Ch12Simulation.ppt

8/14/2019 Ch12Simulation.ppt

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Slide 2008 Thomson South-Western. All Rights Reserved

Slides by

JOHNLOUCKSSt. EdwardsUniversity


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Chapter 12Simulation

Advantages and Disadvantages of UsingSimulation

Modeling

Random Variables and Pseudo-Random Numbers

Time Increments Simulation Languages

Validation and Statistical Considerations

Examples


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Simulation

Simulation is one of the most frequently employed

management science techniques. It is typically used to model random processes that

are too complex to be solved by analyticalmethods.


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Advantages of Simulation

Among the advantages of simulation is the ability to

gain insights into the model solution which may beimpossible to attain through other techniques.

Also, once the simulation has been developed, itprovides a convenient experimental laboratory to

perform "what if" and sensitivity analysis.


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Disadvantages of Simulation

A large amount of time may be required to develop

the simulation. There is no guarantee that the solution obtained will

actually be optimal.

Simulation is, in effect, a trial and error method of

comparing different policy inputs. It does not determine if some input which was not

considered could have provided a better solution forthe model.


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Simulation Modeling

One begins a simulation by developing a

mathematical statement of the problem. The model should be realistic yet solvable within

the speed and storage constraints of the computersystem being used.

Input values for the model as well as probabilityestimates for the random variables must then bedetermined.


7/49 7Slide 2008 Thomson South-Western. All Rights Reserved

Random Variables

Random variable values are utilized in the model

through a technique known as Monte Carlosimulation.

Each random variable is mapped to a set of numbersso that each time one number in that set is generated,

the corresponding value of the random variable isgiven as an input to the model.

The mapping is done in such a way that thelikelihood that a particular number is chosen is the

same as the probability that the corresponding valueof the random variable occurs.



Pseudo-Random Numbers

Because a computer program generates random

numbers for the mapping according to some formula,the numbers are not truly generated in a randomfashion.

However, using standard statistical tests, the

numbers can be shown to appear to be drawn from arandom process.

These numbers are called pseudo-random numbers.



Time Increments

In a fixed-time simulation model, time periods are

incremented by a fixed amount. For each timeperiod a different set of data from the inputsequence is used to calculate the effects on themodel.

In a next-event simulation model, time periods arenot fixed but are determined by the data valuesfrom the input sequence.



Simulation Programs

The computer program that performs the simulation

is called a simulator. Flowcharts can be useful in writing such a program.

While this program can be written in any generalpurpose language (e.g. BASIC, FORTRAN, C++, etc.)

special languages which reduce the amount of codewhich must be written to perform the simulationhave been developed.

Special simulation languages include SIMSCRIPT,

SPSS, DYNAMO, and SLAM.



Model Verification/Validation

Verification/validation of both the model and the

method used by the computer to carry out thecalculations is extremely important.

Models which do not reflect real world behaviorcannot be expected to generate meaningful results.

Likewise, errors in programming can result innonsensical results.



Model Verification/Validation

Validation is generally done by having an expert

review the model and the computer code for errors. Ideally, the simulation should be run using actual

past data.

Predictions from the simulation model should be

compared with historical results.


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13Slide 2008 Thomson South-Western. All Rights Reserved

Experimental Design

Experimental design is an important consideration in

the simulation process. Issues such as the length of time of the simulation

and the treatment of initial data outputs from themodel must be addressed prior to collecting andanalyzing output data.

Normally one is interested in results for the steadystate (long run) operation of the system beingmodeled.

The initial data inputs to the simulation generally

represent a start-up period for the process and it maybe important that the data outputs for this start-upperiod be neglected for predicting this long runbehavior.


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Experimental Design

For each policy under consideration by the decision

maker, the simulation is run by considering a longsequence of input data values (given by a pseudo-random number generator).

Whenever possible, different policies should be

compared by using the same sequence of input data.


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The price change of shares of Dynogen, Inc.

has been observed over the past 50 trades. Thefrequency distribution is as follows:

Price Change Number of Trades-3/8 4-1/4 2-1/8 8

0 20+1/8 10

+1/4 3+3/8 2+1/2 1

Total = 50

Example: Dynogen, Inc.


