2 0 10 2 1
2 1 2 62 2 0 0 0
2 10 58 1 9 0
3 9 32 0 9 0
4 9 42 1 8 0
5 8 87 3 5 0
6 5 88 3 2 0
7 2 00 4 0 2 10 7 2
.
.
.
6 1 0 71 2 0 2 1
2 10 34 1 7 0 0
3 7 14 0 7 0
4 7 46 1 6 0
5 6 84 3 3 0
6 3 09 0 3 0
7 3 65 2 1 0 10 2 1
Typical results from simulation of current system:
Probability of shortage = 0.25
Average ending inventory = 3.5 units
Effect on Shortages Caused by Policy Variable Changes
Policy Variable
Change Review Reorder Reorder
Period Quantity Point
Increase Increase Decrease No effect in this
case since all values
were below current
reorder point.
Decrease Decrease Increase Decrease would have
to be drastic, say to
a reorder point of
< 2 units. Such
a change would in-
crease shortages.
12. Solution to Exercise 12:
Daily
Probability Cumulative RD
Demand Probability Assignment
0 .18 .18 01-18
1 .39 .57 19-57
2 .29 .86 58-86
3 .09 .95 87-95
4 .05 1.00 96-00
CHAPTER 2. SIMULATION EXAMPLES 10
Lead
Probability Cumulative RD
Time Probability Assignment
0 .135 .135 001-135
1 .223 .358 136-358
2 .288 .646 359-646
3 .213 .859 647-859
4 .118 .977 860-977
5 .023 1.000 978-000
RD for Lead Time RD for Demand Lead Time
Cycle
Lead Time Demand Demand
1 024 0 - - 0
2
330 1 14 0 0
3
288 1 53 1 1
4
073 0 - - 0
5
197 1 24 1 1
6
924 4 53 1
81 2
70 2
18 0 5
Narrow histogram intervals (say 1 time unit) seem to be more descriptive and less blocky than larger
intervals. For a realistic determination many more cycles would need to be simulated. With a large number
of cycles, narrow histogram intervals will probably be favored.
15. Solution to Exercise 15:
Time Between
Probability Cumulative RD
Calls Probability Assignment
15 .14 .14 01-14
20 .22 .36 15-36
25 .43 .79 37-79
30 .17 .96 80-96
35 .04 1.00 97-00
Service Probability Cumulative RD
Time Probability Assignment
5 .12 .12 01-12
15 .35 .47 13-47
25 .43 .90 48-90
35 .06 .96 91-96
45 .04 1.00 97-00
First, simulate for one taxi for 5 days.
Then, simulate for two taxis for 5 days.
Shown on simulation tables
Comparison
Smalltown Taxi would have to decide which is more important—paying for about 43 hours of idle time
in a five day period with no customers having to wait, or paying for around 4 hours of idle time in a five day
period, but having a probability of waiting equal to 0.59 with an average waiting time for those who wait of
around 20 minutes.
CHAPTER 2. SIMULATION EXAMPLES 11
One Taxi
Day Call RD for Time Time Call RD for Service Time Time Time Time Idle Time
between between Time Service Time Service Customer Service Customer of Taxi
Calls Calls Time Begins Wai t s Ends in System
1 1 15 - 0 01 5 0 0 5 5 0
2 01 20 20 53 25 20 0 55 25 0
3 14 15 35 62 25 55 20 80 45 0
4 65 25 60 55 25 80 20 105 45 0
5 73 25 85 95 35 105 20 140 55 0
6 48 25 110 22 15 140 30 155 45 0
.
.
.
20 77 25 444 63 25 470 25 495 50 0
2
.
.
.
Typical results for a 5 day simulation:
Total idle time = 265 minutes = 4.4 hours
Average idle time per call = 2.7 minutes
Proportion of idle time = .11
Total time customers wait = 1230 minutes
Average waiting time per customer = 11.9 minutes
Number of customers that wait = 61 (of 103 customers)
Probability that a customer has to wait = .59
Average waiting time of customers that wait = 20.2 minutes
Two taxis (using common RDs for time between calls and service time)
Taxi 1 Taxi 2
Day Call Time Call Service Time Service Time Time Service Time Time Time Idle Idle
between Time Time Service Time Service Service Time Service Customer Customer Time Time
Calls Begins Ends Begins Ends Wai t s in System Taxi 1 Ta xi 2
1 1 - 0 5 0 5 5 0 5
2 20 20 25 20 25 45 0 25
3 15 35 25 35 25 60 0 25 35
4 25 60 25 60 25 85 0 25 15
5 25 85 35 80 35 120 0 35
6 25 110 15 110 15 125 0 15 50
.
.
.
20 20 480 25 480 25 505 0 25 10
2
.
.
.
Typical results for a 5 day simulation:
Idle time of Taxi 1 = 685 minutes
Idle time of Taxi 2 = 1915 minutes
Total idle time = 2600 minutes = 43 hours
Average idle time per call = 25.7 minutes
Proportion of idle time = .54
Total time customers wait = 0 minutes
Number of customers that wait = 0
17. Solution to Exercise 17:
X = 100 + 10RNN
x
Y = 300 + 15RNN
y
Z =40+8RNN
z
Typical results...
CHAPTER 2. SIMULATION EXAMPLES 12
RNN
x
X RNN
y
Y RNN
z
Z W
1 -.137 98.63 .577 308.7 -.568 35.46 11.49
2
.918 109.18 .303 304.55 -.384 36.93 11.20
3
1.692 116.92 -.383 294.26 -.198 38.42 10.70
4
-.199 98.01 1.033 315.50 .031 40.25 10.27
5
-.411 95.89 .633 309.50 .397 43.18 9.39
.
.
.
19. Solution to Exercise 19:
T =LeadTime
T ∼ N(7, 2
2
)
T =7+2(RNN)(Rounded to nearest integer)
Daily
Probability Cumulative RD
Demand Probability Assignment
0 0.367 0.367 001-367
1 0.368 0.735 368-735
2 0.184 0.919 736-919
3 0.062 0.981 920-981
4 0.019 1.000 982-000
Cycle RNN for Lead Day RD for Demand Lead
Lead Time Time Demand Time
Demand
1 -.82 5 1 127 0
2 313 0
3 818 2
4 259 0
5 064 0 2
2 -.45 6 1 912 2
2 651 1
3 139 0
4 288 0
5 524 1
6 772 2 6
.
.
.
21. Solution to Exercise 21:
Lead Time
Probability Cumulative RD
(Days) Probability Assignment
0 .166 .166 001-166
1 .166 .332 167-332
2 .166 .498 333-498
3 .166 .664 499-664
4 .166 .830 665-830
5 .166 .996 831-996
996-000
(discard)
CHAPTER 2. SIMULATION EXAMPLES 13
Assume 5-day work weeks.
D =Demand
D =5+1.5(RNN)( Rounded to nearest integer)
Week Day Beginning RN N for Demand Ending Order RD for Lead Lost
Inventory Demands Inventory Quantity Lead Time Time Sales
1 1 18 -1.40 3 15 0
2 15 -.35 4 11 0
3 11 -.38 4 7 13 691 4 0
4 7 .05 5 2 0
5 2 .36 6 0 4
2 6 0 .00 5 0 5
7 0 -.83 4 0 4
8 13 -1.83 2 11 0
9 11 -.73 4 7 13 273 1 0
10 7 -.89 4 3 0
.
.
.
Typical results
Average number of lost sales/week = 24/5 = 4.8 units/weeks
22. Solution to Exercise 22:
Material A (200kg/box)
Interarrival
Probability Cumulative RD
Time Probability Assignment
3 .2 .2 1-2
4 .2 .4 3-4
5 .2 .6 5-6
6 .2 .8 7-8
7 .2 1.0 9-0
Box RD for Interarrival Clock
Interarrival Time Time Time
1 1 3 3
2 4 4 7
3 8 6 13
4 3 4 17
.
.
.
14 4 4 60
Material B (100kg/box)
Box
1 2 3 ··· 10
Clock Time 6 12 18 ··· 60
CHAPTER 2. SIMULATION EXAMPLES 14
Material C (50kg/box)
Interarrival
Probability Cumulative RD
Time Probability Assignment
2 .33 .33 01-33
3 .67 1.00 34-00
Box RD for Interarrival Clock
Interarrival Time Time Time
1 58 5 3
2 92 3 6
3 87 3 9
4 31 2 11
.
.
.
.
.
.
.
.
.
.
.
.
22 62 3 60
Clock A B C
Time
Arrival Arrival Arrival
3 1 1
6
1 2
7
2
9 3
11
4
12
2
.
.
.
Simulation table shown below.
Typical results:
Average transit time for box A (
¯
t
A
)
¯
t
A
=
Total waiting time of A +(No.ofboxesofA)(1 minute up to unload)
No. of boxes of A
=
28 + 12(1)
12
=3.33 minutes
Average waiting time for box B (¯w
B
)
¯w
B
=
(Total time B in Queue)
No. of boxes of B
=
10
10
= 1 minute/box of B
Total boxes of C shipped = Value of C Counter = 22 boxes
Clock No. of A No. of B No. of C Queue Time Time Time A Time B A B C
Time
in Queue in Queue in Queue Wei g ht Service Service in Queue in Queue Counter Counter Counter
Begins Ends
3 1 0 1 250
6 0 0 0 0 6 10 3 0 1 1 2
7
1 0 0 200
9 1 0 1 250
11 1 0 2 300
12 0 0 0 350 12 16 5 0 2 2 4
.
.
.
CHAPTER 2. SIMULATION EXAMPLES 15
25. Solution can be obtained from observing those clearance values in Exercise 24 that are greater than
0.006.
26. Degrees =360(RD/100)
Replication 1
RD
Degrees
57 205.2
45
162.0
22
79.2
Range = 205.2
0
− 79.2
0
= 126
0
(on the same semicircle).
Continue this process for 5 replications and estimate the desired probability.
27. Solution to Exercise 27:
V =1.02
2
+(−.72)
2
+ .28
2
=1.7204
T =
−
.18
1.7204
3
= −.2377
28. Solution to Exercise 28:
Cust. RD for IAT AT RD for Serv. No. in TimeServ. Time Serv. Go Into
Arrival Service Time Queue Begins Ends Bank?
1 30 2 2 27 2 1 -
√
2 46 2 4 26 2 0 4 6
3 39 2 6 99 4 0 6 10
4 86 4 10 72 3 0 10 13
5 63 3 13 12 1 0 13 14
6 83 4 17 17 1 0 17 18
7 07 0 17 78 3 1 18 21
8 37 2 19 91 4 1 -
√
9 69 3 22 82 3 0 22 25
10 78 4 26 62 3 0 26 29
Chapter 3
General Principles
Forsolutionscheckthecoursewebsiteatwww.bcnn.net.
16
Chapter 4
Simulation Software
Forsolutionscheckthecoursewebsiteatwww.bcnn.net.
17
Chapter 5
Statistical Models in Simulation
1. Let X be defined as the number of defectives in the sample. Then X is binomial (n = 100,p= .01) with
the probability mass function
p(x)=
100
x
(.01)
x
(.99)
100−x
,x=0, 1,...,100
The probability of returning the shipment is
P (X>2) = 1 − P (X ≤ 2)
=1−
100
0
(.99)
100
−
100
1
(.01)(.99)
99
−
100
2
(.01)
2
(.99)
98
= .0794
2. Let X be defined as the number of calls received until an order is placed. Then, X is geometric (p = .48)
with the probability mass function
p(x)=(.52)
x−1
(.48),x=0, 1, 2 ...
