 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/27k/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

A dispatch cycle contains

N

interarrival times and has mean length

N / μ

. The expected total passenger waiting time in one cycle is

\frac{1}{μ} (1 + 2 + \dots + (N - 1)) = \frac{N ( N - 1 )}{2 μ} .

The renewal-reward average cost rate is therefore

C (N) = \frac{K + c N ( N - 1 ) / ( 2 μ )}{N / μ} = \frac{μ K}{N} + \frac{c}{2} (N - 1) .

Thus the continuous optimum is

2 μ K / c

, and the integer optimum is the positive integer minimizing

C (N)

; equivalently it is the smallest

N

with

N (N + 1) \geq \frac{2 μ K}{c},

with both adjacent values optimal in the equality case.

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