Random Variate Generation ENM 307 SIMULATIONendustri.eskisehir.edu.tr › gurkan.o › ENM307 › icerik › ENM_307_SIM… · Random Variate Generation • Random variate generation

ENM307 2012 Bahar

ENM 307 SIMULATION

Anadolu Üniversitesi,

Endüstri Mühendisliği Bölümü Yrd. Doç. Dr. Gürkan ÖZTÜRK

Random Variate Generation

ENM307 2012 Bahar

Random Variate Generation

• Random variate generation is about procedures for sampling from a variety of widely-used continuous and discrete distributions.

• Widely-used techniques are considered.

• Routines in programming libraries or built into simulation language

• Inverse transform technique, acceptance-rejection technique

• All techniques use a source of uniform [0,1] random numbers R1, R2, …

otherwise,0

10,1)(

xxfR

1x,1

10,

0,0

)( xx

x

xFR

ENM307 2012 Bahar

Inverse-Transform Technique

• Inverse-transform technique can be used to sample from – Exponential

– Uniform

– Weibull

– Triangular

– Empirical

– Etc.

• It is underlying principle for sampling from a wide variety of discrete distributions.

ENM307 2012 Bahar


Exponential Distribution • Probability density function (pdf)

– The parameter can be interpreted as the mean number of occurrence per time unit. – If the interarrival time X1, X2, X3,… had an exponential distribution with rate , then could

be interpreted as the mean number of arrivals per time unit.

– Expected value of Xi , is the mean of interarrival time.

otherwise,0

0,)(

xexf

x

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

f(x)

x

pdfs for several exponential distributions

0.5

1

1.5

2

1)( iXE

ENM307 2012 Bahar


Exponential Distribution

• Cumulative distribution function (cdf)

• The goal is to develop a procedure for generating values X1, X2, X3,… that have an exponential distribution.

• Inverse-transform technique can be utilized for any distribution

• If F-1 can be computed easily then it is most useful

x

dttfxF )()(

00

0,0

x

xdte

x

t

0,0

0,1)(

x

xexF

x

ENM307 2012 Bahar

Inverse-Transform Technique Step-by-step procedure on

Exponential Distribution

• Step 1: Compute the cdf of the desired random variable X. – For the exponential distribution, cdf

• Step 2: Set F(x)=R on the range of X. – For the exponential distribution,

• Step 3: Solve the equation F(x)=R for X in terms of R. – For the exponential distribution, the solution proceeds as follows:

• Step 4: Generate (as needed) random variables R1,R2,… and compute the desired random variates by – For the exponential case

.0,1)( xexF x

.0,1 xRe X

)1ln(1

)1ln(

1

1

RX

RX

Re

Re

X

X

ii RX ln1

)1ln(

1ii RX

)(1

ii RFX

or

ENM307 2012 Bahar

Example 8.1 i 1 2 3 4 5

Ri 0.1306 0.0422 0.6597 0.7965 0.7696

Xi 0.1400 0.0431 1.0779 1.5921 1.4679

=1

111

XeR

)1ln( 11 RX

1400.0

8694.0ln

)1306.01ln(1

X

F(x)=1-e-x

x

R1=1-e-X1

R2

X2 X1 =-ln(1- R1) x0

F(x0 )

0

1

ENM307 2012 Bahar


Uniform Distribution

• Probability density function (pdf)

• Step1: Cumulative distribution function (cdf) is given by

• Step2: Set F(X)=(X-a)/(b-a)=R

• Step3: Solving for X in terms of R yields X=a+(b-a)R

otherwise,0

,1

)(bxa

abxf

bx

bxaab

ax

ax

xF

,1

,

,0

)(

x

f(x)

a b

x

F(x)

