MIT15_053S13_rec07sol

# Solution yes node 12 gets fathomed if x4 is set to 1

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Unformatted text preview: de 12 get pruned (fathomed)? Justify your answer. Solution. Yes, node 12 gets fathomed. If x4 is set to 1, then the LP relaxation becomes: max ax5 + 28 s.t. 10x5 ≤ −3, 0 ≤ x5 ≤ 1, The relaxation is infeasible because of the ﬁrst constraint, so node 12 gets fathomed. Part B.1 Suppose x4 is set to 0 and parameter a = 12. Does node 13 get pruned (fathomed)? Justify your answer. Solution. becomes: No, node 13 does not get fathomed. If x4 = 0 and a = 12, then the LP relaxation max 12x5 + 10 s.t. 10x5 ≤ 5, 0 ≤ x5 ≤ 1, The optimal solution to this relaxation is x5 = 1/2, with an objective value of 16. Because the solution is fractional and has a value better (higher) than the incumbent, the node wont be fathomed (since its possible that a better solution could be found in the subtree). Part C.1 Suppose x4 is set to 0 and parameter a = 8. Does node 13 get pruned (fathomed)? Justify your answer. Solution. becomes: Yes, node 13 gets fathomed. If x4 = 0 and a = 8, then the LP relaxation max 8x5 + 10 s.t. 10x5 ≤ 5, 0 ≤ x5 ≤ 1, The optimal...
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