hw2-solutionnew

hw2-solutionnew - MIT 2.098/6.255/15.093 Optimization...

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Unformatted text preview: MIT 2.098/6.255/15.093 Optimization Methods Prof. J. Vera, Fall 2007 Homework Assignment 2. Solution Problem 1: BT, exercise 3.2 Solution. (a) If feasible solution x is optimal, suppose there exists a feasible direction d such that c d < 0, then for a small enough > 0, x + d is still feasible, but c ( x + d ) = c x + c d < c x contradiction. If c d 0 for every feasible direction d at x , for any point y P and y 6 = x , d = y- x is a feasible direction. c y- c x = c d 0, which implies x is optimal. (b) The proof is almost identical. If feasible solution x is the unique optimal solution, suppose there exists a feasible direction d such that c d 0, then for a small enough > 0, x + d is still feasible, but c ( x + d ) = c x + c d c x contradict with x being the unique optimal solution. If c d > 0 for every feasible direction d at x , for any point y P and y 6 = x , d = y- x is a feasible direction. c y- c x = c d > 0, which implies x is uniquely optimal. / Problem 2: BT, exercise 3.17 Solution. The initial tableau in Phase I is- 5- 1- 1- 3- 1- 2 x 6 = 2 1 * 3 4 1 1 x 7 = 2 1 2- 3 1 1 x 8 = 1- 1- 4 3 1 The final tableau in Phase I is 1 7 2 1 x 1 = 2 1 3 4 1 1 x 7 = 0- 1 *- 7- 1 1 x 3 = 1 1 / 3 1 4 / 3 1 / 3 1 / 3 1 / 3 Drive x 7 out of the basis 1 1 1 1 x 1 = 2 1- 17 1- 2 3 x 2 = 0 1 7 1- 1 x 3 = 1 1 11 / 3 1 / 3 2 / 3- 1 / 3 1 / 3 The initial tableau in Phase II is:- 7 3- 5 x 1 = 2 1- 17 1 x 2 = 0 1 7 x 3 = 1 1 11 / 3 1 / 3 Final tableau is 3 5 82 / 7 x 5 = 2 1 17 / 7 1 x 4 = 0 1 / 7 1 x 3 = 1 / 3- 1 / 3- 4 / 3 1 / Problem 3: BT, exercise 3.19 Solution. (a) The current solution is optimal but the current basis is not. Thus the current solution is a degenerate optimal solution. So we have = 0. Update the tableau using one simplex iteration- 10 + 2 3 2 / 3 x 3 = 4- 1- 3 1- 3 x 4 = 1 + 4 3 1 4 / 3 x 2 = 0 3 1 1 / 3 There are multiple optimal solutions, thus + 2 3 = 0. In addition, we need to have a feasible direction, which requires 3 0. The conditions could be = 0 , + 2 3 = 0, and 0. (b) The optimal cost is- when we have a feasible solution in the current tableau, a nonbasic variable x i with c i < 0 and u i = B- 1 A i 0. We need 0 for problem feasiblity. The variable 2 x 2 cannot satisfy all the conditions for the- cost. For the variable x 1 , the conditions then can be expressed as follows: , 0, and...
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hw2-solutionnew - MIT 2.098/6.255/15.093 Optimization...

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