Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-3/15h/solution

A balanced transportation problem chooses nonnegative shipments minimizing , subject to and , where the total supply equals total demand. For multipliers , its Lagrangian is
Taking its infimum over produces the linear programming duality problem of maximizing subject to . These are the transportation dual potentials.
The transportation simplex algorithm starts with a feasible transportation spanning tree of basic entries, obtainable by filling the north-west available cell with the smaller remaining supply or demand. Set on basic cells; this determines all potentials up to adding a constant to all and subtracting it from all . If all nonbasic reduced costs are nonnegative, feasibility and the dual bound prove optimality. Otherwise let a negative-cost cell enter. It closes a unique cycle in the basic tree. Alternate additions and subtractions around that cycle and choose the largest amount preserving nonnegativity, namely the minimum shipment on the subtraction cells. A cell that reaches zero leaves the basis. The cost changes by the entering reduced cost times the pivot amount. Zero basic cells and an anti-cycling rule can handle degeneracy; all pivots below are nondegenerate.
The north-west initial shipment matrix is
With , its transportation dual potentials are and . The most negative reduced cost is . The cycle is ; the pivot amount is . Hence
Now and . Enter , of reduced cost , on cycle . The amount is , giving
The new potentials are and . Enter , of reduced cost , on cycle . The amount is . Thus
To certify optimality in the original transportation problem, take and . Their entire reduced-cost matrix is
They satisfy dual feasibility, and their value is , equal to the shipment cost. The minimum transportation cost is therefore , with no unproved optimality assertion from the iterations alone.

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