Ch18 Simulation

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18

Simulation

SUPPLEMENT TO

CHAPTER

LEARNING OBJECTIVES

Af ter completing this supplement,

you should be able to:

LO1 Explain what is meant by thetermsimulation, list some ofthe reasons for simulationspopularity as a tool fordecision making, anddescribe the steps insimulation.

LO2 Explain how randomnumbers and randomvariates are generated insimulation.

LO3 Model and solve typicalproblems that require the useof simulationmanuallyand using Excel.

LO4 Describe some simulationsoftware, be able to useArena for simple problems,and describe someapplications of simulation.

SUPPLEMENT OUTLINE

Introduction and SimulationProcess, 2

Random Number and RandomVariate Generation, 4

Some Simulation Models, 13

Some Simulation Software andApplications, 21

Key Terms, 31

Solved Problems, 31

Discussion and ReviewQuestions, 32

Internet Exercises, 33

Problems, 33

Mini-case: Valencia KidneyWaiting Line, 38

Mini-case: Dofasco, 38


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PART EIGHT Waiting-Line Analysis2

LO 1 INTRODUCTION AND SIMULATION PROCESSSimulation, as used in decision making, is a descriptive technique in which a model ofa process is developed and then experiments are conducted on the model to evaluate itsbehaviour under various conditions. Simulation is not an optimizing technique. It does notproduce a solution per se. Instead, simulation enables decision makers to test their solutionson a model that reasonably duplicates a real process. Simulation models enable decision

makers to experiment with decision alternatives usingwhat-ifapproach.Simulation has applications across a broad spectrum of operations management prob-

lems. In some instances, the simulations models are quite modest, while others are rathercomplex. Their usefulness in all cases depends on the degree to which decision makersare able to successfully answer their what-ifquestions.

A list of operations management topics would reveal that many have simulation applica-tions. For instance, simulation is often helpful in process design, facility layout, capacityplanning, line balancing, testing alternative inventory policies, scheduling, waiting lines,and project management.

Generally, analysts use the simulation approach because the assumptions required byan optimizing technique are not reasonably satisfied in a given situation. Waiting lineproblems are a good example. Although waiting line problems are pervasive, the rather

restrictive assumptions of arrival and service distributions in many cases are simply notmet. Very often, analysts will then turn to simulation as a reasonable alternative for obtain-ing descriptive information about the system in question.

Other reasons for the popularity of simulation include:

Many situations are too complex to permit development of a mathematical solution.1.

Simulation models are fairly simple to use and understand.2.

Simulation enables the decision maker to conduct experiments on a model that will3.help in understanding the process behaviour while avoiding the risks of conductingtests in real life.

Extensive computer software packages make it relatively easy to model fairly sophis-4.ticated processes.

Simulation ProcessCertain basic steps are used for all simulation models:

Define the manufacturing or service process and set simulation objective(s).1.

Develop the simulation model.2.

Test the model to be sure that it is working as intended (called model verification) and3.reflects the system being studied (called model validation).

Develop one or more experiments (conditions under which the models behaviour4.will be examined).

Run the simulation and evaluate the results.5.

Repeat steps 4 and 5 until you are satisfied with the results.6.

The first step in simulation is to clearly define the manufacturing or service process tobe modelled. A clear statement of the simulation objective(s) can provide not only guid-ance for model development but also the basis for evaluation of the success or failure ofa simulation study. In general, the goal is to determine how a process will behave undercertain conditions. The more specific a manager is about what he or she is looking for, thebetter the chances that the simulation model will be designed to accomplish that. Towardthat end, the manager must decide on thescopeand level of detai lof the simulation. Thisdetermines the necessary degree of complexity of the model and the information require-ments of the study.

As part of system definition, the type of simulation should be determined: terminating(or transient) and non-terminating (or steady state). A terminating state or event might

simulation A descriptivetechnique that enables a

decision maker to evaluate the

behaviour of a model under

various conditions.


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SUPPLEMENT TO CHAPTER 18 Simulation 3

be when a particular number of jobs have been completed. A terminating point in timemight be the closing of shop at the end of a business day. In a terminating simulation,performance during successive time intervals during the simulation is measured. A non-terminating simulation means that the simulation could theoretically go on indefinitely withno statistical change in behaviour of the system. The modeller must determine a suitablelength of time to run the model. An example of a non-terminating simulation is a modelof a manufacturing operation in which work always continues exactly as it left off at the

end of previous shift.The next step is simulation model development. Typically, this involves deciding on the

structure of the model and using a computer to carry out the simulation. Data gathering isa significant part of model development. The amount and type of data needed are a directfunction of the scope and level of detail of the simulation. The data are needed for bothmodel development and validation.

Once a model is developed, it must beverified, i.e., debugged to ensure that it workscorrectly. This is much easier to do if a model is built in stages and with minimal detail.To help debug the model, most simulation software provide a trace capability in the formof audit trail, screen messages, or graphic animation. A walk-through of the modelinput is always advisable. Model validations main purpose is to determine if the modeladequately depicts real process performance. An analyst usually accomplishes this by

comparing the results of simulation runs with known performance of the process under thesame circumstances. If such a comparison cannot be made because, for example, real-lifedata are difficult or impossible to obtain, an alternative is to employ a test of reasonable-ness (called face validation), in which the judgments and opinions of individuals familiarwith the process or similar processes are relied on for confirmation that the results areplausible and acceptable. Still another aspect of validation is careful consideration of theassumptions of the model and the values of parameters used in testing the model. Again,the judgments and opinions of those familiar with the real-life process and those who mustuse the results are essential. Finally, note that model development and model verification/validation go hand in hand: model deficiencies uncovered during verification/validationprompt model revisions, which lead to the need for further verification/validation effortsand perhaps further revisions.

The fourth step in simulation is developing experiments (conditions under which themodels behaviour will be examined). Experiments are the essence of a simulation; theyhelp answer thewhat-ifquestions posed in simulation studies. For example, should we useone server or two servers, or use order quantity of 5 units, 6 or 7.

The fifth step is to run the simulation model. If a simulation model is deterministic andall parameters are known and constant, only a single run will be needed for eachwhat-ifquestion. But if the model is probabilistic, with parameters subject to random variability,multiple runs will be needed to obtain confidence in the results. In this chapter supplement,probabilistic simulation (also called the Monte Carlo method or simulation) is the focalpoint of the discussion, and comments are limited to it.

Probabilistic simulation is essentially a form of random sampling, with each run repre-senting one observation (for non-terminating simulation) and a sequence of observations(for terminating simulation). Consequently, statistical theory can be used to determine

appropriate sample sizes. In effect, the larger the degree of variability inherent in simulationresults, the greater the number of simulation runs needed to achieve a reasonable level ofconfidence in the results as true indicators of model behaviour. Another variance reductiontechnique is to use common random numbers, i.e., use the same stream of random num-bers in various experiments. For example, in queueing, if we are comparing two differentconfigurations of tellers in a bank, we would want the (random) time of arrival of thenthcustomer to be generated using the same random number for both configurations.

In probabilistic simulation, each random component of the process under study has aprobability distribution. Random samples taken from probability distributions are analogousto observations made on the process itself. Random sampling is accomplished by the useof random numbers and random variates.

Monte Carlo method orsimulation Probabilisticsimulation technique, used

when a process has one or

more random component(s).


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LO 2 RANDOM NUMBER AND RANDOM VARIATEGENERATION

Random number generation is the process of choosing a number in a given range where everynumber in the range has equal probability of being picked. For example, a two-digit randomnumber in the range [00 to 99] is any of the 100 possible 2-digit numbers (note that 0 to 9

are considered as 00 to 09, i.e., two digits), and each has an equal chance of being picked.In this chapter supplement, we will distinguish betweenrandom numbersthat are integers,usually 2 digits, from a range such as [00, 99], and random values (calledrandom variates)from distributions such as Normal and Poisson. The reason for this distinction is that wewill use both random numbers and variates below, with different roles. The objective is togenerate random variates, and we will use random numbers to do so when we are simulatingmanually. Computer software use random numbers too but these are hidden from the user.For example, in Excels Data Analysis module, there is the Random Number Generationprogram that directly generates random variates from common probability distributionssuch as Poisson and Normal. Note that Excel calls random variates random numbers. Wewill first consider how we can obtain random numbers (e.g., two-digit integers).

Obtaining Random NumbersThe random numbers used in probabilistic simulation come from one of two sources: (a) atable of random numbers like Table 18S-1, or (b) computer software such as Excel. In eithercase, the method of generation is usually a recursive equation such as Zi=(aZi-1) modm,i.e., the next random number Ziis the remainder after dividing atimes the current randomnumber Zi-1bym. This process results in random numbers between 0 and m 1. The derivedrandom number can then be transformed to obtain a random number in any desired range,e.g., [00 to 99]. The problem with the recursive equation method is that the resulting sequenceof numbers repeats itself after a certain number of numbers (called a cycle) are generated.Therefore, the parametersaandmshould be chosen carefully. In the ProModel simulationsoftware,a=630,360,016 andm=231 1, which results in a cycle ofm. Because the resultingnumbers are not truly random, they are sometimes referred to as pseudo random numbers.

