# 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 ${P}_{k}\left(t\right)$ is the probability that there are $k$ arrivals in the pooled arrival stram in interval $\left(0,t\right)$. Since the sources are independent we have:

${P}_{k}\left(t\right)=\sum _{l=0}^{k}{P}_{l}^{\left(1\right)}\left(t\right){P}_{k-l}^{\left(2\right)}\left(t\right)$

Since each source is Poissonian:

${P}_{k}\left(t\right)=\sum _{l=0}^{k}\frac{{\left({\lambda }_{1}t\right)}^{l}}{l!}{e}^{-{\lambda }_{1}t}\frac{{\left({\lambda }_{2}t\right)}^{k-l}}{\left(k-l\right)!}{e}^{-{\lambda }_{2}t}$

${P}_{k}\left(t\right)={e}^{-\left({\lambda }_{1}+{\lambda }_{2}\right)t}\frac{1}{k!}\sum _{l=0}^{k}\frac{k!}{l!\left(k-l\right)!}{\left({\lambda }_{1}t\right)}^{l}{\left({\lambda }_{2}t\right)}^{k-l}$

But the inner sum is a binomial sum:

$\sum _{l=0}^{k}\frac{k!}{l!\left(k-l\right)!}{\left({\lambda }_{1}t\right)}^{l}{\left({\lambda }_{2}t\right)}^{k-l}={\left({\lambda }_{1}t+{\lambda }_{2}t\right)}^{k}$

or

${P}_{k}\left(t\right)=\frac{{\left[\left({\lambda }_{1}+{\lambda }_{2}\right)t\right]}^{k}}{k!}{e}^{-\left({\lambda }_{1}+{\lambda }_{2}\right)t}$

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 $1,\dots ,K$.

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

$\lambda ={\lambda }_{1}+{\lambda }_{2}+\dots +{\lambda }_{K}$