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

The Waiting times of a Poisson process have an exponential distribution with the arrival rate. Measuring time in days, the survival function is $P(W>w)=e^{-\alpha w}$. Thereforeor about $4.98\%$.

A day's admission count is a Poisson random variable with mean three. “Less than three” includes zero, one and two, soThus the probability is about $42.32\%$.

For an exponential distribution, the survival function at the median equals one half. Thus $e^{-10\delta }=1/2$, andThe corresponding mean care time is $1/\delta =10/\ln 2\approx 14.43$ days; a ten-day median should not be substituted as the mean.

This is an immigration--death process, equivalently an M-M-infinity queue: admission occurs at rate $\alpha$ regardless of occupancy, and each of $n$ animals is discharged independently at rate $\delta$. Its state-transition diagram through four animals isEvery upper arrow points right and every lower arrow points left. Thus $q_{n,n+1}=\alpha$ and $q_{n,n-1}=n\delta$ for $n\ge 1$, with no discharge from zero. Four is only the limit of the requested sketch, not an absorbing state or a capacity boundary.

Admissions add to the expected value at rate $\alpha$, while the expected aggregate discharge rate is $\delta \mathbb{E}[X(t)]$. ConsequentlyMultiplying this linear ordinary differential equation by the integrating factor $e^{\delta t}$ givesIntegrating from zero, using $N(0)=0$, gives $e^{\delta t}N(t)=\alpha (e^{\delta t}-1)/\delta$. HenceIt starts at zero, has initial slope $\alpha$, and approaches $\alpha /\delta$ as $t\to \infty$, as the admission and discharge rates balance.

Setting the mean equation's derivative to zero givesWhen the actual current occupancy is $n=25$, the next admission has exponential distribution rate five and the next discharge has rate $n\delta =5$. The memoryless property makes these independent competing clocks conditional on the current occupancy. ThereforeThe rate comparison uses the actual stated occupancy. Knowing only the mean occupancy would not justify replacing the random $X$ inside $X\delta /(\alpha +X\delta )$ by its mean.

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.