Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 339 1 a Solution Created 2026-10-03 Updated 2026-10-05
A feasible point lies in the probability simplex, so is a convex combination of the coordinates of . Hence . Taking a coordinate vector with attains equality. ThereforeMore generally, all optimal solutions are exactly the probability vectors supported on the maximizing indices: equality requires for every . Thus ties give an entire optimal face, while a unique maximizing index gives a unique optimizer.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 339 1 b iii Solution Created 2026-10-03 Updated 2026-10-05
For every decision with finite worst-case value, the previous dual represents that value as a maximum over . Maximizing it jointly with yields the robust linear optimization over the probability simplex formulation:Indeed every feasible triple gives a dual lower bound on that decision's worst-case revenue, while dual attainment supplies a triple reaching it. This argument only combines two maximizations after dualization; it does not interchange the original max and min.
If the probability simplex intersects , this linear program is feasible and bounded above by , and hence attains an optimum. Its maximizing is a robust optimizer. If the intersection is empty, every decision has worst-case value and the linear program is infeasible, consistently with the extended-value convention. Full column rank of ensures every simplex decision has a finite inner optimum, but is not needed for the qualified equivalence.
Worst-case linear revenue over a polyhedral uncertainty set is optimized by the linear programHere is the probability simplex. Finite decisions are exactly its intersection with . If this intersection is empty, all decisions have worst-case value and the displayed linear program is infeasible; its supremum over the empty feasible set is also . Full column rank of suffices to make all simplex decisions finite.