i would need some help for the following problem.

Consider the $M/E_r/1$ queue with arrival rate $\lambda$ and mean service time $r/\mu$. Let $P(z)$ be the generating function of the probabilities $p_n$, so

$$P(z)=\sum_{n=0}^{\infty} p_{n}z^{n},|z|\leq 1$$ with normalization equation $$\sum_{n=0}^{\infty} p_n=1$$

We are given that $P(z)=\frac{p_0\mu}{\mu-\lambda(z+\ldots+z^{r-1}+z^{r})}$ (*)

(iii) Show, by partial fraction decomposition of $P(z)$, that the probabilities $p_n$ can be written as $$p_n= \sum_{k=1}^{r} c_k (\frac{1}{z_k})^n, n=0,1,2,...$$ where $z_1,...,z_r$ are the zeroes of the denominator in (*).

This is where i have got so far but still cannot relate how i can get $p_n$ eventually:

$P(z)=\frac{p_0}{1-\frac{\lambda}{\mu}(z+\cdots+z^{r})}$, we let $c_k=\frac{p_0}{1-\frac{\lambda}{\mu}}$ and rewriting $\frac{1}{z+\cdots+z^r}=\frac{1}{(z-z_1)\cdots(z-z_k)}$

we have that \begin{align*} P(z) &=\sum_{k=1}^{r} c_k\frac{1}{(z-z_k)} &=\frac{c_1}{z-z_1}+...+\frac{c_k}{z-z_k}=\frac{1}{(z-z_1)...(z-z_k)} \end{align*}

Hence we get \begin{align*} c_1\prod\limits_{j\neq1}^k(z-z_j)+\cdots+c_k\prod\limits_{j\neq k}^k(z-z_j)=1 \end{align*}

For $z=z_1$, $c_1\prod\limits_{j\neq1}^k(z_1-z_j)=1$ so $c_1=\frac{1}{\prod\limits_{j\neq1}^k(z_1-z_j)}$, iterating n times

For $z=z_k$, $c_k\prod\limits_{j\neq1}^n(z_k-z_j)=1$ so $c_k=\frac{1}{\prod\limits_{j\neq1}^n(z_k-z_j)}$

Thanks in advance


We use the Ansatz (after arguing that all zeros have multiplicity one) \begin{align*} P(z)&=\sum_{k=1}^r\frac{\alpha_k}{(z-z_k)}\\ &=\sum_{k=1}^r\left(-\frac{\alpha_k}{z_k}\right)\frac{1}{1-\frac{z}{z_k}}\\ &=\sum_{k=1}^r\left(-\frac{\alpha_k}{z_k}\right)\sum_{n=0}^\infty\left(\frac{z}{z_k}\right)^n\\ &=\sum_{n=0}^\infty\sum_{k=1}^r\left(-\frac{\alpha_k}{z_k}\right)\left(\frac{1}{z_k}\right)^nz^n\tag{1} \end{align*} and conclude from (1) since \begin{align*} P(z)=\sum_{n=0}^\infty p_nz^n\qquad\qquad|z|\leq 1 \end{align*} that $p_n$ can be represented as \begin{align*} p_n=\sum_{k=1}^rc_k\left(\frac{1}{z_k}\right)^n\qquad\qquad n\geq 0 \end{align*} with $c_k=-\frac{\alpha_k}{z_k}, 1\leq k\leq r$.

  • $\begingroup$ @user304663: You're welcome! :-) $\endgroup$ – Markus Scheuer Mar 19 '17 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.