hw10_sol - Homework 10 Solution Due Exercise 2 page 376...

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Unformatted text preview: Homework 10 Solution Due April 21, 2009 Exercise 2, page 376, parts (a) and (d) Develop Branch and Bound (B&B) tree for each of the following problems. For convenience, use x 1 as the first branching variable at the starting node. (a) Maximize z = 3 x 1 + 2 x 2 , subject to 2 x 1 + 5 x 2 ≤ 9, 4 x 1 + 2 x 2 ≤ 9, x 1 ≥ 0, x 2 ≥ 0 and both are integer valued. (b) Minimize z = 5 x 1 + 4 x 2 , subject to 3 x 1 + 2 x 2 ≥ 5, 2 x 1 + 3 x 2 ≥ 7, x 1 ≥ 0, x 2 ≥ 0 and both are integer valued. Solution (a) The B & B tree is illustrated in FIgure 1. We first consider the LP relaxation of the problem, which we label as LP1: maximize z = 3 x 1 + 2 x 2 subject to 2 x 1 + 5 x 2 ≤ 9 4 x 1 + 2 x 2 ≤ 9 x 1 ≥ , x 2 ≥ The standard form of this LP1 is maximize z = 3 x 1 + 2 x 2 subject to 2 x 1 + 5 x 2 + x 3 = 9 4 x 1 + 2 x 2 + x 4 = 9 x 1 , x 2 , x 3 , x 4 ≥ . We can solve LP1, by considering all basic solutions (since the problem size is small) and then find the best valued one. We can also consider basic solutions one by one and keep checking optimality conditions, as follows. Let us first consider choosing x 1 and x 2 as basic variables. The basis matrix is B = 2 5 4 2 . The basic solution is given by B- 1 [9 9] T . The basis inverse is B- 1 =- 1 16 2- 5- 4 2 , 1 and the resulting basic solution is x 1 = 1 . 69 and x 2 = 1 . 13. This solution is feasible for the LP. Its shadow prices are [ y 1 y 2 ] = [3 2] B- 1 = [3 2]- 1 16 2- 5- 4 2 = 1 8 11 16 . We now compute the reduced cost of the nonbasic variables x 3 and x 4 , which are ¯ c 3 = y 1- 0 = y 1 ¯ c 4 = y 2- 0 = y 2 . Since the reduced costs are nonnegative, the current basis is optimal, and the optimal value is z * = 7 . 31. From the solution x 1 = 1 . 69 and x 2 = 1 . 13, we branch into two possibilities for x 1 : x 1 ≤ 1 and x 1 ≥ 2, as shown in Figure 1. Suppose we follow the branch x 1 ≥ 2 (LP2), where we face the following LP2: maximize z = 3 x 1 + 2 x 2 subject to 2 x 1 + 5 x 2 ≤ 9 4 x 1 + 2 x 2 ≤ 9 x 1 ≥ 2 x 1 , x 2 ≥ . Solving this LP2, we find that the optimal solution is x 1 = 2 and x 2 = 0 . 5, and the optimal value is z * = 7. We now branch from LP2 into two possibilities: x 2 ≥ 1 (LP3) and x 2 ≤ 0 (LP4). LP3 has constraints 2 x 1 + 5 x 2 ≤ 9 , 4 x 1 + 2 x 2 ≤ 9 , x 1 ≥ 2 , x 2 ≥ 1 , which is not feasible, so this branch is fathomed. We go to LP4, which is given by maximize z = 3 x 1 + 2 x 2 subject to 2 x 1 + 5 x 2 ≤ 9 4 x 1 + 2 x 2 ≤ 9 x 1 ≥ 2 x 2 = 0 . Solving LP4, we find that the optimal solution is x 1 = 2 . 25 and x 2 = 0 with the optimal value z * = 6 . 75 . 2 We now branch from LP4 into two possibilities: x 1 ≥ 3 (LP5) and x 1 ≤ 2 (LP6)....
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hw10_sol - Homework 10 Solution Due Exercise 2 page 376...

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