>Does that work if i were also asked to generate it given mean. What i

>am trying to do is to generate jobs in an exponentially distributed

>fashion given the mean.

>like "The inter-arrival time of jobs is exponentially distributed with

>mean of ...".

>> >I was trying to find out the function in 'C' that would return me the

>> >exponential distribution value, given lambda as the argument. Can you

>> >please help me get to know the same.

>> Let uniform() be your favorite random number generator that returns

>> a random float/double in the range 0.0 < uniform() < 1.0.

>> Then I think -lambda*log(uniform()) will be exponentially distributed.

>> At least that is what some old code of my does.

bellenot 3289> maple

|\^/| Maple 6 (SUN SPARC SOLARIS)

._|\| |/|_. Copyright (c) 2000 by Waterloo Maple Inc.

\ MAPLE / All rights reserved. Maple is a registered trademark of

<____ ____> Waterloo Maple Inc.

| Type ? for help.

Quote:> assume(lambda>0);

> int(exp(-x/lambda),x=0..infinity);

lambda~

Quote:> int(exp(-x/lambda)/lambda,x=0..infinity);

1

Quote:> # So the density is exp(-x/lambda)/lambda -- compute the mean

> int(x*exp(-x/lambda)/lambda,x=0..infinity);

bytes used=1000248, alloc=851812, time=0.16

lambda~

Quote:> # Solve for -lambda*log(uniform())

> solve(P=int(exp(-x/lambda)/lambda,x=0..T),T);

-ln(-P + 1) lambda~

So lambda is the mean in this case.

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