Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/18h/d/solution

With integer supplies and demands, every northwest-corner allocation is integer. At each pivot, is the minimum of finitely many current allocations on the minus cells, so it remains integer; adding and subtracting it preserves integrality. The algorithm therefore reaches an integer-valued optimal solution.

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