= Solution
Put $K=N+s$, the total capacity. The customer count is an <M-M-s-K queue>, hence a <birth-death process> on $\{0,1,\ldots,K\}$. Its nonzero transition rates are
$$
q_{n,n+1}=\lambda
\quad(0\leq n<K),
\qquad
q_{n,n-1}=\mu\min(n,s)
\quad(1\leq n\leq K).
$$
The first rate is zero at $K$ because an arrival finding the shop full is rejected. The second rate is the number of busy servers times the individual service rate.
Write $\rho=\lambda/\mu$. The <detailed balance for a birth-death process> recursion is
$$
\frac{\pi_n}{\pi_{n-1}}
=\frac{\lambda}{\mu\min(n,s)}.
$$
Therefore
$$
\boxed{\begin{gathered}
\pi_n=\frac{w_n}{Z},
\qquad
w_n=
\begin{cases}
\rho^n/n!,&0\leq n\leq s,\\
\rho^n/(s!s^{n-s}),&s<n\leq K,
\end{cases}
\end{gathered}}
$$
where
$$
Z=\sum_{n=0}^{s}\frac{\rho^n}{n!}
+\frac{\rho^s}{s!}\sum_{j=1}^{N}\left(\frac\rho s\right)^j.
$$
This is the <stationary distribution of an M-M-s-K queue>.
The state space is finite, and every state communicates with every other through successive births and deaths because $\lambda,\mu>0$. The chain is therefore an <irreducible Markov chain>, so its invariant distribution is unique. Finally the adjacent-state identities
$$
\pi_n\lambda
=\pi_{n+1}\mu\min(n+1,s)
$$
are detailed balance, while all nonadjacent rates vanish in both directions. Part a now shows that the queue is reversible in equilibrium.
Back to article page
