cs433 modeling and simulation lecture 15 random number generator dr. anis koubâa 24 may 2009...

CS433Modeling and Simulation

Lecture 15Random Number Generator

Dr. Anis Koubâa24 May 2009

Al-Imam Mohammad Ibn Saud Islamic UniversityCollege Computer and Information Sciences

http://www.aniskoubaa.net/ccis/spring09-cs433/

Reading

Required Chapter 4: Simulation and Modeling with Arena Chapter 2: Discrete Event Simulation - A First

Course,

Optional Harry Perros, Computer Simulation Technique -

The Definitive Introduction, 2007Chapter 2 and Chapter 3

Goals of Today

Understand the fundamental concepts of

Random Number Generators (RNGs)

Lean how to generate random variate for

standard uniform distributions in the

interval [0,1]

Learn how to generate random variate for

any probability distribution

Outline

Why Random Number Generators Desired Properties of RNGs Lehmer’s Algorithm Period and full-period RNGs Modulus and multiplier selection

(Lehmer) Implementation - overflow

Why Random Number Generators

Random Numbers (RNs) are needed for doing simulations Generate random arrivals (Poisson), random service times

(Exponential) Random numbers must be Independent (unpredictable)

A1 A2Random Inter-Arrival Time: Exp(A)

D1

Random Service Time: Exp()

A3NOW

The Concept

Goal: create random variates for random variables X of a certain distribution defined with its CDF

Approach: Inverse cumulative distribution function

CDF F(x)=u 0 1F x u

Cumulative Distribution Function

Generate a random variable

Determine x such that

0 1F x u

0 1u

1F x u F u x

u1

X1Exp

u2

X1Poisson

X2Exp X2Poisson

Problem Statement of RNGs

Problem: We want to create a function:u = rand();

that produces a floating point number u, where

0<u<1 AND

any value strictly greater than 0 and less than 1 is equally likely to occur (Uniform Distribution in ]0,1[ ) 0.0 and 1.0 are excluded from possible values

Problem Statement of RNGs

This problem can be simply resolved by the following technique: For a large integer m, let the set m = {1, 2, … m-1}

Draw in an integer x m randomly Compute: u = x/m

m should be very large

Our problem reduces to determining how to randomly select an integer in m

Lehmer’s Algorithm

The objective of Lehmer’s Algorithm is to generate a sequence of random integers in m:

x0, x1, x2, … xi, xi+1, …

Main idea: Generate the next xi+1 value based on the last value of random integer xi

xi+1 = g(xi) for some function g(.)

Lehmer’s Algorithm

In Lehmer’s Algorithm, g(.) is defined using two fixed parameters Modulus m : a large, fixed prime integer Multiplier a : a fixed integer in m

Then, choose an initial seed The function g(.) is defined as:

The mod function produces the remainder after dividing the first argument by the second

More precisely: where is the largest integer n such that n≤ x

modg x a x m

0 mx

modu

u v u vv

x

Observations

The mod function ensures a value lower than m is always produced

If the generator produces the value 0, then all subsequent numbers in the sequence will be zero (This is not desired)

Theorem

if (m is prime and initial seed is non-zero)then the generator will never produce the value 0

In this case, the RNG produces values in m = {1, 2, … m-1}

1 modi ix a x m

Observations

The above equation simulates drawing balls from an urn without replacement, where each value in m represents a ball The requirement of randomness is violated because successive

draws are not independent Practical Fact: The random values can be

approximately considered as independent if the number of generated random variates (ball draws) is << m

The Quality of the random number generator is dependent on good choices for a and m

1 modi ix a x m

The Period of a Sequence Consider sequence produced by:

Once a value is repeated, all the sequence is then repeated itself Sequence: where p is the period: number of elements before

the first repeat clearly p ≤ m-1

It can be shown, that if we pick any initial seed x0, we are guaranteed this initial seed will reappear

1 modi ix a x m

0 1, ,..., ,...,i i px x x x i i px x

Full Period Sequences

[LP] Discrete-Event Simulation: A First Course by L. M. Leemis and S. K Park, Prentice Hall, 2006, page. 42 Theorem 2.1.2

If and the sequence is produced by

the Lehmer generator where m is prime,

then there is a positive integer p with such that:

are all different

and

In addition,

0 mx 0 1 2, , ,...x x x

1 modi ix a x m

1p m

0 1 1, ,..., px x x

, 0,1,2,...i i px x i

1 mod 0m p

Full Period Sequences

Ideally, the generator cycles through all values in m to maximize the number of possible values that are generated, and guarantee any number can be produced

The sequence containing all possible numbers is called a full-period sequence (p = m-1)

Non-full period sequences effectively partition m into disjoint sets, each set has a particular period.

