input modeling aslı sencer graduate program in engineering and technology management simulation-4

INPUT MODELİNGAslı Sencer

Graduate Program in

Engineering and Technology Management

Simulation-4

2

STEPS OF İNPUT MODELİNG

1) Collect data from real system of interest Requires substantial time and effort Use expert opinion in case of no sufficient data

2) Identify a probability distribution to represent the input process

Draw frequency distribution, histograms Choose a family of theoretical distribution

3) Estimate the parameters of the selected distribution4) Apply goodness-of-fit tests to evaluate the chosen

distribution and the parameters Chi-square tests Kolmogorov Smirnov Tests

5) If these tests are not justified, choose a new theoretical distribution and go to step 3! If all theoretical distributions fail, then either use emprical distribution or recollect data.

3

STEP 1: DATA COLLECTİON İNCLUDES LOTS OF DİFFİCULTİES

Nonhomogeneous interarrival time distribution; distribution changes with time of the day, days of the week, etc. You can’t merge all these data for distribution fitting!

Two arrival processes might be dependent; like demand for washing machines and dryers. You shouldn’t treat them seperately!

Start and end of service durations might not be clear; You should split the service into well defined processes!

Machines may breakdown randomly; You should collect data for up and down times!

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STEP 2.1: IDENTİFY THE PROBABİLİTY DİSTRİBUTİON

Histogram with Discrete Data

Arrivals per period Frequency

0 121 102 193 174 105 86 77 58 59 3

10 311 1

10 8 5 1 6 0 4 6 2 32 3 5 9 2 0 2 4 2 35 1 8 9 1 9 3 7 4 02 6 3 1 4 5 0 3 3 22 10 0 3 6 0 6 5 7 08 2 3 7 0 2 2 1 0 40 2 4 1 2 5 1 5 3 28 6 3 4 6 11 3 2 8 02 4 2 4 1 3 1 2 1 23 10 0 7 3 5 3 7 3 4

Raw Data

0 1 2 3 4 5 6 7 8 9 10 110

2

4

6

8

10

12

14

16

18

20

Histogram of Arrivals per Period

Frequency

5

STEP 2.1: IDENTİFY THE PROBABİLİTY DİSTRİBUTİON

3 6 9 12 15 18 21 24 27 30 33 360

5

10

15

20

25

Histogram of Component Life

Frequency

Histogram with Continuous DataComponent Life

(days) Frequency[0-3) 23[3-6) 10[6,9) 5

[9-12) 1[12-15) 1[15-18) 2[18-21) 0[21-24) 1[24-27) 1[27-30) 0[30-33) 1[33-36) 1

... ...[42-45) 1

... ...[57-60) 1

... ...[78-81) 1

... ...[144-147) 1

79.919 3.081 0.062 1.961 5.8453.027 6.505 0.021 0.013 0.1236.769 59.899 1.192 34.760 5.009

18.387 0.141 43.565 24.420 0.433144.695 2.663 17.967 0.091 9.003

0.941 0.878 3.148 2.157 7.5790.624 5.380 3.371 7.078 23.9600.590 1.928 0.300 0.002 0.5437.004 31.764 1.005 1.147 0.2193.217 14.382 1.008 2.336 4.562

Raw Data

6

STEP 2.2: SELECTİNG THE FAMİLY OF DİSTRİBUTİONS

The purpose of preparing a histogram is to infer a known pdf or pmf.

This theoretical distribution is used to generate random variables like interarrival times and service times during simulation runs.

Exponential, normal and poisson ditributions are frequently encountered and are not difficult to analyze.

Yet there are beta, gamma and weibull families that provide a wide variety of shapes.

7

Applications of Exponential Distribution

Used to model time between independent events,like arrivals or breakdowns

Inappropriate for modeling process delay times

9

Applications of Poisson Distribution

•Discrete distribution, used to model the number of independent events occuring per unit time, Eg. Batch sizes of customers and items

•If the time betweeen successive events is exponential,then the number of events in a fixed time intervalsis poisson.

12

Applications of Beta Distribution:

•Often used as a rough model in the absence of data•Represent random proportions•Can be transformed into scaled beta sample Y=a+(b-a)X

14

Applications of Erlang Distribution

•Used to represent the time required to complete a taskwhich can be reprsented as the sum of k exponentially distributed durations.

•For large k, Erlang approaches normal distribution.

•For k=1, Erlang is the exponential distribution withrate=1/β.

•Special case of gamma distribution in which α, theshape parameter of gamma distribution is k.

