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Unformatted text preview: MS&E 211 Linear & Nonlinear Optimization Fall 2011 Prof Yinyu Ye Homework Assignment 3: SOLUTIONS Problem 1. Sensitivity Analysis: (22 points) [[2 points each]] You have rented a metal detector for two and a half hours. You can spend your time with it searching for valuables in one of three locations: the sand, the grass, or the hills. The metal detector has three settingsone for each location. The sand setting uses the battery power at only onethird the normal rate. The grass setting uses the battery power at twothirds the normal rate. And the hills setting uses the batter power at full blast. You have only enough battery life to operate the machine at full blast for one hour and twenty minutes. If you estimate that people find about five dollars per hour on the beach, nine dollars per hour on the grass, and ten dollars per hour on the hills, how should you allocate your time to maximize expected earnings? Hint: read all the questions before starting. We choose an LP formulation: gG 5 + 9 + 10 : 1 3 + 2 3 + 4 3 + + 5 2 0 The first line being expected profit in dollars. The second line being a constraint on battery hours. And the third line being a constraint on human hours. = G. = G. = . (b) Standard Form: g 5 9 10 : 1 3 + 2 3 + + = 4 3 + + + = 5 2 0 Simplex form: 5 9 10 0 0 Slackbatt 1/3 2/3 1 1 0 4/3 Slackhrs 1 1 1 0 1 5/2 Get ready to pivot in grass and out slackbatt 5 9 10 0 0 Slackbatt 1/2 1 3/2 3/2 0 2 Slackhrs 1 1 1 0 1 5/2 Pivot: 1/2 0 7/2 27/2 0 18 Grass 1/2 1 3/2 3/2 0 2 Slackhrs 1/2 0 1/2 3/2 1 1/2 Pivot in sand and out slackhours : (c) 0 0 3 12 1 18 Grass 0 1 2 3 1 3/2 Sand 1 0 1 3 2 1 This is optimal. X*= 1 hours grass, and 1 hour sand . (a) (d) We were able to choose to pivot in grass and sand because we knew the optimal basis in advance, and the simplex theory states that all pivots (regardless of their associated reduced costs) retain feasibility. And because we started at a feasible point, only used valid pivots, and ended at optimality conditions, once we get to the final tableau, we knew it was optimal. (e) The primal was: Min cx, st: Ax = b, x 0. So the dual is: Max by, st Ay c gG 4 3 + 5 2 : 1 3 + 5 2 3 + 9 + 10 0 0 (f) We see that if we resolve the problem with 2hrs of battery life, we get a grass and hills optimal solution....
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This note was uploaded on 01/16/2012 for the course MS&E 211 at Stanford.
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