Modulus and Multiplier Selection Criteria

Selection Criteria 1: m to be “as large as possible” m = 2i - 1 where i is the machine precision (is the largest

possible positive integer on a “two’s complement” machine) Recall m must be prime It happens that 231-1 is prime (for a 32 bit machine) Unfortunately, 215-1 and 263-1 are not prime ;-(

Selection Criteria 2: p gives full-period sequence (p = m-1) For a given prime number m, select multiplier a that provide a

full period Algorithm to test if a is a full-period multiplier (m must be

prime):

Modulus and Multiplier Selection Criteria

Criterias m to be “as large as possible” p gives full-period sequence (p = m-1)

p = 1;x = a; // assume, initial seed is x0=1, thus x1=awhile (x != 1) { // cycle through numbers until repeat p++; x = (a * x) % m; // careful: overflow possible}if (p == m-1) // a is a full period multiplierelse // a is not a full period multiplier

Algorithm for finding if p is full-period multiplier

Other Useful Properties

Theorem 2.1.1[LP]: If the sequence x0, x1, x2, … is produced by a Lehmer generator with multiplier a and modulus m, then

Note this is not a good way to compute xi!

Theorem 2.1.4[LP, p. 45]: If a is any full-period multiplier relative to the prime modulus m, then each of the integers

is also a full period multiplier relative to m if and only if the integer i has no prime factors in common with the prime factors of m-1 (i.e., i and m-1 are relatively prime, or co-prime)

0 mod 0,1,2,...iix a x m i

mod 0,1,2,..., 1ima m i m

Other Useful Properties

// Given prime modulus m and any full period multiplier a, // generate all full period multipliers relative to mi = 1;x = a; // assume, initial seed is 1while (x != 1) { // cycle through numbers until repeat if (gcd(i,m-1)==1) // x=aimod m is full period multiplier i++; x = (a * x) % m; // careful: overflow possible}

Generate all full-period multipliers

Implementation Issues Assume we have a 32-bit machine, m=231-1 Problem

Must compute Obvious computation is to compute first,

then do mod operation The multiplication might overflow, especially if m-1 is

large!

First Solution: Floating point solution Could do arithmetic in double precision floating

point if multiplier is small enough Double has 53-bits precision in IEEE floating point May have trouble porting to other machines Integer arithmetic faster than floating point

moda x m a x

Implementation Issues: Mathematical Solutions

Problem: Compute without overflow

General Idea: Perform mod operation first, before multiplication.

Suppose that (not prime) We have: Thus, No

overflow

moda x m

m a q mod mod moda x m a x a q a x q

mod 1x q q

mod 1a x q a q a q m

For the case, m is prime, so let q = quotient; r = remainder

Let

It can be shown that (Page 59, Lemis/Park Textbook)

and

m a q r

modx

x a x q rq

x a xx

q m

moda x m x m x

Implementation Issues: Mathematical Solutions

0 if

1 if -

m

m

xx

x

Random number variantsDiscrete random variablesContinuous random variables

Next

The Concept

Goal: create random variates for random variables X of a certain distribution defined with its CDF

Approach: Inverse cumulative distribution function

CDF F(x)=u 0 1F x u

Cumulative Distribution Function

Generate a random variable

Determine x such that

0 1F x u

0 1u

1F x u F u x

u1

X1Exp

u2

X1Poisson

X2Exp X2Poisson

Generating Discrete Random Variates

Cumulative Distribution Function of X:

F(x) = P(X≤x)

PDF

x

F(x)

1.0

1 2 3 4 5

0.8

0.6

0.4

0.2

x f(x)

1 0.2

2 0.3

3 0.1

4 0.2

5 0.2u = F(x)0.5 = F(2)

x

u

Inverse Distribution Function (idf) of X:F*(u) = min {x: u < F(x)}

x

F(x)

1.0

1 2 3 4 5

0.8

0.6

0.4

0.2

x = F*(u)2 = F(0.5)

x

u

Random variate generation:1. Select u, uniformly distributed (0,1)2. Compute F*(u); result is random variate withdistribution f()

f(x=2)

Discrete Random Variate Generation

Uniform (a,b): equally likely to select an integer in interval [a,b]

;

if 1 return 0;

else return 1;

u random

u p

Bernoulli(p)Bernoulli(p): Return 1 with probability

p, Return 0 with probability

1-p

Geometric(p): Number of Bernoulli

trials until first ‘0’)

;

return log 1.0 log ;

u random

u p

1kf x p p

;

return 1 ;

u random

a u b a

Exponential Random Variates

Exponential distribution with mean

;

return - log 1 ;

u random

u