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Relative Frequency Distribution and

Random Number Mapping

Price Change Relative Frequency Random Numbers

-3/8 .08 00 - 07

-1/4 .04 08 - 11

-1/8 .16 12 - 27

0 .40 28 - 67

+1/8 .20 68 - 87

+1/4 .06 88 - 93

+3/8 .04 94 - 97

+1/2 .02 98 - 99

Total 1.00


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If the current price per share of Dynogen is

23, use random numbers to simulate the price per

share over the next 10 trades.

Use the following stream of random numbers:

21, 84, 07, 30, 94, 57, 57, 19, 84, 84


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Simulation Worksheet

Trade Random Price Stock

Number Number Change Price1 21 -1/8 22 7/82 84 +1/8 233 07 -3/8 22 5/84 30 0 22 5/85 94 +3/8 236 57 0 237 57 0 238 19 -1/8 22 7/89 84 +1/8 23

10 84 +1/8 23 1/8


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Spreadsheet for Stock Price Simulation

A B C D E F

1 Lower Upper Trade Price Stock

2 Random Random Price Number Change Price

3 Number Number Change 1 0.125 23.125

4 0.00 0.08 -0.375 2 -0.125 23.0005 0.08 0.12 -0.250 3 0.000 23.000

6 0.12 0.28 -0.125 4 0.000 23.000

7 0.28 0.68 0.000 5 0.000 23.000

8 0.68 0.88 0.125 6 0.500 23.500

9 0.88 0.94 0.250 7 -0.125 23.375

10 0.94 0.98 0.375 8 0.000 23.375

11 0.98 1.00 0.500 9 -0.125 23.250

12 10 -0.125 23.125

A B C D E F

1 Lower Upper Trade Price Stock

2 Random Random Price Number Change Price

3 Number Number Change 1 0.125 23.125

4 0.00 0.08 -0.375 2 -0.125 23.0005 0.08 0.12 -0.250 3 0.000 23.000

6 0.12 0.28 -0.125 4 0.000 23.000

7 0.28 0.68 0.000 5 0.000 23.000

8 0.68 0.88 0.125 6 0.500 23.500

9 0.88 0.94 0.250 7 -0.125 23.375

10 0.94 0.98 0.375 8 0.000 23.375

11 0.98 1.00 0.500 9 -0.125 23.250

12 10 -0.125 23.125


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Theoretical Results and Observed Results

Based on the probability distribution, theexpected price change per trade can be calculated by:

(.08)(-3/8) + (.04)(-1/4) + (.16)(-1/8) + (.40)(0)

+ (.20)(1/8) + (.06)(1/4) + (.04)(3/8) + (.02)(1/2) = +.005

The expected price change for 10 trades is

(10)(.005) = .05. Hence, the expected stock price after

10 trades is 23 + .05 = 23.05.

Compare this ending price with the spreadsheetsimulation and manual simulation results on the

previous slides.


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Example: Mark Koffs Process

Mark Koff is a specialist at repairing large metal-

cutting machines that use lasertechnology. His repair

territory consists of the cities

of Austin, San Antonio, and

Houston. His day-to-day

repair assignment locations

can be modeled as a Markov

process. The transition matrix is shownon the next slide.


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This Days Next Day's LocationLocation Austin San Antonio Houston

Austin .60 .15 .25

San Antonio .20 .75 .05

Houston .15 .05 .80


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Random Number Mappings

Currently in Currently in Currently in

Austin San Antonio Houston

Next-Day Random Next-Day Random Next-Day Random

Location Numbers Location Numbers Location NumbersAustin 00 - 59 Austin 00 - 19 Austin 00 - 14

San Ant. 60 - 74 San Ant. 20 - 94 San Ant. 15 - 19

Houston 75 - 99 Houston 95 - 99 Houston 20 - 99


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Assume Mark is currently in Houston.

Simulate where Mark will be over the next 16 days.

What percentage of time will Mark be in each of the

three cities?