a b

0

1

ENM307 2012 Bahar


Weibull Distribution


• Step1: The is given by F(X)=1-e-(x/) , x0

• Step2: Set 𝐹 𝑋 = 1 − 𝑒−

𝑥

𝛼

𝛽

= 𝑅

• Step3: Solving for X in terms of R yields

𝑋 = 𝛼 − ln 1 − 𝑅 1/𝛽

otherwise,0

0,)(

)/(1 xexxf

ax

ENM307 2012 Bahar


Triangular Distribution


• Cumulative distribution function

• For 0 x 1,

• For 1 x 2,

otherwise,0

21,2

10,

)( xx

xx

xf

2,1

21,2

)2(1

10,2

0,0

)(2

2

x

xx

xx

x

xF

2

2xR

2

)2(1

2xR

1,)1(22

0,2

21

21

RR

RRX

0.0

0.2

0.4

0.6

0.8

1.0

-1 0 1 2 3

0.0

0.2

0.4

0.6

0.8

1.0

0 1 2 3

ENM307 2012 Bahar


Empirical Continuous Distribution-Example 8.2

• Five observations of fire-crew response time (in minutes) to incoming alarms: – 2.76 1.83 0.80 1.45 1.24

Arrange the data from smallest to largest x(1) x(2) x(3) … x(n)

Smallest value is considerd as 0, so x(0)=0

n

xx

nini

xxa

iiii

i/1/)1(/

)1()()1()(

i Interval x(i-1) x x(i)

Probability 1/n

Cumulative Probability,

i/n

Slope ai

1 0.0 x 0.80 0.2 0.2 4

2 0.80 x 1.24 0.2 0.4 2.25

3 1.24 x 0.45 0.2 0.6 1.05

4 1.45 x 1.83 0.2 0.8 1.90

5 1.83 x 2.76 0.2 1.0 4.65

n

iRaxRFX ii

)1()(ˆ

)1(

1

0; 0

0.8; 0.2

1.24; 0.4

1.45; 0.6

1.83; 0.8

2.76; 1 3; 1

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3 4

ENM307 2012 Bahar


Empirical Continuous Distribution-Example 8.2 i Interval

x(i-1) x x(i)

Probability 1/n

Cumulative Probability, i/n

Slope ai

1 0.0 x 0.80 0.2 0.2 4

2 0.80 x 1.24 0.2 0.4 2.25

3 1.24 x 0.45 0.2 0.6 1.05

4 1.45 x 1.83 0.2 0.8 1.90

5 1.83 x 2.76 0.2 1.0 4.65

0; 0

0.8; 0.2

1.24; 0.4

1.45; 0.6

1.83; 0.8

2.76; 1 3; 1

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 0.5 1 1.5 2 2.5 3 3.5

R1

X1

66.1

)60.071.0(90.145.1

)/)14(( 14)14(1

nRaxX

ENM307 2012 Bahar

Continuous distributions without Closed-Form Inverse

• A number of continuous distributions do not have closed form expression for their cdf or its inverse: – For example

• Normal

• Gamma

• Beta

– Inverse transform technique for random variete generation is not available for these distributions.

– An approximation for inverse transform can be used.

– Simple approximation to the inverse cdf of standard normal distribution, (Schmeiser [1979])

1795.0

)1()(

135.0135.01 RR

RFX

R Approximate Inverse Exact Inverse

0.01 -2.3263 -2.3373

0.10 -1.2816 -1.2813

0.25 -0.6745 -0.6713

0.50 0.0000 0.0000

0.75 0.6745 0.6713

0.90 1.2816 1.2813

0.99 2.3263 2.3373

ENM307 2012 Bahar

Discrete Distributions

• All discrete distributions can be generated via inverse transform tecnique, – Numerically , through table-lookup procedure

– Algebraically, final generation scheme being in terms of a formula

• Other techniques are sometime used for certain distributions, e.g. – Convolution technique for binomial distribution.

ENM307 2012 Bahar

Discrete Distribution

An Empirical Discrete distribution Example 8.4.