Fortunately, we dont have to generate random numbers and will only use them. In

manual simulation we will use Table 18S-1 (or Table 18S-2 for Normal variates), and incomputer simulation we will bypass random numbers and use Excel to generate randomvariates directly. If in Excel it is necessary to first generate random numbers, this can bedone by using the function @rand, which will generate a real number greater or equal to0 and less than 1. To generate a random (integer) number between 00 and 99, we can usethe formula =round(rand()*100-.5,0).

The numbers in Table 18S-1 have two digits for convenience, but the digits can be used singly,in pairs, or in whatever grouping a given problem calls for. Each digit ranges from 0 to 9.

random number An integer,usually 2 digits, chosen

randomly from a range of

integers such as [00, 99]

random variate A randomvalue for a distribution such as

Poisson or Normal

Table 18S-1

Random numbers

1 2 3 4 5 6 7 8 9 10 11 12

1 18 20 84 29 91 73 64 33 15 67 54 07

2 25 19 05 64 26 41 20 09 88 40 73 34

3 73 57 80 35 04 52 81 48 57 61 29 35

4 12 48 37 09 17 63 94 08 28 78 51 23

5 54 92 27 61 58 39 25 16 10 46 87 17

6 96 40 65 75 16 49 03 82 38 33 51 20

7 23 55 93 83 02 19 67 89 80 44 99 72

8 31 96 81 65 60 93 75 64 26 90 18 59

9 45 49 70 10 13 79 32 17 98 63 30 05

10 01 78 32 17 24 54 52 44 28 50 27 68

11 41 62 57 31 90 18 24 15 43 85 31 97

12 22 07 38 72 69 66 14 85 36 71 41 58


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For any size grouping of digits (e.g., two-digit numbers), every possible outcome (e.g.,34, 89, 00) has the same probability of appearing. This implies that there are no discerniblepatterns in sequences of numbers to enable one to predict numbers further in the sequence.Therefore, the numbers can be read across rows and up or down columns.

When using the table, it is important to avoid always starting in the same spot. For ourpurposes, the starting point will be specified so that everyone obtains the same results.

Generating Random VariatesTo generate most random variates, theinverse transformation methodis used:

Construct the Cumulative probability distribution F (from which a random variate is1.desired).

Generate a random real number2. ugreater or equal to 0 and less than 1.

Determine the value3. xsuch thatF(x) =u.

Below, we will illustrate the inverse transformation method for some common probabilitydistributions. In each case we will first perform the random variate generation manually,and then use Excel.

Generating Discrete Random Variates

The manager of a machine shop is concerned about machine breakdowns. He has made adecision to simulate breakdowns for a 10-day period. Historical data on breakdowns overthe last 100 days are given in the following table:

Number of Breakdowns Frequency0

1

2

3

4

5

0 . . . . . . . . . . . . 10

1 . . . . . . . . . . . . 30

2 . . . . . . . . . . . . 25

3 . . . . . . . . . . . . 20

4 . . . . . . . . . . . . 10

5 . . . . . . . . . . . . 5

100

Simulate breakdowns for a 10-day period. Read two-digit random numbers fromTable 18S-1, starting at the top of column 1 and reading down.

Example S-1

SolutionDevelop cumulative probabilities for breakdowns:a.

Convert frequencies into relative frequencies by dividing each frequency by thei.sum of the frequencies. Thus, 10 becomes 10/100=.10, 30 becomes 30/100=.30,and so on.

Develop cumulative relative frequencies (i.e., cumulative probabilities) by successiveii.

summing. The results are shown in the following table:

Number ofBreakdowns Frequency

RelativeFrequency

CumulativeProbability

0 10 .10 .10

1 30 .30 .40

2 25 .25 .65

3 20 .20 .85

4 10 .10 .95

5 5 .05 1.00

100 1.00


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b. Assign random-number intervals to correspond to the cumulative probabilities for break-downs. (Note: Use two-digit numbers because the cumulative probabilities are givento two decimal places.) You want a 10-percent probability of obtaining the event 0breakdowns in our simulation. Therefore, you must designate 10 percent of the possiblerandom numbers as corresponding to that event. There are 100 two-digit numbers, sowe can assign the 10 numbers 01 to 10 to that event.

Similarly, assign the numbers 11 to 40 to one breakdown, 41 to 65 to two break-

downs, 66 to 85 to three breakdowns, 86 to 95 to 4 breakdowns and 96 to 00 tofive breakdowns.

Note that 00 is assigned as if it was 100, not 0. This makes the interval maxi-mums correlate well to the cumulative Probabilities.

Number ofBreakdowns Frequency

RelativeFrequency

CumulativeProbability

CorrespondingRandom Numbers

0 10 .10 .10 01 to 10

1 30 .30 .40 11 to 40

2 25 .25 .65 41 to 65

3 20 .20 .85 66 to 85

4 10 .10 .95 86 to 955 5 .05 1.00 96 to 00

100 1.00

c. Obtain the random numbers from Table 18S-1, column 1, as specified in the problem:18 25 73 12 54 96 23 31 45 01

d. Convert the random numbers into numbers of breakdowns on each day, starting fromday 1:18 falls in the interval 11 to 40 and corresponds, therefore, to one breakdown onday 1.25 falls in the interval 11 to 40 and corresponds to one breakdown on day 2.73 corresponds to three breakdowns on day 3.

12 corresponds to one breakdown on day 4.54 corresponds to two breakdowns on day 5.96 corresponds to five breakdowns on day 6.23 corresponds to one breakdown on day 7.31 corresponds to one breakdown on day 8.45 corresponds to two breakdowns on day 9.01 corresponds to no breakdowns on day 10.

The following table summarizes these results:

DayRandomNumber

Simulated Numberof Breakdowns

1 18 1

2 25 1

3 73 3

4 12 1

5 54 2

6 96 5

7 23 1

8 31 1

9 45 2

10 01 0

17


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The mean number of breakdowns for this 10-day simulation is 17/10=1.7 breakdowns perday. Compare this to theexpectednumber of breakdowns based on the historical data:

0(.10)+1(.30)+2(.25)+3(.20)+4(.10)+5(.05)=2.05 per day.The two (1.7 and 2.05) are close but because of random variability, not equal. As the length

of simulation increases, however, the sample mean should approach the population mean.

Using Excel to Generate Discrete Random Variates While the inverse transformtion method

can be used to generate discrete random variates in Excel using rand() and VLookup func-tions, there is a more direct way. Excels Data Analysis module contains the Random NumberGeneration program that can directly generate Discrete random variates, as well as some othervariates presented below. The Data Analysis module should be in the top right-hand cornerof the Data folder (see the screenshot below). If it is not, you can add it in Excel 2010 as fol-lows:, click File, then click Options, next click Add-ins, then click Analysis Toolpak on thetop, next click Go in the bottom, next tick the Analysis Toolpak, and finally click OK.

To use the Random Number Generation program in Excel, first we need to enter thedata (the numbers of breakdowns and the probability of each value) in the worksheet. Thenwe click on Data Analysis; in the small window that opens we find and click on RandomNumber Generation, and finally click OK. If we want 10 Discrete random variates fromthe breakdown distribution (in cells A4 to B9), arranged in column D starting with cellD1, we fill the Random Number Generation window as follows:


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Finally, clicking OK will result in the following:

The problem with using Random Number Generation program is that it does not recal-culate the numbers automatically when the recalculate key F9 is pushed. The recalculationis useful when we want to quickly replicate the simulation. Also, it is useful when DataTables (explained later) are used to rerun the simulation for various values of an inputvariable. Therefore, we need to briefly look at the generation of a Discrete random variateusing rand() and Vlookup.

First we need to enter the values and the probabilities, and compute the cumulativeprobabilities:

In the above spreadsheet, the formula in cell C4 is =B4, the formula in cell C5is =B5+C4, and the formula in cell C6 is =B6+C5, etc. Now, we will add theMinimum of Interval in column D as follows: In cell D4 we enter 0, in cell D5 weenter =C4, and copy it down. In effect, we are just shifting the cumul. Prob. valuesone row down:


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Now, we just copy column A to column E:

Suppose we wish to generate 10 random values from the breakdown distribution and putthem in cells G1 to G10. All we have to do is to enter =VLOOKUP(RAND(),D$4:E$9,2)in cell G1 and copy it down. What this formula does is that first rand() will generate a realnumber greater or equal to 0 and less than 1, then VLookup will take the random number to therange D4 to E9, and will find the Interval where the random number fits (in the left column ofthe range) and will look up the corresponding value in the right column of the range; 2 in theVLOOKUP formula means that it should look up the second column in the given range:

Generating Poisson Random Variates Generation of a Poisson random variate requiresthe mean of the distribution. Given the mean, we can obtain the cumulative probabilitiesfor the Poisson distribution from Appendix B, Table C. Then, the inverse transformationmethod can be used as in the Discrete case above. However, we should obtain three-digit

random numbers from Table 18S-1 because the Poisson cumulative probabilities are givenin three digits. Example S-2 illustrates this.

