Individuals arrive in a shop in the manner of a Poisson process with intensity $\lambda$, and they await service in the order of their arrival. Their service times are independent, identically distributed random variables $S_{1}, S_{2}, \ldots$. For $n \geqslant 1$, let $Q_{n}$ be the number remaining in the shop immediately after the $n$th departure. Show that

$$Q_{n+1}=A_{n}+Q_{n}-h\left(Q_{n}\right)$$

where $A_{n}$ is the number of arrivals during the $(n+1)$ th service period, and $h(x)=$ $\min \{1, x\}$.

Show that

$$\mathbb{E}\left(A_{n}\right)=\rho, \quad \mathbb{E}\left(A_{n}^{2}\right)=\rho+\lambda^{2} \mathbb{E}\left(S^{2}\right),$$

where $S$ is a typical service period, and $\rho=\lambda \mathbb{E}(S)$ is the traffic intensity of the queue.

Suppose $\rho<1$, and the queue is in equilibrium in the sense that $Q_{n}$ and $Q_{n+1}$ have the same distribution for all $n$. Express $\mathbb{E}\left(Q_{n}\right)$ in terms of $\lambda, \mathbb{E}(S), \mathbb{E}\left(S^{2}\right)$. Deduce that the mean waiting time (prior to service) of a typical individual is $\frac{1}{2} \lambda \mathbb{E}\left(S^{2}\right) /(1-\rho)$.