• Distribution of number of Shipments, X

• The probability mass function (pmf) – P(0)=P(X=0)=0.5

– P(1)=P(X=1)=0.3

– P(2)=P(X=2)=0.2

• Cumulative distribution function

x p(x) F(x)

0 0.50 0.50

1 0.30 0.80

2 0.20 1.00

x

x

x

x

xF

21

218.0

105.0

00

)(

x

F(x)

0 1 2 3

R1=0.73

X1=1

0.18.02

8.05.01

5.00

R

R

R

X

ENM307 2012 Bahar


Uniform Distribution

• Discrete uniform distribution on {1,2,…,k} – pmf

– Cdf

– xi=i and ri=p(1)+p(2)+…+p(xi)=F(xi)=1/k for i=1,2,…,k

kxk

xp ,...,2,1,1

)(

xk

kxkk

k

xk

xk

x

xF

,1

1,1

32,2

21,1

1,0

)(

k

irR

k

ir ii

11

RkXRkiRkiRki 11

ENM307 2012 Bahar


Geometric Distribution

• Geometric distribution with – pmf

– Cdf

– Inverse transform

...,2,1,0,)1()( xppxp x

1

1

0

)1(1)(

)1(1

)1(1)1()(

x

xx

j

j

pxF

p

ppppxF

)()1(1)1(1)1( 1 xFpRpxF xx

xx pRp )1(1)1( 1

)1ln()1ln()1ln()1( pxRpx

)1ln(

)1ln(1

)1ln(

)1ln(

p

Rx

p

R

1

)1ln(

)1ln(

p

RX

ENM307 2012 Bahar

Acceptance-Rejection Technique

• X uniformly distributed between 1/4 and 1 – Step1: Generate a random number R

– Step2a: if R1/4, accept X=R then go to Step 3.

– Step2b: if R<1/4, reject R then return to Step 1.

– Step3: if another uniform vaiete on [1/4,1] is needed, go to Step 1 else stop.

ENM307 2012 Bahar


Poisson Distribution

• A Poisson random variable, N, with mean >0 has – pmf

– N=n

,...2,1,!

)()(

nn

enNPnp

n

t=0 t=1 t=2

A1 A2 A3 A… An An+1

12121 1 nnn AAAAAAA

1

11

1

1

1

11 1

1

11

lnlnlnln

ln1

1ln1

n

i

i

n

i

i

n

i

i

n

i

i

n

i

n

i

ii

n

i

i

n

i

i

ReR

RRRR

RR

ENM307 2012 Bahar



• The procedure for generating a Poisson random variate, N, is given by following steps – Step1: Set n=0, P=1

– Step2: Generate a random number Rn+1 and replace P by P.Rn+1

– Step3: if P<e- , then N=n. Otherwise, reject n, increase n by one, and return to step 2.

• Example 8.8 : =0.2 First compute e-=e-2=0.8187. – Step1: n=0, P=1

– Step2: R1=0.4357, P=1. R1=0.4357

– Step3: since P=0.4357<e- =0.8187, accept N=0.

– Step1-3: (R1=0.4146, leads to N=0)

ENM307 2012 Bahar



– …

– Step1: n=0, P=1

– Step2: R1=0.8353, P=1. R1=0.8353

– Step3: since P=0.8353e- =0.8187, reject, and return step 2 with n=1.

– Step2: R2=0.9952, P=R1. R2=0.8313

– Step3: since P=0.8313e- =0.8187, reject, and return step 2 with n=2.

– Step2: R3=0.8004, P=R1. R2.R3=0.6654

– Step3: since P=0.8313<e- =0.8187, accept, N=2.

n Rn+1 P Accept/reject Result

0 0.4357 0.4357 P<e- accept N=0

0 0.4146 0.4146 P<e- accept N=0

0 0.8353 0.8353 Pe- reject

1 0.9952 0.8313 Pe- reject

2 0.8004 0.6654 P<e- accept N=2

ENM307 2012 Bahar



=4 buses per hour Pe-4 0.0183

When is large, say ≥ 15 the rejection technique becomes quite expensive.

Z is approximately normal distributed with mean zero and variance 1.

Then possion variate can be generated

n Rn+1 P Accept/reject Result

0 0.4357 0.4357 Pe- reject

1 0.4146 0.1806 Pe- reject

2 0.8353 0.1508 Pe- reject

3 0.9952 0.1502 Pe- reject

4 0.8004 0.1202 Pe- reject

5 0.7945 0.0955 Pe- reject

6 0.1530 0.0146 P<e- accept N=6

NZ

5.0 ZN

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Random Variate Generation ENM 307 SIMULATIONendustri.eskisehir.edu.tr › gurkan.o › ENM307 › icerik › ENM_307_SIM… · Random Variate Generation • Random variate generation