The average number of lost-time accidents at a large plant has been determined from histori-cal records to be two per day. Moreover, it has been determined that this accident rate canbe well approximated by a Poisson distribution. Simulate five days of accident experiencefor the plant. Read random numbers from columns 1 and 2 of Table 18S-1.

Example S-2

SolutionFirst obtain the cumulative Poisson probability distribution from Appendix B, Table C fora mean of 2.0, and determine the intervals:

x

Cumulative

Probability

Random Number

Intervals0 . . . . . . . . . . . . .135 001 to 135

1 . . . . . . . . . . . . .406 136 to 406

2 . . . . . . . . . . . . .677 407 to 677

3 . . . . . . . . . . . . .857 678 to 857

4 . . . . . . . . . . . . .947 858 to 947

5 . . . . . . . . . . . . .983 948 to 983

6 . . . . . . . . . . . . .995 984 to 995

7 . . . . . . . . . . . . .999 996 to 999

8 . . . . . . . . . . . . 1.000 000


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Next obtain three-digit random numbers from Table 18S-1. Reading from column 1 and2 as instructed, you find 182, 251, 735, 124, and 549.

Finally, convert the above random numbers into number of lost-time accidents using theestablished set of intervals above. Because 182 falls in the second interval, it correspondsto one accident on day 1. The second random number (251) falls in the same interval,indicating one accident on day 2. The number 735 falls between 678 and 857, which cor-responds to three accidents on day 3; 124 corresponds to 0 accidents on day 4; and 549

corresponds to two accidents on day 5.

Using Excel to Generate Poisson Random Variates In the Random Number Generation win-dow, choose Poisson as the Distribution and enter the mean (2) for Lambda. For example,the following will generate 5 Poisson random variates with mean=2 in cells A1 to A5:

Generating Normal Random Variates There are a number of ways to generate a Normalrandom variate of a given mean and standard deviation, but perhaps the simplest is touse Table 18S-2, a table of standard Normal random variates, i.e., a Normal distributionwith mean 0 and standard deviation 1.00. Like all such tables, the numbers are arranged

randomly, so that when they are read in any sequence they exhibit randomness. Numbersobtained from Table 18S-2 can be converted to the desired variates by multiplying themby the standard deviation and adding this amount to the mean. That is:

Simulated value Mean Standard Normalrandom number Standard= + deviation (18S-1)

In effect, the standard Normal random variate is a zvalue, which indicates how far aparticular value is above or below the distribution mean.

Table 18S-2

Standard Normal random

variates

1 2 3 4 5 6 7 8 9 10

1 1.46 -0.09 -0.59 0.19 -0.52 -1.82 0.53 -1.12 1.36 -0.44

2 -1.05 0.56 -0.67 -0.16 1.39 -1.21 0.45 -0.62 -0.95 0.27

3 0.15 -0.02 0.41 -0.09 -0.61 -0.18 -0.63 -1.20 0.27 -0.504 0.81 1.87 0.51 0.33 -0.32 1.19 2.18 -2.17 1.10 0.70

5 0.74 -0.44 1.53 -1.76 0.01 0.47 0.07 0.22 -0.59 -1.03

6 -0.39 0.35 -0.37 -0.52 -1.14 0.27 -1.78 0.43 1.15 -0.31

7 0.45 0.23 0.26 -0.31 -0.19 -0.03 -0.92 0.38 -0.04 0.16

8 2.40 0.38 -0.15 -1.04 -0.76 1.12 -0.37 -0.71 -1.11 0.25

9 0.59 -0.70 -0.04 0.12 1.60 0.34 -0.05 -0.26 0.41 0.80

10 -0.06 0.83 -1.60 -0.28 0.28 -0.15 0.73 -0.13 -0.75 -1.49


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Example S-3

SolutionThe first three values are: 1.46, -1.05, and 0.15. The simulated values are:

For 1.46: 30+1.46(4)=35.84 minutes.

For-1.05: 30-1.05(4)=25.80 minutes.For 0.15: 30+0.15(4)=30.60 minutes.

Using Excel to Generate Normal Random Variates In the Random Number Generationwindow, choose Normal as the Distribution and enter its mean and standard deviation.For example, the following will generate 3 Normal random variates with mean =30 andstandard deviation=4 in cells A1 to A3:

Alternatively, we can use the formula =NORMINV(rand(),mean,stddev). Forexample, to generate a random value from the above Normal distribution in a cell, weenter =NORMINV(rand(),30,4) in that cell.

Generating Continuous Uniform Random Variates The Continuous Uniform distributionU[a, b] has equally probable values anywhere betweenaandb, a


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Solution a=10 minutes, b=15 minutes,b-a=5 minutes

Obtain the random numbers: 15, 88, 57, and 28.a.

Convert to Continuous Uniform variates:b.

Random Number ComputationSimulated Value

(minutes)

15 . . . . . . . . . . . . 10 +5(.15) = 10.75

88 . . . . . . . . . . . . 10 +5(.88) = 14.40

57 . . . . . . . . . . . . 10 +5(.57) = 12.85

28 . . . . . . . . . . . . 10 +5(.28) = 11.40

Using Excel to Generate Continuous Uniform Variates In the Random Number Generationprogram, choose Uniform as the Distribution and enter the extreme valuesaandbafterBetween and and, respectively. For example, the following will generate four Continu-ous Uniform random variates between 10 and 15 in cells A1 to A4:

Alternatively, we can use the formula =a+rand()*(b-a). For example, to gen-

erate a random value from the above Continuous Uniform distribution in a cell, weenter =10+rand()*5 in that cell.

Generating Exponential Random Variates An Exponential distribution is portrayed inFigure 18S-1. The probability is fairly high that the random variable will assume a valueclose to zero, and it decreases as the value of the random variable increases. The probabil-ity that the Exponential random variable will take on a value greater than some specifiedvalueT, given mean of 1/, is:

P(t>T) =e-T (18S-3)

FIGURE 18S-1

An Exponential distri bution

0 T

P(t >T) = u

(t)

t


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To simulate an Exponential value manually, we obtain a real random numberubetween0 and 1, set this equal to the probabilityP(t>T), and solve Formula 18S-3 forT.The resultis a random variate from the Exponential distribution with mean of 1/. This concept isillustrated in Figure 18S-1.

We can obtain an expression for Tby taking the natural logarithm of both sides of theequation. Thus, withP(t>T) =u, we have

ln(u) =ln(e-T

)

The natural logarithm of a power of eis equal to the power itself, so

ln(u) =-T

Then

T u= 1

ln( ) (18S-4)

As for the Continuous Uniform case above, a random number can be obtained fromTable 18S-1 and converted tousimply by placing a decimal point to the left of it. This isdemonstrated in the following example.

Times between breakdowns of a piece of equipment can be described by an Exponentialdistribution with mean of five hours. Simulate the time between two pairs of breakdowns.Read two-digit random numbers from column 3 of Table 18S-1.

Example S-5

SolutionThe mean, 1/, is 5 hours. The random numbers are 84 and 05. Using formula 18S-4, thesimulated times are:

For 84:T=5[ln(.84)]=-5[-0.1744]=0.872 hours.For 05:T=5[ln(.05)]=-5[-2.9957]=14.979 hours.

Note that the smaller the value of the random number, the larger the simulated value of T.Using Excel to Generate Exponential Random Variates In Excel use the formula=-5*ln(rand()).

LO 3 SOME SIMULATION MODELSReal problems involve many components or steps, with process flows that may be compli-cated. In some cases, it is helpful to construct a process flowchart (similar to a process flowdiagram). If the model is complicated, one needs to use a specialized simulation softwaresuch as Arena. Otherwise, Excel can be used. Below, we illustrate some simple models ofoperations decision making using Excel.

An Inventory Control ModelWhile we solved most inventory control problems analytically in Chapter 12, we madesome simplifying assumptions. For example, we calculated EOQ and ROP separately. Ifa more exact solution is required, simulation should be used. Simulation is also used toillustrate the behaviour of the system over time so that the decision maker can gain insight/confidence in it.

The manager of a truck dealership wants to acquire some insight into how a proposedpolicy for reordering trucks from the manufacturer might result in shortage. Under thenew policy, two trucks are to be ordered whenever the number of trucks on hand at the

Example S-6


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Solution a.

end of the day plus number of trucks on order is two or fewer. Assume that purchase leadtime is only two full days. According to the dealers records, the probability distributionfor daily demand (i.e., its sales) is:

Demand,x P(x)

0 . . . . . . . . . .50

1 . . . . . . . . . .40

2 . . . . . . . . . .10

Draw a flowchart that describes a 10-day simulation. Assume a beginning inventory ofa.four trucks and that all shortage will be back-ordered.

Manually simulate the inventory system for 10 days. Use two-digit random numbers fromb.Table 18S-1, column 11, reading down. Estimate the probability of shortage.