Use the following random numbers:93, 63, 26, 16, 21, 26, 70, 55, 72, 89, 49, 64, 91, 02, 52, 69



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Simulation Worktable

Starting in HoustonRandom Day's Random Day's

Day Number Location Day Number Location1 93 Houston 9 72 San Ant.2 63 Houston 10 89 San Ant.3 26 Houston 11 49 San Ant.4 16 San Ant. 12 64 San Ant.5 21 San Ant. 13 91 San Ant.

6 26 San Ant. 14 02 Austin7 70 San Ant. 15 52 Austin8 55 San Ant. 16 69 San Ant.


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Repeat the simulation with Mark currently

in Austin. Use the following random numbers:

13, 08, 60, 13, 68, 40, 40, 27, 23, 64, 36, 56, 25, 88, 18, 74

Compare the percentages with those found withMark starting in Houston.



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Simulation Worksheet

Starting in AustinRandom Day's Random Day's

Day Number Location Day Number Location1 13 Austin 9 23 San Ant.

2 08 Austin 10 64 San Ant.3 60 San Ant. 11 36 San Ant.4 13 Austin 12 56 San Ant.5 68 San Ant. 13 25 San Ant.

6 40 San Ant. 14 88 San Ant.7 40 San Ant. 15 18 Austin8 27 San Ant. 16 74 San Ant.


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Simulation Summary

Starting in Houston

Austin = 2/16 = 12.50%

San Antonio = 11/16 = 68.75%

Houston = 3/16 = 18.75%

Starting in Austin

Austin = 4/16 = 25%

San Antonio = 12/16 = 75%Houston = 0/16 = 0%


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Partial Spreadsheet w/Variable Look-up Table

LRN = Lower Random NumberURN = Upper Random Number

NDL = Next-Day Location

A B C D E F G H I

1

2

3 LRN URN NDL LRN URN NDL LRN URN NDL

6 0.00 0.60 Aus. 0.00 0.20 Aus. 0.00 0.15 Aus.

7 0 .6 0 0.7 5 S.A. 0 .20 0 .9 5 S.A. 0 .15 0 .2 0 S.A.

8 0.75 1.00 Hou. 0.95 1.00 Hou. 0.20 1.00 Hou.

Aus. S.A. Hou.

Current Location

A B C D E F G H I

1

2

3 LRN URN NDL LRN URN NDL LRN URN NDL

6 0.00 0.60 Aus. 0.00 0.20 Aus. 0.00 0.15 Aus.

7 0 .60 0 .75 S.A. 0.20 0 .95 S.A. 0.1 5 0 .20 S.A.

8 0.75 1.00 Hou. 0.95 1.00 Hou. 0.20 1.00 Hou.

Aus. S.A. Hou.

Current Location


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Partial Spreadsheet with Simulation Table

=IF(A13=$A$2,VLOOKUP(C13,$A$6:$C$8,3),

IF(A13=$D$2,VLOOKUP(C13,$D$6:$F$8,3),

VLOOKUP(C13,$G$6:$I$8,3)))

A B C D E F

11

12

13

1415

16

17

18

19

Next-Day

S.A.

S.A.

S .A . 0.61 S .A .

N u mb e r

0.33

Current

Location

Random

S.A.

0.45

S.A.

Location

A us . 0. 64 S .A .

S .A . 0.87 S .A .

S.A.S.A.

S.A.

0.30

0.71S.A.


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Example: Wayne International Airport

Wayne International Airport primarily serves

domestic air traffic. Occasionally, however,

a chartered plane from abroad will arrive

with passengers bound for Wayne's two

great amusement parks, Algorithmlandand Giffith's Cherry Preserve.

Whenever an international plane

arrives at the airport the two customs

inspectors on duty set up operations toprocess the passengers.


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Incoming passengers must first have their

passports and visas checked. This is handled by

one inspector. The time required to check

a passenger's passports and visas can be

described by the probability distributionon the next slide.


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Time Required toCheck a Passenger'sPassport and Visa Probability

20 seconds .20

40 seconds .4060 seconds .30

80 seconds .10


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After having their passports and visas checked,

the passengers next proceed to the second customsofficial who does baggage inspections. Passengers form

a single waiting line with the official inspecting

baggage on a first come, first served basis. The time

required for baggage inspection follows the probability

distribution shown on the next slide.



