Selecting Input Probability Distribution

Simulation Machine Simulation can be considered as an Engine with input and output as follows: Simulation Engine Input Output

Realizing Simulation Input Analysis: is the analysis of the random variables involved in the model such as: –The distribution of IAT –The distribution of Service Times Simulation Engine is the way of realizing the model, this includes: –Generating Random variables involved in the model –Performing the requiring formulas. Output Analysis is the study of the data that are produced by the Simulation engine.

Input Analysis collect data from the field Analyze these data Two ways to analyze the data: –Build Empirical distribution and then sample from this distribution. –Fit the data to a theoretical distribution ( such as Normal, Exponential, etc.) See Chapter 6 of Text for more distributions.

How to select an Input Probability distribution 1.Hypothesize a family of distributions. 2.Estimate the parameters of the fitted distributions 3.Determine how representative the fitted distributions are Repeat 1-3 until you get a fitted distribution foe the collected data. Otherwise go with an empirical distribution.

Hypothesizing a Theoretical Distribution To Fit a Theoretical Distribution Need a good background of the theoretical distributions (Consult your Text: Section 6.2) Histogram may not provide much insight into the nature of the distribution. Need Summary statistics

Summary Statistics Mean Median Variance  2 Coefficient of Variation (cv =  ) for continuous distributions Lexis ration    for discrete distributions Skewness index

Summary Stats. Cont. If the Mean and the Median are close to each others, and low Coefficient of Variation, we would expect a Normally distributed data. If the Median is less than the Mean, and  is very close to the Mean (cv close to 1), we expect an exponential distribution. If the skewness ( close to 0) is very low then the data are symmetric.

n Consider the following data Example

Example Cont. Mean Median Standard Deviation Skewness Range Minimum Maximum

Example Continue n We might take these data and construct a histogram The given summary statistics and the histogram suggest a Normal Distribution

Empirical Distribution

Disadvantages of Empirical distribution The empirical data may not adequately represent the true underlying population because of sampling error The Generated RV’s are bounded To overcome these two problems, we attempt to fit a theoretical distribution.

Estimation of Parameters of the fitted distributions Suppose we hypothesized a distribution, then use the Maximum Likelihood Estimator (MLE) to estimate the parameters involved with the hypothesized distribution. Suppose that  is the only parameter involve in the distribution then construct (for example the mean 1/ in the exponential distribution) Let L(  f   (X 1 )  f  (X 2 )  f  (X n ) Find  that maximize L(  ) to be the required parameter. Example: the exponential distribution. Do in class

Determine how representative the fitted distributions are Goodness of Fit (Chi Squared method)

Goodness of Fit (Chi Square method) 1.Divide the range of the fitted distribution into k (k<30) intervals [a 0, a 1 ), [a 1, a 2 ), … [a k-1, a k ] Let N j = the number of data that belong to [a j-1, a j ) 2.Compute the expected proportion of the data that fall in the j th interval using the fitted distribution call them p j 3.Compute the Chi-square

Chi-square cont. Note that np j represents the expected number of data that would fall in the jth interval if the fitted distribution is correct. If Where r is the number of parameters in the distribution (in Exponential dist. r = 1 which is ) Then do not reject distribution with significance (1-  )100%.

Example: Consider the following data: 0.01, 0.07, 0.03, 0.23, 0.04, 0.10, 0.31, 0.10, 0.31, 1.17, 1.50, 0.93, 1.54, 0.19, 0.17, 0.36, 0.27, 0.46, 0.51, 0.11, 0.56, 0.72, 0.39, 0.04, 0.78 Suppose we hypothesize an exponential distribution, Use Chi-square test by dividing the range into 5 subintervals.

The estimate of =2.5 Since k = 5, we have p i =0.2 For the exponential distribution Therefore

Therefore chi-square = 0.4 From the tables of chi-square we can accept the hypothesis With significance level 5%

Probability, pDegrees of Freedom The Chi-square table