(a) The probability that the first order will come on the fourth call is
p(4) = .0675
(b) The number of orders, Y, in eight calls is binomial
(n =8,p= .48) with the probability mass function
p(y)=
8
y
(.48)
y
(.52)
8−y
,y =0, 1,...,8
The probability of receiving exactly six orders in eight calls is
p(6) = .0926
(c) The number of orders, X, in four calls is binomial (n =4,p = .48) with
probability mass function
p(x)=
4
x
(.48)
x
(.52)
8−x
,x=0, 1, 2, 3, 4
18
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 19
The probability of receiving one or fewer orders in four calls is
P (X ≤ 1) =
4
0
(.52)
4
+
4
1
(.48)(.52)
3
= .3431
3. Let X be defined as the number of women in the sample never married
P (2 ≤ X ≤ 3) = p(2) + p(3)
=
20
2
(.18)
2
(.82)
18
+
20
3
(.18)
3
(.82)
17
= .173 + .228 = .401
4. Let X be defined as the number of games won in the next two weeks. The random variable X is described
by the binomial distribution:
p(x)=
5
x
(.55)
x
(.45)
5−x
P (3 ≤ X ≤ 5) = p(3) + p(4) + p(5)
=
5
3
(.55)
3
(.45)
2
+
5
4
(.55)
4
(.45) +
5
5
.55
5
= .337 + .206 + .050 = .593
5. Solution to Exercise 5:
(a) Using the geometric probability distribution, the desired probability is given by
p(.4) = (.6)
3
(.4) = .0864
(b) Using the binomial distribution, the desired probability is given by
P (X ≤ 2) =
5
i=0
5
i
(.4)
i
(.6)
5−i
= .07776 + .2592 + .3456
= .68256
6. X = X
1
+ X
2
∼ Erlang with Kθ =1. SinceK =2,θ =1/2
F (2) = 1 −
1
i=0
e
−2
2
i
/i!=0.406
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 20
P (X
1
+ X
2
> 2) = 1 − F (2) = .594
7. The geometric distribution is memoryless if
P (X>s+ t|X>s)=P (X>t)
where s and t are integers and X is a geometrically distributed random variable. The probability of a failure
is denoted by q and
P (X>s)=
∞
j=s+1
q
j−1
p = q
s
,
P (X>t)=q
t
, and
P (X>s+ t)=q
s+t
;so,
P [(X>s+ t)|X>s]=
q
s+t
/q
s
= q
t
which is equal to P (X>t).
8. The number of hurricanes per year, X, is Poisson (α =0.8) with the probability mass function
p(x)=e
−0.8
(0.8)
x
/x!,x=0, 1,...
(a) The probability of more than two hurricanes in one year is
P (X>2) = 1 − P (X ≤ 2)
=1− e
−0.8
− e
−0.8
(0.8) − e
−0.8
(0.8
2
/2)
= .0474
(b) The probability of exactly one hurricane in one year is
p(1) = .3595
9. The number of arrivals at a bank teller’s cage, X, is Poisson (α =1.2) with the probability mass function
p(x)=e
−1.2
(1.2)
x
/x!,x=0, 1, 2,...
(a) The probability of zero arrivals during the next minute is
p(0) = .3012
(b) The probability of zero arrivals during the next two minutes (α =2.4) is p(0) = 0.0907.
10. Using the Poisson approximation with the mean, α,givenby
α = np = 200(.018) = 3.6
The probability that 0 ≤ x ≤ 3 students will drop out of school is given by
F (3) =
3
x=o
e
α
α
x
x!
= .5148
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 21
11. Let X be the number of calls received. The variance and mean are equal. Thus,
σ
2
= α =4
and the standard deviation is
σ =2
Then using the Poisson distribution
P (X> 6) = 1 − .889 = .111
12. Let X be defined as the lead time demand. Then, X is Poisson (α = 6) with cumulative distribution
function
F (x)=
x
i=0
e
−6
(6)
i
/i!
The order size at various protection levels is given by:
Order Size Protection(%) F (x)
6 50 .606
8 80 .847
9 90 .916
10 95 .957
11 97 .979
11 97.5 .979
12 99 .991
13 99.5 .996
15 99.9 .999
13. A random variable, X, has a discrete uniform distribution if its probability mass function is
p(x)=1/(n +1) R
X
= {0, 1, 2,...n}
(a) The mean and variance are found by using
n
i=0
i =[n(n +1)]/2and
n
i=0
i
2
=[n(n + 1)(2n +1)]/6
E(X)=
n
i=0
x
i
p(x
i
)=
n
i=0
ip(i)
=[1/(n +1)]
n
i=0
i = n/2
V (X)=E(X
2
) − [E(X)]
2
=
n
i=0
x
2
i
p(x
i
) − (n/2)
2
=(n
2
+2n)/12
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 22
(b) If R
X
= {a, a +1,a+2,...,b}, the mean and variance are
E(X)=a +(b − a)/2=(a + b)/2
V (X)=[(b − a)
2
+2(b − a)]/12
14. Let X be defined as the lifetime of the satellite. Then, X is exponential (λ = .4) with cumulative
distribution function
F (x)=1− e
−.4x
,x≥ 0
(a) The probability of the satellite lasting at least five years is
P (X ≥ 5) = 1 − F (5) = .1353
(b) The probability that the satellite dies between three and six years is
P (3 ≤ X ≤ 6) = F (6) −F (3) = .2105
15. Let X be the number of hours until a crash occurs. Using the exponential distribution, the desired
probability is given by
F (48) − F (24) = [1 − e
−
1
36
(48)
] − [1 − e
−
1
36
(24)
]
= e
−2/3
− e
−4/3
= .513 − .264 = .249
16. Let X be defined as the number of ball bearings with defects in a random sample of 4000 bearings.
Then, X is binomial (n = 4000,p=1/800) with probability mass function
p(x)=
4000
x
(1/800)
x
(1 − (1/800))
n−x
,x=0, 1, 2,...,4000
The probability that the random sample yields three or fewer ball bearings with defects is
P (X ≤ 3) = p(0) + p(1) + p(2) + p(3)
= .2649
Also, X can be approximated as Poisson (λ = 4000/800) with a probability mass function
p(x)=e
−5
(5)
x
/x!,x=0, 1, 2,...
The probability that the random sample yields three or fewer ball bearings with defects is
P (X ≤ 3) = p(0) + p(1) + p(2) + p(3)
= .2650
17. An exponentially distributed random variable, X, that satisfies
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 23
P (X ≤ 3) = .9P (X ≤ 4),
can be specified by letting
1 − e
−3λ
= .9(1 − e
−4λ
)
By letting z = e
−λ
,
0=z
3
− .90z
4
− .10, or
z = .6005 and λ = .51
18. Let X be the number of accidents occuring in one week. The mean is given by
α =1
The probability of no accidents in one week is given by
p(0) =
e
−1
α
0
0!
= .368
The probability of no accidents in three successive weeks is given by
[p(0)]
3
= .368
3
= .05
19. Let X be defined as the lifetime of the component. Then X is exponential (λ =1/10, 000 hours) with
cumulative distribution function
F (x)=1− e
−x/10000
,x>0
Given that the component has not failed for s =10, 000 or s =15, 000 hours, the probability that it lasts
5000 more hours is
P (X ≥ 5000 + s|X>s)=P (X ≥ 5000) = .6065
In both cases, this is due to the memoryless property of the exponential distribution.
20. Let X be defined as the lifetime of the battery. Then, X is exponential (λ =1/48) with cumulative
distribution function
F (x)=1− e
−x/48
,x>0
(a) The probability that the battery will fail within the next twelve months, given that it has operated for
sixty months is
P (X ≤ 72|X>60) = P (X ≤ 12)
= F (12) = .2212
due to the memoryless property.
(b) Let Y be defined as the year in which the battery fails, Then,
P (Y =oddyear)=(1− e
−.25
)+(e
−.50
e
−.75
)+...
P (Y =evenyear)=(1− e
−.50
)+(e
−.75
− e
−1
)+...
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 24
So,
P (Y =evenyear)=e
−.25
P (Y =oddyear),
P (Y =evenyear)+P (Y =oddyear)=1, and
e
−.25
P (Y =oddyear)=1− P (Y =oddyear)
The probability that the battery fails during an odd year is
P (Y =oddyear)=1/(1 + e
−.25
)=.5622
(c) Due to the memoryless property of the exponential distribution, the remaining expected lifetime is 48
months.
21. Service time, X
i
, is exponential (λ =1/50) with cumulative distribution function
F (x)=1− e
−x/50
,x> 0
(a) The probability that two customers are each served within one minute is
P (X
1
≤ 60,X
2
≤ 60) = [F (60)]
2
=(.6988)
2
= .4883
(b) The total service time, X
1
+ X
2
, of two customers has an Erlang distribution (assuming independence)
with cumulative distribution function
F (x)=1−
1
i=0
[e
−x/50
(x/50)
i
/i!],x>0
The probability that the two customers are served within two minutes is
P (X
1
+ X
2
≤ 120) = F (120) = .6916
22. A random variable, X, has a triangular distribution with probability density function
f(x)=
[2(x − a)]/[(b − a)(c − a)],a≤ x ≤ b
[2(c − x)] /[(c − b)(c − a)],b≤ x ≤ c
Thevarianceis
V (X)=E(X
2
) − [E(X)]
2
E(X)=(a + b + c)/3
E(X
2
)=
2
(b − a)(c − a)
b
a
x
2
(x − a)dx
+
2
(c − b)(c − a)
c
b
x
2
(c − x)dx
=[1/6(c − a)][c(c
2
+ cb + b
2
) − a(b
2
+ ab + a
2
)]
V (X)=[(a + b + c)
2
/18] − [(ab + ac + bc)/6]
23. The daily use of water, X,isErlang(k =2,θ = .25) with a cumulative distribution function
F (x)=1−
2−1
i=0
[e
−x/2
(x/2)
i
/i!],x>0
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 25
The probability that demand exceeds 4000 liters is
P (X>4) = 1 − F (4) = .4060
24. Let X
i
be defined as the lifetime of the ith battery and X = X
1
+ X
2
+ X
3
.ThenX is Erlang
(k =3,θ =1/36) with cumulative distribution function
F (x)=1−
3−1
i=0
[e
−x/12
(x/12)
i
/i!],x>0
The probability that three batteries are sufficient is
P (X>30) = 1 − F (30) = .5438
25. Let X represent the time between dial up connections. The desired probability is Erlang distributed
with
Kθ =1/15 and X =30
The probability that the third connection occurs within 30 seconds is given by
F (30) = 1 −
2
i=0
e
−
1
15
(30)
[
1
15
30]
i
i!
= .323
and its complement gives the desired probability, or 1 − .323 = .677.
26. Let X represent the life of a single braking system. Using the Erlang distribution, the probability of a
crash within 5,000 hours is given by
F (5, 000) = 1 −
1
i=0
e
−2(8,000)(5,000)
[2(1/8, 000)(5, 000)]
i
i!
= i − e
−5/4
− e
−5/4
(5/4)
=1− .2865 − .3581 = .3554
The complement gives the desired probability, or,
p(no crash) = .6446
27. Let X represent the time until a car arrives. Using the Erlang distribution with
Kθ =4andX =1
the desired probability is given by
F (1) = 1 −
2
i=o
e
−4(1)
[4(1)]
i
i!
= .762
28. Let X be defined as the number of arrivals during the next five minutes. Then X is Poisson (α =2.5)
with cumulative distribution function
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 26
F (x)=
x
i=0
e
−2.5
(2.5)
i
/i!,x=0, 1,...
The probability that two or more customers will arrive in the next five minutes is
P (X ≥ 2) = 1 − F (1) = .7127
29. Let X be defined as the grading time of all six problems. Then X is Erlang (k =6,θ =1/180) with
cumulative distribution function
F (x)=1−
6−1
i=0
[e
−x/30
(x/30)
i
/i!],x>0
(a) The probability that grading is finished in 150 minutes or less is
P (X ≤ 150) = F (150) = .3840
(b) The most likely grading time is the mode = (k − 1)/kθ = 150 minutes.