15

Applications of Gamma Distribution

•Used to represent time required to complete a task

•Same as Erlang distribution when the shape parameterα is an integer.

16

Applications of Johnson Dist.Flexible domain being bounded or unbounded allows it to fit many data sets.If δ>0, the domain is boundedIf δ<0, the domain is unbounded

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Applications of Lognormal DistributionUsed to represent quantities which is the product of large number of random quantitiesUsed to represent task times which are skewed to right. If X~LOGN( ), then lnX ~NORM(μ,σ)

ll ,

19

Applications of Weibull Distribution

•Widely used in reliability models to represent lifetimes.

•If the system consists of large number of parts that failindependently, time between successive failures can beWeibull.

•Used to model nonnegative task times that are skewed to left.

•It turns out to be exponential distribution when =1.

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Applications of Continuous Empirical Distribution

•Used to incorporate empirical data as an alternative totheoretical distribution, when there are multimodes,significant outliers, etc.

21

Applications of Discrete Empirical Distribution

•Used for discrete assignments such as job type, visitation sequence or batch size

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STEP 3: ESTİMATE THE PARAMETERS OF THE SELECTED DİSTRİBUTİON A theoretical distribution is specified by its parameters that

are obtained from the whole population data.Ex: Let V,W,X,Y,Z be random variables, thenV~N(µ,σ2), where µ is the mean and σ2 is the variance.W~Poisson (λ), where λ is the meanX~Exponential (β), where β is the mean Y~Triangular (a,m,b), where a, m,b are the minimum,mod and the maximum of the dataZ~Uniform (a,b), where a and b are the minimum and maximum of the data

These parameters are estimated by using the point estimators defined on the sample data

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STEP 3: ESTİMATE THE PARAMETERS OF THE SELECTED DİSTRİBUTİON Sample mean and the sample variance are the point estimators for

the population mean and population variance

Let Xi; i=1,2,...,n iid random variables (raw data are known) , then

the sample mean and sample variance s2 are calculated as

10 8 5 1 6 0 4 6 2 32 3 5 9 2 0 2 4 2 35 1 8 9 1 9 3 7 4 02 6 3 1 4 5 0 3 3 22 10 0 3 6 0 6 5 7 08 2 3 7 0 2 2 1 0 40 2 4 1 2 5 1 5 3 28 6 3 4 6 11 3 2 8 02 4 2 4 1 3 1 2 1 23 10 0 7 3 5 3 7 3 4

Discrete Raw Data79.919 3.081 0.062 1.961 5.8453.027 6.505 0.021 0.013 0.1236.769 59.899 1.192 34.760 5.009

18.387 0.141 43.565 24.420 0.433144.695 2.663 17.967 0.091 9.003

0.941 0.878 3.148 2.157 7.5790.624 5.380 3.371 7.078 23.9600.590 1.928 0.300 0.002 0.5437.004 31.764 1.005 1.147 0.2193.217 14.382 1.008 2.336 4.562

Continuous Raw Data

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STEP 3: ESTİMATE THE PARAMETERS OF THE SELECTED DİSTRİBUTİON If the data are discrete and have been grouped in a

frequency distribution, i.e., the raw data are not known, then

where k is the number of distinct values of X and fj; j=1,2,...,k is the observed frequency of the value Xj of X.

Arrivals per period Frequency Arrivals per period Frequency

0 12 6 7

1 10 7 5

2 19 8 5

3 17 9 3

4 10 10 3

5 8 11 1

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STEP 3: ESTİMATE THE PARAMETERS OF THE SELECTED DİSTRİBUTİON If the data are discrete or continuous and have been

grouped in class intervals, i.e., the raw data are not known, then

where fj; j=1,2,...,c is the observed frequency of the jth

class interval and mj is the midpoint of the jth interval.Component Life

(days) FrequencyComponent Life

(days) FrequencyComponent Life

(days) Frequency

[0-3) 23 [21-24) 1 ... ...

[3-6) 10 [24-27) 1 [57-60) 1

[6,9) 5 [27-30) 0 ... ...

[9-12) 1 [30-33) 1 [78-81) 1

[12-15) 1 [33-36) 1 ... ...

[15-18) 2 ... ... [144-147) 1

[18-21) 0 [42-45) 1

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STEP 3: ESTİMATE THE PARAMETERS OF THE SELECTED DİSTRİBUTİON

The minimum, mod (i.e., data value with the highest frequency) and maximum of the population data are estimated from the sample data as

Xt is the data value that has the highest frequency.

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STEP 4: GOODNESS OF FİT TEST

Goodness of fit tests (GFTs) provide helpful guidance for evaluating the suitability of the selected input model as a simulation input.