Enter the model for this problem in Excel and run the simulation for 1,000 days. Esti-c.mate the probability of shortage.

UpdateinventoryI = I + 2

I = Amount of End-of-Day inventory

StartI = 4Day = 0

Stop

Yes

No

No

Day = 10?

I + on order


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The only negative Ending Inventory is in Day 7. Therefore, estimate for probabilityof shortage=1/10=.10. However, this is probably not very accurate. If we simulateanother 10 days, we may get a different answer, e.g., no shortage. Therefore, we mustsimulate longer (e.g., 1,000 days) and possibly run many replications (e.g., 10) andaverage their results.

The top part of the worksheet is shown below. The reorder decision components havec.

been explicitly defined here. Therefore, there are separate columns for Quantity OnOrder, Order?, and Day of Arrival. On the other hand, we have used the RandomNumber Generator program of Excel to generate the demand in Column D (so we didnot need to generate random numbers first).

The formula for Units Receid in cell B17 is =IF(A17=H14,$C$9,0), i.e., ifthe day number equals the Day of Arrival two full days before, then a shipment isreceived. In this case, H14 will be evaluated as 0 by Excel. The formula for Begin-ning Stock Level in cell C17 is =B17 +E16, i.e., it equals Units Receid plusEnding Stock the previous day. The formula for Ending Stock in cell E17 is=C17 D17, i.e., the beginning Stock Level minus Demand. The formula forQuantity On Order in cell F17 is =IF(G16 =Yes,F16 +C$9 B17,F16 B17), i.e., if Order? on previous day is Yes, then Quantity On Order =QuantityOn Order the previous day+order quantity any amount received today; else Quan-tity On Order =Quantity On Order the previous day any amount received today.The formula for Order? in cell G17 is =IF(E17 +F17>$C$10, No, Yes),i.e., if Ending Stock plus Quantity On Order are larger than Reorder Point, thenOrder? =No, else Order? =Yes. The formula for Day of Arrival in cellH17 is =IF(G17=Yes,A17 +C$11+1, ), i.e., if Order? is Yes then Dayof Arrival =todays Day No.+lead time+1, else it is blank. The +1 is becausethe lead time is 2 full days and we are ordering in the end of the current day and willreceive it the morning of day of arrival.

Finally, the formula in cell E12 is =COUNTIF(E17:E1016,


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A Waiting Line ModelWhile we presented many queueing models in Chapter 18, we made some simplifyingassumptions. For example, for most models we assumed that the distributions of inter-arrival and service times are Exponential, and that only steady state results are desired.Also, some performance measures are difficult to obtain analytically, e.g., maximum waittime and maximum number of customers in the queue. In these cases, simulation should

be used. Simulation will also illustrate the behaviour of the system over time so that thedecision maker can gain insight/confidence in it.

The time between mechanics requests for tools in a large plant is Normally distributedwith a mean of 10 minutes and a standard deviation of 1 minute. The time to fill requestsis also Normal with a mean of 9 minutes per request and a standard deviation of 1 minute.Mechanics waiting time represents a cost of $2 per minute, and servers time representsa cost of $1 per minute.

1. Manually simulate nine mechanic requests and their service times, and determine themechanics waiting time, assuming one server. Would it be economical to add anotherserver? Explain. Use Table 18S-2, column 8 for requests and column 9 for service.

2. Enter the model for this problem in Excel and simulate 1,000 arrivals. Does your answerto partachange?

Example S-7

Solution 1. i. Obtain standard Normal random variates from Table 18S-2 and convert to times[see columns (a) and (b) in the following table for requests for tool (i.e., arrivals)and columns (f) and (g) for service]. For example, the first Time Between Arrivals=10-1.12(1)=8.88 and the first Service Time=9+1.36(1)=10.36.

CUSTOMER ARRIVALS SERVICE

(a) (b) (c) (d) (e) (f) (g) (h)

Random

Number

TimeBetween

Arrivals

Arrival

Time

(e c)Wait

Time

ServiceStart

Time

Random

Number

Service

Time

(e +g)Service

End Time-1.12 8.88 8.88 .00 8.88 1.36 10.36 19.24

-.62 9.38 18.26 .98 19.24 -.95 8.05 27.29

-1.20 8.80 27.06 .23 27.29 .27 9.27 36.56

-2.17 7.83 34.89 1.67 36.56 1.10 10.10 46.66

.22 10.22 45.11 1.55 46.66 -.59 8.41 55.07

.43 10.43 55.54 .00 55.54 1.15 10.15 65.69

.38 10.38 65.92 .00 65.92 -.04 8.96 74.88

-.71 9.29 75.21 .00 75.21 -1.11 7.89 83.10

-.26 9.74 84.95 .00 84.95 .41 9.41 94.36

4.43 82.60

ii. Determine Arrival Times [column (c)] by adding to previous Arrival Time to theTime Between Arrivals in column (b).

iii. Use Arrival Times for Service Start Times unless service is sti l l in progresson a previous request.In that case, determine how long the arrival must wait(e-c). Column (h) values are the sum of Service Start Time and Service Time[column (g)].

iv.Total waiting time is 4.43 minutes (see the table).


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v.The total cost is:

Waiting cost (including service time): (4.43+82.60) minutes at $2 per minute=$174.06

Server cost: 94.36 minutes at $1 per minutes=94.36

$268.42

vi. Usually, a second simulation with two servers would be needed (e.g., for 18 arriv-als). However, in this case it is apparent that a second server would increase servercost by about $94 but could not eliminate more than approximately $8.86 of waitingcost. Hence, the second server would not be justified.

2.The top part of the worksheet is shown below. The column headings are identical to themanual simulation table above, except for the Mechanic No. column (to keep trackof numbers of arrivals). Also, we have used the Random Number Generator program ofExcel for Time Between Arrivals and Service Time, so we did not need to generaterandom numbers first.

The formula for Arrival Time in cell C9 is =C8+B9, i.e., the previous ArrivalTime plus the current Time Between Arrivals. The formula for Wait Time in cell

D9 is =MAX(G8,C9)-C9, i.e., the maximum of Service End Time of the previousmechanic and the Arrival Time of current mechanic minus the Arrival Time of

current mechanic. The formula for Service Start Time in cell E9 is =C9+D9, i.e.,the Arrival time plus the Wait Time. The formula for Service End Time in cell G9is =E9+F9, i.e., Service Start Time plus Service Time. The formula in cell D3 is=SUM(D9:D1008), i.e., sum of the Wait Time of 1,000 mechanics, and the formulain cell F3 is =SUM(F9:F1008), i.e., the sum of Service Time of 1,000 mechanics.

The answer to the question of the need for the second server does not change becausethe average wait time is almost the same as in part 1: 401/1000 =.401 minutes permechanic4.43/9=.492 minutes per mechanic from part 1.

It is interesting to compare the maximum wait times of part 1 and 2. In part 1, the maxi-mum wait time is 1.67 minutes, whereas in part 2 the maximum wait time is approximately

5 minutes (not shown). This is expected because a longer simulation run allows for moreextreme cases to occur.


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Optimization in SimulationSimulation is usually used to compare two or more configurations of the system, basedon some measure of performance. Choosing the best configuration is sometimes calledoptimization. We have to run the simulation for each configuration, and choose the configu-ration with the best measure. However, to make the right decision, the simulation resultsshould be accurate enough. For this reason, we need to have long runs, and therefore need

to use the computer. Excel has a program called Data Table that will run the simulationfor different values of a decision variable. In the following example, we will use Exceland Data Table to determine the optimal order quantity of the single period inventorycontrol model of Chapter 12.

Suppose you are responsible for ordering soft drinks for an event such as a sports game.1Onecanister can serve 100 drinks. Demand is Exponentially distributed with mean of five can-isters. Suppose that unmet demand costs $40 per canister (or $.40 per drink) for lost profit,and returning excess soft drink to the bottler costs $10 per canister (or $.10 per drink).Use Excel to determine the order quantity that will minimize the expected total cost. Run2,000 replications of the event for each order quantity between 5 and 10.

1A. F. Seila et al,Applied Simulation Modeling, Australia: Thomson Learning, 2003, p. 49.

Example S-8

Solution In the following worksheet, the formula for Demand in cell B13 is =-C$6*LN(RAND()),i.e., negative of the mean of the Exponential distribution times the natural logarithm of a(real) random number greater or equal to 0 and less than 1. The formula for Shortage Costin cell C13 is =IF(B13 >C$9,B13 C$9,0)*C$3, i.e., if Demand is larger than OrderQuantity, shortage=Demand Order Quantity, else shortage=0, and multiply the shortageby the unit shortage cost. The formula for Excess Cost in cell D13 is=IF(C$9>B13,C$9 B13,0)*C$4, i.e., if Order Quantity is larger than Demand, excess =Order Quantity Demand, else excess=0, and multiply the excess by the unit excess cost. The formula forTotal Cost in cell E13 is =C13+D13, i.e., the sum of the Shortage and Excess Costs.The formula for Avg. Total Cost in cell F2 is =AVERAGE(E13:E2012), i.e., the averageof Total Cost of 2,000 games.

