Clients arrive at 1 server according to a Poisson process with intensity $a$ and leave according to a Poisson process with intensity $b$. However if there are already $N$ clients waiting, a new client will also wait with probability $1/(N+1)$ or leave with probability $N/(N+1)$.
- How do I show that the stationary distribution of the number of clients present is Poisson distributed with expectation $a/b$?
- What is the average number of clients that arrive per unit time?
What I know:
Our state space is $\mathbb{N}_0$, the number of clients present.
I thought that a state $i$ can only transition to $i-1,i,i+1$ with intensity $b,-(a+b),a$ resp. for $i<N$. For $i\geq N$ the intensity to $i+1$ is $a/(N=1)$, to $i$ is $-a/(N+1)-b$ and to $i-1$ remains $b$. These are the entries of the transition rate matrix $G$.
The transition rate matrix G is given by
$ g_{ij} = \left\{
\begin{array}{lr}
v_ip_{ij} & : i \neq j\\
-v_i & : i=j
\end{array}
\right.$
where the residence time in state $i$ is distributed exponentially with parameter $v_i$ and $p_{ij}$ are the transition probabilities.
Further the stationary distribution is $P=(P_j)_j$ where $P_j=\lim_{t\rightarrow\infty}p_{ij}(t)$. And $P$ is the solution of $v_jx_j=\sum_{k\neq j}g_{kj}x_k$.
How can I use these to find 1. and 2.?