 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-3/25i/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

(λ + α) π_{0 C} = β π_{0 W}

and

β π_{0 W} = α π_{0 C} + μ π_{1 W}

. At every

i \geq 1

they are

(λ + α) π_{i C} = λ π_{i - 1, C} + β π_{iW}, (μ + β) π_{iW} = α π_{i C} + μ π_{i + 1, W} .

The zero-level equations give

π_{0 C} = β π_{0 W} / (λ + α)

and

π_{1 W} = β λ π_{0 W} / [μ (λ + α)]

. Substituting the latter in the level-one

C

(π_{1 C}, π_{1 W}) = \frac{β π _{0 W}}{λ + α} (\frac{λ ( μ + β )}{μ ( λ + α )}, \frac{λ}{μ}) .

Solve the

W

π_{i + 1, W}

, then substitute it into the next

C

equation. This yields the required row-vector recurrence with

B = (\frac{λ μ - β α}{μ ( λ + α )} \frac{β ( β + μ )}{μ ( λ + α )} - \frac{α}{μ} \frac{β + μ}{μ}) .

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