Lecture18

Lecture18 - C&O 355 Mathematical Programming Fall 2010...

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Mathematical Programming Fall 2010 Lecture 18 N. Harvey
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Topics Network Flow Max Flow / Min Cut Theorem Total Unimodularity Proof of Max Flow / Min Cut Theorem
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Network Flow Let D=(N,A) be a directed graph. Every arc a has a “capacity” c a ¸ 0. (Think of it as an oil pipeline) Want to send oil from node s to node t through pipelines Oil must not leak at any node, except s and t : flow in = flow out. How much oil can we send? For simplicity, assume no arc enters s and no arc leaves t . 3 1 8 7 1 5 1 2 2 2 3 2 4 1 2 s t
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Schematic diagram of the railway network of the Western Soviet Union and Eastern European countries, with a maximum flow of value 163,000 tons from Russia to Eastern Europe, and a cut of capacity 163,000 tons indicated as ‘The bottleneck’. [Schrijver, 2005] Harris and Ross [1955]
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Delbert Ray Fulkerson Let D=(N,A) be a digraph, where arc a has capacity c a . Definition: For any U µ N, the cut ± + (U) is: The capacity of the cut is: Theorem: The maximum amount of flow from s to t equals the minimum capacity of a cut ± + (U), where s 2 U and t 2 U Furthermore, if c is integral then there is an integral flow that achieves the maximum flow.
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LP Formulation of Max Flow Variables: x a = amount of flow to send on arc a Constraints: Flow through each arc can not exceed its capacity.
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This note was uploaded on 06/16/2011 for the course CO 355 taught by Professor Harvey during the Winter '10 term at Waterloo.

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Lecture18 - C&O 355 Mathematical Programming Fall 2010...

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