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# lecture 16 - Math 482(Lecture 16 Max flow and min cut This...

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Math 482 (Lecture 16): Max flow and min cut This lecture discusses Sections 6.1 and 6.2 of the textbook. * We have studied the primal-dual method and shown how it applies to the shortest path problem. * We will now study the max-flow problem at greater depth. It is efficiently solved by the Ford-Fulkerson algorithm. * Later we will comment on how the Ford-Fulkerson algorithm can be seen from the point of view of primal-dual methods. We have aleady introduced the max flow problem: max v s.t. Af+dv=0, f≤ b, g≥0 where d is defined by d s =-1, d t =1 and d i =0 if i is not s,t. In class exercise: (a) Write this LP out for the wiki problem . (b) In eq (5.21) of the textbook, it is mentioned that one can replace Af+dv=0 with Af+dv≤ 0. Why? Solution: (a) is straightforward and discussed in class. (b): the textbook says "since a deficit in the flow balance at any node implies a surplus at some other". Indeed, convince yourself that it must be an equality in the rows for s and t. Then a strict inequality means some of the goods being flowed are lost at a vertex. But some must be gained elsewhere if at the end "t" receives everything "s" sent. The min cut problem Definition: An s-t cut is a partition (A,B) of the nodes V into A and B such that s A and t B. The capacity of an s-t cut is C(A,B)=Σ b(i,j) where the sum is over all arcs (i,j) with i A and j&isin B. In class exercise: What is the minimum cut in the wiki problem ? Answer: 15 = the max flow In class exercise: "Clearly" the max flow is at most the capacity of any cut. Why?

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lecture 16 - Math 482(Lecture 16 Max flow and min cut This...

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