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Time Required ForBaggage Inspection Probability

No Time .251 minute .60

2 minutes .103 minutes .05



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Time Required toCheck a Passenger's RandomPassport and Visa Probability Numbers

20 seconds .20 00 - 1940 seconds .40 20 - 59

60 seconds .30 60 - 89

80 seconds .10 90 - 99


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Time Required For RandomBaggage Inspection Probability Numbers

No Time .25 00 - 24

1 minute .60 25 - 84

2 minutes .10 85 - 94

3 minutes .05 95 - 99


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Next-Event Simulation Records

For each passenger the following informationmust be recorded:

When his service begins at the passport control

inspection The length of time for this service When his service begins at the baggage inspection The length of time for this service


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Time Relationships

Time a passenger begins service

by the passport inspector

= (Time the previous passenger started passport

service)+ (Time of previous passenger's passport service)

l l


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Time Relationships

Time a passenger begins service

by the baggage inspector

( If passenger does not wait in line for baggage

inspection)= (Time passenger completes service

with the passport control inspector)

(If the passenger does wait in line for baggage

inspection)= (Time previous passenger completes

service with the baggage inspector)

E l W I i l Ai


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Time Relationships

Time a customer completes service

at the baggage inspector

= (Time customer begins service with baggage

inspector) + (Time required for baggage inspection)

E l W I i l Ai


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A chartered plane from abroad lands at

Wayne Airport with 80 passengers. Simulate the

processing of the first 10 passengers through customs.

Use the following random numbers:

For passport control:

93, 63, 26, 16, 21, 26, 70, 55, 72, 89

For baggage inspection:

13, 08, 60, 13, 68, 40, 40, 27, 23, 64


E l W I i l Ai


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Simulation Worksheet (partial)

Passport Control Baggage Inspections

Pass. Time Rand. Service Time Time Rand. Service Time

Num Begin Num. Time End Begin Num. Time End

1 0:00 93 1:20 1:20 1:20 13 0:00 1:202 1:20 63 1:00 2:20 2:20 08 0:00 2:20

3 2:20 26 :40 3:00 3:00 60 1:00 4:00

4 3:00 16 :20 3:20 4:00 13 0:00 4:00

5 3:20 21 :40 4:00 4:00 68 1:00 5:00

E l W I t ti l Ai t


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Simulation Worksheet (continued)

Passport Control Baggage Inspections

Pass. Time Rand. Service Time Time Rand. Service Time

Num Begin Num. Time End Begin Num. Time End

6 4:00 26 :40 4:40 5:00 40 1:00 6:007 4:40 70 1:00 5:40 6:00 40 1:00 7:00

8 5:40 55 :40 6:20 7:00 27 1:00 8:00

9 6:20 72 1:00 7:20 8:00 23 0:00 8:00

10 7:20 89 1:00 8:20 8:20 64 1:00 9:20

E l W I t ti l Ai t


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Explanation

For example, passenger 1 begins being served bythe passport ontrol inspector immediately. His servicetime is 1:20 (80 seconds) at which time he goesimmediately to the baggage inspector who waves him

through without inspection.Passenger 2 begins service with passport inspector

1:20 minutes (80 seconds) after arriving there (as this iswhen passenger 1 is finished) and requires 1:00 minute

(60 seconds) for passport inspection. He is wavedthrough baggage inspection as well.

This process continues in this manner.

E l W I t ti l Ai t


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Question

How long will it take for the first 10 passengersto clear customs?

Answer

Passenger 10 clears customs after 9 minutes and

20 seconds.

E l W I t ti l Ai t


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Question

What is the average length of time a customerwaits before having his bags inspected after he clearspassport control? How is this estimate biased?

E l W I t ti l Ai t


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Answer

For each passenger calculate his waiting time:

(Baggage Inspection Begins) - (Passport ControlEnds)

= 0+0+0+40+0+20+20+40+40+0 = 120 seconds.120/10 = 12 seconds per passenger

This is a biased estimate because we assume that

the simulation began with the system empty. Thus,the results tend to underestimate the average waitingtime.


E d f Ch t 12


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End of Chapter 12

Documents

Ch12Simulation.ppt