= Solution
Immediately after departure $n$, either $Q_n>0$ and one of those customers starts the next service, or $Q_n=0$ and the next arrival both initiates service and is not counted among those waiting after it. During that service $A_n$ further customers arrive. Thus the <embedded departure chain of an M-G-1 queue> obeys
$$
\boxed{Q_{n+1}=A_n+Q_n-h(Q_n),\qquad h(x)=\min\{1,x\}.}
$$
Conditional on the next service time $S_{n+1}=s$, the <Poisson process> property gives
$$
A_n\mid S_{n+1}=s\sim\operatorname{Poisson}(\lambda s).
$$
The <law of total expectation> and <law of total variance> therefore give
$$
\boxed{\mathbb EA_n=\lambda\mathbb ES=\rho,}
$$
and
$$
\boxed{\mathbb EA_n^2
=\mathbb E[\lambda S+(\lambda S)^2]
=\rho+\lambda^2\mathbb ES^2.}
$$
In equilibrium let $Q$ have the common distribution of $Q_n,Q_{n+1}$ and put $H=\mathbf1_{\{Q>0\}}=h(Q)$. The fresh pair $(A_n,S_{n+1})$ is independent of $Q_n$. Taking expectations in the recursion gives
$$
0=\rho-\mathbb EH,
\qquad\text{so}\qquad
\mathbb P(Q>0)=\rho.
$$
Because $QH=Q$ and $H^2=H$, stationarity of the second moment gives
$$
\begin{aligned}
0
&=\mathbb E[(Q-H+A)^2-Q^2]\\
&=-2\mathbb EQ+\rho
+2\rho(\mathbb EQ-\rho)
+\rho+\lambda^2\mathbb ES^2.
\end{aligned}
$$
Solving,
$$
\boxed{\mathbb EQ
=\rho+\frac{\lambda^2\mathbb ES^2}{2(1-\rho)}.}
$$
In stationarity, level crossing together with <Poisson arrivals see time averages> identifies this as the mean number $L$ in the shop. The <Little law> gives $L=\lambda(W_q+\mathbb ES)$, where $W_q$ is waiting time before service. Since $\rho=\lambda\mathbb ES$,
$$
\boxed{W_q=\frac{\lambda\mathbb ES^2}{2(1-\rho)}.}
$$
This is the <Pollaczek-Khinchine mean formula>.
Back to article page
