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Max Flow – Min Cut Problem

Max Flow – Min Cut Problem. Directed Graph Applications. Shortest Path Problem (Shortest path from one point to another) Max Flow problems (Maximum material flow through conduit) Liquid flow through pipes Parts through assembly line Current through electrical network, etc. Flow Networks.

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Max Flow – Min Cut Problem

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  1. Max Flow – Min Cut Problem

  2. Directed Graph Applications • Shortest Path Problem (Shortest path from one point to another) • Max Flow problems (Maximum material flow through conduit) • Liquid flow through pipes • Parts through assembly line • Current through electrical network, etc

  3. Flow Networks • Directed graph G = (V, E) • Only one Source(s) and one Sink(t) • Weight on each edge = Capacity of the edge • if (u, v) ÎE then Capacity is non-negative, i.e. c(u, v) ≥ 0 • if (u, v) E then Capacity is assumed zero, i.e. c(u, v) = 0

  4. Flow Networks and Flows • flow(f) in Flow Network(G) is a real-valued function f: V x V → R • f(u, v) is flow from vertex u to vertex v. • flow f(u, v) can be positive, negative or zero Constraints on flow : • Capacity constraint : f(u, v) ≤ c(u, v), for all u, v Î V • Skew symmetry constraint : f(u, v) = - f(v, u)

  5. j1 j2 j3 i jn k4 Flow Networks and Flows (Contd.) • Flow conservation constraint : Total net flow at vertex must equal 0. ∑jf(i, j) - ∑kf(k, i) = 0 for all i Î V – {s, t} flow in equals flow out k1 k2 k3

  6. 1 (8/10) (7/8) (1/1) s t (5/6) (6/10) 2 Maximum Flow • We refer a flow f as maximum if it is feasible and maximize ∑kf(s, k). Where f(s, k) is flow out of source s. • Problem: • Objective: To find a maximum flow (flow/capacity)

  7. Factories s1 10 Warehouses 3 12 t1 s2 15 5 6 8 t2 s3 20 14 13 7 t3 s4 11 18 2 s5 Multiple source – Multiple sink Network Convert this problem to single source - single sink.

  8. Factories s1 10 3 Warehouses 12 t1 s2 15 5 Super-sink 6 Super-source 8 t2 t s3 20 14 s  13 7  t3 s4 11 18    2  s5   Conversion to flow Network • Add a super – source with infinite weighted edges emanting out to original sources • Add a super – sink with edges of infinite weight from original sinks

  9. persons tasks 1 5 2 6 3 7 4 8 Another Example : Find a feasible match • Assign persons to tasks such that each task is assigned to a person and each person is assigned to a task in feasible manner.

  10. persons tasks 1 5 1 1 1 1 2 6 1 1 s t 1 3 7 1 4 8 Transformation to maximum flow problem

  11. Ford-Fulkerson Algorithm for Max Flow • Key ingredients:- • Residual Networks • Augmenting Paths • Cut • Limitations :- • Flow should be integral or rational • On each iteration residual capacity should be integral

  12. 1 (7/8) (8/10) f (u, v) / c (u, v) (1/1) u u v v s t (5/6) (6/10) 2 1 1 2 c (u, f) – f (u, v) 8 7 1 s t f (u, v) 4 1 2 6 5 The Residual Network Flow Network (flow / capacity) Residual Network Residual capacity r (u ,v)

  13. 1 1 2 8 7 1 s t 4 1 1 2 2 6 5 8 8 1 s t 4 2 6 6 Augmenting Paths • An augmenting path is a path from s to t in the residual network. • The Residual capacity of augmented path P is d(P) = min {r(i, j):(i, j) P} • Augmentation along P • Add d(P) in each arc along P inflow network • Modify residual capacities in residual network r(u, v) = r(u, v) - d(P)andr(v, u) = r(v, u) + d(P)for u, v P

  14. 1 (8/8) (9/10) (1/1) s t (6/6) (7/10) 2 Cut An (S,T)-cut in a flow network G = (V,E) is a partition of vertices V into two disjoint subsets S and T such that s  S, t  T e.g., S = { s, 1 } and T = { 2, t }. The capacity of a cut (S,T) is CAP(S,T) = uSvT c(u,v)

  15. The Ford Fulkerson Maximum Flow Method 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);  := d(P); send  units of flow along P; update the residual capacities; end end {the flow x is now maximum}.

  16. Proof of Correctness of the Algorithm Assume that all data are integral. Lemma: At each iteration all residual capacities are integral. Proof: By assumption it is true at beginning. Assume it is true after the first k-1 augmentations, and consider augmentation k along path P. The residual capacity D of P is the smallest residual capacity on P, which is integral. After updating, we modify residual capacities by 0, or D, and thus residual capacities stay integral.

  17. Proof of finiteness of Algorithm 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. So, the sum of the residual capacities of arcs out of s keeps decreasing, and is bounded below by 0. Number of augmentations is O(nU), where U is the largest capacity in the network.

  18. 1 (8/8) (9/10) (1/1) s t (6/6) (7/10) 2 1 1 9 8 1 s t 3 2 7 6 Indication of Optimum Flow There is no augmenting path in the residual network.

