Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 27J / 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

Conditional on the next service time $S_{n+1}=s$, the Poisson process property givesThe law of total expectation and law of total variance therefore giveand

In equilibrium let $Q$ have the common distribution of $Q_n,Q_{n+1}$ and put $H=1_{\{Q>0\}}=h(Q)$. The fresh pair $(A_n,S_{n+1})$ is independent of $Q_n$. Taking expectations in the recursion givesBecause $QH=Q$ and $H^2=H$, stationarity of the second moment givesSolving,

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 $L=\lambda (W_q+\mathbb{E}S)$, where $W_q$ is waiting time before service. Since $\rho =\lambda \mathbb{E}S$,This is the Pollaczek-Khinchine mean formula.