To compute the feasible flow of minimum cost in minimum-cost multi-commodity-flow problem and express the problem as a linear program.

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Chapter 29.2, Problem 7E

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To compute the feasible flow of minimum cost in minimum-cost multi-commodity-flow problem and express the problem as a linear program.

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rate(F, C) = cap(C): To prove this, it is sufficient to prove that everyedge u, v crossing from U to V has flow in F at capacity (Fu,v = cu,v) andevery edge v, u crossing back from V to U has zero flow in F. These give thatrate(F, C) = u∈Uv∈V [Fu,v − Fv,u] = u∈Uv∈V [cu,v − 0] = cap(C).

QUESTION 9 We are given a connected graph G with costs ci, ) on edges. Assume all costs are positive integers and that there are no ties (ie. no two edges have the sanvo cost costs are squarod (meaning that edge () now costs c'. ) = (c(, ))) the Minimum Spanning Tree (MST) remains the same tree. O True O False QUESTION 10 e are given a connected graph G with costs c(i, ) on edges. Assume all costs are positive integers and that there are no ties (1.e. no two edges have the same ce sts are squared (meaning that edge (3) now costs c'( 1) (ci, ) the Shortest Path Tree rooted at node A (SPT) remains the same tree. True False QUESTION 11

T/F 1).In the aggregate analysis different operations may have different amortized costs,while in the accounting method all operations have the same amortized cost. 2).BFS finds the shortest distance to a node from the starting point in unweighted graphs. 3).Given a graph G. If the edge e is not part of any MST of G, then it must be the maximum weight edge on some cycle in G. 4).Assume that no two men have the same highest-ranking woman. If the women carried out the proposal to men, then the Gale-Shapley algorithm will contain a matching set where every man gets their highest-ranking woman. 5).If a path P is the shortest path from u to v and w is a node on the path, then the part of the path from u to w is also the shortest path from u to w.