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lecture24 - Taking Stock IE170 Algorithms in Systems...

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IE170: Algorithms in Systems Engineering: Lecture 24 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University March 28, 2007 Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 1 / 24 Taking Stock Last Time Transitive Closure (Fast) Flows in Networks This Time Flows, Flows, Flows Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 2 / 24 Flows Flows in Networks G = ( V, E ) directed. Each edge ( u, v ) E has a capacity c ( u, v ) 0 If ( u, b ) E c ( u, v ) = 0 We will typically have a special source vertex s V , a sink vertex t V , and we will assume there exists paths from s v t v V The combination of all of these things ( G, s, t, c ) is known as a flow network . Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 3 / 24 Flows Net Flows A net flow is a function f : V × V R | V |×| V | that satisfies three conditions: 1 Capacity Constraints: f ( u, v ) c ( u, v ) 2 Skew Symmetry: f ( u, v ) = - f ( v, u ) u V, v V 3 Flow Conservation: v V f ( u, v ) = 0 u V \ { s, t } Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 4 / 24
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Flows More Flow An important value we will be worried about is the value of flow f = | f | = v V f ( s, v ) : The total flow out of the source. The Maximum Flow Problem Given G = ( V, E ) . source node s V , sink node t V , edge capacities c . Find a flow whose value is maximum. Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 5 / 24 Flows Lemma, Lemma, Lemma Recall Shorthand f ( X, Y ) = x X y Y f ( x, y ) . 1 f ( X, X ) = 0 X V 2 f ( X, Y ) = - f ( Y, X ) X, Y V 3 Let X, Y, Z V be such that X Y = , then f ( X Y, Z ) = f ( X, Z ) + f ( Y, Z ) f ( Z, X Y ) = f ( Z, X ) + f ( Z, Y ) 4 | f | = f ( V, t ) Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 6 / 24 Flows Cuts A cut of a (flow) network G = ( V, E ) is a partition of V into S and T = V \ S such that s S and t T For flow f , net flow across a cut is f ( S, T ) and the cuts capacity is c ( S, T ) = u S v T c ( u, v ) A minimum cut of G is a cut whose capacity is minimum Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 7 / 24 Flows A Simple Upper Bound Flow Across Cuts Lemma For any cut ( S, T ) , f ( S, T ) = | f | Coronary :-) The value of any flow is no more than the capacity of any cut | f | = f ( S, T ) = u S v T f ( u, v ) u
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