1.
Table of Contents
Chapter 12 (Computer
Simulation: Basic Concepts)
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consent of McGraw-Hill Education.
12.1
The Essence of Computer Simulation (Section 12.1) 12.2
Example 1: A Coin-Flipping Game (Section 12.1) 12.3–12.7
Example 2: Heavy Duty Company (Section 12.1) 12.8–12.14
A Case Study: Herr Cutter’s Barber Shop (Section 12.2) 12.15–12.23
Analysis of the Case Study (Section 12.3) 12.24–12.31
Outline of a Major Computer Simulation Study (Section 12.4) 12.32–12.34

2.
The Essence of Computer
Simulation
• A stochastic system is a system that evolves over time
according to one or more probability distributions.
• Computer simulation imitates the operation of such a system
by using the corresponding probability distributions to
randomly generate the various events that occur in the
system.
• Rather than literally operating a physical system, the computer
just records the occurrences of the simulated events and the
resulting performance of the system.
• Computer simulation is typically used when the stochastic
system ivolved is too complex to be analyzed satisfactorily by
analytical models.
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consent of McGraw-Hill Education.
12.2

3.
Example 1: A Coin-Flipping
Game
Rules of the game:
1. Each play of the game involves repeatedly flipping an unbiased coin until the
difference between the number of heads and tails tossed is three.
2. To play the game, you are required to pay $1 for each flip of the coin. You are
not allowed to quit during the play of a game.
3. You receive $8 at the end of each play of the game.
• Examples:
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12.3
HHH 3 flips You win $5
THTTT 5 flips You win $3
THHTHTHTTTT 11 flips You lose $3

4.
Computer Simulation of Coin-
Flipping Game
• A computer cannot flip coins. Instead it generates a sequence
of random numbers.
• A number is a random number between 0 and 1 if it has been
generated in such a way that every possible number within
the interval has an equal chance of occurring.
• An easy way to generate random numbers is to use the
RAND() function in Excel.
• To simulate the flip of a coin, let half the possible random
numbers correspond to heads and the other half to tails.
• 0.0000 to 0.4999 correspond to heads.
• 0.5000 to 0.9999 correspond to tails.
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consent of McGraw-Hill Education.
12.4

5.
Computer Simulation of Coin-
Flipping Game
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consent of McGraw-Hill Education.
12.5
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A B C D E F G
Coin-Flipping Game
Required Difference 3
Cash At End of Game $8
Summary of Game
Number of Flips 7
Winnings $1
Random Total Total
Flip Number Result Heads Tails Stop?
1 0.9016 Tails 0 1
2 0.0515 Heads 1 1
3 0.4396 Heads 2 1
4 0.6349 Tails 2 2
5 0.8241 Tails 2 3
6 0.7694 Tails 2 4
7 0.6820 Tails 2 5 Stop
8 0.9856 Tails 2 6 NA
9 0.2223 Heads 3 6 NA

6.
Data Table for 14 Replications of Coin-
Flipping Game
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12.6
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I J K L M
Data Table for Coin-Flipping Game
(14 Replications)
Number
Play of Flips Winnings
7 $1
1 19 -$11
2 5 $3
3 3 $5
4 11 -$3
5 9 -$1
6 5 $3
7 5 $3
8 3 $5
9 3 $5
10 5 $3
11 5 $3
12 5 $3
13 9 -$1
14 3 5
Average 6.43 $1.57
Select the
whole table
(J6:L20),
before
choosing
Table from
the Data
menu.

7.
Data Table for 1000 Replications of Coin-
Flipping Game
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12.7
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1001
1002
1003
1004
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1007
1008
I J K L M
Data Table for Coin-Flipping Game
(1000 Replications)
Number
Play of Flips Winnings
5 $3
1 3 $5
2 9 -$1
3 9 -$1
4 5 $3
5 5 $3
6 9 -$1
7 5 $3
8 7 $1
9 5 $3
10 11 -$3
995 5 $3
996 7 $1
997 7 $1
998 7 $1
999 17 -$9
1000 19 -$11
Average 8.83 -$0.83

