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Published byGuadalupe Modrell Modified about 1 year ago

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Max Flow Problem Given network N=(V,A), two nodes s,t of V, and capacities on the arcs: u ij is the capacity on arc (i,j). Find non-negative flow f ij for each arc (i,j) such that for each node, except s and t, the total incoming flow is equal to the total outgoing flow (flow conservation), such that f ij is at most the capacity u ij on arc (i,j); The total amount of flow going out of s (which is equal to the total amount of flow coming into t ) is maximized

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3 t s the black numbers next to an arc is its capacity

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3 t s C ts = -1 Set costs all other arcs at 0 The minimum cost flow circulation (Af=0) maximises the s-t flow

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3 t s the black numbers next to an arc is its capacity Push flow over the path s-1-4-t The bottlenecks on this path are the edges {1,4} and {4,t}. So we can send flow 2 along this path

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3 t s the black number next to an arc is its capacity the green number next to an arc is the flow on it We can push another extra flow of 1 on the path s-1-3-t. The bottleneck is now {s,1} with remaining capacity 1.

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3 t s the black number next to an arc is its capacity the green number next to an arc is the flow on it Since {4,t} is on its capacity, the only path remaining through which we can send extra flow is s t. Here {4,3} is the bottleneck and thus we can send extra flow of 1

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3 t s the black number next to an arc is its capacity the green number next to an arc is the flow on it There are no s-t paths left on which we can send extra flow. So is this the maximal flow?

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the residual graph 3 t s blue arcs (i,j) are forward arcs (f ij __0) the blue number is the residual capacity of a blue arc the green number is the capacity of a green arc the black number is the original capacity of the arc
__

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the residual graph 3 t s red arcs form an s-t-path in the residual graph and therefore a flow-augmenting path in the original network 3

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the residual graph 3 t s red arcs form an augmenting path Augment the flow by the minimum capacity of a red arc, i.e, 1 3

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the residual graph 3 t s Augment the flow by the minimum capacity of a red arc, i.e, 1: - increase the flow by 1 on all arcs corresponding to forward red arcs - decrease the flow by 1 on all arcs corresponding to backward red arcs 3

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the residual graph 3 t s Augment the flow by the minimum capacity of a red arc, i.e, 1: - increase the flow by 1 on all arcs corresponding to forward red arcs - decrease the flow by 1 on all arcs corresponding to backward red arcs Construct the new residual graph 3 1

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3 t s An s-t cut is defined by a set S of the nodes with s in S and t not in S.

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3 t s An s-t cut is defined by a set S of the nodes with s in S and t not in S. S={s,1,2}

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3 t s An s-t cut is defined by a set S of the nodes with s in S and t not in S Size of cut S is the sum of the capacities on the arcs from S to N\S. S={s,1,2}

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3 t s An s-t cut is defined by a set S of the nodes with s in S and t not in S Size of cut S is the sum of the capacities on the arcs from S to N\S. C(S)=u 13 +u 14 +u 24 = 2+2+4=8

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3 t s C(S 1 )=u s1 +u 24 = 2+4=6 S1S1

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S1S1 S2S2

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3 t s S2S2 Max Flow ≤ Min Cut

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 Max Flow ≤ Min Cut f s1 +f s2 ≤ Min Cut ≤ 5

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t s the residual graph Max Flow = Min Cut f s1 +f s2 =2+3=5 = C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2

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t s the residual graph Max Flow = Min Cut f s1 +f s2 =2+3=5 = C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 Theorem: Max Flow = Min Cut

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t s Proof sketch of Max Flow=Min Cut f s1 +f s2 =2+3=5 = C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 There is no s-t-path in the residual graph!!! S 2 ={s,1,2,4} the set of nodes reachable from S

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 f s1 +f s2 =2+3=5 = Proof sketch of Max Flow=Min Cut from S 2 ={s,1,2,4} to {3,t} f 13 = u 13 = 2 f 43 = u 43 = 1 f 4t = u 4t = 2

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 f s1 +f s2 =2+3=5 = from S 2 ={s,1,2,4} to {3,t} f 13 = u 13 = 2 f 43 = u 43 = 1 f 4t = u 4t = 2 Full capacity of cut is used Proof sketch of Max Flow=Min Cut

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 f s1 +f s2 =2+3=5 = from S 2 ={s,1,2,4} to {3,t} f 13 = u 13 = 2 f 43 = u 43 = 1 f 4t = u 4t = 2 Full capacity of cut is used from {3,t} to {S 2 ={s,1,2,4} f 32 = 0 f 34 = 0 Proof sketch of Max Flow=Min Cut

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 f s1 +f s2 =2+3=5 = from S 2 ={s,1,2,4} to {3,t} f 13 = u 13 = 2 f 43 = u 43 = 1 f 4t = u 4t = 2 Full capacity of cut is used from {3,t} to {S 2 ={s,1,2,4} f 32 = 0 f 34 = 0 nothing flows back across the cut Proof sketch of Max Flow=Min Cut

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3 t s C(S 2 )=u 13 +u 43 +u 4t = 2+1+2=5 S2S2 f s1 +f s2 =2+3=5 = from S 2 ={s,1,2,4} to {3,t} f 13 = u 13 = 2 f 43 = u 43 = 1 f 4t = u 4t = 2 from {3,t} to {S 2 ={s,1,2,4} f 32 = 0 f 34 = 0 Proof sketch of Max Flow=Min Cut Capacity of the cut is equal to the flow from s to t

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