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The larger a transportation problem (that is, as the problem has more rows and more columns), the smaller the fraction of all possible routes that will be filled in a solved problem. Explain.
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Operations Management
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The number of filled cells in a solved transportation problem is rows plus columns minus one. For a problem with three origins and four destinations, the number of filled cells will be no more than 3 + 4 - 1 = 6, which is 6/12 or one-half of all 3 ∗ 4 = 12 possible routes. If a problem has six origins and eight destinations, the number of filled cells will be no more than 6 + 8 - 1 = 13, which is 13/48 of all 6 ∗ 8 = 48 possible routes. There are four times as many routes, but the number of filled cells barely doubles. In general, (R + C - 1) / R ∗ C falls as R and/or C rise.
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