8.
Corrective Maintenance versus
Preventive Maintenance
• The Heavy Duty Company has just purchased a large machine
for a new production process.
• The machine is powered by a motor that occasionally breaks
down and requires a major overhaul. Therefore, a second
standby motor is kept, and the two motors are rotated in use.
• The breakdowns always occur on the fourth, fifth, or sixth day
that the motor is in use. Fortunately, it takes fewer than three
days to overhaul a motor, so a replacement is always ready.
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consent of McGraw-Hill Education.
12.8
Cost of a Replacement Cycle that Begins with a Breakdown
Replace a Motor $2,000
Lost production during replacement 5,000
Overhaul a motor 4,000
Total $11,000

9.
Probability Distribution of
Breakdowns
Day
Probability of
a Breakdown
Corresponding
Random Numbers
1, 2, 3 0
4 0.25 0.0000 to 0.2499
5 0.5 0.2500 to 0.7499
6 0.25 0.7500 to 0.9999
7 or more
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consent of McGraw-Hill Education.
12.9

10.
Computer Simulation of
Corrective Maintenance
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12.10
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A B C D E F G H I J K
Heavy Duty Company Corrective Maintenance Simulation
Random Time Since Last Cumulative Cumulative Distribution of
Breakdown Number Breakdown Day Cost Cost Time Between Breakdowns
1 0.5267 5 5 $11,000 $11,000 Number
2 0.9685 6 11 $11,000 $22,000 Probability Cumulative of Days
3 0.8465 6 17 $11,000 $33,000 0.25 0 4
4 0.4735 5 22 $11,000 $44,000 0.5 0.25 5
5 0.4525 5 27 $11,000 $55,000 0.25 0.75 6
6 0.9994 6 33 $11,000 $66,000
7 0.4022 5 38 $11,000 $77,000 Breakdown Cost $11,000
8 0.1556 4 42 $11,000 $88,000
9 0.9489 6 48 $11,000 $99,000
10 0.2082 4 52 $11,000 $110,000
26 0.0688 4 132 $11,000 $286,000
27 0.7892 6 138 $11,000 $297,000
28 0.6054 5 143 $11,000 $308,000
29 0.1216 4 147 $11,000 $319,000 Average Cost per Day
30 0.0506 4 151 $11,000 $330,000 $2,185

11.
Preventive Maintenance
Options
• Preventive maintenance would involve scheduling the motor to be
removed (and replaced) for an overhaul at a certain time, even if a
breakdown has not occurred.
• The goal is to provide maintenance early enough to prevent a breakdown.
• Scheduling the overhaul enables removing and replacing the motor at a
convenient time when the machine is not in use, so no production is lost.
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consent of McGraw-Hill Education.
12.11
Cost of a Replacement Cycle that Begins without a Breakdown
Replace a motor on overtime $3,000
Lost production during replacement 0
Overhaul a motor before a breakdown 3,000
Total $6,000

12.
Replace Motor After 3 Days
• Cost of a replacement cycle is $6,000. (Replacement always
occurs without a breakdown).
• Replacement cycle occurs every three days.
• E(Cost per day) = ($6,000) / (3 days) = $2,000 per day.
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consent of McGraw-Hill Education.
12.12

