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

random variate generation
SMART_READER_LITE
LIVE PREVIEW

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.


slide-1
SLIDE 1

Chapter 8 Random-Variate Generation

Banks, Carson, Nelson & Nicol Discrete-Event System Simulation

slide-2
SLIDE 2

2

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

slide-3
SLIDE 3

Exponential Distribution

[Inverse-transform]

 Histogram for 200 Ri:

(empirical) (theoretical)

3

slide-4
SLIDE 4

Exponential Distribution

[Inverse-transform]

 Histogram for 200 Xi :

(empirical) (theoretical)

4

slide-5
SLIDE 5

5

Inverse-transform Technique

 The concept:

 For cdf function: r = F(x)  Generate r from uniform (0,1)  Find x:

x = F-1(r)

r1 x1

r = F(x)

slide-6
SLIDE 6

6

Exponential Distribution

[Inverse-transform]

 Exponential Distribution:

 Exponential cdf:  To generate X1, X2, X3 …

r = F(x) = 1 – e-lx

for x 0

Xi = F-1(Ri) =

  • (1/l) ln(1-Ri)

[Eq’n 8.3]

i i

R X ln 1 l  

slide-7
SLIDE 7

Exponential Distribution

[Inverse-transform]

 Generate using the graphical view:

7

Figure: Inverse-transform technique for exp(l = 1)

slide-8
SLIDE 8

8

Other Distributions

[Inverse-transform]

 Examples of other distributions for which inverse

cdf works are:

 Uniform distribution  Weibull distribution

(v = 0)

R a b a X b a U X ) ( ) , ( ~    

      

 

  • therwise

x e x x f

a x

, , ) (

) / ( 1

 

 

/ 1

)] 1 ln( [ R X    

( )

( ) 1 ,

x

F X e x

 

  

slide-9
SLIDE 9

Other Distributions

[Inverse-transform]

 Triangular distribution

9

  • therwise

x x x x x f 2 1 1 , 2 . ) (           

1 2 1 2 1 , ) 1 ( 2 2 , 2             R R R R X 

slide-10
SLIDE 10

10

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  Assign the probability 1/n to each interval

where

(n) (2) (1)

x x x    

(i) 1)

  • (i

x x x  

          

 

n i R a x R F X

i i

) 1 ( ) ( ˆ

) 1 ( 1

n x x n i n i x x a

i i i i i

/ 1 / ) 1 ( /

) 1 ( ) ( ) 1 ( ) (  

     

slide-11
SLIDE 11

11

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

i Interval (Hours) Probability 1/n Cumalative Probability, i/n Slope, a i 1 0.0 ≤ x ≤ 0.80 0.2 0.20 4.00 2 0.8 ≤ x ≤ 1.24 0.2 0.40 2.20 3 1.24 ≤ x ≤ 1.45 0.2 0.60 1.05 4 1.45 ≤ x ≤ 1.83 0.2 0.80 1.90 5 1.83 ≤ x ≤ 2.76 0.2 1.00 4.65

slide-12
SLIDE 12

Empirical Continuous Dist’n

[Inverse-transform]

12

Consider R1 = 0.71: (i-1)/n = 0.6 < R1 < i/n = 0.8 X1 = x(4-1) + a4(R1 – (4-1)/n) = 1.45 + 1.90(0.71-0.60) = 1.66

slide-13
SLIDE 13

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 ci-1 < R ≤ ci (ci : cumulative frequency) Where

13

) ( ) ( ˆ

1 ) 1 ( 1   

   

i i i

c R a x R F X

1 ) 1 ( ) (  

  

i i i i i

c c x x a

slide-14
SLIDE 14

14

Empirical Continuous Dist’n

[Inverse-transform]

 Example: Suppose the data collected for100 broken-

widget repair times are:

i Interval (Hours) Frequency Relative Frequency Cumulative Frequency, c i Slope, a i 1 0.25 ≤ x ≤ 0.5 31 0.31 0.31 0.81 2 0.5 ≤ x ≤ 1.0 10 0.10 0.41 5.0 3 1.0 ≤ x ≤ 1.5 25 0.25 0.66 2.0 4 1.5 ≤ x ≤ 2.0 34 0.34 1.00 1.47

Consider R1 = 0.83: c3 = 0.66 < R1 < c4 = 1.00 X1 = x(4-1) + a4(R1 – c(4-1)) = 1.5 + 1.47(0.83-0.66) = 1.75

slide-15
SLIDE 15

15

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:  Method - Given R, the generation

scheme becomes:

. 1 8 . 8 . 5 . 5 . , 2 , 1 ,          R R R x  

Consider R1 = 0.73: F(xi-1) < R <= F(xi) F(x0) < 0.73 <= F(x1) Hence, x1 = 1

x p(x) F(x)

0.50 0.50 1 0.30 0.80 2 0.20 1.00

slide-16
SLIDE 16

16

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)  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.

Procedures: Step 1. Generate R ~ U[0,1] Step 2a. If R >= ¼, accept X=R. Step 2b. If R < ¼, reject R, return to Step 1

Generate R Condition Output R’ yes no

slide-17
SLIDE 17

17

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 Condition R <= l(t) Output E ’~ t yes no

slide-18
SLIDE 18

18

NSPP

[Acceptance-Rejection]

 Example: Generate a random variate for a NSPP Procedures:

Step 1. l* = max l(t) = 1/5, t = 0 and i = 1. Step 2. For random number R = 0.2130, E = -5ln(0.213) = 13.13 t = 13.13 Step 3. Generate R = 0.8830 l(13.13)/l*=(1/15)/(1/5)=1/3 Since R>1/3, do not generate the arrival Step 2. For random number R = 0.5530, E = -5ln(0.553) = 2.96 t = 13.13 + 2.96 = 16.09 Step 3. Generate R = 0.0240 l(16.09)/l*=(1/15)/(1/5)=1/3 Since R<1/3, T1 = t = 16.09, and i = i + 1 = 2

t (min) Mean Time Between Arrivals (min) Arrival Rate l(t) (#/min) 15 1/15 60 12 1/12 120 7 1/7 180 5 1/5 240 8 1/8 300 10 1/10 360 15 1/15 420 20 1/20 480 20 1/20

Data: Arrival Rates

slide-19
SLIDE 19

19

Summary

 Principles of random-variate generate via

 Inverse-transform technique  Acceptance-rejection technique

 Important for generating continuous and discrete

distributions