(c) The expected grading time is
E(X)=1/θ = 180 minutes
30. Let X be defined as the life of a dual hydraulic system consisting of two sequentially activated hydraulic
systems each with a life, Y , which is exponentially distributed (λ = 2000 hours). Then X is Erlang (k =
2,θ =1/4000) with cumulative distribution function
F (x)=1
2−1
i=0
[e
−x/2000
(x/2000)
i
/i!],x>0
(a) The probability that the system will fail within 2500 hours is
P (X ≤ 2500) = F (2500) = .3554
(b) The probability of failure within 3000 hours is
P (X ≤ 3000) = F (3000) = .4424
If inspection is moved from 2500 to 3000 hours, the probability that the system will fail increases by .087.
32. Letting X represent the lead time in 100’s of units, the Erlang distribution with
β = K =3,θ=1, and X =2
will provide the probability that the lead time is less than 2 with
F (2) = 1 −
2
i=o
e
−6
6
i
i!
= .938
The complement gives the desired probability, or
P (Lead Time ≥ 2) = 1 − .938 = .062
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 27
33. Let X be the lifetime of the card in months. The Erlang distribution gives the desired probability where
β = K =4,Kθ=4(1/16) =
1
4
, and X =24
Then
F (24) = 1 −
3
i=o
e
6
6
i
i!
=1− .151 = .849
The complement gives the desired probability, or
P (X ≥ 2years)=1− .849 = .151
34. Let X be defined as the number on a license tag. Then X is discrete uniform (a = 100,b = 999) with
cumulative distribution function
F (x)=(x − 99)/900,x= 100, 101,...,999
(a) The probability that two tag numbers are 500 or higher is
[P (X ≥ 500)]
2
=[1− F (499)]
2
= .5556
2
= .3086
(b) Let Y be defined as the sum of two license tag numbers. Then Y is discrete triangular which can be
approximated by
F (y)=
(y −a)
2
/[(b − a)(c − a)],a≤ y ≤ b
1 − [(c − y)
2
/[(c − a)(c − b)]],b≤ y ≤ c
where a = 2(100) = 200, c = 2(999) = 1998, and b = (1998 + 200)/2 = 1099.
The probability that the sum of the next two tags is 1000 or higher is
P (Y ≥ 1000) = 1 − F (999) = .6050
35. A normally distributed random variable, X, with a mean of 10, a variance of 4, and the following
properties
P (a<X<b)=.90 and |µ − a| = |µ − b|
exists as follows
P (X<b)=P (X>a)=.95 due to symmetry
Φ[(b − 10)/2] = .95 b =13.3
1 − Φ[(a − 10)/2] = .95 a =6.7
36. Solution to Exercise 36:
Normal (10, 4)
F (8) − F (6) = F
8 − 10
2
− F
6 − 10
2
= F (−1) − F (−2) = (1 − .84134) − (1 − .97725)
= .13591
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 28
Triangular (4, 10, 16)
F (8) − F (6) =
(8 − 4)
2
(10 − 4)(16 − 4)
−
(6 − 4)
2
(10 − 4)(16 − 4)
=1/6=.1667
Uniform (4, 16)
F (8) − F (6) =
(8 − 4)
16 − 4
−
(6 − 4)
16 − 4
=1/6=.1667
37. Letting X be the random variable
Z =
x − u
σ
2.33 =
x − 20
2
x =24.66 (1%)
1.645 =
x − 20
2
x =23.29 (5%)
1.283 =
x − 20
2
x =22.57 (10%)
38. Let X be defined as I.Q. scores. Then X is normally distributed (µ = 100,σ = 15).
(a) The probability that a score is 140 or greater is
P (X ≥ 140) = 1 − Φ[140 − 100)/15] = .00383
(b) The probability that a score is between 135 and 140 is
P (135 ≤ X ≤ 140) = Φ[(140 − 100)/15] − Φ[(135 − 100)/15]
= .00598
(c) The probability that a score is less than 110 is
P (X<110) = Φ[(110 − 100)/15] = .7475
39. Let X be defined as the length of the ith shaft, and Y as the linkage formed by i shafts. Then X
i
is
normally distributed.
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 29
(a) The linkage, Y , formed by the three shafts is distributed as
Y ∼ N
3
i=1
µ
i
,
3
i=1
σ
2
i
Y ∼ N (150,.25)
(b) The probability that the linkage is larger than 150.2 is
P (Y>150.2) = 1 − Φ[(150.2 − 150)/.5] = .3446
(c) The probability that the linkage is within tolerance is
P (149.83 ≤ Y ≤ 150.21) = Φ[(150.21 − 150)/.5] − Φ[(149.83 − 150)/.5]
= .2958
40. Let X be defined as the circumference of battery posts. Then X is Weibull (γ =3.25,β =1/3,α = .005)
with cumulative distribution function
F (x)=1− exp[−((x − 3.25)/.005)
1/3
] ,x≥ 3.25
(a) The probability of a post having a circumference greater than 3.40 is
P (X>3.40) = 1 − F (3.40) = .0447
(b) The probability of a post not meeting tolerance is
1 − P (3.3 <X<3.5) = 1 − F (3.5) + F (3.3) = .9091
41. Let X be defined as the time to failure of a battery. Then X is Weibull (γ =0,β =1/4,α =1/2) with
cumulative distribution function
F (x)=1− exp[−(2x)
1/4
],x≥ 0
(a) The probability that a battery will fail within 1.5 years is
P (X<1.5) = F (1.5) = .7318
(b) The mean life of a battery is
E(X)=(1/2)Γ(4 + 1) = 12 years
The probability of a battery lasting longer than twelve years is
P (X>12) = 1 − F (12) = .1093
(c) The probability that a battery will last from between 1.5 and 2.5 years is
P (1.5 ≤ X ≤ 2.5) = F (2.5) − F (1.5) = .0440
42. Let X be the demand for electricity. Suppose
1000 = a<median = 1425 ≤ b =Mode
so that the probability that the demand is less than or equal to 1425 kwh is given by
F (1425) = 0.5=
(1425 − 1000)
2
(b − 1000)(1800 − 1000)
=
425
2
(b − 1000(800)
implying b = 1451.56 kwh. Since 1451.56 ≥ 1425 we have Mode = 1451.56.
43. Letting X represent the time to failure
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 30
(a) E(X) = 100Γ(1 + 2) = 1000Γ(3) = 2000 hours
(b) F (3000) = 1 − exp
−
3000
1000
1
2
F (3000) = 1 − e
−1.732
= .823
44. Let X be defined as the gross weight of three axle trucks. Then X is Weibull (γ =6.8,β =1.5,α=1/2)
with cumulative distribution function
F (x)=1− exp[−((x − 6.8)/.5)
1.5
],x≥ 6.8
The weight limit, a, such that .01 of the trucks are considered overweight is
P (X>a)=1− F (a)=.01
exp[−((a − 6.8)/.5)
1.5
]=.01
a =8.184 tons
45. Let X be defined as the car’s gas mileage. Then X is triangular (a =0,c= 50) with an expected value,
E(X), equal to 25.3 miles per gallon.
The median can be determined by first finding the mode, b, by setting
E(X)=(a + b + c)/3=25.3
b =25.9 miles per gallon,
then, determining which interval of the distribution contains the median by setting
F (b)=(x − a)
2
/[(b − a)(c − a)],a≤ x ≤ b
to compute F (25.9) = .518, so the median is in the interval (0,25.9). The median is then computed by
finding x such that F (x)=.50, or median = 25.45 miles per gallon.
46. Let T represent the time to complete the route. Then T ∼ N(µ
T
,σ
2
T
)
(a) µ
T
=
i
µ
i
= 38 + 99 + 85 + 73 + 52 + 90 + 10 + 15 + 30 = 492 minutes
(b) σ
2
T
=
i
σ
2
i
= 16 + 29 + 25 + 20 + 12 + 25 + 4 + 4 + 9 = 144 minutes
2
and σ
T
=12minutes
Φ(z)=Φ
x−µ
σ
T
=Φ
480−492
12
=Φ(−1) = .3413
P (X>480) = 1 − .3413 = .6587
(c) P (X>2) = 1 − P (X<2) = 1 −
2
x=0
6
x
(.6587)
x
(.3413)
6−x
=1− .108 = .892
(d) P (456 <X<504) = F (504) − F (456)
=Φ
504−496
12
− Φ
456−496
12
=Φ(2/3) − Φ(−31/3) = .7476 − .0001 = .7475
47. 1 − F (600) = exp[−(600/400)
1/2
]=e
−(1.5)1/2
= e
−1.22
= .295
48. R(x)=1− F (x)=
2
i=0
e
−.0001(32,000)
[(.001)(32,000)]
i
i!
= .2364
CHAPTER 5. STATISTICAL MODELS IN SIMULATION 31
49. Solution to Exercise 49.
(a)
E(X
2
)=
92
85
x
2
(2)(x − 85)
119
dx +
102
95
x
2
(2)(102 − x)
170
dx = 3311.75 + 5349.41 = 8661.16
E(X)=(a + b + c)/3 = (85 + 92 + 102)/3=93
V (X)=E(X
2
) − [E(X)]
2
= 8661.16 − (93)
2
=12.16
◦
F
2
(b)
0.5=1−
(102 − x)
2
170
(102 − x)
2
=85
x =92.8
◦
F
(c) Mode = b =92
◦
F
50. (a) E(X)=1.8+1/3Γ(2+1)=1.8+1/3(2) = 2.47 × 10
3
hours
F (2.47) = 1 − exp
−
2.47 − 1.80
.33
1/2
=1− exp[−(2)
1/2
]=.757
P (X>2.47) = 1 − .757 = .243
(b)
.5=1− exp
−
x − 1.8
.33
1/2
, where x =median
.5=exp
−
x − 1.8
.33
1/2
n .5=−
x − 1.8
.33
1/2
x =1.96 × 10
3
hours
51.
F (4) = 1 −
1
i=0
e
−2(1/4)4
[2(1/4)(4)]
i
i!
=1−
1
i=0
e
−2
2
i
i!
= .594
P (X>4) = 1 − .594 = .406
Chapter 6
Queueing Models
For Maple procedures that help in evaluating queueing models see the course web site at www.bcnn.net.
1. The tool crib is modeled by an M/M/c queue (λ =1/4,µ =1/3,c = 1 or 2). Given that attendants are
paid $6 per hour and mechanics are paid $10 per hour,
Mean cost per hour = $10c + $15L
assuming that mechanics impose cost on the system while in the queue and in service.
CASE 1: one attendant - M/M/1 (c =1,ρ= λ/µ = .75)
L = ρ/(1 − ρ) = 3 mechanics
Mean cost per hour = $10(1) + $15(3) = $55 per hour.
CASE 2: two attendants - M/M/2 (c =2,ρ= λ/cµ = .375)
L = cρ +
(cρ)
c+1
P
0
/
c(c!)(1 − ρ)
2
= .8727,
where
P
0
=
c−1
n=0
(cρ)
n
/n!
+[(cρ)
c
(1/c!)(1/(1 − ρ))]
−1
= .4545
Mean cost per hour = $10(2) + $15(.8727) = $33.09 per hour
It would be advisable to have a second attendant because long run costs are reduced by $21.91 per hour.
2. A single landing strip airport is modeled by an M/M/1 queue (µ =2/3). The maximum arrival rate, λ,
such that the average wait, w
Q
, does not exceed three minutes is computed as follows:
w
Q
= λ/[µ(µ − λ)] ≤ 3
or
λ = µ/[1/µw
Q
+1]≤ .4444 airplanes per minute.
Therefore, λ
max
= .4444 airplanes per minute.
