 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/7h/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

x_{1} - 2 x_{2} \leq 2, - x_{1} + 2 x_{2} \leq 2.

Introduce slacks

s_{1}, s_{2}, s_{3}

x_{2}

enter and

s_{2}

leave, then let

x_{1}

enter and

s_{3}

leave. The final dictionary is

x_{1} = \frac{2}{3} + \frac{1}{9} s_{2} - \frac{2}{9} s_{3}, x_{2} = \frac{4}{3} - \frac{4}{9} s_{2} - \frac{1}{9} s_{3},

x_{1} + x_{2} = 2 - \frac{1}{3} s_{2} - \frac{1}{3} s_{3} .

Thus

(x_{1}, x_{2}) = (\frac{2}{3}, \frac{4}{3}), max (x_{1} + x_{2}) = 2 .

For sufficiently small perturbations, the same two constraints remain active:

- x_{1} + 2 x_{2} = 2 + ϵ_{1}, 4 x_{1} + x_{2} = 4 + ϵ_{2} .

Solving gives

x_{1} = \frac{6 + 2 ϵ _{2} - ϵ _{1}}{9}, x_{2} = \frac{12 + ϵ _{2} + 4 ϵ _{1}}{9},

and therefore the perturbed optimal value is

2 + \frac{ϵ _{1} + ϵ _{2}}{3} .

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