Past exam of the mathematics course of the University of Cambridge 2009 ib Paper 2 9H Solution Created 2026-09-24 Updated 2026-10-14
Label the towns from left to right, and the treatment plants as upper-left, upper-right, lower-left and lower-right. Model each treatment capacity as a sink arc to the sea. Incoming untreated sewage can be forwarded, so the numbers inside the plant circles constrain treatment, not total throughput.
The sum of treatment capacities is , so no feasible flow network can handle more than 62. Here is a feasible routing attaining that upper bound. Send town inflowsEvery town-to-plant flow respects its directed pipe capacity. Send an additional five units from to through their bidirectional pipe, whose capacity is twelve, and two from to through their bidirectional pipe, whose capacity is fourteen. Use neither horizontal plant-to-plant pipe.
Now receives and treats all of it. Plant receives , forwards five and treats fourteen. Plant receives and treats all of it. Plant receives , forwards two and treats twelve. Thus flow conservation holds at every plant and all treatment capacities are saturated. The routing proves the upper bound is achievable:This is one optimal assignment, not a uniqueness assertion. The total bound comes from the cut consisting of all treatment-to-sea arcs; the explicit routing supplies a matching maximum flow certificate.