13.
Replace Motor After 4 Days
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12.13
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A B C D E F G H I J K L M
Heavy Duty Company Preventive Maintenance Simulation (Replace After 4 Days)
Random Time Until Scheduled Time Event That Cumulative Cumulative Distribution of
Cycle Number Breakdown Until Replacement Concludes Cycle Day Cost Cost Time Between Breakdowns
1 0.0298 4 4 Breakdown 4 $11,000 $11,000 Number
2 0.7147 5 4 Replacement 8 $6,000 $17,000 Probability Cumulative of Days
3 0.3191 5 4 Replacement 12 $6,000 $23,000 0.25 0 4
4 0.7966 6 4 Replacement 16 $6,000 $29,000 0.5 0.25 5
5 0.4653 5 4 Replacement 20 $6,000 $35,000 0.25 0.75 6
6 0.3701 5 4 Replacement 24 $6,000 $41,000
7 0.1245 4 4 Breakdown 28 $11,000 $52,000 Breakdown Cost $11,000
8 0.1751 4 4 Breakdown 32 $11,000 $63,000 Replacement Cost $6,000
9 0.8580 6 4 Replacement 36 $6,000 $69,000
10 0.7298 5 4 Replacement 40 $6,000 $75,000 Replace After 4 days
11 0.0324 4 4 Breakdown 44 $11,000 $86,000
12 0.9562 6 4 Replacement 48 $6,000 $92,000
13 0.2773 5 4 Replacement 52 $6,000 $98,000
14 0.2109 4 4 Breakdown 56 $11,000 $109,000
15 0.9327 6 4 Replacement 60 $6,000 $115,000
16 0.4827 5 4 Replacement 64 $6,000 $121,000
17 0.3666 5 4 Replacement 68 $6,000 $127,000
18 0.3375 5 4 Replacement 72 $6,000 $133,000
19 0.2818 5 4 Replacement 76 $6,000 $139,000
20 0.6351 5 4 Replacement 80 $6,000 $145,000
21 0.4333 5 4 Replacement 84 $6,000 $151,000
22 0.9768 6 4 Replacement 88 $6,000 $157,000
23 0.1488 4 4 Breakdown 92 $11,000 $168,000
24 0.4420 5 4 Replacement 96 $6,000 $174,000
25 0.2325 4 4 Breakdown 100 $11,000 $185,000
26 0.7521 6 4 Replacement 104 $6,000 $191,000
27 0.8106 6 4 Replacement 108 $6,000 $197,000
28 0.2376 4 4 Breakdown 112 $11,000 $208,000
29 0.3160 5 4 Replacement 116 $6,000 $214,000 Average Cost per Day
30 0.8749 6 4 Replacement 120 $6,000 $220,000 $1,833

14.
Replace Motor After 5 Days
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12.14
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A B C D E F G H I J K L M
Heavy Duty Company Preventive Maintenance Simulation (Replace After 5 Days)
Random Time Until Scheduled Time Event That Cumulative Cumulative Distribution of
Cycle Number Breakdown Until Replacement Concludes Cycle Day Cost Cost Time Between Breakdowns
1 0.6401 5 5 Breakdown 5 $11,000 $11,000 Number
2 0.4804 5 5 Breakdown 10 $11,000 $22,000 Probability Cumulative of Days
3 0.8561 6 5 Replacement 15 $6,000 $28,000 0.25 0 4
4 0.0251 4 5 Breakdown 19 $11,000 $39,000 0.5 0.25 5
5 0.5282 5 5 Breakdown 24 $11,000 $50,000 0.25 0.75 6
6 0.5269 5 5 Breakdown 29 $11,000 $61,000
7 0.3786 5 5 Breakdown 34 $11,000 $72,000 Breakdown Cost $11,000
8 0.9322 6 5 Replacement 39 $6,000 $78,000 Replacement Cost $6,000
9 0.7445 5 5 Breakdown 44 $11,000 $89,000
10 0.3389 5 5 Breakdown 49 $11,000 $100,000 Replace After 5 days
11 0.1712 4 5 Breakdown 53 $11,000 $111,000
12 0.6093 5 5 Breakdown 58 $11,000 $122,000
13 0.3909 5 5 Breakdown 63 $11,000 $133,000
14 0.3067 5 5 Breakdown 68 $11,000 $144,000
15 0.7306 5 5 Breakdown 73 $11,000 $155,000
16 0.9599 6 5 Replacement 78 $6,000 $161,000
17 0.8690 6 5 Replacement 83 $6,000 $167,000
18 0.9593 6 5 Replacement 88 $6,000 $173,000
19 0.1577 4 5 Breakdown 92 $11,000 $184,000
20 0.1414 4 5 Breakdown 96 $11,000 $195,000
21 0.6092 5 5 Breakdown 101 $11,000 $206,000
22 0.5343 5 5 Breakdown 106 $11,000 $217,000
23 0.7374 5 5 Breakdown 111 $11,000 $228,000
24 0.9481 6 5 Replacement 116 $6,000 $234,000
25 0.2882 5 5 Breakdown 121 $11,000 $245,000
26 0.1733 4 5 Breakdown 125 $11,000 $256,000
27 0.1046 4 5 Breakdown 129 $11,000 $267,000
28 0.0690 4 5 Breakdown 133 $11,000 $278,000
29 0.4880 5 5 Breakdown 138 $11,000 $289,000 Average Cost per Day
30 0.1791 4 5 Breakdown 142 $11,000 $300,000 $2,113