3. The Port of Trop is modeled by an M/M/1/4 queue (λ =7,µ =8,a =7/8,N = 4). The expected
number of ships waiting or in service, L,is
L =
a[1 − (N +1)a
N
+ Na
N+1
]
(1 − a
N+1
)(1 − a)
=1.735 ships
32
CHAPTER 6. QUEUEING MODELS 33
since λ = µ and system capacity is N = 4 ships.
4. String pulling at City Hall is modeled by an M/M/2 queue (λ =1/10,µ=1/15,ρ= .75).
(a) The probability that there are no strings to be pulled is
P
0
=
c−1
n=0
(cρ)
n
/n!
+[(cρ)
c
(1/c!)/(1 − ρ)]
−1
= .1429
(b) The expected number of strings waiting to be pulled is
L
Q
=
(cρ)
c+1
P
0
/
c(c!)(1 − ρ)
2
=1.929 strings
(c) The probability that both string pullers are busy is
P (L(∞) ≥ 2) =
(cρ)
2
P
0
/ [c!(1 − ρ)] = .643
(d) If a third string puller is added to the system, (M/M/3 queue, c =3,ρ= .50), the measures of performance
become
P
0
= .2105,L
Q
= .2368,P(L(∞) ≥ 3) = .2368
5. The bakery is modeled by an M/G/1 queue (µ =4,σ
2
= 0). The maximum arrival rate, λ, such that the
mean length of the queue, L
Q
, does not exceed five cakes is
L
Q
=[λ
2
/2µ
2
(1 − λ/µ)] ≤ 5cakes
λ
2
+40λ − 160 ≤ 0
λ ≤ 3.6643 cakes per hour.
6. The physical examination is modeled as an M/G/1 queue. The arrival rate is λ =1/60 patient per
minute. The mean service time is 15 + 15 + 15 = 45 minutes, so the service rate is µ =1/45 patient per
minute. Thus, ρ = λ/µ =3/4. The variance of the service time is σ
2
=15
2
+15
2
+15
2
= 675 minutes, the
sum of the variance of three exponentially distributed random variables, each with mean 15. Applying the
formula for L
Q
for the M/G/1 queue we obtain
L
Q
=
ρ
2
(1 + σ
2
µ
2
)
2(1 − ρ)
=1
1
2
patients.
7. The tool crib is modeled as an M/G/1 queue with arrival rate λ = 10 per hour, service rate µ =60/4=15
per hour, and service-time variance σ
2
=(2/60)
2
=(1/30)
2
hours. Thus, ρ = λ/µ =2/3. The wages for
non-productive waiting in line amounts to 15w
Q
per mechanic’s visit to the tool crib. Since there are λ =10
visits per hour on average, the average cost per hour of having mechanics delayed is λ($15w
Q
) = $15L
Q
,
using L
Q
= λw
Q
. Applying the formula for L
Q
for the M/G/1 queue we obtain
L
Q
=
ρ
2
(1 + σ
2
µ
2
)
2(1 − ρ)
=0.833 mechanics.
Thus, the average cost per hour is $15L
Q
= $12.50.
CHAPTER 6. QUEUEING MODELS 34
8. The airport is modeled as an M/G/1 queue with arrival rate λ =30/60 = 0.5 per minute, service rate
µ =60/90 = 2/3 per minute, and service-time variance σ
2
= 0. The runway utilization is ρ = λ/µ =3/4.
Applying the formulas for the M/G/1 queue we obtain
L
Q
=
ρ
2
(1 + σ
2
µ
2
)
2(1 − ρ)
=1.125 aircraft
w
Q
=
L
Q
λ
=2.25 minutes
w = w
Q
+
1
µ
=3.75 minutes
L =
λ
µ
+ L
Q
=1.875 aircraft.
9. The machine shop is modeled by an M/G/1 queue (λ =12/40 = .3/hour, µ =1/2.5=.4/hour,
ρ = .75,σ
2
=1).
(a) The expected number of working hours that a motor spends at the machine shop is
w = µ
−1
+[λ(µ
−2
+ σ
2
)]/[2(1 − ρ)] = 6.85 hours
(b) The variance that will reduce the expected number of working hours, w, that a motor spends in the shop
to 6.5 hours is calculated by solving the equation in (a) for σ
2
:
σ
2
=[(w −µ
−1
)(2(1 − ρ))]/λ − µ
−2
σ
2
= .4167 hours
2
.
10. The self-service gasoline pump is modeled by an M/G/1 queue with (λ = 12/hour, µ = 15/hour,
ρ = .8,σ
2
=1.333
2
min
2
= .0222
2
hour
2
. The expected number of vehicles in the system is
L = ρ +[ρ
2
(1 + σ
2
µ
2
)]/[2(1 − ρ)] = 2.5778 vehicles.
11. The car wash is modeled by an M/G/1 queue (λ =1/45,µ=1/36, ρ = .8,σ
2
= 324).
(a) The average time a car waits to be served is
w
Q
=90minutes
(b) The average number of cars in the system is
L =2.8cars
(c) The average time required to wash a car is
1/µ =36minutes.
12. The cotton spinning room is modeled by an M/M/c/10/10 queue with (λ =1/40,µ =1/10,N = K =
10). Given that operators are paid $10 per hour, and idle looms cost $40 per hour, the mean cost per hour
of the system is
Mean cost per hour = $10c + $40L
The table below is generated for various levels of c.
CHAPTER 6. QUEUEING MODELS 35
cL
Q
Lw
Q
(min) K − L Cost
1 5.03 6.02 50.60 3.98 $250.80
2 1.46 3.17 8.55 6.83 146.80
3 0.32 2.26 1.65 7.74 120.40
4 0.06 2.05 0.30 7.95 122.00
5 0.01 2.01 0.05 7.99 130.40
(a) The number of operators that should be employed to minimize the total cost of the room is three,
resulting in a total cost of $120.40.
(b) Four operators should be employed to ensure that, on the average, no loom should wait for more than
one minute for an operator (i.e., to ensure w
Q
≤ 1 min.). In this case, a loom will only have to wait an
average of w
q
=0.3 min. = 18 seconds for a cost of $122.00.
(c) Three operators should be employed to ensure that an average of at least 7.5 looms are running at all
times (i.e., to ensure K −L ≥ 7.5looms)
13. Given an M/M/2/10/10 queue (λ =1/82,µ =1/15,c =2,K =10,N = 10), the average number of
customers in the queue is L
Q
=0.72. The average waiting time of a customer in the queue is
W
Q
= L
Q
/λ
e
=0.72/0.09567 = 7.526 time units.
The value of λ such that L
Q
= L/2 is found by trial and error to be
λ =0.0196
14. Assuming Figure 6.6 represents a single-server LIFO system, the time in system, W
i
,oftheith customer
can be found to be W
1
=2,W
2
=5,W
3
=9,W
4
=3,W
5
=4,so
N
i=1
W
i
= 23.
Also,
λ = N/T =5/20 = 0.25
w =
N
i=1
w
i
/N =4.6 time units
L =(1/T )
∞
i=0
iT
i
=1.15 customers
Note that:
L =1.15 = (.25)(4.6) =
λ w
Allowing T −→ ∞,andN −→ ∞, implies that
L −→ L,
λ −→ λ,and w −→ w,and
L =
λ w becomes L = λw
The total area under the L(t) function can be written as:
T
0
L(t)dt =
N
i=1
W
i
Note that LIFO did not change the equations.
15. (a) Assume Figure 6.6 is for a FIFO system with c = 2 servers. As before, N =5andT = 20,
so
λ = N/T =0.25 customer/time unit. The solution for this system is given by Figure 6.8. Hence,
CHAPTER 6. QUEUEING MODELS 36
W
1
=2,W
2
=8− 3=5,W
3
=10− 5=5,W
4
=14− 7 = 7, and W
5
=20− 16 = 4. To show
L =
λ w,one
proceeds as in Exercise 14.
(b) Assume Figure 6.6 is for LIFO system with c = 2 servers. The solution is identical to that of Exercise
11.
16. (d) The values of µ
1
,µ
2
,andp needed to achieve a distribution with mean E(X) = 1 and coefficient of
variation cv = 2 can be determined as follows:
Note that
E(X)=p/µ
1
+(1− p)/µ
2
and
(cv)
2
=[2p(1 − p)(1/µ
1
− 1/µ
2
)
2
]/[E(X)]
2
+1
By choosing p =1/4 arbitrarily, the following equations can be simultaneously solved
1/(4µ
1
)+3/(4µ
2
)=1and3/8(1/µ
1
− 1/µ
2
)
2
+1=4
Solving the left equation for µ
1
yields
µ
1
= µ
2
/(4µ
2
− 3)
Substituting µ
1
into the right equation and solving for µ
2
yields
µ
2
=1/(1 −
2/2) = 3.4142
µ
1
=3.4142/[4(3.4142 − 3)] = .3204
17. In Example 6.18, the milling machine station is modeled by M/M/c/K/K queue (λ =1.20,µ=1/5,K =
10). A table comparing the relevant parameters of the system for c =1, 2, and 3 is given below:
c =1 c =2 c =3
L
Q
5.03 1.46 0.32
L − L
Q
0.994 1.708 1.936
ρ
0.994 0.854 0.645
As more servers are hired, the average server utilization, ρ, decreases; but the average queue length, L
Q
,
also decreases.
18. Modeling the system as an M/M/c/12/12 queue we need λ
e
to obtain ρ = λ
e
/(cµ), where λ =1/20 and
µ =1/5. Results are given in the table below:
cλ
e
ρ
1 0.200 0.999
2 0.374 0.934
3 0.451 0.752
19. The lumber yard is modeled by a M/M/c/N/K queue (λ =1/3,µ=1,N = K = 10).
(a) Assume that unloading time is exponentially distributed with mean 1/µ = 1 hour. Also assume that
travel time to get the next load of logs and return is exponentially distributed with mean 1/λ = 3 hours. The
exponential distribution is highly variable (mean=std.dev.) and therefore it may be reasonable for travel
times provided the trucks travel varying distances and/or run into congested traffic conditions. On the other
hand, actual unloading times are probably less variable than the exponential distribution.
(b) With one crane to unload trucks, c =1.
CHAPTER 6. QUEUEING MODELS 37
The average number of trucks waiting to be unloaded is
L
Q
=6trucks.
The average number of trucks arriving at the yard each hour is
λ
e
=1.0 trucks/hour.
The fraction of trucks finding the crane busy upon arrival is
1 − P
0
= .997 = 99.7%
The long run proportion of time the crane is busy is
ρ =1.0
(c) With two cranes to unload trucks, c =2.
A table comparing one crane and two cranes follows:
one crane two cranes
c 12
L
Q
6.0 2.47
λ
e
1.0 1.88
busy
0.997 0.844
ρ
1.0 0.94
(d) The value of a truckload is $200 and the cost of a crane is $50 per hour independent of utilization. The
cost per hour is $50 (number of cranes) - $200 (number of arrivals per hour), or cost per hour = $50c−$200λ
e
.
Cost ($) per hour Cost ($) per hour
cλ
e
Exercise 19(d) Exercise 19(e)
1 1.000 -150.00 90.00
2 1.883 -276.60 -177.80
3 2.323 -314.60 -286.20
4 2.458 -291.60 -284.80
5 2.493 -248.60 -247.40
Three cranes should be used because the value of logs received per hour is $314.60 more than the cost of
three cranes, and is higher than with any other option.
(e) In addition to the above costs, the cost of an idle truck and driver is $40 per hour. Then,
cost = $50c + $40L
Q
− $200λ
e
and three cranes should be installed as shown in the table above, since the value of the logs is $286.20 more
than the combined cost of three cranes and L
Q
= .71 idle trucks and drivers on the average.
