Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 26I (Applied Probability)

The model is a general-service infinite-server queue, usually denoted $M/G/\infty$: arrivals form a Poisson process, each wandering time is a service time with distribution that of $S$, and trapping ends service. Infinitely many servers mean that mosquitoes wander concurrently without waiting or delaying one another. General service times are not necessarily exponential, so the count alone need not be a Markov chain.

An arrival at time $u\in [0,t]$ is still present at $t$ exactly when its independent service time exceeds $t-u$. Thinning the marked Poisson process with this retention probability gives a Poisson count with parameterThusThis reasoning allows atoms in the service-time distribution; the strict survival inequality correctly treats a customer whose service has ended.

The tail integral identity gives ${\int }_0^{\infty }\mathbb{P}(S>v)dv=\mathbb{E}S$. Therefore the limiting and stationary occupancy distribution isIt is finite under the stated finite-mean assumption.

Only arrivals at times $u\le t$ remain after the window closes. Such an arrival survives until $t+v$ exactly when its service time exceeds $t+v-u$. The same independent Poisson thinning givesAt $v=0$ this agrees with part (b), while for large $v$ the parameter tends to zero.

After closing, the count can only decrease. Thus $V(t)>v$ is the event that at least one mosquito remains at $t+v$. The zero probability of a Poisson count yields the clearance time after closing an infinite-server queue:The clearance time has an atom at zero of mass $e^{-m_t}$, corresponding to an empty room when the window closes.

For equilibrium, let $t\to \infty$ in part (e). Monotone convergence of the tail integral and the finite mean giveand consequentlyThis limiting argument uses the equilibrium age and residual-service distribution produced by the queue, not just a fresh independent copy of $S$ for each mosquito present. Replacing their residual service by fresh service would generally be wrong unless the service law is memoryless.