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hw8sols - C O250/CM340 I NTRODUCTION TO O PTIMIZATION HW 6...

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CO250/CM340 I NTRODUCTION TO O PTIMIZATION - HW 6 SOLUTIONS Exercise 1 (25 marks) . (a) If there is a topological ordering y and a directed cycle i 1 i 2 , i 2 i 3 , . . . , i k - 1 i k , i k i 1 then y i 1 > y i 2 + 1 > y i 3 + 2 > · · · > y i k + k - 1 > y i 1 + k , a contradiction. (b) (P 0 ) is the following. max x a + x b + x c + x d + x e + x f + x g subject to x a - x b = 0 x d + x c - x a = 0 - x d - x e = 0 x f + x b - x c - x g = 0 x e + x g - x f = 0 x z 0 ( z ∈ { a, b, c, d, e, f, g } ) . (c) This is feasible since x z = 0 for all z ~ E is a feasible solution. The same works for the arbitrary (P). (d) For (P 0 ): for an arbitrary number b 0 we can let x f = x g = b , and x z = 0 otherwise. This is feasible. The objective function is 2 b , so it is unbounded. For (P): If there is a directed cycle, set x e = b for all edges e in the cycle, and x e = 0 otherwise, where b 0 . Then this is a feasible solution, because each constraint with an edge of the cycle corresponds to a vertex v . There must be exactly one arc of the cycle going into v and one going out. The variables corresponding to these arcs have value b each and have opposite signs in the constraint corresponding to v , and all other arcs involved with v have value 0. So the constraint is satisfied. (For a vertex not involved in the cycle, the constraint says 0 = 0 ). The value of this solution is b times the number of arcs in the cycle, so it is unbounded. (e) The dual is (D): min 0 subject to y i - y j 1 (for all e = ij ~ E ) y i free (for all i V ) The dual of (P 0 ) is the following. 1
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2 max 0 subject to y 1 - y 2 1 y 4 - y 1 1 y 2 - y 4 1 y 2 - y 3 1 y 5 - y 3 1 y 4 - y 5 1 y 5 - y 4 1 y i free ( i ∈ { 1 , . . . , 5 } ) . (f) A topological ordering satisfies exactly the constraints given in the dual (D) since y i y j + 1 implies y i - y j 1 . So it is a feasible solution of (D). (And vice versa.) (g) If there is a topological ordering then by (f), (D) is feasible. Its objective function is 0. So by the Weak Duality theorem, any feasible solution of (P) is at most 0. So (P) is not unbounded. But if there is a directed cycle then (P) is unbounded by (d). So there is no directed cycle in the digraph. (h) If (P) has a feasible solution of value greater than 0, then we can choose any arc e = ij with x e > 0 , and by the constraint for vertex j in (P), we see that x e 1 > 0 for some arc e 1 of the form ji 1 . Then similarly, x e 2 > 0 for some arc of the form i 1 i 2 , and so on. Eventually the we must reach a vertex already seen, and complete a cycle of ~ G . So ~ G has a directed cycle. (i) If ~ G has no directed cycle, then by (h), (P) has no feasible solution of value greater than 0. But we saw in (c) that (P) has a feasible solution of value 0. Hence 0 is the optimal value of (P). Hence, by strong duality, 0 is the optimal value of (D), and in particular, (D) must be feasible. We noticed in (f) that the statement to be proved there was iff, that is, a feasible solution to (D) is a topological ordering. So
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