  19. 1 (8/8) (9/10) (1/1) s t (6/6) (7/10) 2 Indication of Optimum Flow (Contd.) Flow across the network is equal to the capacity of some cut (Max Flow Min Cut Theorem)

  20. S T Weak Duality Theorem for the Max Flow Problem Theorem: If f is any feasible flow and if (S,T) is an (s,t)-cut, then the flow | f| from source to sink in the network is at most CAP(S,T). Proof: We define the flow across the cut (S,T) to be f(S,T) = iSjT f(i, j) - iSjT f(j, i)

  21. 1 (8/8) (9/10) (1/1) s t (6/6) (7/10) 2 1 (8/8) (9/10)  S (1/1) s t (6/6) (7/10)  T 2  S/T cut 1 (8/8) (9/10) (1/1) s t (6/6) (7/10) 2 Flowsacross different cuts If S = {s}, then the flow across (S, T) is 9 + 6 = 15 If S = {s,1}, then the flow across (S, T) is 8 + 1 + 6 = 15 If S = {s,2}, then the flow across (S, T) is 9 + 7 – 1 = 15

  22. S T More on Flows Across Cuts Claim: Let (S,T) be any s-t cut. Then f(S,T) = | f | = flow into t. Proof: Add the conservation of flow constraints for each node i Î S - {s} to the constraint that the flow leaving s is |f|. The resulting equality is f(S,T) = |f|. • Sjf(i, j) - Skf(k, i) = 0 for each i  {S} - s • Sjf(s, j) = | f |

  23. More on Flows Across Cuts Claim: The flow across (S,T) is at most the capacity of a cut. Proof: If i S, and j  T, then f(i, j)  c(i, j). If i T, and j  S, then f(i, j)  0. f(S,T) = iSjT f(i, j) - iSjT f(j, i) CAP(S,T) = iSjT c(i, j) - iSjT 0

  24. Max Flow Min Cut Theorem Theorem: (Optimality conditions for max flows). The following are equivalent. 1. A flow x is maximum. 2. There is no augmenting path in G(x). 3. There is an s-t cutset (S, T) whose capacity is the flow value of x. Corollary:(Max-flow Min-Cut). The maximum flow value is the minimum value of a cut. Proof of Theorem: 1 Þ 2. (not 2 Þ not 1) Suppose that there is an augmenting path in G(x). Then x is not maximum.

  25. Max Flow Min Cut Theorem Contd. 3 Þ 1. Let v = Fx(S, T) be the flow from s to t. By assumption, v = CAP(S, T). By weak duality, the maximum flow is at most CAP(S, T). Thus the flow is maximum. 2 Þ 3. Suppose there is no augmenting path in G(x). Claim: Let S be the set of nodes reachable from s in G(x). Let T = N\S. Then there is no arc in G(x) from S to T.

  26. Saturated .t .s 0 Reachable from s Not Reachable from s S T Max Flow Min Cut Theorem Contd. Thus iÎ S and j Î T Þf(i, j) = c(i, j) iÎ T and jÎ S Þ f(i, j) = 0. There is no arc from S to T in G(x) It follows that Fx(S,T) = iSjT f(i, j) - iSjT f(j, i) = iSjT c(i, j) - iSjT 0 = CAP(S,T)

  27. Limitation of Algorithm : example 1 1 M M s t 1 M M 2 • Termination even with integral flow and few nodes can take • large no. of steps. • Depends on path selection and capacity of flow network. Î Flow M >> 1 Flow Network 27

  28. After 1st augmentation 1 M-1 M 1 s t 1 M-1 M 1 2 Residual Network 28

  29. After 2nd augmentation 1 M-1 M-1 1 1 s t 1 M-1 M-1 1 1 2 29

  30. After 3rd augmentation 1 M-2 M-1 2 1 s t 1 M-2 M-1 1 2 2 30

  31. After 2M augmentations 1 M M s t 1 M M 2 31

  32. Limitation of Algorithm (Non termination for irrational flow) Flow Network 1 m m r s 2 3 t 1 m m 1 m m 4 m >> 1 1/2 < r < 1 32

  33. After 1st augmentation 1 S 2 3 t r 1 1 1 2 3 1 1 4 Flow in Flow Network Residual Network 33 33

  34. After 2nd augmentation 1 1 r r r 2 3 2 3 1-r 1-r r 4 4 Residual Network Flow in Flow Network 34 34 34

  35. After 3rd augmentation 1 1 r r 2 3 2 3 1 r 1-r r 4 Residual Network Flow in Flow Network 35 35 35 35

  36. 1 2r -1 1-r 1 - r 2 3 r 1 4 After 4th augmentation 1 1 - r 2 3 1 - r 1-r 4 Flow in Flow Network Residual Network 36 36 36 36 36

  37. 1 1 2r -1 1-r 1-r 1-r 2 3 2 3 1 1 - r r 1-r 4 4 After 5th augmentation Flow in Flow Network Residual Network 37 37 37 37 37 37

  38. Non termination in irrational flow 1 1 ai+1 ai+1 - (ai - ai+1) = ai+3 3 2 3 2 ai ai - ai+1 = ai+2 4 4 And this goes on ……………. 38

  39. Summary and Extensions 1. Augmenting path theorem 2. Ford-Fulkerson Algorithm 3. Duality Theory. 4. Computational Speedups. 39

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