Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 38 3 a Solution Created 2026-10-03 Updated 2026-10-07
Label the vertices (left), (upper middle), (lower middle), (upper right), (lower right). Use the flow balance convention outgoing minus incoming equals supply. The printed initial flows, in the order , areThey obey all capacity constraints and the flow balances . The four strictly interior edges form a spanning tree. The remaining edges are at a bound: at zero, and at their upper bounds. Thus this is already a feasible network simplex tree basis; no artificial feasibility phase is needed. Its total cost is .
For a tree, choose network dual potentials with and zero network reduced cost on each tree edge. InitiallyThe nonbasic lower-bound edge has network reduced cost . Increase its flow and adjust around the graph cycle , reversing and . The available step is , so leaves at zero. This givesThe tree is now , and the potentials are .
Now is at its upper bound but has network reduced cost , so decreasing it improves cost. Its reverse residual network edge enters along the cycle . The forward edges increase; decrease. The step isThus leaves at its upper bound, andThe new tree has potentials .
The upper-bound edge now has network reduced cost . Enter its reverse direction along , decreasing and and increasing . The step is , so leaves at zero. The resulting flow isThe tree is , with potentials . The nonbasic edges have network reduced costs at its upper bound, at its lower bound, and at its lower bound. All have the correct signs, so there is no improving network simplex pivot. The next part proves these signs certify optimality.
