Ii next we wish to find a vector y which satisfiies

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(ii) Next, we wish to find a vector y which satisfiies the complementary slackness condition for x = [0 , 3 , 2 , 1] T . As before, we first check that this is indeed a feasible solution for the primal. Next, we must find y that satisfies the second, third, and fourth dual constraint with equality: 2 y 1 - y 2 = 5 - y 1 + 3 y 2 = - 5 3 y 1 - 2 y 2 = 8 . Solving y 1 , y 2 using the first two of the equality constraints above, we obtain y 1 = 2 , y 2 = - 1. Also note that this solution also satisfies the fourth dual constraint with equality (i.e., it satisfies all three equality constraints above). We still need to check that this solution is feasible for the dual problem, by checking whether it satisfies the first dual constraint: 6(2) - 6( - 1) = 12 - 6 = 6 4 . So, y = [1 , - 2] T is a feasible solution for the dual problem and it satisfies the com- plementary slackness condition. Therefore, by the complementary slackness theorem (Theorem 19.1), x = [0 , 3 , 2 , 1] T is optimal for the primal problem. (c) Consider y = [2 , - 1] T that we obtain in part (b) above. We saw that y is feasible for the dual problem. Further observe that the dual objective value corresponding to y is: 7(2) + ( - 1) = 13 , and the primal objective value corresponding to x = [0 , 3 , 2 , 1] T from part (b)(ii) is: 4(0) + 5(3) - 5(2) + 8(1) = 13 . Since x and y are feasible for the primal and dual problems, respectively, and they have the same objective function value, then by certificate of optimality (Corollary 16.7), both are optimal for the primal and dual problems, respectively. In particular, we have found an optimal solution for the dual, namely y = [2 , - 1] T . Grading scheme: 4
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