20. The tool crib is modeled by an M/M/c/N/K queue (λ =1.20, µ =1.3,N = K =10,c = 1 or 2). As in
Exercise 1,
mean cost per hour = $6c + $10L
Case 1: one attendant (c =1)
L
Q
=2.82
λ
e
=0.311
CHAPTER 6. QUEUEING MODELS 38
L =3.75
Mean cost per hour = $6(1) + $10(3.75) = $43.50
Case 2: two attendants (c =2)
L
Q
=0.42
L =1.66
Mean cost per hour = $6(2) + $10(1.66) = $28.60
A second attendant reduces mean costs per hour by $43.50 - $28.60 = $14.90.
21. For an M/G/∞ queue with λ = 1000/hour and 1/µ = 3 hours,
P
n
=
e
−λ/µ
(λ/µ)
n
n!
If c is the number of parking spaces, the probability we need more than c spaces is
∞
n=c+1
P
n
=1−
c
n=0
P
n
By trial and error we find that c = 3169 spaces makes this probability < 0.001.
22. If the overall arrival rate increases to λ = 160/hour, then λ
1
= .4λ =64,λ
2
= .6λ = 96, and λ
3
=
λ
1
+ λ
2
= 160. The offered load at service center 2 is λ
2
/µ
2
=96/20 = 4.8, so we need at least c = 5 clerks.
At service center 3, λ
3
/µ
3
= 160/90 = 1.8, so we need at least c = 2 clerks.
23. The system can be approximated as an M/M/c queue with arrival rate λ = 24 per hour and service rate
µ =1/2 per minute = 30 per hour. Currently c = 1 server (copy machine), but the proposal is for c =2
servers. The steady-state probability that the line reaches outside the store is
p =
∞
n=5
P
n
=1−
4
n=0
P
n
For the M/M/1 queue p ≈ 0.33, while for the M/M/2 queue p ≈ 0.01. Thus, adding another copier
substantially reduces the likelihood of having a line reach outside the store.
24. The system can be approximated as an M/M/c/N queue. In both system designs the capacity is N =7
cars. Currently there are c = 4 servers (stalls), and the proposal is to change to c = 5 stalls. The arrival
rate is λ = 34 cars per hour, so the rate at which cars are lost is λP
7
.
Theexpectedservicetimeis
3(0.2) + 7(0.7) + 12(0.1) = 6.7minutespercar
implying a service rate of approximately µ = 9 cars per hour. Clearly the service time is not exponentially
distributed, but we are approximating it as exponentially distributed with the same mean.
When c =4wehaveλP
7
≈ (34)(0.14) = 4.8 cars per hour lost, but when c =5wehaveλP
7
≈
(34)(0.08) = 2.7 cars per hour lost.
Chapter 7
Random-Number Generation
1. Place 10 slips of paper into a hat, where each slip has one of the integers 0, 1, 2,...,9 written on it. Draw
two slips of paper (one-at-a-time, with replacement), and let the resulting numbers be F, S.Thenset
R =0.F S
This procedure generates random numbers on the interval [0, 0.99].
2. Video gambling games, military draft, assigning subjects to treatments in a pharmaceutical experiment,
state lotteries and pairing teams in a sports tournament.
3. Let X = −11 + 28R.
4. Solution to Exercise 4:
X
0
=27,a=8,c=47,m= 100
X
1
=(8× 27 + 47)mod 100 = 63,R
1
=63/100 = .63
X
2
=(8× 63 + 47)mod 100 = 51,R
2
=51/100 = .51
X
3
=(8× 51 + 47)mod 100 = 55,R
3
=55/100 = .55
5. None. A problem would occur only if c =0also.
6. Solution to Exercise 6:
X
0
= 117,a=43,m =1, 000
X
1
= [43(117)]mod 1, 000 = 31
X
2
= [43(31)]mod 1, 000 = 333
X
3
= [43(333)]mod 1, 000 = 319
X
4
= [43(319)]mod 1, 000 = 717
7. Solution to Exercise 7:
R
(i)
.11 .54 .68 .73 .98
i/N .20 .40 .60 .80 1.0
i/N −R
(i)
.09 – – .07 .02
R
(i)
− (i − 1)/N .11 .34 .28 .13 .18
39
CHAPTER 7. RANDOM-NUMBER GENERATION 40
D
+
=max
1≤i≤N
(i/N − R
(i)
)=.09
D
−
=max
1≤i≤N
(R
(i)
− (i − 1)/N )=.34
D =max(D
+
,D
−
)=.34
The critical value, D
α
, obtained from Table A.8 is
D
.05
= .565
since D<D
.05
, the hypothesis that there is no difference between the true distribution of {R
1
,R
2
,...,R
5
}
and the uniform distribution on [0, 1] cannot be rejected on the basis of this test.
8. Let ten intervals be defined each from (10i − 9) to (10i)wherei =1, 2,...,10. By counting the numbers
that fall within each interval and comparing this to the expected value for each interval, E
i
= 10, the
following table is generated:
Interval O
i
(O
i
− E
i
)
2
/E
i
(01-10) 9 0.1
(11-20) 9 0.1
(21-30) 9 0.1
(31-40) 6 1.6
(41-50) 17 4.9
(51-60) 5 2.5
(61-70) 10 0.0
(71-80) 12 0.4
(81-90) 7 0.9
(91-00) 16 3.6
100 14.2= χ
2
0
From Table A.6, χ
2
.05,9
=16.9. Since χ
2
0
<χ
.05,9
, then the null hypothesis of no difference between the
sample distribution and the uniform distribution is not rejected.
9. The numbers are given a “+” or a “−” depending on whether they are followed by a larger or smaller
number:
+ − + −−−++++− + − +++−−+++−−−−
+++− + − + −−+ − + − + −−−++−−++
There are a = 27 runs in this sequence.
For N = 50,
µ
a
=(2N − 1)/3=33, and
σ
2
a
=(16N − 29)/90 = 8.5667
Z
0
=(a − µ
a
)/σ
a
= −2.05
z
α/2
= z
.025
=1.96
Since Z
0
< −z
.025
, the null hypothesis of independence can be rejected.
CHAPTER 7. RANDOM-NUMBER GENERATION 41
10. A “+” sign is used to denote an observation above the mean (.495) and a “−” sign will denote an
observation below the mean.
+ − + −−+ −−−−−−++− + − + − + − + − ++−
++++−−−+ −−+ −−++−−+++− + −−+
n
1
=24,n
2
=26, and b =31
µ
b
=[(2n
1
n
2
)/N ]+1/2=25.46
σ
2
b
=[2n
1
n
2
(2n
1
n
2
− N )]/[N
2
(N − 1)] = 12.205
Z
0
=(b − µ
b
)/σ
b
=1.586
z
α/2
= z
.025
=1.96
Since −z
.025
<Z
0
<z
.025
, the null hypothesis of independence cannot be rejected.
11. The lengths of runs up and down are
1, 1, 1, 3, 4, 1, 1, 1, 3, 2, 3, 4, 3, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2
E(Y
i
)=[2/(i + 3)!][N(i
2
+3i +1)− (i
3
+3i
2
− i − 4),i≤ N − 2
E(Y
i
)=2/N !,i= N −1
E(Y
1
)=(2/24)[50(5) − (−1)] = 20.917
E(Y
2
)=(2/120)[50(11) − 14] = 8.933
E(Y
3
)=(2/720)[50(19) − 48] = 2.506
E(Y
≥4
)=µ
a
− E(Y
1
) − E(Y
2
) − E(Y
3
)
=(2n − 1)/3 − (20.917 + 8.933 + 2.506) = 0.644
Run Length Observed Runs Expected Runs
[O
i
−E(Y
i
)]
2
E(Y
i
)
(i)(O
i
) E(Y
i
)
1 14 20.917 2.2874
2 6 8.933
3 5 2.506 0.0696
≥ 4 2 0.644
χ
2
0
=2.3570
χ
2
.05,1
=3.84
since χ
2
0
<χ
2
.05,1
, the null hypothesis of independence cannot be rejected. Notice that we grouped run
lengths i =2, 3, ≥ 4 together into a single cell with O
i
=13andE(Y
i
)=12.083.
12. Solution to Exercise 12:
The sequence is as follows:
+ − + −−+ −−−−−++− + − + − + − + − ++−
++++−−−+ −−+ −−++−−+++− + −−+
CHAPTER 7. RANDOM-NUMBER GENERATION 42
Run length, i 1234
Observed Runs, O
i
19822
n
1
=24andn
2
=26
w
1
=2(24/50)(26/50) = .4992
w
2
=(24/50)
2
(26/50) + (24/50)(26/50)
2
= .2496
w
3
=(24/50)
3
(26/50) + (24/50)(26/50)
3
= .1250
E(I)=24/26 + 26/24 = 2.00
E(A)=50/2.00 = 25
E(Y
1
) = 50(.4992)/2.00 = 12.48
E(Y
2
) = 50(.2496)/2.00 = 6.24
E(Y
3
) = 50(.1250)/2.00 = 3.125
Run Length Observed Runs Expected Runs
[O
i
−E(Y
i
)]
2
E(Y
i
)
(i)(O
i
) E(Y
i
)
1 19 12.48 3.41
2 8 6.24 .50
3 2 3.125
≥ 4 2 3.155 .83
4.74
χ
2
.05,2
=5.99
Therefore, do not reject the hypothesis of independence on the basis of this test. Notice that we grouped
run lengths i =3, ≥ 4 together into a single cell with O
i
=4andE(Y
i
)=6.28.
13. Solution to Exercise 13:
ρ
14
=(1/8)[(.48)(.61) + (.61)(.37) + (.37)(.37) + (.37)(.99) + (.99)(.09)
+(.09)(.55) + (.55)(.60) + (.60)(.19)] − .25 = −.0495
σ
ρ
14
= .1030
Z
0
= −.0495/.1030 = −0.48
Since −z
.025
<Z
0
<z
.025
, the null hypothesis of independence cannot be rejected on the basis of significant
autocorrelation.
14. Solution to Exercise 14:
Gap Length Relative
Classes Frequency Frequency S
N
(x) F (x) |F (X) − S
N
(x)|
0-3 33 .3000 .3000 .3439 .0439
4-7 23 .2091 .5091 .5695 .0604
8-11 23 .2091 .7182 .7176 .0006
12-15 15 .1364 .8546 .8146 .0400
16-19 7 .0636 .9182 .8784 .0398
20-23 5 .0455 .9637 .9202 .0435
24-27 1 .0091 .9728 .9497 .0231
28-31 0 0 .9728 .9657 .0071
32-35 2 .0182 .9910 .9775 .0135
36-39 1 .0091 1.0 .9852 .0148
110
CHAPTER 7. RANDOM-NUMBER GENERATION 43
D =max|F (x) − S
N
(x)| = .0604, and
D
α
= D
.05
= .136. Since D<D
.05
, the null hypothesis of independence cannot be rejected on the basis of
this test.
15. Solution to Exercise 15:
(a)
P (4 different digits) = (.9)(.8)(.7) = .5040
P (exactly one pair) = (
4
2
)(.1)(.9)(.8) = .4320
P (two pairs) = (
3
2
)(.1)(.9)(.1) = .0270
P (triplet) = (
4
3
)(.1)(.1)(.9) = .0360
P (4 like digits) = (.1)(.1)(.1) = .0010
(b)
P (5 different digits) = (.9)(.8)(.7)(.6) = .3024
P (exactly one pair) = (
5
2
)(.1)(.9)(.8)(.7) = .5040
P (2 different pairs) = 15(.1)(.9)(.1)(.8) = .1080
P (triplet and pair) = (
5
3
)(.1)(.9)(.1) = .0090
P (exactly one triplet) = (
5
3
)(.1)(.1)(.9)(.8) = .0720
P (4 like digits) = (
5
4
)(.1)(.1)(.1)(.9) = .0045
P (5 like digits) = (.1)(.1)(.1)(.1) = .0001
16. Solution to Exercise 16:
Combination Observed Expected
(O
i
−E
i
)
2
E
i
iO
i
E
i
4 different digits 565 504 7.3829
1 pair 392 432 3.7037
2 pairs 17 27 3.7037
3 like digits 24 36
4 like digits 2 1 3.2703
Totals 1000 1000 18.0606
χ
2
.05,3
=7.81 <χ
2
0
=18.0806
Reject the null hypothesis of independence based on this test. Notice that we grouped “3 like digits” and “4
like digits” into a single cell with O
i
=26andE
i
= 37.
