Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 27K (Applied Probability)

An M-M-1 queue has Poisson process arrivals of rate $\lambda$, independent exponential service times of rate $\mu$, and one server. The two letters $M$ indicate the memoryless property of the interarrival and service-time distributions. Its queue length is a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. When $\rho =\lambda /\mu <1$, its normalized stationary distribution isthe powers ${\rho }^n$ quoted in the question form the corresponding unnormalized invariant measure.

Let $N$ be the number of visits to the server made by one customer. Then $N$ has the geometric distribution with success parameter $\delta$, and its service periods $E_1,E_2,\dots$ are independent exponential random variables of rate $\mu$. For $s\ge 0$, the Laplace transform of the total service time $T=E_1+\cdots +E_N$ isThus

At each service completion, a successful departure reduces the queue length by one and occurs at rate $\delta \mu$ while the queue is nonempty; feedback completions leave its length unchanged. The queue-length process is therefore the birth-death process of an ordinary $M/M/1$ queue with service rate $\delta \mu$. It is positive recurrent exactly when $\lambda <\delta \mu$, and then, with $\rho =\lambda /(\delta \mu )$,

The Poisson output theorem for a feedback queue, equivalently the Burke theorem applied to the effective queue-length process with service rate $\delta \mu$, shows that successful departures in equilibrium form a Poisson process of rate $\lambda$.

The process of all entries into the queue is different. Its long-run rate isImmediately after any queue-entry event the server is busy, so the conditional probability of another entry in the next interval of length $h$ isthe next entry can be external or can be a service completion followed by feedback. A Poisson process of the displayed long-run rate would instead give $(\lambda /\delta )h+o(h)$. These coefficients are unequal when $\lambda <\delta \mu$, sinceHence the combined arrival process is not Poissonian.