The values in cells F3 to F8 are the Avg. Total Cost for Order Quantities 5 to 10,respectively. These are determined by Excels Data Table program. Data Table was used asfollows: select cells E2 to F8, then click on Data folder, next click What-if Analysis, thenclick Data Table; in the small window enter $C$9 (the location of the Order Quantity) infront of Column input cell:, and click OK:

It can be seen from cells F2 to F8 that Order Quantity =8 canisters has the lowest Avg.Total Cost. However, even 2,000 replications may not be enough for an accurate decision.If the whole experiment is repeated (press the F9 key to recalculate), in some instancesother Order Quantities, in particular 7 and 9, will have a lower Avg. Total Cost than 8. Toget a more accurate result, one should increase the number of games.


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A Maintenance ModelWhile we solved some maintenance problems analytically in the Supplement to Chapter 15,we made some simplifying assumptions. For example, we calculated the preventive main-tenance interval approximately, and we assumed that individual units will not fail againbefore group replacement time. If a more exact solution is required, simulation should beused. Simulation is also used to illustrate the behaviour of the system over time so that thedecision maker can gain insight/confidence in it.

A milling machine has two different bearings that fail.2The distribution of the life ofeach bearing is identical and is shown below. When a bearing fails, the machine stops,

a mechanic is called, and he installs a new bearing. Each bearing costs $32. It takes anaverage of 27 minutes to change the bearing. Downtime for the machine costs $10 perminute. The engineering staff has proposed a new policy: replace both bearings when-ever one fails. This will take the mechanic an average of 37 minutes. Simulate 1,000failure/replacement cycles for each bearing for the current and for both bearings for theproposed replacement policy and determine the lower-cost policy.

2Based on J. Banks et al, Di screte-event System Simul ati on, 5th ed, 2010, New Jersey: Prentice-Hall,pp. 6567.

Example S-9


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Bearing Life (Hours) Prob

1,000 .08

1,100 .15

1,200 .26

1,300 .19

1,400 .13

1,500 .09

1,600 .06

1,700 .03

1,800 .01

Solution The individual replacement policy is modelled in Excel as shown in the worksheet below.Each cycle is the length of life of a new bearing. These are created using the DiscreteDistribution of the Random Number Generation program of Excel. The formula in cellH13 is =C4*2000, i.e., the cost of replacing 2,000 bearings. The formula in cell H14 is=C5*2,000, i.e., the cost of downtime of 2,000 individual failures The formula in cellH15 is =H13 +H14. The formula in cell H16 is =SUM(B13:B1012), i.e., the total

lives of Bearing 1s, and the formula in cell H17 is =SUM(C13:C1012), i.e., the totallives of Bearing 2s. The formula in cell H18 is =H16+H17. Finally, the formula in cellH19 is =H15/H18*1000, i.e., the total cost divided by total life hours times 1,000, orexpected cost per 1,000 hours of work.

The group replacement policy is modelled in Excel as shown in the worksheet below.It is very similar to the individual replacement policy worksheet above, but for each cyclethe minimum length of life of the two bearings (1 and 2) is computed (called First Failure)and used for both bearings, because both are replaced at that time. Therefore, the formulain cell D13 is =MIN(B13:C13). The formula in cell I14 is =C5*1000, i.e., the totaldowntime cost of joint replacement (note only 1,000 downtimes). The formulas in cells


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I16 and I17 are=SUM(D13:D1012), i.e., the total lives of the first failures. The otherformulas are the same as the individual policy.

Because the cost per 1,000 hours $184.02


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the resources used in each operation are entered (these affect the times). The analyst needsto specify the level of resources to use and run the simulation for a certain length of timeto see its effects on entities (e.g., percentage of time spent in an operation) and resources(e.g., percentage of time idle, costs). Important data can also be represented graphically(called dashboard monitors). The simulation is run several times for each different levelof resources and the best level of resources is chosen. For more details, see http://www.simprocess.net/solutions/er_model.html.

Level 2 or 3

Sign In

Level 1

Entrance Door

AmbulanceAlways Level 1

Transferto room

Level-1

Level 1

Not Level-1

Not Level 1

Emergency room

Registration

Treatment

Releaseand

Admission

Triage

Source: http://www.simprocess.net/solutions/models_html/EmergencyRoom.

Extend.Extend is a product of Imagine That! Software company. Modelling in Extendis similar to SIMPROCESS. Just click on the library menu and select a type of opera-tion (Generator, Queue Activity, Plotter, etc.) to include in the model. Then, connect itto another operation by left-clicking in the middle of the right edge of the icon and drag-ging the line to the input activity. For further information, see http://www.extendsim.com/downloads/manuals/support_manuals_dl.html. An Extend model for bank staff ing

is shown below.

Source:Bank Line Tutorial in Extend Demo Software, downloaded from http://www.extendsim.com/prods_demo.html. Screen capture courtesy of Imagine That Inc. of San Jose, CA.


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The front-axle line used a closed-loop conveyor. At Station 1 (the loading station), partsand kits are loaded onto pallets. Pallets progress through subsequent stations to reach theunloading station, where the fully assembled axles are unloaded from the pallets. The fin-ished axle assemblies leave the system while the empty pallets are moved to the loadingstation to receive new parts.

Visteon formed a project team that used WITNESS to model and simulate the productionof the front axle line. Most stations used a machine with varying operation times (a rangeof 8 seconds to 33 seconds per axle). In addition, there were machine failures (a range of

11 minutes to 11,000 minutes for mean time between failures) and repair (a range of 1minute to 5 minutes for mean time to repair). Also, a small percentage of axles had to betorn down for repair at various stages during the process.

Although there were more pallets circulating in the front axle line than stations, toofew pallets resulted in stations starving, but too many pallets resulted in blocking.So, the team wanted to identify the optimal number of pallets. Also, the team wanted toknow the bottleneck station and if a given station-time reduction would help increase theline capacity. As a bottleneck was relieved and capacity of the line increased, another sta-tion would become the bottleneck. After each improvement, simulation was run again toinvestigate the best way to increase the capacity next. At the start of the study, capacity ofthe line was 57 axles per hour. By the end of project 1.5 years later, capacity was increasedto 75 axles per hour (a 30 percent increase).

Simulation was also used to show that the line capacity could not easily be increased muchmore. Therefore, Ford decided to set up another front axle line to meet the excess demand.

ProModel.ProModel is a product of the ProModel corporation of Utah (http://www.promodel.com). The following is a simple application of ProModel. Gasification convertscoal into carbon monoxide and hydrogen, which can then be converted into chemicals orput in other uses. Eastman Chemicals used ProModel to determine how many gasifiersto use in a gasification plant in order to produce the required output.3It is also possibleto use a gasifier as standby. For example, when 4 gasifiers are used, if the reliability

3http://www.promodel.com/solutions/manufacturing/Eastman-Chemical-web.pdf.

FIGURE 18S-2

Visteons front axle l ineSource:G. Pfeil, et al.,Visteons Sterling PlantUses Simulation-basedDecision Support inTraining, Operations, andPlanning, Interfaces30(1),JanuaryFebruary 2000,pp. 115133.

STA.1

LOADING

STATION

STA.36

UNLOADING

STATION

STA.39

OFF-LINE

WASH

REPAIR BAY #3 REPAIR BAY #2

REPAIR BAY #1

STA. 2 STA. 11 STA. 12 STA. 16 STA. 17

STA. 16

STA. 19

STA. 20

STA. 21STA. 22STA. 24STA. 25STA. 26STA. 31STA. 36STA. 37

WITNESS.WITNESS is a product of the Lanner Group, based in the UK (http://www.lanner.com). The following describes an application of WITNESS.Visteon is a subsidiary

of Ford Motor Company. In the mid-1990s, the Sterling Heights plant (in Michigan) wasthe only source of Fords axles. With the introduction of new cars, trucks, and SUVs, Fordneeded to substantially increase production of the front axle line (see Figure 18S-2).


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(percentage uptime) of each gasifier is 90 percent, then the output will be approximately92 percent of planned output, whereas if there is an additional standby gasifier, the outputwill be about 98 percent of planned output.

Source: http://www.promodel.com/solutions/manufacturing/Eastman-Chemical-web.pdf.