17. Solution to Exercise 17(c): a =1+4k −→ k = 1237.5 which is not an integer. Therefore, maximum
period cannot be achieved.
18. Solution to Exercise 18:
X
1
=[7× 37 + 29] mod 100 = 88
R
1
= .88
X
2
=[7× 88 + 29] mod 100 = 45
R
2
= .45
X
3
=[7× 45 + 29] mod 100 = 44
R
3
= .44
19. Use m =25
X
1
=[9× 13 + 35] mod 25 = 2
CHAPTER 7. RANDOM-NUMBER GENERATION 44
X
2
=[9× 2 + 35] mod 25 = 3
X
3
=[9× 3 + 35] mod 25 = 12
21. Solution to Exercise 21:
X
1
= [4951 × 3579] mod 256 = 77
R
1
=77/256 = .3008
23. Solution to Exercise 23:
Case (a) Case (b) Case (c) Case (d)
iX
i
X
i
X
i
X
i
07878
113818
215 7 8
35
47
Inferences:
Maximum period, p = 4, occurs when X
0
is odd and a =3+8k where k = 1. Even seeds have the minimal
possible period regardless of a.
24. X
1,0
= 100,X
2,0
= 300,X
3,0
= 500
The generator is
X
1,j+1
= 157 X
1,j
mod 32363
X
2,j+1
= 146 X
2,j
mod 31727
X
3,j+1
= 142 X
3,j
mod 31657
X
j+1
=(X
1,j+1
− X
2,j+1
+ X
3,j+1
) mod 32362
R
j+1
=
X
j+1
32363
, if X
j+1
> 0
32362
32363
=0.999 , if X
j+1
=0
The first 5 random numbers are
X
1,1
= [157 × 100] mod 32363 = 15700
X
2,1
= [146 × 300] mod 31727 = 12073
X
3,1
= [142 × 500] mod 31657 = 7686
X
1
= [15700 − 12073 + 7686] mod 32362 = 11313
R
1
= 11313/32363 = 0.3496
X
1,2
= 5312
X
2,2
= 17673
X
3,2
= 15074
X
2
= 2713
R
2
=0.0838
X
1,3
= 24909
X
2,3
= 10371
X
3,3
= 19489
X
3
= 1665
R
3
=0.0515
X
1,4
= 27153
X
2,4
= 22997
CHAPTER 7. RANDOM-NUMBER GENERATION 45
X
3,4
= 13279
X
4
= 17435
R
4
=0.5387
X
1,5
= 23468
X
2,5
= 26227
X
3,5
= 17855
X
5
= 15096
R
5
=0.4665
29. Two results that are useful to solve this problem are
(c + d)modm = c mod m + d mod m
and that if g = h mod m,thenwecanwriteg = h − km for some integer k ≥ 0. The last result is true
because, by definition, g is the remainder after subtracting the largest integer multiple of m that is ≤ h.
(a) Notice that
X
i+2
= aX
i+1
mod m
= a[aX
i
mod m]modm
= a[aX
i
− km]modm (for some integer k ≥ 0)
= a
2
X
i
mod m − akm mod m
= a
2
X
i
mod m (since akm mod m =0).
(b) Notice that
(a
n
X
i
)modm = {(a
n
mod m)+[a
n
− (a
n
mod m)]}X
i
mod m
= {(a
n
mod m)X
i
mod m} + {[a
n
− (a
n
mod m)]X
i
mod m}
= {(a
n
mod m)X
i
mod m} + {kmX
i
mod m} (for some integer k ≥ 0)
=(a
n
mod m)X
i
mod m.
(c) In this generator a = 19, m = 100 and X
0
= 63. Therefore, a
5
mod 100 = 19
5
mod 100 = 99. Thus,
X
5
= (99)(63) mod 100 = 37.
Chapter 8
Random-Variate Generation
1. Solution to Exercise 1:
Step 1.
cdf = F (x)=
e
2x
/2, −∞ <x≤ 0
1 − e
−2x
, 0 <x<∞
Step 2. Set F(X)=R on −∞ <X<∞
Step 3. Solve for X to obtain
X =
1/2ln2R 0 <R≤ 1/2
−1/2ln(2− 2R)1/2 <R<1
2. Solution to Exercise 2:
Step 1.
cdf = F (x)=
1 − x + x
2
/4, 2 ≤ x<3
x − x
2
/12 − 2, 3 <x≤ 6
Step 2. Set F(X)=R on 2 ≤ X ≤ 6
Step 3. Solve for X to obtain
X =
2+2
√
20≤ R ≤ 1/4
6 − 2
√
3 − 3R 1/4 <R≤ 1
Thetruemeanis(a + b + c)/3=(2+3+6)/3=11/3.
3. Triangular distribution with a =1,b=4,c = 10. Total area = 1 = base × height/2 = 9h/2, so h =2/9
Step 1: Find cdf F (x) = total area from 1 to x.
For 1 ≤ x ≤ 4,f(x)/h =(x − 1)/(4 − 1) by similar triangles so
F (x)=(x − 1)f(x)/2=(x − 1)
2
/27
For 4 <x≤ 10,f(x)/h =(10− x)/(10 − 4) by similar triangles so
F (x)=1− (10 − x)f (x)/2=1− (10 − x)
2
/54.
Step 2: Set F(X)=R on 1 ≤ X ≤ 10.
Step 3: Solve for X.
46
CHAPTER 8. RANDOM-VARIATE GENERATION 47
X =
1+
√
27R, 0 ≤ R ≤ 9/27
10 −
54(1 − R), 9/27 <R≤ 1
4. Triangular distribution with a =1,c =10andE(X) = 4. Since (a + b + c)/3=E(X), the mode is at
b = 1. Thus, the height of the triangular pdf is h =2/9. (See solution to previous problem. Note that the
triangle here is a right triangle.)
Step 1: Find cdf F (x) = total area from 1 to x.
=1− (total area from x to 10).
By similar triangles, f(x)/h =(10− x)/(10 − 1), so
F (x)=1− (10 − x)f(x)/2=1− (10 − x)
2
/81, 1 ≤ x ≤ 10.
Step 2: Set F(X)=R on 1 ≤ X ≤ 10.
Step 3: X =10−
81(1 − R), 0 ≤ R ≤ 1
5. Solution to Exercise 5:
X =
6(R − 1/2) 0 ≤ R ≤ 1/2
32(R − 1/2) 1/2 ≤ R ≤ 1
6. X =2R
1/4
, 0 ≤ R ≤ 1
7. Solution to Exercise 7:
F (x)=x
3
/27, 0 ≤ x ≤ 3
X =3R
1/3
, 0 ≤ R ≤ 1
8. Solution to Exercise 8:
Step 1:
F (x)=
x/3, 0 ≤ x ≤ 2
2/3+(x − 2)/24, 2 <x≤ 10
Step 2: Set F(X)=R on 0 ≤ X ≤ 10.
Step 3:
X =
3R, 0 ≤ R ≤ 2/3
2 + 24(R − 2/3) = 24R − 14, 2/3 <R≤ 1
9. Use Inequality (8.14) to conclude that, for R given, X will assume the value x in R
X
= {1, 2, 3, 4} provided
F (x − 1) =
(x − 1)x(2x − 1)
180
<R≤
x(x + 1)(2x +1)
180
= F (x)
By direct computation, F(1) = 6/180 = .033, F (2) = 30/180 = .167, F(3) = 42/180 = .233, F (4) = 1.
Thus, X can be generated by the table look-up procedure using the following table:
x
1 2 3 4
F (x) .033 .167 .233 1
CHAPTER 8. RANDOM-VARIATE GENERATION 48
R
1
=0.83 −→ X =4
R
2
=0.24 −→ X =4
R
3
=0.57 −→ X =4
10. Weibull with β =2,α= 10. By Equation (9.6)
X = 10[−ln(1 − R)]
0.5
11. The table look-up method for service times:
Input Output Slope
ir
i
x
i
a
i
1 0 15 244.89
2 .0667 30 112.53
3 .2000 45 89.98
4 .3667 60 128.59
5 .6000 90 150.00
6 .8000 120 450.11
7 .9333 180 1799.10
8 1.0000 300 —
12. The table look-up method for fire crew response times, assuming 0.25 ≤ X ≤ 3:
Input Output Slope
ir
i
x
i
a
i
1 0 .25 3.29
2 .167 .80 2.65
3 .333 1.24 1.26
4 .500 1.45 2.28
5 .667 1.83 5.60
6 .833 2.76 1.44
7 1.000 3.00 —
13. By Example 8.5, 17R generates uniform random variates on {1, 2,...,17},thus
X =7+17R
generates uniform random variates on {8, 9,...,24}.
15. The mean is (1/p) − 1=2.5, so p =2/7 . By Equation (9.21),
X = −2.97 ln(1 − R) − 1
17. Use X = −3.7lnR.
19. Generate X =8[−ln R]
4/3
CHAPTER 8. RANDOM-VARIATE GENERATION 49
If X ≤ 5, set Y = X.
Otherwise, set Y =5.
(Note: for Equation 8.6, it is permissible to replace 1 − R by R.)
20.
Method 1: Generate X
1
∼ U(0, 8) and X
2
∼ U(0, 8).
Set Y =min(X
1
,X
2
).
Method 2: The cdf of Y is
F (y)=P (Y ≤ y)=1− P (Y>y)
=1− P (X
1
>y,X
2
>y)
=1− (1 − y/8)
2
, 0 ≤ y ≤ 8
by independence of X
1
and X
2
.
F (Y )=1− (1 − Y/8)
2
= R
implies
Y =8− 8
√
1 − R, 0 ≤ R ≤ 1.
21. Assume X
i
is exponentially distributed with mean 1/λ
i
,where1/λ
1
= 2 hours and 1/λ
2
= 6 hours.
Method 1 is similar to that in Exercise 20.
Method 2: The cdf of Y is
F (y)=P (Y ≤ y)=1− P (Y>y)
=1− P (X
1
>y,X
2
>y)
=1− e
−λ
1
y
e
−λ
2
y
=1− e
−(λ
1
+λ
2
)y
Therefore Y is exponential with parameter λ
1
+ λ
2
=1/2+1/6=2/3.
Generate Y = −1.5lnR.
Clearly, method 2 is twice as efficient as method 1.
22. Generate R
1
,R
2
,...R
n
.
Set X
i
=
1ifR
i
≤ p
0ifR
i
>p.
Compute X =
n
i=1
X
i
23. Solution to Exercise 23:
Step 1: Set n =0
Step 2: Generate R
Step 3: If R ≤ p,setX = n, and go to step 4.
CHAPTER 8. RANDOM-VARIATE GENERATION 50
If R>p,incrementn by 1 and return to step 2.
Step 4: If more geometric variates are needed, return to step 1.
28. Recall that one can obtain exponentially distributed variates with mean 1 using the inverse cdf trans-
formation
X = F
−1
(1 − R)=−ln(1 − R).
The reverse transformation (known as the probability-integral transformation) also works: If X is exponen-
tially distributed with mean 1, then
R = F (X)=1− e
−X
is uniform (0, 1). ThisgetsusfromX to R; we then use the inverse cdf for the triangular distribution to go
from R to a triangularly distributed variate.
