Help on cancelling out or substituting for n in the equation for P0 of an M/M/c system

Question

I am currently trying to derive the queueing equations for an M/M/c queueing system. The method I am using is to consider to take the Markov chain and use the rate in = rate out principle to first find P0 and Pn. I have found the equations for the two different cases of Pn, but for P0 and all other performance measures I keep getting the equation for a general term in terms of n (the state of the system). The method I am using to find P0 is to express all probabilities in terms of P0, equate the sum of all probabilities to 1, take P0 as a common factor and split the remaining sum into two geometric sequences. Any ideas on what I am doing wrong?

Answer 1

Let $X(t)$ be the number of customers in the system at time $n$. Then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on $\mathbb N\cup \{0\}$ with transition rates $$ q_{ij} = \begin{cases} \lambda,& j=i+1\\ j\mu,& j=i-1\text{ and } j<c\\ c\mu,& j=i-1\text{ and } j\geqslant c\\ 0,& \text{otherwise}. \end{cases} $$ Since $X(t)$ is a birth-death process, assuming that $\lambda<c\mu$ we have $$ \sum_{n=1}^\infty \prod_{i=1}^n\frac{q_{i-1,i}}{q_{i,i-1}} = \sum_{n=1}^{c-1}\prod_{i=1}^n\frac\lambda{i\mu} + \sum_{n=c}^\infty\left(\prod_{i=1}^{c-1}\frac\lambda{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{n-c+1}<\infty, $$ so the process is positive recurrent and there exists a unique stationary distribution $\pi$ such that $\mathbb P(X(t)=j) = \pi(j)$ when $X(0)\sim\pi$. Since $q_{ij}>0$ only when $|i-j|=1$, there is rate conservation between each pair of neighboring states, yielding the detailed balance equations \begin{align} \lambda\pi_{j-1} &= j\mu\pi_j,\quad 1\leqslant j\leqslant c-1\\ \lambda\pi_{j-1} &= c\mu\pi_j,\quad j\geqslant c. \end{align} By recursion this gives $$ \pi_j = \prod_{i=1}^j \frac{\lambda}{i\mu}\pi_0,\quad 1\leqslant j\leqslant c-1 $$ and $$ \pi_j = \left(\prod_{i=1}^{c-1} \frac{\lambda}{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{j-c+1}\pi_0,\quad j\geqslant c. $$ Since $\sum_{j=0}^\infty \pi_0=1$, we have $$ \pi_0 = \left[ \sum_{i=1}^{c-1}\prod_{i=1}^j \frac{\lambda}{i\mu} + \sum_{n=c}^\infty\left(\prod_{i=1}^{c-1} \frac{\lambda}{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{n-c+1}\right ]^{-1} $$ and hence $$ \small \pi_j = \begin{cases} \sum_{i=1}^{c-1}\prod_{i=1}^j \frac{\lambda}{i\mu}\left[ \sum_{i=1}^{c-1}\prod_{i=1}^j \frac{\lambda}{i\mu} + \sum_{n=c}^\infty\left(\prod_{i=1}^{c-1} \frac{\lambda}{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{n-c+1}\right ]^{-1},& 0\leqslant j\leqslant c-1\\ \left(\prod_{i=1}^{c-1} \frac{\lambda}{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{j-c+1}\left[ \sum_{i=1}^{c-1}\prod_{i=1}^j \frac{\lambda}{i\mu} + \sum_{n=c}^\infty\left(\prod_{i=1}^{c-1} \frac{\lambda}{i\mu}\right)\left(\frac\lambda{c\mu} \right)^{n-c+1}\right ]^{-1},& j\geqslant c. \end{cases} $$