15.
Herr Cutter’s Barber Shop
• Herr Cutter is a German barber who runs a one-man barber shop.
He opens his shop at 8:00 AM each weekday morning. His customers
arrive randomly at an average rate of two customers per hour. He
requires an average of 20 minutes for each haircut.
• As his business has increased, his customers now often wait awhile
(sometimes over half-an-hour). His loyal customers are willing to
wait, but new customers are much less likely to return if they have
to wait.
• An article in The Barber’s Journal states
• In a well-run barber shop, loyal customers will tolerate an average
wait of 20 minutes, while new customers will tolerate only a 10
minute average wait. (With longer waits, they typically take their
business elsewhere in the future.)
Question: Should Herr Cutter hire a new associate to share the
workload?
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consent of McGraw-Hill Education.
12.15

16.
Probability Distributions
• The time required to give a haircut varies between 15 and 25
minutes. His best estimate is that the times between 15 and
25 minutes are equally likely.
• Estimated distribution of service times: The Uniform distribution
over the interval from 15 to 25 minutes.
• To generate a random observation from the uniform
distribution, with values between a and b equally likely,
=a + (b–a)*RAND()
• For a uniform distribution between 15 and 25 minutes,
= 15 + 10*RAND()
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12.16

17.
Probability Distributions
• The barber shop has random arrivals of customers, averaging
two per hour.
• Estimated distribution of interarrival times: An exponential
distribution with a mean of 30 minutes.
• To generate a random observation from the exponential
distribution,
= –(mean) * LN(RAND())
• For a mean of 30 minutes,
= –30*LN(RAND())
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consent of McGraw-Hill Education.
12.17

18.
Building Blocks of a Simulation
Model
1.A description of the components of the system, including how
they are assumed to operate and interrelate.
2.A simulation clock.
3.A definition of the state of the system.
4.A method for randomly generating the (simulated) events that
occur over time.
5.A method for changing the state of the system when an event
occurs.
6.A procedure for advancing the time on the simulation clock.
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12.18

19.
Building Blocks for Herr
Cutter’s Simulation Model
1.A description of the components of the system, including how
they are assumed to operate and interrelate.
• The components are the customers, the queue, and Herr Cutter
as the server.
2.A simulation clock.
• t = Amount of simulated time that has elapsed so far
3.A definition of the state of the system.
• N(t) = Number of customers in the barber shop at time t.
4.A method for randomly generating the (simulated) events that
occur over time.
• The interarrival and service times are generated using the
inverse transformation method.
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consent of McGraw-Hill Education.
12.19

20.
Building Blocks for Herr
Cutter’s Simulation Model
5. A method for changing the state of the system when an event occurs.
• Reset N(t) = N(t) + 1 if an arrival occurs at time t
• Reset N(t) = N(t) – 1 if a service completion occurs at time t
6. A procedure for advancing the time on the simulation clock.
• The Next-Event Time-Advance Procedure
The Next-Event Time-Advance Procedure
a) Observe the current time t and randomly generated times of the next occurrence of
each event type that can occur (service or arrival in Herr Cutter’s problem).
Determine which event will occur next.
b) Advance the time on the simulation clock to the time of this next event.
c) Updated the system by determining its new state as a result of this event and by
randomly generating the time until the next occurrence of any event type that can
occur from this new state (if not previously generated). Also record desired
information about the performance of the system. Then return to the first step.
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consent of McGraw-Hill Education.
12.20

21.
AComputerSimulationofHerr Cutter’sBarberShop
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consent of McGraw-Hill Education.
12.21
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A B C D E F G H I
Herr Cutter's Barber Shop
(exponential)
Mean Interarrival Time 30 minutes
(uniform)
Min Service Time 15 minutes
Max Service Time 25 minutes
Average Time in Line (Wq) 14.1 minutes
Average Time in System (W) 34.0 minutes
Time Time Time Time Time
Customer Interarrival of Service Service Service in in
Arrival Time Arrival Begins Time Ends Line System
1 43.8 43.8 43.8 23.1 66.9 0.0 23.1
2 65.6 109.4 109.4 19.7 129.1 0.0 19.7
3 26.0 135.4 135.4 20.3 155.7 0.0 20.3
4 12.1 147.6 155.7 20.1 175.8 8.2 28.2
5 6.8 154.4 175.8 18.3 194.1 21.4 39.7
96 25.3 3,082.6 3,082.6 20.1 3,102.7 0.0 20.1
97 23.7 3,106.3 3,106.3 17.2 3,123.5 0.0 17.2
98 31.7 3,138.0 3,138.0 19.7 3,157.8 0.0 19.7
99 24.3 3,162.3 3,162.3 22.7 3,185.0 0.0 22.7
100 12.9 3,175.2 3,185.0 15.0 3,200.0 9.8 24.8