Chapter 9
Input Modeling
12. Solution to Exercise 12:
ln
¯
X − 1.255787
20
i=1
ln X
i
=21.35591
1/M =5.319392
θ =0.3848516
β =2.815
13. Solution to Exercise 13:
j
β
j
20
i=1
X
βj
i
20
i=1
X
β
i
j ln X
i
20
i=1
X
β
i
j(ln X
i
)
2
f(
β
j
) f
(
β
j
)
β
j+1
0 2.539 1359.088 2442.221 4488.722 1.473 -4.577 2.861
1
2.861 2432.557 4425.376 8208.658 .141 -3.742 2.899
2
2.899 2605.816 4746.920 8813.966 .002 -3.660 2.899
3
2.899 2607.844 4750.684 8821.054 .000 -3.699 2.899
β =2.899
α =5.366
14. H
0
: Data are uniformly distributed
R
(i)
.0600 .0700 ··· .4070 ··· .8720 ··· .9970
1/3
.0333 .0667 ··· .4333 ··· .7333 ··· 1.0000
1/3−R
(i)
——··· .0653 ··· — ··· .0030
R
(i)
− (i − 1)/30 .0600 .0367 ··· .0070 ··· .1720 ··· .0303
D
+
= .0653
D
−
= .1720
D =max(.0653,.1720) = .1720
D
.05,30
= .24 >D= .1720
Therefore, do not reject H
0
16. Solution to Exercise 16:
(a) α =
¯
X =1.11
51
CHAPTER 9. INPUT MODELING 52
x
i
O
i
p
i
E
i
(O
i
−E
i
)
2
E
i
0 35 .3296 32.96 .126
1
40 .3658 36.58 .320
2
13 .2030 20.30 2.625
3
6 .0751 7.51
4 4 .0209 2.09
5 1 .0046 .46
≥6 1 .0010 .10 .333
Totals 100 1.0000 100 3.404 = χ
2
0
χ
2
.05,2
=5.99
Therefore, do not reject H
0
. Notice that we have grouped cells i =3, 4, 5 ≥ 6 together into a single cell with
O
i
=12andE
i
=10.16.
(b) α =1
x
i
O
i
p
i
E
i
(O
i
−E
i
)
2
E
i
0 35 .3679 36.79 .087
1
40 .3679 36.79 .280
2
13 .1839 18.39 1.580
3
6 .0613 6.13
4 4 .0153 1.53
5 1 .0031 .31
≥6 1 .0006 .06 1.963
Totals 100 1.0000 100 3.910 = χ
2
0
χ
2
.05,3
=7.81
Therefore, do not reject H
0
. Notice that we have grouped cells 3, 4, 5 ≥ 6 into a single cell with O
i
=12and
E
i
=8.03.
17. Solution to Exercise 17:
H
0
= Data are exponentially distributed
λ =
¯
X =1.206
S =1.267
i
O
i
(O
i
−E
i
)
2
E
i
1 8 .013
2
11 .853
3
9 .053
4
5 1.333
5
10 .333
6
7 .213
Totals 50 2.798=χ
2
0
χ
2
.05,4
=9.49
Therefore, do not reject H
0
CHAPTER 9. INPUT MODELING 53
18. Using the Arena Input Analyzer, the Kolmogorov-Smirnov statistic for normality is 0.0985, which
corresponds to a p-value greater than 0.15. The chi-square test statistic with 5 intervals (yielding 2 degrees
of freedom) is 4.85, which corresponds to a p-value of 0.09. With 7 intervals (yielding 4 degrees of freedom),
the chi-square statistic is 5.98, corresponding to a p-value of 0.21. These statistics show no strong evidence
against the hypothesis of normality, although the chi-square statistic with 2 degrees of freedom could be
interpreted as rejecting the hypothesis of normality.
19. H
0
= Data are normally distributed
µ =
¯
X =99.222
σ
2
= S
2
= 103.41
Number of
χ
2
0
χ
2
.05,k−3
Decision
Cells (k)
10 3.2 14.1 Do not reject H
0
8 1.2 11.1 Do not reject H
0
5 1.0 5.99 Do not reject H
0
20. H
0
: Data are normally distributed
µ =
¯
X =4.641
σ
2
= S
2
=2.595
Number of
χ
2
0
χ
2
.05,k−3
Decisions
Cells (k)
10 5.6 14.1 Do not reject H
0
8 1.52 11.1 Do not reject H
0
5 .6 5.99 Do not reject H
0
21. H
0
: Data are exponentially distributed
λ =1/
¯
X =1/9.459 = .106
i
O
i
(O
i
−E
i
)
2
E
i
1 7 .8
2
3 .8
3
5 0.0
4
5 0.0
5
5 0.0
6
6 .2
7
5 0.0
8
7 .8
9
4 .2
10
3 .8
Totals 50 3.6=χ
2
0
χ
2
.05,8
=15.5
Therefore, do not reject H
0
22. H
0
: Data are Poisson distributed
α =
¯
X = .48
CHAPTER 9. INPUT MODELING 54
x
i
O
i
p
i
E
i
(O
i
−E
i
)
2
E
i
0 31 .6188 30.94 .0001
1
15 .2970 14.85 .0015
2
3 .0713 3.565
≥3 1 .0129 .645 .0140
Totals 50 1.0000 50.00 .0120 = χ
2
0
χ
2
.05,1
=3.84
Therefore, do not reject H
0
. Notice that we grouped cells i =2, 3intoasinglecellwithO
i
=4andE
i
=4.21.
Note: In Section 9.4.1 it was stated that there is no general agreement regarding the minimum size of E
i
and that values of 3, 4 and 5 have been widely used. We prefer E
i
> 5. If we follow our suggestion in this
case, the degrees of freedom will equal zero, which results in an undefined tabular value of χ
2
. The concern
is that a very small E
i
will result in an undue contribution to χ
2
0
.WithE
i
=4.21 this is certainly not a
cause for concern. Thus, combining cells as shown is appropriate.
23. Solution to Exercise 23:
a) The data seem positively dependent.
b) The sample correlation is ρ =0.9560.
c) To fit a bivariate normal distribution we need the sample means, sample variances, and sample correlation.
Sample mean µ Sample Variance σ
2
Milling Time 17.7 (6.7)
2
Planning Time 13.1 (3.6)
2
Obtain ρ from part (b).
26. For an AR(1) process
µ =
X =20
φ = ρ =0.48
σ
2
ε
= σ
2
=(1−
φ
2
)(3.93)
2
(1 − 0.48
2
)=11.89
For an EAR(1) process
λ =1/
X =0.05
φ = ρ =0.48
A histogram and q-q plot suggest that AR(1) is a better fit since the distribution appears more normal than
exponential.
27. Both exponential and lognormal models look feasible for this data (the Arena Input Analyzer gives p-
values > 0.15 for the Kolmogorov-Smirnov test in both cases). Since many transactions in a bank are routine
and brief, but there are occasional very long transaction times, an exponential model can be justified.
Chapter 10
Verification and Validation of
Simulation Models
1. Solution to Exercise 1:
(a) System: µ
0
=22.5
Model:
¯
Y =(18.9+22.0+...+20.2)/7=20.614
S
Y
=1.36
Test for significance (H
0
: E(Y )=µ
0
)
t
0
=(20.614 − 22.5)/(1.36/
√
7) = −3.67
For α =0.05,t
6,0.025
=2.45
Since |t
0
| > 2.45, reject null hypothesis
(b) Power of the test
δ =2/1.36 = 1.47
For α =0.05 and n =7,δ(1.47) = 0.10
Power = 1 − 0.10 = 0.90
Sample size needed for β ≤ 0.20
Assume that σ =1.36
Then for α =0.05 and δ =1.47,n= 6 observations
2. Solution to Exercise 2:
(a) System: µ
0
=4
Model:
¯
Y =(3.70 + 4.21 + ...+4.05)/7=4.084
S
y
=0.2441
Test for significance (H
0
: E(Y )=µ
0
)
t
0
=(4.084 − 4)/(0.2441/
√
7) = 0.91
For α =0.01,t
6,0.005
=3.71
55
CHAPTER 10. VERIFICATION AND VALIDATION OF SIMULATION MODELS 56
Since |t
0
| < 3.71, do not reject null hypothesis
(b) Sample size needed for β ≤ 0.10
δ =0.5/0.2441 = 2.05
for α =0.01 and δ =2.05,n= 7 observations.
Then, assuming that the population standard deviation is 0.2441, the current power of the test is 0.90.
3. Solution to Exercise 3:
(a)Testforsignificance(H
0
: µ
d
=0)
Letting d
i
= y
i
− z
i
,
¯
d =3.35,S
d
=1.526
t
0
=3.35/(1.526/
√
4) = 4.39
For α =0.05,t
3,0.025
=3.18
Since |t
0
| > 3.18, reject the null hypothesis.
(b) Sample size needed for β ≤ 0.20
δ =2/1.526 = 1.31
For α =0.05,β ≤ 0.20 and δ =1.31
n = 8 observations.
Chapter 11
Output Analysis for a Single Model
For additional solutions check the course web site at www.bcnn.net.
3. The 95% confidence interval based on only 5 replications is [1.02, 16.93], which is much wider than the
interval based on all 10 replications. From the ensemble averages across five replications, and upper and
lower confidence limits, it is not possible to detect a trend in the data.
6. It was assumed that orders could be partially fulfilled before backlogging occurred.
(a) For the (50,30) policy, the average monthly cost over 100 months,
¯
Y
r.
, for replication r (r =1, 2, 3, 4), is
given by
¯
Y
1·
= $233.71,
¯
Y
2·
= $226.36,
¯
Y
3·
= $225.78,
¯
Y
4·
= $241.06.
By Equation (12.39), the point estimate is
¯
Y
..
= $231.73 and by Equation (12.40), S
2
=($7.19)
2
.
An approximate 90% confidence interval is given by
$231.73 ± t
0.05,3
($7.19)/
√
4, (t
0.05,3
=2.353) or [$223.27, $240.19]
(b) The minimum number of replications is given by
R =min{R>R
0
: t
α/2,R−1
S
0
/
√
R ≤ $5} =8
where R
0
=4,α=0.10,S
0
=$7.19 and = $5.
The calculation proceeds as follows:
R ≥ (z
.05
S
0
/)
2
=[1.645(7.19)/5]
2
=5.60
R
6 7 8
t
.05,R−1
1.94 1.90 1.86
t
.05,R−1
S
0
/
2
7.78 7.46 7.15
Thus, four additional replications are needed.
7. Solution to Exercise 7:
(a) The following estimates were obtained for the long-run monthly cost on each replication.
¯
Y
1·
= $412.11,
¯
Y
2·
= $437.60,
¯
Y
3·
= $411.26,
¯
Y
4·
= $455.75,
¯
Y
··
= $429.18,S = $21.52
57
CHAPTER 11. OUTPUT ANALYSIS FOR A SINGLE MODEL 58
An approximate 90% c.i. for long-run mean monthly cost is given by
$429.18 ± 2.353($21.52)/
√
4, or
[$403.86, $454.50]
(b) With R
0
=4,α=0.10,S
0
= $21.52, and = $25 the number of replications needed is
min{R ≥ R
0
: t
α/2,R−1
S/
√
R<$25} =5
Thus, one additional replication is needed to achieve an accuracy of = $25.
To achieve an accuracy of = $5, the total number of replications needed is
min{R ≥ R
0
: t
.05,R−1
S
0
/
√
R<5} =53.
The calculations for = $5 are as follows:
R ≥ [z
.05
S
0
/]
2
=[1.645(21.52)/5]
2
=50.12
R
51 52 53
t
.05,R−1
1.675 1.674 1.674
[t
.05,R−1
S
0
/]
2
52.9 52.9 52.9
Therefore, for = $5, the number of additional replications is 53 − 4 = 49.