Arena.Arena is a product of Rockwell Automation (http://www.arenasimulation.com).An application is as follows. Dofasco used Arena to model its primary steelmaking process.The steps of this production line are (1) iron making and de-sulfurizing, (2) steelmaking, (3)ladle metallurgy facility, (4) vacuum degassing (if pure steel is needed), and (5) continuous

casting (see the following figure, from left to right). Dofasco managers wanted to know ifrelatively small expenditures in some critical parts of the process would result in produc-tivity improvements, or whether major expenditures would be needed. After constructionof the model in Arena, which used 87 different steps/activities, the model was validatedagainst previous years actual performance (e.g., the quantity of slabs made, number ofladles fed into the continuous caster, etc.). The results showed that only simultaneouslyincreasing the uptimes of major steps of the process would increase productivity. In otherwords, there were no unique bottlenecks. Later, the model was also used to evaluate themajor changes to the process.

oxygen

oxygen

Off Gas

Source:http://www.informs-sim.org/wsc06papers/255.pdf


25/38


How to Use ArenaDownload a free trial version of Arena from http://www.arenasimulation.com (you have toregister first). Unzip the file and install it. Arena is relatively simple to use. After startingArena, the initial screen looks like the following:

The trial version has access to only Basic Processes (see the left pane). We will model asingle server queueing model, Example 2 of Chapter 18, in Arena. The time between arriv-als and service times both have Exponential distributions with mean of 4 and 3 minutes,respectively. We wish to determine the average and maximum wait time (in queue), andaverage and maximum numbers in the queue. Click on Create icon in the left pane and

drag it to the main window:


26/38


This is the arrival process. In the bottom pane, we need to change the Value into 4and the Units into minutes. Note that the distribution of arrivals is already Expo, i.e.,Exponential:

Now, with Create 1 selected we click on Process icon in the left pane and drag it to themain pane and drop it to the right of Create 1:


27/38


In the bottom pane for Process 2, we need to change the Action into Seize Delay Release(or else no queue will be developed), Delay Type into Expression (because Exponential isnot one of the distributions we can choose directly here), Units into Minutes, and Expres-sion into EXPO(3) (we need to type in the 3):

Now double-click on Process 2 because we need to assign a resource to it (to seizethe arrivals):

In the Process window above, click Add...:


28/38


In the Resources window above, click OK:

Resource 1 will appear in the middle (see above). Now, click OK again (the Processwindow will close). Then, in the main window, with Process 2 selected click on Disposeicon in the left pane, and drag it and drop it to the right of Process 2. Then click on Run(on the top) and next click Setup:


29/38


Run Setup window will open:

In the Run Setup window, change the Replication Length from Infinite to 24 (becausewe want to run it for 24 hours), and change the Base Time Units to Minutes. Then, clickOK (Run Setup window will close). In the main window, click Go under the Run (in thetop). Simulation starts. A simple animation shows that a paper (representing a customer)comes out of Create 1, possibly queues above Process 2, and exits in Dispose 2. The numbersunder Create and Dispose are the number of entities generated and disposed, respectively.The number under Process is the number of customers in the queue and the paper iconsabove it represent the queue:


30/38


At the end of simulation, the following window will open. Click Yes:

The Report will Preview. Move to the second page:

These are the results of simulation: Average wait time is 7.5226, maximum wait timeis 34.022 minutes, average number in the queue is 1.8841, and maximum number in thequeue is 9. Finally, we need to end the simulation run (click on End under Run at the top)and exit Arena (FileExit) after saving the model.

@RISK.@RISK is an Excel add-in software that performs risk analysis using MonteCarlo (probabilistic) simulation. @RISK is a product of Palisade Corporation (http://www.palisade.com/risk). BC Hydro used it to evaluate the uncertainties surrounding its energyconservation program.4BC Hydro plans to meet most of its future electricity needs throughdemand management, i.e., energy conservation by its customers. Around 60 projects wereconsidered, such as compact fluorescent light promotions, subsidies for energy-efficientappliances, variable-speed motor promotions for home furnaces, and promotional activity

4http://www.palisade.com/cases/bchydro.asp.


31/38


aimed at motivating customers to use less energy. The uncertainties were analyzed on acase-by-case basis, and a probability distribution for the aggregate conservation-savingsforecast was developed. BC Hydro used estimates from experts in various industries toprovide input to the @RISK model. The displays assisted the experts to visualize how theirpredictions influenced the final outcome of the decision-making process. Also, interrela-tionships among key uncertainties were explored.

Key TermsMonte Carlo method, 3random numbers, 4

random variates, 4simulation, 2

Solved ProblemsThe number of customers who arrive during each hour at a car repair shop can be described by

a Poisson distribution. Assuming that the average number of customers per hour is three during

the first four hours of a day, simulate customer arrivals for the first four hours. Read three-digit

random numbers from Table 18S-1, columns 4 and 5, going down.

Obtaind. cumulative probabilities of the Poisson distribution from Appendix B, Table C, for themean specified (3). Then, determine the random number intervals.

xCumulativeProbability

Random NumberIntervals

0 . . . . . . . . . . .050 001 to 050

1 . . . . . . . . . . .199 051 to 199

2 . . . . . . . . . . .423 200 to 423

3 . . . . . . . . . . .647 424 to 647

4 . . . . . . . . . . .815 648 to 815

5 . . . . . . . . . . .916 816 to 916

6 . . . . . . . . . . .966 917 to 966

7 . . . . . . . . . . .988 967 to 988

8 . . . . . . . . . . .996 989 to 996

9 . . . . . . . . . . .999 997 to 999

10 . . . . . . . . . 1.000 000Obtain three-digit random numbers 299, 642, 350, and 091.e.

Convert the random numbers to simulated values. Note where each number falls in the random-f.number interval list. For instance, 299 falls between 200 and 423. Interpret this to mean thattwo customers arrive in the first hour. Similarly, 642 is interpreted to mean that three customersarrive in the second hour, 350 implies two customers in the third hour, and 091 implies onecustomer in the fourth hour.

To summarize, the number of customers per hour for the four-hour simulation is:g.

HourNumber ofArrivals

1 . . . . . . . . . . 2

2 . . . . . . . . . . 3

3 . . . . . . . . . . 2

4 . . . . . . . . . . 1

Jobs arrive at a workstation at fixed intervals of one hour. Processing time is approximately Normal

and has a mean of 56 minutes per job and a standard deviation of 4 minutes per job. Using the

fifth row of the table of standard Normal random variates (Table 18S-2), simulate the processing

times for four jobs and determine the amount of operator idle time and job waiting time. Assume

that the first job arrives at time =0.

Solution

Problem 2

Problem 1


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Obtain thea. random numbers from the table: 0.74, -0.44, 1.53, and-1.76.

Convert the random numbers to simulated processing times:b.

RandomNumber Computation Simulated Time

0.74 . . . . . . . . . . 56 +4(0.74) = 58.96-0.44 . . . . . . . . . 56 +4(-0.44) = 54.24

1.53 . . . . . . . . . . 56 +4(1.53) = 62.12

-1.76 . . . . . . . . . 56 +4(-1.76) = 48.96

Note that three of the times are less than the inter-arrival times for the jobs (i.e., 1 hour), mean-

ing that the operator will be idle after those three jobs. One time exceeds the one-hour interval,

so the next job must wait, and possibly the job following it if the waiting plus processing time

exceeds 60 minutes.

Calculate waiting and idle times:c.

Job

Number

Arrives

at

Processing Time,

t(minutes)

60 -tOperator Idle

(minutes)

t-60Next Job Waits

(minutes)

1 0 58.96 1.04

2 60 54.24 5.76

3 120 62.12 2.12

4 180 48.96 8.92*

15.72 2.12

*60-2.12-48.96=8.92.

Solution

What is simulation (as used in operations management)? (LO1)1.

What are some of the primary reasons for the widespread use of simulation in practice?2.

(LO1)

What are some of the ways managers can use simulation? (LO1)3.

What is the difference between random numbers and random variates? (LO2)4.

How can Excel be used for generating random variates? (LO2)5.

How can random values from each of the following distributions be generated using random6.

number tables? (LO2)

Discretea.

Normalb.

Continuous Uniformc.

Poissond.

Exponentiale.

Respond to the following comment: I ran the simulation several times, and each run gave me7.

a different result. Therefore, the technique does not seem to be useful. I need one answer!

(LO13)

Name three popular simulation software programs. (LO4)8.

Discussion andReview Questions


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Internet ExercisesVisit http://www.promodel.com/solutions/logistics/, watch the video, and briefly summarize how1.ProModels simulation helped the Salt Lake 2002 Winter Olympics. (LO4)

Visit http://www.lanner.com/en/case-studies.cfm, pick a case study, and summarize how2.

WITNESS is being used. (LO4)

Visit http://www.palisade.com/cases/ctt.asp?caseNav=byProduct, pick a case study, and sum-3.

marize how @RISK is being used. (LO4)

Google IIE/RA contest problems, pick a simulation contest problem, and determine what4.

topic it is on. Note: This is an annual contest sponsored and managed by Rockwell Interna-

tional (Arena). (LO3)

Visit http://arenasimulation.com/Tools_Resources_Video_Library.aspx, click on What is5.

Simulation?, register with Arena, watch the video, and summarize what, why, and when of

simulation. (LO1)

Visit http://arenasimulation.com/Tools_Resources_Video_Library.aspx, click on Emergency6.

