Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 5 / f / Solution

Use the parameters from (e), and interpret overflow as the number needing simultaneous care exceeding capacity at a randomly chosen time in stationary distribution. The occupancy of the immigration--death process is not fixed at its mean. Detailed balance givesNormalizing yields ${\pi }_0=e^{-25}$, so stationary occupancy has a Poisson distribution with mean 25. Thus integer capacity $K$ must satisfyThe adjacent tails areSince the tail decreases with $K$, the minimum capacity is $\boxed{32\text{animals}}$. The small excess above $0.1$ at 31 matters; a normal approximation rounded too early can choose the wrong integer.

This is a stationary occupancy criterion. If the event is instead an offered arrival finding at least $K$ animals already present in this unrestricted process, Poisson arrivals see time averages gives $P(X\ge K)$, whose minimum qualifying capacity is 33. That uses a different event definition. From an initially empty centre, occupancy at a finite time is Poisson distributed with mean $N(t)\le 25$, so the stationary criterion also bounds the fixed-time overflow probability during initial filling. A probability of ever exceeding capacity over an unlimited future is a different question: this irreducible recurrent process will eventually exceed every finite capacity with probability one. If instead arrivals are refused when all places are occupied, that changes the model to an loss network, whose blocking probability is not the unrestricted occupancy tail calculated here.