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Maximum Flow Problem

Maximum Flow Problem. (Thanks to Jim Orlin & MIT OCW). 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 S j x ij - S k x ki = 0 for each i  s,t

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Maximum Flow Problem

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  1. Maximum Flow Problem (Thanks to Jim Orlin & MIT OCW)

  2. The Max Flow Problem • G = (N,A) • xij = flow on arc (i,j) • uij = capacity of flow in arc (i,j) • s = source node • t = sink node Maximize v Subject to Sj xij - Sk xki= 0 for each i  s,t Sj xsj = v 0£ xij£ uij for all (i,j)ÎA.

  3. 1 8, 7 10 8 , 1, 1 s t 6, 5 10, 6 2 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. A max flow problem. Capacities and a non-optimum flow.

  4. warehouses warehouses retailers 6 1 6 5 6 2 7 7 4 3 8 6 5 4 9 5 4 5 The feasibility problem: find a feasible flow Is there a way of shipping from the warehouses to the retailers to satisfy demand?

  5. 6 6 5 6 6 5 7 7 s 4 4 t 5 6 6 5 4 5 5 4 Transformation to a max flow problem warehouses warehouses retailers 1 6 2 7 3 8 4 9 5 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. persons tasks 1 5 2 6 3 7 4 8 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?

  7. 1 1 1 1 1 1 1 s t 1 1 Transformation to a maximum flow problem persons tasks 5 2 6 3 7 4 8 Does the maximum flow from s to t have 4 units?

  8. 1 8, 7 10 8 , 1, 1 x s t i ij j u ij 6, 5 10, 6 2 u - x ij ij i j x 1 1 2 ij 8 7 1 s t 4 1 2 6 5 The Residual Network We let rij denote the residual capacity of arc (i,j) The Residual Network G(x)

  9. 1 1 2 8 7 1 s t 4 1 1 2 2 6 5 8 8 1 s t 4 2 6 6 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 d(P) = min{rij : (i,j)  P}. • To augment along P is to send d(P) units of flow along each arc of the path. We modify x and the residual capacities appropriately. • rij := rij - d(P) and rji := rji + d(P) for (i,j)  P.

  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); •  := d(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 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.

  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. • 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.

  13. 1 8, 8 10 9 , 1, 1 s t 6, 6 10, 7 2 reachable from s in G(x) 1 8 1 1 9 s t 3 6 7 not reachable from s in G(x) 2 How do we know when a flow is optimal? METHOD 1. There is no augmenting path in the residual network.

  14. 1 8, 8 10 9 , 1, 1 s t 6, 6 10, 7 2 Method 2: Cut Duality Theory An (s,t)-cut in a network G = (N,A) is a partition of N into two disjoint subsets S and T such that s  S and t  T, e.g., S = { s, 1 } and T = { 2, t }. The capacity of a cut (S,T) is CAP(S,T) = iSjTuij

  15. 1 8, 8 10 9 , 1, 1 s t 6, 6 10, 7 2 Weak Duality Theorem for the Max Flow Problem • Theorem.If x is any feasible flow and if (S,T) is an (s,t)-cut, then the flow v(x) from source to sink in x is at most CAP(S,T). • PROOF. We define the flow across the cut (S,T) to be • Fx(S,T) = iSjT xij - iSjT xji If S = {s}, then the flow across (S, T) is 9 + 6 = 15.

  16. 1 8, 8 10 9 , 1, 1 s t 6, 6 10, 7 2 1 8, 8 10 9 , 1, 1 s t 6, 6 10, 7 2 Flows Across Cuts 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.

  17. i  S i  T j  T j  S i j i j xij -xij More on Flows Across Cuts • Claim:Let (S,T) be any s-t cut. Then Fx(S,T) = v = 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 v. The resulting equality is Fx(S,T) = v. Sj xij - Sk xki= 0 for each i  S\sSj xsj = v

  18. 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 xij  uij. If i  T, and j  S, then xij  0. Fx(S,T) = iSjT xij - iSjT xji Cap(S,T) = iSjT uij - iSjT 0

  19. 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.

  20. Continuation of the proof. • 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. • Thus i Î S and j Î T Þ xij = uij • i Î T and j Î S Þ xij = 0.

  21. 1 2 s t 3 4 Set S Set T Final steps of the proof • Thus i Î S and j Î T Þ xij = uij • i Î T and j Î S Þ xij = 0. There is no arc from S to T in G(x) x12 = u12 x43 = 0 If follows that Fx(S,T) = iSjT xij - iSjT xji = iSjT uij - iSjT 0 = CAP(S,T)

  22. Review • Corollary. If the capacities are finite integers, then the Ford-Fulkerson Augmenting Path Algorithm terminates in finite time with a maximum flow from s to t. • Corollary. If the capacities are finite rational numbers, then the Ford-Fulkerson Augmenting Path Algorithm terminates in finite time with a maximum flow from s to t. (why?) • Corollary. To obtain a minimum cut from a maximum flow x, let S denote all nodes reachable from s in G(x). • Remark. This does not establish finiteness if uij =  or if capacities may be irrational.

  23. 1 M M s t 1 M M 2 A simple and very bad example

  24. After 1 augmentation 1 M-1 M 1 s t 1 M-1 M 1 2

  25. After two augmentations 1 M-1 M-1 1 1 s t 1 M-1 M-1 1 1 2

  26. After 3 augmentations 1 M-2 M-1 2 1 s t 1 M-2 M-1 1 2 2

  27. And so on

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

  29. An even worse example • There are examples that takes an infinite number of augmentations on irrational data, and does not converge to the correct flow. • But we shall soon see how to solve max flows in a polynomial number of operations, even if data can be irrational.

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

  31. Application 6.4 Scheduling on Uniform Parallel Machines Job(j) 1 2 3 4 Processing 1.5 3 4.5 5 Time Release 2 0 2 4 Time Due Date 5 4 7 9 Suppose there are 2 parallel machines 2 1 4 3 No schedule is possible unless preemption is allowed

  32. Application 6.4 Scheduling on Uniform Parallel Machines Job(j) 1 2 3 4 Processing 1.5 3 4.5 5 Time Release 2 0 2 4 Time Due Date 5 4 7 9 Suppose there are 2 parallel machines 2 1 4 3 4 3 A feasible schedule with preemption allowed

  33. 0-2 1 t s 2-4 4-5 5-7 7-9 Transformation into a maximum flow problem red arcs: the amount of processing of job j during t time units is at most t. green arcs: the amount of processing during t time units is at most Mt. 2 4 1.5 4 2 3 2 4.5 1 3 5 4 4 4 2 blue arcs: the amount of processing of job j over all time units is pj.

  34. 0-2 1 t s 2-4 4-5 5-7 7-9 A maximum flow There is a feasible schedule if there is a maximum flow that saturates each of the arcs out of s. 1 .5 4,2 1.5 2 1 4,4 2 3 2 4.5 2,2 .5 3 5 2 4,4 1 2 4 4,2 2 Flow decomposition can be used to transform flows into schedules

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