Department Simulation Demo, register with Arena if necessary, watch the video, and sum-

marize how Arena can help with the operations of an emergency department. (LO4)

Visit http://arenasimulation.com/Solutions_Solutions.aspx, click on an application area, find7.

a case study, and summarize how Arena is being used. (LO4)

ProblemsThe number of jobs received by a small machine shop is to be simulated for an eight-day1.period. The shop manager has collected the following data for the number of jobs received

daily: (LO2)

Number of Jobs Frequency

2 or less 0

3 10

4 50

5 80

6 40

7 168 4

9 or more 0

200

Use the third column of Table 18S-1 and read two-digit numbers, going down. Determine

the average number of jobs per day for the eight-day simulation period.

Jack sells insurance on a part-time basis. His records on the number of policies sold per week2.

over a 50-week period are: (LO2)

Number Sold per Week Frequency

0 8

1 15

2 17

3 7

4 3

50

Simulate three five-week periods. Use Table 18S-1, column 6 for the first simulation, column

7 for the second, and column 8 for the third. In each case, read two-digit numbers, beginning


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at thebottomof the column and going up.For each simulation, determine the percentage of

weeks during which two or more policies are sold.

After a careful study of requests for a special tool at a large tool crib, an analyst concluded3.

that demand for the tool can be adequately described by Poisson distribution with mean of two

requests per hour. Simulate demand for a 12-hour period for this tool crib using Table 18S-1.

Read three-digit numbers from columns 5 and 6 combined, starting at the topand reading

down(e.g., 917, 264, 045). (LO2)

The number of lost-time accidents per month at a forestry company can be described using a4.

Poisson distribution that has mean of four accidents. Using the last two columns of Table 18S-1

(e.g., 540, 733, 293), simulate the number of accidents for a 12-month period. (LO2)

The time a surgeon spends performing a particular surgery can be modelled using a Normal5.

distribution that has mean of 20 minutes and standard deviation of 2 minutes. Using the table

of standard Normal random variates (Table 18S-2), simulate the times the surgeon might spend

on the next seven surgeries. Use column 4 of the table; start at the bottom of the column and

read up. (LO2)

The time between job arrivals at a workstation tends to be Normally distributed with a mean6.

of 15 minutes and a standard deviation of 1 minute. Job processing time is also Normally

distributed with a mean of 14 minutes and a standard deviation of 2 minutes. (LO3)

Using Table 18S-2, simulate the arrival and processing of five jobs. Use column 4 of thea.table for job inter-arrival times and column 3 for processing times. Start each column at

row 4 and go down. Find the total time jobs wait for processing.

The company is considering the use of a new equipment that would result in processingb.

times that is Normal with mean of 13 minutes and standard deviation of 1 minute. Job

waiting represents a cost of $3 per minute, and the new equipment would represent an

additional cost of $.50 per minute. Would the equipment be cost justif ied? (Note: Use the

same arrival times and the same random numbers for processing times.)

Daily usage of sugar in a small bakery can be described by a Continuous Uniform distribution7.

with endpoints of 30 kg and 50 kg. Simulate daily usage for a 10-day period. Read four-digit

numbers from Table 18S-1, columns 5 and 6, going up from the bottom. (LO2)

Weekly usage of a specific spare part in a maintenance storeroom can be described by Pois-8.

son distribution with mean of 2.8 parts per week. Lead time to replenish the spare part is twoweeks (constant). Simulate the total usage of the part during lead time 6 times (dont overlap

the lead times), and then determine the frequency of lead time demand (i.e., what percentage

of times was the demand equal to 2, 3, 4, etc.?). Read three-digit numbers from Table 18S-1,

columns 8 and 9, going down. (LO2 & 3)

(9. Excel exerci se.) Repeat Problem 8 for 100 lead-time periods. Hint: To make a frequency

distribution, use the Histogram tool in Data Analysis. (LO2 & 3)

10. A machine shop breaks an average of .6 unit of a tool per day. The number of breakages10.

(i.e., demand) can be described by Poisson distribution. The number of days required to obtain

replacement is almost constant (3 days). Because ordering is at the end of a day and receiving

is at the beginning of another day, the order will arrive 4 days after the order date. Four units

are ordered whenever the stock on hand and on order is two or less. Initially we have three

units on hand and none on order. All shortage is back-ordered. (LO3)

Draw a flowchart to describe this process.a.

Simulate this problem for a 12-day period. Read three-digit numbers from Table 18S-1,b.

columns 5 and 6, going down (e.g., 917, 264), for tool breakage. What is the probability

of shortage?

(11. Excel exercise.) Repeat Problem 10 for 100 days. What is the probability of shortage?

(LO3)

The time between arrivals of a part to a workstation varies Uniformly between 10 and 2012.

minutes. Process time is Normal with mean of 15 minutes and standard deviation of 2 min-

utes. (LO3)


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Simulate waiting times and processing for nine parts. Read three-digit numbers going downa.

columns 9 and 10 of Table 18S-1 for inter-arrivals (e.g., 156, 884, 576). Use column 8 of

Table 18S-2 for processing time. Calculate total waiting time.

If management can reduce the range of inter-arrival times to between 13 and 17 minutes,b.

what would the impact be on part waiting times? (Use the same service times and the

same random numbers for inter-arrival times from part a.) Round inter-arrival times to

two decimal places.

At a call centre for a cell phone company, customer service associates are employed to respond13.

to customer calls and complaints.5During a non-peak period, on average 6 customers call per

hour (Poisson), i.e., the inter-arrival times are Exponential with mean of 10 minutes. The length

of calls has a triangular distribution with minimum of 2 minutes, maximum of 10 minutes,

and a most likely value of 5 minutes. Use Arena to model this problem and simulate for 200

hours to answer the following question: How many associates should the company employ if

the policy is both of the following: (LO4)

Average wait time in the phone queue should not be larger than 2 minutes.a.

Maximum number of calls waiting should be no more than 5.b.

A product requires that a hole be drilled into a metal block and that a cylindrical shaft be14.

inserted into it.6The probability distribution of shaft radius is Normal with mean of 1 inch and

standard deviation of .001 inch. Similarly, the probability distribution of hole radius is Normalwith a mean of 1.003 inches and standard deviation of .001 inch. The clearance between a

hole and a shaft is the difference in their radii. The objective is to determine how frequently

interference (i.e., negative clearance) will occur. Simulate 1,000 pairs. (LO3)

An analyst found that the length of telephone conversations in an office could be described by15.

an Exponential distribution with a mean of four minutes. Reading two-digit random numbers

from Table 18S-1, column 6, simulate the length of f ive calls and calculate the simulated aver-

age time. Why is the simulated average different from the mean of four minutes? (LO2)

The length of time between calls for service of a certain piece of equipment can be described16.

by an Exponential distribution with a mean of 40 minutes. Service time can be described by

a Normal distribution with mean of eight minutes and standard deviation of two minutes.

Simulate the queuing system for four breakdowns. For breakdowns, read two-digit numbers

from Table 18S-1, column 7; for service times, read numbers from Table 18S-2, column 8.Calculate the total wait time. (LO3)

17. A simple project has only two activities: A and B. A is an immediate predecessor of B. A has

a triangular distribution with minimuma=10 days, modem=20 days, and maximumb=50

days. B also has a triangular distribution with minimum =15 days, mode=25 days, and maxi-

mum=60. Simulate 100 replications of the project in Excel to determine the distribution of

the project completion time. What is a reasonable point estimate for the project duration?

Hint: Excel does not provide random triangular distribution variates. We need to generate

a (real) random numberUgreater or equal to 0 and less than 1; if U(m a)/(b a), then

random variate=a+ U b a m a ( )( ); else random variate=b- ( ) ( )( )1 U b a b m

See e.g., Generating Triangular-distributed random variates section in http://en.wikipedia.

org/wiki/Triangular_distribution. (LO3)

A service operation consists of three steps. The first step can be described by a Continuous18.Uniform distribution that ranges between five and nine minutes. The second step can be

described by a Normal distribution with a mean of seven minutes and a standard deviation of

one minute, and the third step can be described by an Exponential distribution with a mean

of five minutes. Simulate three services using two-digit numbers from Table 18S-1, row 4 for

step 1; Table 18S-2, row 6 for step 2; and two-digit numbers from column 4 of Table 18S-1

for step 3. Determine the simulated time for each of the three services. (LO3)

5C. Harrell et al, Simulation Using ProModel, 2000, Boston: McGraw-Hill, pp. 397399.6Based on F. S. Hillier and G. J. Lieberman,Operations Research, 2nd ed, 1974, San Francisco: Holden-Day,p. 653.

*


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A project consists of f ive major activities, as illustrated in the diagram below. Activity times19.

are Normally distributed with means and standard deviations as shown in the following table.

Note that there are two paths through the project: A-B-D-End and A-C-E-End. Project duration

is defined as the larger sum of times along the two paths. Simulate 10 times for each activity.