10. Ten initial replications were made. The estimated profit is $98.06 with a standard deviation of S
0
=
$12.95.
For α =0.10 and absolute precision of = $5.00, the sample size is given by
min{R ≥ 10 : t
α/2,R−1
(12.95)/
√
R<$5}
R
t
α/2,R−1
S
0
/
√
R
19 5.15
20
5.01
21
4.87
Thus, 21 replications are needed. Based on 21 replications the estimated profit is:
¯
Y = $96.38,S = $13.16
and a 90% c.i. is given by
$96.38 ± t
.05,20
S/
√
21
or $96.38 ± $4.94.
If = $0.50 and α =0.10, then the sample size needed is approximately 1815.
13. The table below summarizes the results from each replication:
Response Time (hrs.) Average Utilization
for Job Type at each Station
Replications 1 2 3 4 1 2 3 4
1 146.6 88.82 82.81 42.53 0.509 0.533 0.724 0.516
2 146.4 89.79 80.45 46.48 0.517 0.537 0.772 0.569
3 144.4 88.40 81.59 45.01 0.468 0.516 0.692 0.491
4 144.3 88.00 82.13 47.17 0.486 0.489 0.673 0.496
5 144.9 88.29 82.53 43.26 0.471 0.473 0.627 0.461
¯
Y
..
145.3 88.66 81.90 44.89 0.465 0.510 0.698 0.507
S 1.103 .697 .932 1.998 0.022 0.028 0.054 0.049
CHAPTER 11. OUTPUT ANALYSIS FOR A SINGLE MODEL 59
A 97.5% c.i. for utilization at each work station is given by
Station 1, [.463,.518]
Station 2, [.475,.544]
Station 3, [.631,.765]
Station 4, [.457,.556]
Note that by the Bonferroni inequality, Equation (12.20), the overall confidence level is 90% or greater.
A 95% c.i. for mean total response time (hrs.) of each job type is given by
Job type 1, [143.6, 147.0]
Job type 2, [87.57, 89.75]
Job type 3, [80.44, 83.36]
Job type 4, [41.77, 48.01]
Note that the overall confidence level is 80% or greater.
Chapter 12
Comparison and Evaluation of
Alternative System Designs
For additional solutions check the course web site at www.bcnn.net.
2. Using common random numbers, the following results were obtained:
Policy
Rep.
(50,30) (50,40) (100,30) (100,40)
1 $233.71 $226.21 $257.73 $261.90
2
$226.36 $232.12 $252.58 $257.89
3
$225.78 $221.02 $266.48 $258.16
4
$241.06 $243.95 $270.61 $270.51
¯
Y
·i
$231.73 $230.83 $261.85 $262.12
S
i
$7.19 $9.86 $8.19 $5.89
To achieve an overall α
E
=0.10, compute 97.5% confidence intervals (c.i.) for mean monthly cost for each
policy by using
¯
Y
·i
± t
.0125,3
S
i
/
√
4, (t
.0125,3
=4.31 by interpolation)
Policy c.i.
(50,30) $231.73 ± $15.49
(50,40) $230.83 ± $21.25
(100,30) $261.85 ± $17.65
(100,40) $262.12 ± $12.69
The overall confidence level is at least 90%.
To obtain confidence intervals which do not overlap, policies (50,30) and (50,40) should be estimated with
an accuracy = ($231.73 − $230.83)/2=$.45, and policies (100,30) and (100,40) with = ($262.12 −
$261.85)/2=$.135.
An estimate for R is given by
R>
z
α/2
S
i
&
2
with z
.0125
=2.24
60
CHAPTER 12. COMPARISON AND EVALUATION OF ALTERNATIVE SYSTEM DESIGNS 61
Policy R (replications)
(50,30) 1281
(50,40) 2411
(100,30) 18,468
(100,40) 9551
The above number of replications might take excessive computer time and thus be too expensive to run. A
better technique would be to compute c.i.’s for the differences.
At a 90% level, policies (50,30) and (50,40) appear to be better than the other two. A 90% c.i. for the
difference between the (50,30) and (50,40) policies is given by
$.9025 ± t
.05,3
× 6.250/
√
4or[−$6.451, $8.256].
Since this interval includes zero, no significant difference is detected.
3. Using common random numbers, the following results were obtained for 4 replications:
Policy
Rep
(50,30) (50,40) (100,30) (100,40) D
1 $412.11 $405.69 $419.57 $398.78 $6.91
2
$437.60 $409.54 $429.82 $410.60 -$1.06
3
$411.26 $399.30 $470.17 $416.37 -$17.07
4
$455.75 $418.01 $466.55 $438.95 -$20.94
¯
Y
i
$429.18 $408.14 $446.53 $416.18 -$8.04=
¯
D
S
i
$21.52 $7.82 $25.60 $16.86 $13.17 = S
D
It appears that the (50,40) policy dominates the other three policies. A 90% c.i. was computed for the
mean difference in cost between the (50,40) and (100,40) policies. The differences, sample mean difference
and sample standard deviation are given in the table above. It is clear that a 90% c.i. will contain zero.
Thus, there is no significant difference between the 2 policies. The 90% c.i. is −$8.04 ± $15.47. A complete
analysis would compute c.i.’s for all differences, perhaps discard clearly inferior policies, and then replicate
the remaining ones to determine the best policy.
6. Using common random numbers, 21 replications were made for different ordering sizes. The table below
summarizes the results:
Estimate of Estimated Standard
Q (cards) Mean Profit ($) Deviation ($)
250 85.05 51.17
300 96.38 13.16
350 101.4 20.89
356 101.8 20.92
357 101.9 20.88
360 101.9 21.00
375 101.5 21.71
400 99.91 22.83
CHAPTER 12. COMPARISON AND EVALUATION OF ALTERNATIVE SYSTEM DESIGNS 62
Based on Exercise 11.10, a 90% c.i. for mean total profit at Q = 300 was $96.38 ± $4.94. To obtain an
accuracy of =$5.00 at α =0.10 additional replications should be made for Q in the range 350 to 400.
Confidence intervals for differences could be computed to determine a range of Q significantly better than
other Q.
9. Use c
i
>λ
i
/µ
i
applied one station at a time.
Station 1
Station 1 receives types 1, 2 and 4 arrivals. Therefore,
Arrival rate λ
1
= .4(.25) + .3(.25) + .1(.25) = .20 per hour
Mean service time
1
µ
1
=
.4
.8
(20) +
.3
.8
(18) +
.1
.8
(30) = 20.5 hours
c
1
>λ
1
/µ
1
= .20(20.5) = 4.1,c
1
= 5 servers.
Station 2
If station 1 is stable (i.e. has 5 or more servers), then departures occur at the same rate as arrivals. Station
2 receives type 1 arrivals from station 1 and type 3 arrivals from the outside. Therefore,
Arrival rate λ
2
= .4(.25) + .2(.25) = .15 per hour
Mean service time
1
µ
2
=
.4
.6
(30) +
.2
.6
(20) = 26.67 hours
c
2
>λ
2
/µ
2
= .15(26.67) = 4.00,c
2
=5servers
Station 3
Station 3 receives types 1, 2, and 3 arrivals. Therefore,
Arrival rate λ
3
= .4(.25) + .3(.25) + .2(.25) = .225 per hour
Mean service time
1
µ
3
=
.4
.9
(75) +
.3
.9
(60) +
.2
.9
(50) = 64.44 hours
c
1
>λ
3
/µ
3
= .225(64.44) = 14.50,c
3
=15servers
Station 4
Station 4 receives all arrivals. Therefore,
Arrival rate λ
4
= .25 per hour
Mean service times
1
µ
4
= .4(20) + .3(10) + .2(10) + .1(15) = 14.5hours
c
4
>λ
4
/µ
4
= .25(14.5) = 3.63,c
4
=4servers
For c
1
=5,c
2
=5,c
3
= 15, and c
4
= 4 the following results are obtained for one replication with T
0
= 200
hours and T
E
= 800 hours.
Jobs
Average Response Time (hours)
Type 1 170.3
Type 2
106.8
Type 3
106.6
Type 4
56.44
All jobs
126.8
Station Estimated Server Utilization
1 .754
2
.751
3
.828
4
.807
CHAPTER 12. COMPARISON AND EVALUATION OF ALTERNATIVE SYSTEM DESIGNS 63
Additional replications should be conducted and standard errors and confidence intervals computed. In
addition, initialization bias should be investigated. Since λ
4
/c
4
µ
4
was calculated to be 3.63/4=.9075 and
ρ
4
= .807, it appears that significant bias may be present for T
0
= 200 hours and T
E
= 800 hours.
13. Let S be the set-up time, which is exponentially distributed with mean 20. Let P
j
bethetimetoprocess
the jth application, which is normally distributed with mean 7 and standard deviation 2. For a particular
design point, x, we generate n replications of total processing time as follows:
for i =1ton
do
generate S
for j =1tox
do
generate P
j
enddo
Y
i
= S + P
1
+ P
2
+ ···+ P
x
enddo
15. Because the samples across design points are dependent, MS
E
/S
xx
is a biased estimator of the variance
of
β
1
, and the degrees of freedom are not n − 2.
18. Let m be the number of buffer spaces (m = 50 in this problem). Since x
1
+ x
2
+ x
3
= m, x
3
is
determined once x
1
and x
2
are specified. Thus, what we really need are all assignments to x
1
and x
2
such
that x
1
+ x
2
≤ m. Clearly there are m + 1 possible assignments for x
1
; specifically, 0, 1, 2,...,m.Ifx
1
is
assigned value , then there are m +1− possible assignments for x
2
; specifically, 0, 1, 2,...,m− .Ifwe
sum over the possible assignments for x
1
we obtain
m
)=0
(m +1− )=
(m +1)(m +2)
2
which is 1326 when m = 50.
The scheme we will develop for sampling (x
1
,x
2
,x
3
) will first sample x
1
,thenx
2
given the value of x
1
,and
finally compute x
3
= m − x
2
− x
1
.
Let n =(m +1)(m +2)/2, the number of possible outcomes for (x
1
,x
2
,x
3
), all equally likely. The marginal
probability that x
1
= m is 1/n,since(m, 0, 0) is the only way it can happen. The marginal probability that
x
1
= m −1is2/n since (m −1, 1, 0) and (m −1, 0, 1) are the only ways it can happen. Arguing this way we
can show that
P (x
1
= j)=
m − j +1
n
for j =0, 1, 2,...,m. Thus, we can use one of the general methods for sampling from discrete distributions
to sample x
1
.
Now given x
1
, we can show that the marginal distribution of x
2
is discrete uniform on {0, 1,...,m−x
1
},a
distribution that is easy to sample. And finally, x
3
= m − x
2
− x
1
.
19. For this problem the true optimal solution can be computed analytically: x
∗
=2.611 years, giving an
expected cost of $11,586. This solution is obtained by minimizing the expected cost, which can be written
as
2000x +
∞
0
20000 I(y ≤ 1)
e
−y/x
x
dx
where I is the indicator function.
CHAPTER 12. COMPARISON AND EVALUATION OF ALTERNATIVE SYSTEM DESIGNS 64
20. For this problem the true optimal solution can be computed analytically: x
∗
=2.611 years, giving an
expected cost of $11,586. This solution is obtained by minimizing the expected cost, which can be written
as
2000x +
∞
0
20000 I(y ≤ 1)
e
−y/x
x
dx
where I is the indicator function.
21. There are two optimal solutions, x
∗
=9, 10, with objective function value approximately 0.125.
Chapter 13
Simulation of Manufacturing and
Material Handling Systems
Forsolutionscheckthecoursewebsiteatwww.bcnn.net.
65
Chapter 14
Simulation of Computer Systems
Forsolutionscheckthecoursewebsiteatwww.bcnn.net.
66
