Queuing Systems I: Theory

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Exercise. This is the solution to exercise 2.1 in the book.

Solution. Let’s consider the case when K = 2 and Pk(t) is the probability that there are k arrivals in the pooled arrival stram in interval (0,t). Since the sources are independent we have:

Pk(t) = l=0kP l(1)(t)P kl(2)(t)

Since each source is Poissonian:

Pk(t) = l=0k(λ1t)l l! eλ1t(λ2t)kl (k l)! eλ2t

Pk(t) = e(λ1+λ2)t 1 k! l=0k k! l!(k l)!(λ1t)l(λ 2t)kl

But the inner sum is a binomial sum:

l=0k k! l!(k l)!(λ1t)l(λ 2t)kl = (λ 1t + λ2t)k

or

Pk(t) = [(λ1 + λ2)t]k k! e(λ1+λ2)t

Hence the arrival stream pooled together from 2 independent Poissonian sources is also Poissonian. Iteratively one can show that pooled streams 1,2,3 are also Poissonian, up to streams 1,,K.

As shown above for K = 2 the Poisson parameter for the pool of sources is the sum of parameters for each source:

λ = λ1 + λ2 + + λK