GFTs check the discrepancy between the emprical and the selected theoretical distribution to decide whether the sample is taken from that theoretical distribution or not.

The role of sample size, n: If n is small, GFTs are unlikely to reject any theoretical

distribution, since discrepancy is attributed to the sampling error!

If n is large, then GFTs are likely to reject almost all distributions.

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STEP 4: GOODNESS OF FİT TESTSCHİ SQUARE TEST Chi square test is valid for large sample sizes and for both

discrete and continuous assumptions when parameters are estimated with maximum likelihood.

Hypothesis test:

Ho: The random variable X conforms to the theoretical distribution with the estimated parameters

Ha: The random variable does NOT conform to the theoretical distribution with the estimated parameters

We need a test statistic to either reject or fail to reject Ho. This test statistic should measure the discrepency between the theoretical and the emprical distribution.

If this test statistic is high, then Ho is rejected, Otherwise we fail to reject Ho! (Hence we accept Ho)

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STEP 4: GOODNESS OF FİT TESTSCHİ SQUARE TEST

Test statistic:Arrange n observations into a set of k class intervals or cells. The test statistic is given by

where Oi is the observed frequency in the ith class interval and Ei is the expected frequency in the ith class interval.

where pi is the theoretical probability associated with the ith class, i.e., pi =P(random variable X belongs to ith class).

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STEP 4: GOODNESS OF FİT TESTSCHİ SQUARE TEST

Recommendations for number of class intervals for continuous data

It is suggested that . In case it is smaller, then that class should be combined with the adjacent classes. Similarly the corresponding Oi values should also be combined and k should be reduced by every combined cell.

Sample Size,

n

Number of Class Intervals

k

20 Do not use chi-square test

50 5-10

100 10 to 20

>100 to n/5

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STEP 4: GOODNESS OF FİT TESTSCHİ SQUARE TEST Evaluation

Let α =P(rejecting Ho when it is true); the significance level is 5%.

If probability of the test statistic < α, reject Ho and the distribution

otherwise, fail to reject Ho.

follows the chi-square distribution with k-s-1 degress

of freedom, where s is the number of

estimated parameters.

Reject HoFail to Reject Ho

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CHİ-SQUARE DİSTRİBUTİON TABLE

α(k-s-1)

𝜒𝛼 ,𝑘− 𝑠− 12

33

STEP 4: GFT - CHİ SQUARE TEST EX: POİSSON DİSTRİBUTİON

Consider the discrete data we analyzed in step 2.Ho: # arrivals, X~ Poisson (λ=3.64)Ha: owλ is the mean rate of arrivals, =3.64

The following probabilities are found by using the pmf

P(0)=0.026 P(6)=0.085

P(1)=0.096 P(7)=0.044

P(2)=0.174 P(8)=0.020

P(3)=0.211 P(9)=0.008

P(4)=0.192 P(10)=0.003

P(5)=0.140 P(>11)=0.001

34

STEP 4: GFT - CHİ SQUARE TEST EX: POİSSON DİSTRİBUTİON Calculation of the chi-square test statistic with k-s-1=7-1-

1=5 degrees of freedom and α=0,05.

So, Ho is rejected!

35

STEP 4: GFT - CHİ SQUARE TEST EX: ARENA İNPUT ANALYZER

Distribution Summary

Distribution: Normal

Expression: NORM(225, 89)

Square Error: 0.037778

Chi Square Test

Number of intervals = 12

Degrees of freedom = 9

Test Statistic = 1.22e+004

Corresponding p-value < 0.005

Data Summary

Number of Data Points = 27009

Min Data Value = 1

Max Data Value = 1.88e+003

Sample Mean = 225

Sample Std Dev = 89

Histogram Summary

Histogram Range = 0.999 to 1.88e+003

Number of Intervals = 40

Fit all summaryFunction Sq Error-----------------------Normal 0.0506Gamma 0.0625Beta 0.0639Erlang 0.0673Weibull 0.079Lognormal 0.0926Exponential 0.286Triangular 0.311Uniform 0.36

Reject Normal distribution at 5% significance level!

36

STEP 4: GFT - CHİ SQUARE TEST EX: ARENA İNPUT ANALYZERDistribution Summary

Distribution: Lognormal

Expression: 2 + LOGN(145, 67.9)


Chi Square Test



Test Statistic = 207


Data Summary


Min Data Value = 2


Sample Mean = 146

Sample Std Dev = 79.5

Histogram Summary

Histogram Range = 2 to 6.01e+003


Reject Lognormal distribution at 5% significance level!

