GE330_lect15 - Lecture 15 Max-Flow Min-Cut Maximal Flow...

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Unformatted text preview: Lecture 15 Max-Flow - Min-Cut Maximal Flow Algorithm March 18, 2009 Maximum Flow Problem: Example Given a network G = ( N, A ) maximize X { j :( s,j ) ∈ A } x sj subject to X { j :( i,j ) ∈ A } x ij- X { j :( j,i ) ∈ A } x ji = 0 for all i 6 = s ≤ x ij ≤ c ij for all ( i, j ) ∈ A. NOTE: if some links are bidirectional, we label capacities in both directions. 1 Max-Flow Algorithm This is an iterative method for finding the maximum flow • At each iteration, the algorithm is searching for a path from the source node to the sink node along which it can send a positive flow • The algorithm terminates when such a path cannot be found • The path is referred to as “breakthrough” path • After a breakthrough, the amount of flow sent along a path is re- moved from the link capacities, the resulting capacities are “residual” (correspondingly, the resulting network is referred to as residual network) • Typical such iteration has the following steps: • Start with the source node • Among one-hop neigboring nodes j [link ( s, j ) ∈ A ] find the node with the largest “residual capacity” • Label that node by the ammount of flow and s . • Repeat the process from the labeled node until a breakthrough • If no breakthrough, then “backtrack” 2 Example: Directed Links Iteration 1: 3 Iteration 2: Iteration 3: 4 Iteration 4: The maximum flow is the total flow sent: f 1 + f 2 + f 3 = 5 + 5 + 5 = 15 . 5 Example: Double Capacity Labels Iterations 1 and 2: 6 Iteration 3: Iteration 4: 7 Iteration 5: Iteration 6: The maximum flow is the total flow sent in the iterations: f 1 + f 2 + f 3 + f 4 + f 5 = 20 + 10 + 10 + 10 + 10 = 60 ....
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This note was uploaded on 08/31/2010 for the course IESE GE 330 taught by Professor Nedich during the Spring '09 term at University of Illinois at Urbana–Champaign.

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GE330_lect15 - Lecture 15 Max-Flow Min-Cut Maximal Flow...

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