co227_f19_a10_sln.pdf - CO 227 Fall 2019 Assignment 10 Solutions 1(a min s.t 6y1 17y2 12y3 \u2212y1 3y2 y3 2y1 \u2212 y2 6y3 y1 free y2 \u2265 0 y3 \u2264 0 y4 8y4

# co227_f19_a10_sln.pdf - CO 227 Fall 2019 Assignment 10...

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CO 227 Fall 2019: Assignment 10 Solutions 1. (a) min 6 y 1 + 17 y 2 + 12 y 3 + 8 y 4 s.t. - y 1 + 3 y 2 + y 3 + 2 y 4 - 3 2 y 1 - y 2 + 6 y 3 + y 4 4 y 1 free , y 2 0 , y 3 0 , y 4 0 (b) x 1 = 0 or - y 1 + 3 y 2 + y 3 + 2 y 4 = - 3 x 2 = 0 or 2 y 1 - y 2 + 6 y 3 + y 4 = 4 y 1 = 0 or - x 1 + 2 x 2 = 6 Note: this condition is optional. y 2 = 0 or 3 x 1 - x 2 = 17 y 3 = 0 or x 1 + 6 x 2 = 12 y 4 = 0 or 2 x 1 + x 2 = 8 (c) For (8 , 7) T , we see that both x 1 , x 2 6 = 0, so both constraints in the dual are tight. We have x 1 + 6 x 2 > 12, so this implies that y 3 = 0. Also, 2 x 1 + x 2 > 8, so y 4 = 0. This leaves us with the equations - y 1 + 3 y 2 = - 3 and 2 y 1 - y 2 = 4. Solving this gives the dual solution ( 9 5 , - 2 5 , 0 , 0) T , which is not feasible since x 2 < 0. Hence the solution is not optimal. For (4 , 5) T , both dual constraints are tight. We have 3 x 1 - x 2 < 17 , x 1 +6 x 2 > 12 , 2 x 1 + x 2 > 8, so this implies that y 2 = y 3 = y 4 = 0. We then have - y 1 = - 3 and 2 y 1 = 4, which results in y 1 = 3 and y 1 = 2. This is a contradiction, so no dual solution exists in this case. Hence the solution is not optimal. For (2 , 4) T , both dual constraints are tight. We have 3 x 1 - x 2 < 17 , x 1 + 6 x 2 > 12, so this implies that y 2 = y 3 = 0. This gives us the equations #### You've reached the end of your free preview.

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Unformatted text preview: -y 1 + 2 y 4 =-3 , 2 y 1 + y 4 = 4. Solving this gives the dual solution ( 11 5 , , ,-2 5 ) T , which is feasible. Hence the solution is optimal. 2. (a) (D) min { y T b : A T y ≥ c, y ≥ } (D’) min { 0 : A T y ≥ c, y ≥ } (b) Since (P) is unbounded, (D) is infeasible. Since both (D) and (D’) have the same set of constraints, (D’) is infeasible. (c) Since (D’) is infeasible, (P’) is either infeasible or unbounded. However, y = 0 is a feasible solution to (P’). Therefore, (P’) must be unbounded. (d) Since (P’) is unbounded, there exists a series of feasible solution d such that c T d → ∞ . This means that there exists a feasible solution for which c T d > 0. Such a solution satsifies Ad ≤ , d ≥ , c T d > 0....
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