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**Unformatted text preview: **Fall 2010 Optimization I (ORIE 3300/5300) Assignment 12 Solution Problem 1 (a) Since the variables are binary, on the left branch we place the restriction x = 0 and on the right branch x = 1. (b) The branch-and-bound tree is as follows in Figure b. The optimal solution is x = (1 , , 0) with value z = 14. (1) Original Problem z = 19.6, x = (0, 0, F) (2) z = 13, x = (0, 0, 1) current incumbent (3) z = 18, x = (F, F, 0) (4) z = 14, x = (1, 0, 0) new incumbent (5) infeasible x 3 = 0 x 3 = 1 x 1 = 0 x 1 = 1 Figure 1: The branch-and-bound tree for problem 1(b) (c) The branch-and-bound tree is as follows in Figure c. The optimal solution is x = (1 , , 0) with value z = 14, as in part (b). Note however, that the decision tree here is much larger. (1) Original Problem z = 19.6, x = (0, 0, F) (9) z = 13, x = (0, 0, 1) worse than incumbent (2) z = 18, x = (F, F, 0) (6) z = 16.5, x = (F, 1, 0) (3) z = 17.3, x = (F, 0, 0) x 3 = 0 x 3 = 1 x 2 = 0 x 2 = 1 (4) infeasible (5) z = 14, x = (1, 0, 0) current incumbent (7) infeasible (8) z = 9, x = (1, 1, 0) worse than incumbent x 1 = 0 x 1 = 0 x 1 = 1 x 1 = 1 Figure 2: The branch-and-bound tree for problem 1(c) Grading scheme: 1 Fall 2010...

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