Random-Variate Generation Banks, Carson, Nelson & Nicol - PowerPoint PPT Presentation

Chapter 8 Random-Variate Generation Banks, Carson, Nelson & Nicol Discrete-Event System Simulation

Purpose & Overview  Develop understanding of generating samples from a specified distribution as input to a simulation model.  Illustrate some widely-used techniques for generating random variates.  Inverse-transform technique  Acceptance-rejection technique 2

Exponential Distribution [Inverse-transform]  Histogram for 200 R i : (empirical) (theoretical) 3

Exponential Distribution [Inverse-transform]  Histogram for 200 X i : (empirical) (theoretical) 4

Inverse-transform Technique  The concept:  For cdf function: r = F(x)  Generate r from uniform (0,1)  Find x: r = F(x) x = F -1 (r) r 1 x 1 5

Exponential Distribution [Inverse-transform]  Exponential Distribution:  Exponential cdf: r = F(x) 1 – e - l x for x  0 =  To generate X 1 , X 2 , X 3 … F -1 (R i ) X i = -(1/ l) ln(1-R i ) = [Eq ’ n 8.3] 1   X ln R l i i 6

Exponential Distribution [Inverse-transform]  Generate using the graphical view: Figure: Inverse-transform technique for e xp( l = 1) 7

Other Distributions [Inverse-transform]  Examples of other distributions for which inverse cdf works are:  Uniform distribution     X ~ U ( a , b ) X a ( b a ) R  Weibull distribution         1 ( / ) x a (v = 0) x e , x 0     f ( x )   0 , otherwise   x   ( )   F X ( ) 1 e , x 0       1 / X [ ln( 1 R )] 8

Other Distributions [Inverse-transform]  Triangular distribution    . 0 1 x x       ( ) 2 , 1 2 f x x x   0 otherwise 1    0 R  2 R , 2    X    1  2 2 ( 1 R ) ,   R 1 2 9

Empirical Continuous Dist ’ n [Inverse-transform]  When theoretical distribution is not applicable  To collect empirical data:  Resample the observed data  Interpolate between observed data points to fill in the gaps  For a small sample set (size n ):  Arrange the data from smallest to largest     x x x (1) (2) (n)   x x x  Assign the probability 1/n to each interval (i - 1) (i)    ( i 1 ) ˆ        1 X F ( R ) x a R  ( 1 ) i i   n   x x x x where     ( i ) ( i 1 ) ( i ) ( i 1 ) a   i i / n ( i 1 ) / n 1 / n 10

Empirical Continuous Dist ’ n [Inverse-transform]  Five observations of fire-crew response times (in mins.):  2.76 1.83 0.80 1.45 1.24 Cumalative Interval Probability Probability, i (Hours) 1/n i/n Slope, a i 0.0 ≤ x ≤ 0.80 1 0.2 0.20 4.00 0.8 ≤ x ≤ 1.24 2 0.2 0.40 2.20 1.24 ≤ x ≤ 1.45 3 0.2 0.60 1.05 1.45 ≤ x ≤ 1.83 4 0.2 0.80 1.90 1.83 ≤ x ≤ 2.76 5 0.2 1.00 4.65 11

Empirical Continuous Dist ’ n [Inverse-transform] Consider R 1 = 0.71: (i-1)/n = 0.6 < R 1 < i/n = 0.8 X 1 = x (4-1) + a 4 (R 1 – (4-1)/n) = 1.45 + 1.90(0.71-0.60) = 1.66 12

Empirical Continuous Dist ’ n [Inverse-transform]  For a large sample set :  Summarize into frequency distribution with small number of intervals. (equal intervals) x (i-1) < x ≤ x (i) : i th interval  Fit a continuous empirical cdf to the frequency distribution. if c i-1 < R ≤ c i (ci : cumulative frequency) ˆ      1 X F ( R ) x a ( R c )   ( i 1 ) i i 1 Where  x x   ( i ) ( i 1 ) a  i c c  i i 1 13

Empirical Continuous Dist ’ n [Inverse-transform]  Example: Suppose the data collected for100 broken- widget repair times are: Interval Relative Cumulative Slope, i (Hours) Frequency Frequency Frequency, c i a i 0.25 ≤ x ≤ 0.5 1 31 0.31 0.31 0.81 0.5 ≤ x ≤ 1.0 2 10 0.10 0.41 5.0 1.0 ≤ x ≤ 1.5 3 25 0.25 0.66 2.0 1.5 ≤ x ≤ 2.0 4 34 0.34 1.00 1.47 Consider R 1 = 0.83: c 3 = 0.66 < R 1 < c 4 = 1.00 X 1 = x (4-1) + a 4 (R 1 – c (4-1) ) = 1.5 + 1.47(0.83-0.66) = 1.75 14

Discrete Distribution [Inverse-transform]  Example: Suppose the number of shipments, x, on the loading dock of IHW company is either 0 , 1 , or 2  Data - Probability distribution: x p(x) F(x) 0 0.50 0.50 1 0.30 0.80 2 0.20 1.00  Method - Given R, the generation scheme becomes:   0 , 0 . 5 R      x 1 , 0 . 5 R 0 . 8  Consider R 1 = 0.73:    2 , 0 . 8 R 1 . 0 F(x i-1 ) < R <= F(x i ) F(x 0 ) < 0.73 <= F(x 1 ) Hence, x 1 = 1 15

Acceptance-Rejection technique  Useful particularly when inverse cdf does not exist in closed form, a.k.a. thinning  Illustration: To generate random variates, X ~ U( 1/4, 1 ) Generate R Procedures: no Step 1. Generate R ~ U[0,1] Condition Step 2a. If R >= ¼, accept X=R. yes Step 2b. If R < ¼, reject R, return to Step 1 Output R ’  R does not have the desired distribution, but R conditioned ( R ’ ) on the event { R  ¼ } does.  Efficiency: Depends heavily on the ability to minimize the number of rejections. 16

NSPP [Acceptance-Rejection]  Non-stationary Poisson Process (NSPP): a Possion arrival process with an arrival rate that varies with time  Idea behind thinning:  Generate a stationary Poisson arrival process at the fastest rate, l * = max l (t)  But “ accept ” only a portion of arrivals, thinning out just enough to get the desired time-varying rate Generate E ~ Exp( l *) t = t + E no Condition R <= l (t) yes Output E ’ ~ t 17

NSPP [Acceptance-Rejection]  Example: Generate a random variate for a NSPP Procedures: Data: Arrival Rates Step 1. l * = max l (t) = 1/5 , t = 0 and i = 1 . Mean Time Step 2. For random number R = 0.2130 , Between Arrival Rate l (t) E = -5ln(0.213) = 13.13 t Arrivals (min) (min) (#/min) t = 13.13 0 15 1/15 Step 3. Generate R = 0.8830 60 12 1/12 l (13.13)/ l *=(1/15)/(1/5)=1/3 120 7 1/7 Since R>1/3 , do not generate the arrival 180 5 1/5 Step 2. For random number R = 0.5530 , 240 8 1/8 300 10 1/10 E = -5ln(0.553) = 2.96 360 15 1/15 t = 13.13 + 2.96 = 16.09 420 20 1/20 Step 3. Generate R = 0.0240 480 20 1/20 l (16.09)/ l *=(1/15)/(1/5)=1/3 Since R<1/3, T 1 = t = 16.09 , and i = i + 1 = 2 18

Summary  Principles of random-variate generate via  Inverse-transform technique  Acceptance-rejection technique  Important for generating continuous and discrete distributions 19