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# Set6 - Introduction to Maximum Flows The Max Flow Problem G...

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Introduction to Maximum Flows

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2 The Max Flow Problem G = (N,A) x ij = flow on arc (i,j) u ij = capacity of flow in arc (i,j) s = source node t = sink node Maximize v Subject to Σ j x ij - Σ k x ki = 0 for each i s,t Σ j x sj = v 0 x ij u ij for all (i,j) A.
3 Maximum Flows We refer to a flow x as maximum if it is feasible and maximizes v. Our objective in the max flow problem is to find a maximum flow. s 1 2 t 10 , 8 8, 7 1, 1 10, 6 6, 5 A max flow problem. Capacities and a non- optimum flow.

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4 The feasibility problem: find a feasible flow retailers 1 2 3 4 5 6 7 8 9 warehouses 6 7 6 5 Is there a way of shipping from the warehouses to the retailers to satisfy demand? 6 5 4 5 4
5 Transformation to a max flow problem 1 2 3 4 5 6 7 8 9 warehouses retailers There is a 1-1 correspondence with flows from s to t with 24 units (why 24?) and feasible flows for the transportation problem. 6 5 4 5 4 6 7 6 5 6 5 4 5 4 s 6 7 6 5 t

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6 The feasibility problem: find a matching Is there a way of assigning persons to tasks so that each person is assigned a task, and each task has a person assigned to it? 1 2 3 4 5 6 7 8 persons tasks
7 Transformation to a maximum flow problem Does the maximum flow from s to t have 4 units? 1 2 3 4 5 6 7 8 persons tasks s 1 1 1 1 t 1 1 1 1

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8 The Residual Network s 1 2 t 10 , 8 8, 7 1, 1 10, 6 6, 5 s 1 2 t 2 1 1 4 1 8 5 6 7 i j x u ij ij i j u - x x ij ij ij We let r ij denote the residual capacity of arc (i,j) The Residual Network G(x)
9 A Useful Idea: Augmenting Paths An augmenting path is a path from s to t in the residual network. The residual capacity of the augmenting path P is δ (P) = min{r ij : (i,j) P}. To augment along P is to send δ ( P) units of flow along each arc of the path. We modify x and the residual capacities appropriately. r ij := r ij - δ (P) and r ji := r ji + δ (P) for (i,j) P. s 1 2 t 2 1 1 4 1 8 5 6 7 s 1 2 t 2 1 4 8 6 6 8

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10 The Ford Fulkerson Maximum Flow Algorithm Begin x := 0; create the residual network G(x); while there is some directed path from s to t in G(x) do begin let P be a path from s to t in G(x); := δ (P); send units of flow along P; update the r's; end end {the flow x is now maximum}. Ford- Fulkerson Algorithm Animation
11 Proof of Correctness of the Algorithm Assume that all data are integral. Lemma : At each iteration all residual capacities are integral. Proof . It is true at the beginning. Assume it is true after the first k-1 augmentations, and consider augmentation k along path P. The residual capacity of P is the smallest residual capacity on P, which is integral. After updating, we modify residual capacities by 0, or , and thus residual capacities stay integral.

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12 Theorem. The Ford-Fulkerson Algorithm is finite Proof . The capacity of each augmenting path is at least 1. The augmentation reduces the residual capacity of some arc (s, j) and does not increase the residual capacity of (s, i) for any i.
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