22.
Evolution of Number of Customers
Over First 100 Minutes
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12.22
20 40 60 80 100
1
2
3
Elapsed time (in minutes)
Number of
customers in
the system

23.
Simulating the Barber Shop
with an Associate
• The only difference occurs when the next-event time-advance
procedure is determining which event occurs next.
• Instead of just two possibilities, there are now three:
1. A departure because Herr Cutter completes a haircut.
2. A departure because the associate completes a haircut.
3. An arrival.
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12.23

24.
Analysis of the Case Study:
Financial Factors
• Revenue = $15 per haircut
• Average tip = $2 per haircut
• Cost of maintaining the shop = $50 per working day
• Salary of an associate = $120 per working day
• Commission for an associate = $5 per haircut given by the
associate.
• In addition to his salary and commission, the associate would
keep his own tips. Otherwise, the revenue would go to Herr
Cutter.
• The shop is open from 8:00 AM to 5:00 PM, so it admits
customers for nine hours.
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consent of McGraw-Hill Education.
12.24

25.
Analysis of Continuing without
an Associate
• The current distribution of interarrival times has a mean of 30
minutes. Therefore, he averages 18 customers per working
day.
• After subtracting the cost of maintaining the shop, his average
net income per working day is
Net daily income = ($15 + $2) (18 customers) –
$50
= $306 – $50
= $256
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consent of McGraw-Hill Education.
12.25

26.
Features of the Queueing
Simulator
1. Can run computer simulations of various kinds of basic queueing systems.
2. Can have any number of servers up to a maximum of 25.
3. Can use any of the following probability distributions for either interarrival times
or service times:
a) Constant time (also called the degenerate distribution).
b) Exponential distribution.
c) Translated exponential distribution (the sum of a constant time and a time from
an exponential distribution).
d) Uniform distribution.
e) Erlang distribution.
4. Provides estimates of various key measures of performance:
L = Expected number of customers in the system, including those being served.
Lq = Expected number of customers in the queue.
W = Expected waiting time in the system (includes service time) for a customer.
Wq = Expected waiting time in the queue (excluding service time) for a customer.
Pn = Probability of exactly n customers in the system (for n = 1, 2, … , 10).
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consent of McGraw-Hill Education.
12.26

27.
Queueing Simulator for Herr
Cutter (Without Associate)
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consent of McGraw-Hill Education.
12.27
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B C D E F G H
Data Results
Number of Servers = 1 Point
Estimate Low High
Interarrival Times L = 1.358 1.332 1.385
Distribution = Exponential Lq = 0.689 0.666 0.712
Mean = 30 W = 40.582 39.983 41.180
5 Wq = 20.577 19.980 21.174
Service Times P0 = 0.330 0.326 0.335
Distribution = Uniform P1 = 0.310 0.307 0.313
Minimum Value = 15 P2 = 0.183 0.180 0.185
Maximum Value = 25 P3 = 0.0492 0.0920 0.0963
P4 = 0.0451 0.0433 0.0469
Length of Simulation Run P5 = 0.0206 0.0192 0.0220
Number of Arrivals = 10,000 P6 = 0.00950 0.00849 0.01050
P7 = 0.00432 0.00360 0.00503
P8 = 0.00219 0.00163 0.00274
P9 = 0.000876 0.000540 0.001210
P10 = 0.000372 0.000165 0.000579
95% Confidence Interval
Run Simulation

28.
Testing the Validity of the
Simulation Model
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12.28
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A B C D E F G
Template for the M/G/1 Queueing Model
Data Results
 0.0333 (mean arrival rate) L = 1.344
 20 (expected service time) Lq = 0.678
 2.887 (standard deviation)
s = 1 (# servers) W = 40.356
Wq = 20.356
 0.666
P0 = 0.334