37

STEP 4: GFT - CHİ SQUARE TEST EX: ARENA İNPUT ANALYZERDistribution Summary

Distribution: Weibull

Expression: 0.999 + WEIB(94.7, 0.928)


Chi Square Test



Test Statistic = 838


Data Summary


Min Data Value = 1


Sample Mean = 108

Sample Std Dev = 135

Histogram Summary

Histogram Range = 0.999 to 1.47e+003


Reject Weibull distribution at 5% significance level!

38

STEP 4: GOODNESS OF FİT TESTSDRAWBACKS OF CHİ-SQUARE GFT The Chi-square test uses the estimates of the

parameters obtained from the sample that decreases the degrees of freedom.

Chi-square test requires the data to be placed in class intervals in the continuous distributions where these classes are arbitrary and affects the value of the chi-square test statistic.

The distribution of the chi-square test statistic is known approximately and the power of the test (probability of rejecting an incorrect theoretical distribution) is sometimes low.

Hence other GFTs are also needed!

39

STEP 4: GOODNESS OF FİT TESTSKOLMOGOROV-SMİRNOV TEST

Useful when the sample sizes are small and when no parameters are estimated from the sample data.

Compares the cdf of the theoretical distribution, F(x) with the emprical cdf, SN(x) of the sample of N observations.

Hypothesis test:

Ho: Data follow the selected pdf

Ha: Data do NOT follow the selected pdf

Test Statistic:

The largest deviation, D between F(x) and SN(x).

40


Steps of K-S Test:

1. Rank the data so that

2. Calculate the maximum discrepancy D between F and SN,

𝑋(1) ≤ 𝑋(2) ≤ ⋯ ≤ 𝑋(𝑁)

𝐹൫𝑋(𝑖)൯= 𝑃(𝑋≤ 𝑋(𝑖))

𝑆𝑁൫𝑋(𝑖)൯= # 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒𝑑 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠≤ 𝑋(𝑖)𝑁 = 𝑖𝑁

41


If F is discrete , where

If F is continuous

𝐷+ = max0≤𝑖≤𝑁൛𝑆𝑁൫𝑋ሺ𝑖ሻ൯− 𝐹൫𝑋ሺ𝑖ሻ൯ൟ= max0≤𝑖≤𝑁൜𝑖𝑁− 𝐹൫𝑋ሺ𝑖ሻ൯ൠ 𝐷− = max0≤𝑖≤𝑁൛𝐹൫𝑋ሺ𝑖ሻ൯− 𝑆𝑁൫𝑋ሺ𝑖−1ሻ൯ൟ= max0≤𝑖≤𝑁൜𝐹൫𝑋ሺ𝑖ሻ൯− 𝑖 − 1𝑁 ൠ

𝐷= 𝑚𝑎𝑥ሼ𝐷+,𝐷−ሽ

𝐷= max0≤𝑖≤𝑁ห𝐹൫𝑋ሺ𝑖ሻ൯− 𝑆𝑁൫𝑋ሺ𝑖ሻ൯ห

42


3. Evaluation 𝐼𝑓 𝐷> 𝐷∝,𝑁,𝑡ℎ𝑒𝑛 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻𝑜 𝐼𝑓 𝐷≤ 𝐷∝,𝑁,𝑡ℎ𝑒𝑛 𝑓𝑎𝑖𝑙 𝑡𝑜 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻𝑜

43

STEP 4: GOODNESS OF FİT TESTSEXAMPLE: KOLMOGOROV-SMİRNOV TEST

Consider the data: 0.44, 0.81, 0.14, 0.05, 0.93

Ho: Data are uniform between (0,1)Ha: ow

i 1 2 3 4 5

0.05

0.14

0.44

0.81

0.93

0.05

0.14

0.44

0.81

0.93

0.20

0.40

0.60

0.80

1.00

0.15

0.26

0.16

- 0.07

0.05

- 0.04

0.21

0.13

𝑋ሺ𝑖ሻ

𝑆𝑁൫𝑋ሺ𝑖ሻ൯= 𝑖/𝑁

𝐹൫𝑋ሺ𝑖ሻ൯= 𝑋ሺ𝑖ሻ

𝑖/𝑁− 𝐹൫𝑋ሺ𝑖ሻ൯ 𝐹൫𝑋ሺ𝑖ሻ൯= (𝑖 − 1)/𝑁

Since D=0.26 < = 0.565

Ho is not rejected!Data are uniform between

(0,1)

𝐷0.05,5

input modeling aslı sencer graduate program in engineering and technology management simulation-4

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