CS 6998 - 3 : Solutions to Problem Set # 1 Etienne
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چکیده
Proof. We check that the flow S routing all supply through the first edge is both optimal and Nash. Since S sends no supply through the second edge, S is optimal if the marginal cost of the first edge is at most that of the second: a ≤ (i+ 1)−1 a(i+ 1) ≤ 1 [ a(i+ 1)x ] (1) ≤ 1 d dx ( xax ) (1) ≤ d dx (x) (0) c1(1) ≤ c2(0), as required. Thus S is optimal. To check S is Nash, we check that the latency of the first edge is at most the latency of the second:
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تاریخ انتشار 2008