Past exam of the mathematics course of the University of Cambridge 2019 ib Paper 4 20H c Solution Created 2026-09-24 Updated 2026-09-29
Assume the capacities are integers and begin with the zero flow. Inductively, if every edge flow is an integer, then every forward residual capacity and reverse residual capacity is an integer. The bottleneck on an augmenting path is therefore a positive integer. Augmenting by it changes every affected edge flow by an integer, so all edge flows remain integral.
Each augmentation raises the flow value by at least one. On the other hand, every feasible flow satisfiesand the right side is a finite integer because the network is finite. There can therefore be only finitely many augmentations. When the algorithm stops, no augmenting path remains, and the source-reachable cut in the residual network has capacity equal to the current flow value. The max-flow min-cut theorem proves that the terminating flow is maximum. This proves both termination and the Integrality of the Ford-Fulkerson algorithm.