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ENM 307 SIMULATION

Anadolu Üniversitesi,

Endüstri Mühendisliği Bölümü 2012-2013, Spring

Zehra Kamışlı Öztürk

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, …

Random Variate Generation

≤≤

=otherwise,0

10,1)(

xxfR

>

≤≤

<

=

1x,1

10,

0,0

)( xx

x

xFR

Inverse-Transform Technique

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

◦ Uniform

◦ Weibull

◦ Triangular

◦ Empirical

◦ Et

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


Exponential Distribution• Probability density function (pdf)

◦ The parameter λ 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 λ, then λ 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

xλλ

pdfs for several exponential distributions

f(x)

0,0

0,2

0,4

0,6

0,8

1,1

1,3

1,5

1,7

1,9

2,1

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,7 1,9 2,1 2,3 2,5 2,7 2,9 3,1

0.511.52

λ

1)( =iXE


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

tλλ

<

≥−=

−

0,0

0,1)(

x

xexF

xλ

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)( ≥−=−

xexFxλ

.0,1 ≥=−−

xReXλ

)1ln(1

)1ln(

1

1

RX

RX

Re

Re

X

X

−−=

−=−

−=

=−

−

−

λ

λ

λ

λ

iiRX ln

1

λ−=)1ln(

1iiRX −−=

λ

)(1iiRFX

−=

or

Example 8.1i 1 2 3 4 5

Ri

0.130

6

0.042

2

0.659

7

0.796

5

0.769

6X

i 0.1400 0.0431 1.0779 1.5921 1.4679

λ =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


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


Weibull Distribution• Probability density function (pdf)

• Step1: The is given by F(X)=1-e-(x/α)β , x≥0

• 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 ββ

βα

β


Triangular Distribution• Probability density function (pdf)

• 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

2x

R =

2

)2(1

2x

R−

−=

≤≤−−

≤≤=

1,)1(22

0,2

21

21

RR

RRX

0,0

0,2

0,4

0,7

0,9

1,1

-0,8 0,0 0,8 1,5 2,3 3,0

0,0

0,2

0,4

0,7

0,9

1,1

0 1 1 2 2


Empirical Continuous Distribution• If the modeler has been unableto find a theoretical distribution that

provides a good model for the input data, then it may be necessary to use the emprical distribution of the data.

• Finite number of values !

• Otherwise, interpolate between the observed data

points to fill in the gaps !


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 ≤ 1.45 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()(! )1(

1

0,0

0,3

0,5

0,8

1,0

0,0 0,8 1,5 2,3 3,0


Empirical Continuous Distribution-Example 8.2i 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

0,3

0,5

0,8

1,0

0,00 0,75 1,50 2,25 3,00

R1

X1

66.1

)60.071.0(90.145.1

)/)14(( 14)14(1

=

−+=

−−+= − nRaxX

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,

1795.0

)1()(

135.0135.01 RRRFX

−−≈=

−

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

Discrete Distributions

• All discrete distributions can be generated via inverse transform technique, ◦ 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.

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


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

)(

ΜΜ

k

irR

k

ir

ii=≤<

−=−

1

1 RkXRkiRkiRki =⇒+≤≤⇒<≤−⇒ 11


Geometric Distribution

• Geometric distribution with ◦ pmf

◦ Cdf

◦ Inverse transform

...,2,1,0,)1()( =−= xppxpx

{ }

1

1

0

)1(1)(

)1(1

)1(1)1()(

+

+

=

−−=

−−

−−=−=∑

x

xx

j

j

pxF

p

ppppxF

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

xx=−−≤<−−=−

+

xxpRp )1(1)1( 1

−<−≤−+

)1ln()1ln()1ln()1( pxRpx −<−≤−+

)1ln(

)1ln(1

)1ln(

)1ln(

p

Rx

p

R

−

−<≤−

−

−

−

−

−= 1

)1ln(

)1ln(

p

RX

Acceptance-Rejection Technique

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

◦ Step2a: if R≥1/4, 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.


Poisson Distribution

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

◦ pmf

◦ N=n

,...2,1,!

)()( ====−

nn

enNPnp

nα

α

t=0 t=1 t=2

A1 A2 A3 A… An An+1

121211 +++++<≤+++

nnnAAAAAAA ΛΛ

∏∏

∏∑∏ ∑

∑∑

+

==

−

+

=

+

== =

+

==

>≥

=>−≥=

−<≤−

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

α

α

αα



• 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-α , then N=n. Otherwise, reject n, increase n by one, and return to step 2.

• Example 8.8 : α =0.2 First compute e-α=e-2=0.8187.

◦ Step1: n=0, P=1 ◦ Step2: R1=0.4357, P=1. R1=0.4357

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

◦ Step1-3: (R1=0.4146, leads to N=0)



◦ … ◦ Step1: n=0, P=1 ◦ Step2: R1=0.8353, P=1. R1=0.8353

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

◦ Step2: R2=0.9952, P=R1. R2=0.8313

◦ Step3: since P=0.8313 ≥e-α =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-α =0.8187, accept, N=2.

n Rn+1 P Accept/reject Result

0 0.4357 0.4357 P<e-α accept N=0

0 0.4146 0.4146 P<e-α accept N=1

0 0.8353 0.8353 P ≥e-α reject

1 0.9952 0.8313 P ≥e-α reject

2 0.8004 0.6654 P<e-α accept N=2

enm 307 simulation - eskisehir.edu.trendustri.eskisehir.edu.tr/zkamisli/enm307/icerik...•random...

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