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hw5solutions

# hw5solutions - c Î 1 â‰¥ In the above derivation...

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ORF 307 Homework 5 Solutions Exercise 1. The adjacency matrix A (including the extra link) is: 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 - 1 0 1 - 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 - 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 0 0 - 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 0 0 - 1 0 - 1 0 1 0 - 1 0 - 1 0 0 0 0 0 0 0 0 - 1 1 0 0 0 0 0 0 0 - 1 1 0 0 0 0 0 0 0 0 0 0 0 - 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 1 - 1 - 1 0 0 0 1 0 0 0 0 0 0 0 0 - 1 0 0 0 0 0 1 1 0 0 0 The max flow problem is the same as we’ve done: min - t s.t. A T x - et = 0 0 x u Where [A -e] is the above adjacency matrix and u is a vector of capacities given in the problem. See the Matlab file for how to solve this problem. One solution to the problem is: x = (4 . 0 , 7 . 0 , 4 . 0 , 0 . 0 , 4 . 4 , 0 . 0 , 2 . 0 , 2 . 4 , 3 . 7 , 3 . 0 , 1 . 0 , 0 . 2 , 0 . 7 , 7 . 2 , 3 . 0 , 0 . 7 , 2 . 5 , 1 . 2 , 11 . 0) The dual variables associated with the flow constraints are: z = (0 , 0 , 1 , 1 , 0 , 1 , 1 , 1 , 1 , 1) The dual is derived as such: The Lagrangian of this problem is: L ( x, λ 1 , λ 2 , ν ) = c T x + λ T 1 ( Ax - b ) - λ T 2 x + ν T Dx with λ 1 , λ 2 0 The Lagrange dual function is: g ( λ 1 , λ 2 , ν ) = inf x D L ( x, λ 1 , λ 2 , ν ) = inf x D c T x + λ T 1 ( Ax - b ) - λ T 2 x + ν T Dx 1

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= inf x D ( c T + λ T 1 A - λ T 2 + ν T D ) x - λ T 1 b = - λ T 1 b if c + A T λ 1 - λ 2 + D T ν = 0, = -∞ otherwise Thus the dual problem is: max g ( λ 1 , λ 2 , ν ) s.t. λ 1 , λ 2 0 This is equivalent to: max - λ T 1 b s.t. c + A T λ 1 - λ 2 + D T ν = 0 λ 1 , λ 2 0 Note that λ 2 is like a surplus variable and we switch max to min so this problem is equivalent to: min λ T 1 b s.t.
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Unformatted text preview: c Î» 1 â‰¥ In the above derivation A=I,b=u,D=our adjacency matrix above, and c=[0,. ..,0,-1]. Not-ing the structure in D (every column has one 1 and one -1), we can write our problem as: min y T u s.t. y 9 , 1 + z 9-z 1 â‰¥ 1 y i,j + z i-z j â‰¥ y â‰¥ Using the multipliers given above from Matlab, our minimum cut is U=1,2,5 and V=3,4,6,7,8,9. Notice that the cut could have gone through arc 2 or through both arcs 5 and 17. 2...
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hw5solutions - c Î 1 â‰¥ In the above derivation...

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