Use column 1 of Table 18S-2 for activity A, column 2 for activity B, column 3 for activity

C, column 4 for activity D, and column 5 for activity E. Determine the project duration for

each of the 10 sets, and prepare a frequency distribution of project duration. Use categories

of 25 to less than 30, 30 to less than 35, 35 to less than 40, 40 to less than 45, and 45 or more.Determine the proportion of time that a simulated duration of less than 40 days occurred. How

might this information be used? (LO3)

A

B D

End

C E

Activity Mean (days)Standard

Deviation (days)

A 10 2

B 12 2

C 15 3

D 14 2

E 8 1

20.A drug manufacturer has recently accepted an order from its best customer for 950 ouncesof a new drug and wants to plan its production schedule to meet the promised delivery date.7

The order has to be produced in 1 or more batches. There are three sources of uncertainty: (a)

time required to produce a batch (Normal with mean of 8 days and standard deviation of 1

day rounded to the nearest integer), (b) the yield of a batch, in ounces (triangular distribution

[600, 1000, 1100]), and (c) the probability that a batch passes final inspection (prob=.99). If

the batch fails inspection, it will be discarded. Use simulation and Excel to decide how many

days prior to the due date the production should begin. (LO3)

21. A newsvendor must decide how many papers to buy from the newspaper publisher each

weekday.8Each weekday she will order the same quantity. A newspaper costs her $.55

and sells for $1. Daily demand during each weekday is Normal with mean of 136 and

standard deviation of 27 newspapers. Any unsold newspapers will be sold to a recycler

for $.03 each. A shortage results in loss of profit. Simulate demand for 1,000 days using

Excel and determine the number of newspapers the newsvendor should buy each morning

*

7Based on W. L. Winston and S. C. Albright, Pract ical M anagement Science, 2nd ed, 2001, Pacific Grove,California: Duxbury, pp. 626630.8W. D. Kelton et al, Simulation wi th Arena, 4th ed, 2007, Boston: McGraw-Hill, p. 38.

*


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to minimize expected total shortage and excess cost. Assume that she can buy only in

multiples of 10. (LO3)

22.The inter-arrival times to a single server queue are Exponentially distributed with a mean of

1.6 minutes.9Service times have a Continuous Uniform distribution between .27 and 2.29

minutes. Download a trial version of Arena and use it to perform 100 hours of simulation to

determine the maximum customer wait time in the queue. (LO3)

23. A technical support centre is staffed by two people, Able and Baker.10

Able is more experi-enced than Baker, and hence provides faster service. If both are idle, Able takes the call. If

both are busy, the caller waits. The distributions of inquiry inter-arrival times during peak

periods, Ables service times, and Bakers service times are all Exponential with means of

2.25, 3.35, and 4.25 minutes, respectively. Model this problem in Excel, run the simulation

for 1,000 calls, and determine the average wait time (in the queue). Should another person be

assigned to the centre? (LO3)

24. An appliance retail chain keeps inventories of its most popular fridges in its warehouse.11

Consider one such model and size. The company uses the fixed interval inventory model for

replenishing the fridge from the manufacturer. The order interval used is 1 week (5 work-

days). The distribution of daily demand is Normal with mean of 2 and standard deviation

of 1 (but non-negative) and lead time from the manufacturer is Exponential with mean of

1.5 workdays (but minimum=1 day and maximum=4 days; 1 day LT =next-day arrival).

Currently the order up to level used for this fridge is 13 units. Model this problem in Excel,

run the simulation for 1,000 days, and determine the average daily number of fridges

short. (LO3)

25. A multi-user mainframe computer includes two disk drives that are prone to failure.12If a

disk drive fails, users lose their files, which need to be restored. Restoration is achieved by

copying copies of the files, held on magnetic tapes, onto the disk backup. This restoration

is inconvenient and so a new operation policy is being considered. At the moment, the disk

drives are repaired and restored as and when they fail. The proposal is to introduce a joint

repair system. The life of a disk drive is Normally distributed with a mean of 4 months and

a standard deviation of 1 month. Under the current policy, it costs $50 to repair and restore

a failed disk drive. The joint repair and restore policy would operate as follows: when either

unit fails, the failed unit is repaired and restored, and the other unit is cleaned to bring it to

a state equivalent to having been just repaired. This will cost $75 in total. Is the new jointrepair and restore policy cost effective? Use Excel to simulate 200 individual failures and

100 joint repairs. (LO3)

26. An airport hotel has 100 rooms.13On any given night the hotel takes up to 105 reserva-

tions because of the possibility of no-shows. Past records indicate that the number of daily

reservations is uniformly distributed over the integer range [96, 105]. The no-shows have a

Normal distribution with mean of 2.5 and standard deviation of 1, but rounded to the near-

est integer. Develop and run a simulation model in Excel for 100 days to find the expected

number of rooms used per night and the percentage of nights when more than 100 rooms

are claimed. (LO3)

*

*

**

*

*

9W. D. Kelton et al, Simulation with Arena, 4th ed, 2007, Boston: McGraw-Hill, pp. 4344.10Based on J. Banks et al, Di screte-event System Simulation, 5th ed, 2010, New Jersey: Prentice-Hall, pp. 5152.11J. Banks et al,Di screte-event System Simulation, 5th ed, 2010, New Jersey: Prentice-Hall, pp. 6162.12M. Pidd,Computer Simulati on in Management Science, 1984, Chichester, UK: J ohn Wiley, pp. 2224.13W. L. Winston,Operat ions Research, 2nd ed, 1991, Boston: PWS-Kent Publishing, p. 1159.


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MINI-CASE

Valencia Kidney Waiting Line

Some staff researchers at the Valencia Regional HealthAuthority of Spain are wondering what will happen to thewaiting line for kidney transplants in their region. They believethat the waiting line is likely to decrease. During the 1997

to 1999 period (1,095 days), the (new) demand for a kidney

transplant was 531, with the number of (new) patients per day

distributed approximately according to Poisson distribution.

Hence, the average number of arrivals to the queuing system

is estimated to be 531/1,095=0.485 patients per day. There

were 241 double donations and 82 single donations during

the same time period. Hence, the average number of kidneys

donated per day was (2412+821)/1,095=.515. This can

be used to estimate the average number of patients receiving a

kidney each day (there was no instance of a patient receiving

two kidneys as transplants). The number of kidney donations

per day can also be approximated by Poisson distribution. There

were 446 people in the queue as of December 31, 1999. What

is the probability that the waiting line will decrease after 365

days? Hint: (a) First in Excel compute the cumulative Poisson

distribution with means of .485 and .515 using =POISSON.

DIST(x, mean, cumulative). For example, to get the cumulative

probability that Xtakes the values up to and including 2 use

POISSON.DIST(2, .485, TRUE) for the arrival distribution, (b)

Use the cumulative Poisson distributions of part ato simulate

365 days of arrivals and services, keeping track of the length

of the queue (initially it is 446 people long), (c) use the Data

Table to obtain 100 values for the end-of-the-year number of

people in the queue, and (d) count the numbers that are smaller

than 446.

Source: Based on J. J. Abellan, et al., Predicting the Behaviourof the Renal Transplant Waiting List in the Pais Valencia (Spain)Using Simulation Modelling, Proceedings of the 2004 WinterSimulation Conference, pp. 1969974. http://www.informs-sim.org/wsc04papers/263.pdf

MINI-CASE

DofascoThe superintendent of blast furnaces at Dofasco has to decide

whether or not to recommend that an engineering modificationcosting $600,000 be made to the lance desulfurizing plant. The

lance desulfurizing plant is part of the process for converting

iron into steel. Molten iron from the blast furnaces is transported

in special rail cars, called torpedoes, to the lance desulfurizing

plant where sulfur content of torpedoes is reduced. After des-

ulfurizing, a locomotive is called to haul the torpedoes to the

melt shop where the iron is converted to steel.

The desulfurizing plant consists of two side-by-side ports,

each capable of desulfurizing a torpedo. However, the lance used

at each port tends to break down, and repairs take up to a week.

Management is concerned that lance breakdowns could cause a

bottleneck, which would hold up a large number of torpodeos,causing the blast furnaces to shut down. There were 23 torpedoes

circulating between the blast furnaces, desulfurizing plant, and

melt shop. The proposed modification would install a backup

system, which would solve the problem.

There are four blast furnaces. Each needs approximately

two torpedoes for each discharge. The discharge time for each

blast furnace is Normally distributed with a mean of 200 min-

utes and standard deviation of 20 minutes. The processingtime of each desulfurizing port is approximately Normally

distributed with a mean of 15 minutes and standard deviation

of 3 minutes. The wait time for a locomotive to haul a torpedo

out of a port is approximately Exponential with a mean of

8 minutes.

Model this problem in Excel, assuming that a lance has bro-

ken down for a week (i.e., just use one server). What is the impli-

cation for the number of torpedoes stuck in the lance queue?

Management wont like to see more than five to six torpedoes

in the lance line. Hint: The distribution of sum of two or more

independent Normal distributions tends to the Continuous Uni-

form distribution. Therefore, one can assume that inter-arrival

times are Uniform.

Source: P. C. Bell,Management Science/Operat ionsResearch, 1999, Cincinnati: South-Western College Publishing,pp. 203206.

Documents

Ch18 Simulation