29.
Analysis of the Option of Adding an
Associate
• Adding an associate would
• reduce the average waiting time before a haircut begins to less than 10
minutes.
• gradually attract new business until this average waiting time reaches 10
minutes.
Queueing Simulator Results:
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consent of McGraw-Hill Education.
12.29
Mean of
Interarrival Times
Point
Estimate of Wq
95 Percent Confidence
Interval for Wq
20 minutes 3.33 minutes 3.05 to 3.61 minutes
15 minutes 8.10 minutes 6.98 to 9.22 minutes
14 minutes 10.80 minutes 9.51 to 12.08 minutes
14.2 minutes 9.83 minutes 8.83 to 10.84 minutes
14.3 minutes 9.91 minutes 8.76 to 11.05 minutes

30.
Queueing Simulator for Herr
Cutter (With an Associate)
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12.30
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B C D E F G H
Data Results
Number of Servers = 2 Point
Estimate Low High
Interarrival Times L = 2.126 2.090 2.163
Distribution = Exponential Lq = 0.719 0.689 0.748
Mean = 14.3 W = 30.212 29.833 30.591
5 Wq = 10.211 9.834 10.588
Service Times P0 = 0.163 0.160 0.166
Distribution = Uniform P1 = 0.266 0.262 0.270
Minimum Value = 15 P2 = 0.233 0.230 0.235
Maximum Value = 25 P3 = 0.1541 0.1518 0.1564
P4 = 0.0877 0.0855 0.0898
Length of Simulation Run P5 = 0.0467 0.0448 0.0487
Number of Arrivals = 100,000 P6 = 0.02417 0.02264 0.02570
P7 = 0.01282 0.01162 0.01401
P8 = 0.00634 0.00546 0.00722
P9 = 0.003208 0.002530 0.003890
P10 = 0.001546 0.001076 0.002017
95% Confidence Interval
Run Simulation

31.
Analysis of the Option of
Adding an Associate
• Herr Cutter believes 14.3 minutes provides an adequate and
conservative estimate of the mean interarrival rate with an
associate.
• This corresponds to a mean arrival rate of 37.8 customers per
day.
• Net daily income = 37.8 ($15) (shop revenue)
+ 18.9 ($2) (his tips)
– $50 (shop maintenance)
– $120 (associate’s salary)
– $18.9 ($5)(associate’s commission)
Total $340.30
• This compares to Herr Cutter’s current net daily income of
$256.
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consent of McGraw-Hill Education.
12.31

32.
Outline of a Major Computer
Simulation Study
• Step 1: Formulate the Problem and Plan the Study
• What is the problem that management wants studied?
• What are the overall objectives for the study?
• What specific issues should be addressed?
• What kinds of alternative system configurations should be
considered?
• What measures of performance of the system are of interest to
management?
• What are the time constraints for performing the study?
• Step 2: Collect the Data and Formulate the Simulation Model
• The probability distributions of the relevant quantities are needed.
• Generally it will only be possible to estimate these distributions.
• A simulation model often is formulated in terms of a flow diagram.
• Step 3: Check the Accuracy of the Simulation Model
• Walk through the conceptual model before an audience of all the key
people.
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consent of McGraw-Hill Education.
12.32

33.
Outline of a Major Computer
Simulation Study
• Step 4: Select the Software and Construct a Computer Program
• Classes of software
• Spreadsheet software (e.g., Excel, Crystal Ball)
• A general purpose programming language (e.g., C, FORTRAN, Pascal, etc.)
• A general purpose simulation language (e.g., GPSS, SIMSCRIPT, SLAM,
SIMAN)
• Applications-oriented simulators
• Step 5: Test the Validity of the Simulation Model
• If the real system is currently in operation, performance data should
be compared with the corresponding output generated by the
simulation model.
• Conduct a field test to collect data to compare to the simulation
model.
• Have knowledgeable personnel check how the simulation results
change as the configuration of the simulated system is changed.
• Watch animations of simulation runs.
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consent of McGraw-Hill Education.
12.33

34.
Outline of a Major Computer
Simulation Study
• Step 6: Plan the Simulations to Be Performed
• Determine length of simulation runs.
• Keep in mind that the simulation runs do no produce exact
values. Each simulation run can be viewed as a statistical
experiment that is generating statistical observations of the
performance of the system.
• Step 7: Conduct the Simulation Runs and Analyze the Results
• Obtain point estimates and confidence intervals to indicate the
range of likely values for the measures.
• Step 8: Present Recommendations to Management
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consent of McGraw-Hill Education.
12.34