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# h5sol - CO 227 Homework assignment 5 Fall `10 Page 1 CO 227...

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CO 227 - Homework assignment 5 Fall ‘10 Page 1 CO 227 - Fall ‘10 Homework assignment #5: (Due at the beginning of the class, 10:30AM, on Friday, Nov 19th) Instructions: Please show all your work and justify your answers. Answers without proper justification will not be considered. Please list who you collaborated with. Please staple your homework and write your name and student id in pen. Total value = 108 points Question 1 - Duality (16 points) Consider the following LP: max (4 , 1 , 6 , 7 , 2) x s.t. 2 0 8 1 0 3 1 4 5 3 5 0 0 0 1 x 5 1 6 x 0 (1) (a) (5 points) Find the dual of (1) Solution: min (5 , 1 , 6) y s.t. 2 3 5 0 1 0 8 4 0 1 5 0 0 3 1 y 4 1 6 7 2 y 1 0 , y 2 , y 3 0 (-1pt for each mistake) (b) (11 points) Notice that ¯ x = (0 , 5 , 1 , 0 , 0) is feasible for (1). Argue, using the knowledge about duality and the dual you found in part (a) that problem (1) is unbounded. (I do NOT want you to use the simplex method to do this question!) Solution: Note that the second constraint of the dual is y 2 ≤ − 1, which contradicts the fact that y 2 0. As a result we conclude that the dual is infeasible. Now the possible outcomes for the primal are: infeasible, unbounded or has an optimal solution. Since ¯ x is feasible for the primal, then the primal cannot be infeasible. Now if the primal has an optimal solution, then strong duality must hold, so there must exist an optimal solution to the dual. But that is impossible, since the dual is infeasible. Therefore, the only option left is that the primal must be unbounded. (4 pts for finding out dual is infeasible) (1 pt for noticing primal is feasible) (2 pts for mentioning possible outcomes of LP) (4 pts for arguing why the LP cannot have an optimal solution)

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CO 227 - Homework assignment 5 Fall ‘10 Page 2 Question 2 - Duality and complementary slackness (13 points) Consider the following LP: min x 1 2 x 2 + x 3 +8 x 4 s.t. x 1 x 2 9 2 x 1 +5 x 2 6 x 3 3 4 x 2 +10 x 3 7 x 4 = 5 x 1 0 x 2 0 (2) (a) (5 points) Find the dual of (2). Solution: max 9 y 1 +3 y 2 +5 y 3 s.t. y 1 +2 y 2 1 y 1 +5 y 2 4 y 3 2 6 y 2 +10 y 3 = 1 7 y 3 = 8 y 1 0 y 2 0 (-1pt for each mistake) (b) (8 points) Write down the complementary slackness conditions for (2). Solution: The complementary slackness conditions are as follows: x 1 x 2 = 9 or y 1 = 0 2 x 1 + 5 x 2 6 x 3 = 3 or y 2 = 0 y 1 + 2 y 2 = 1 or x 1 = 0 y 1 + 5 y 2 4 y 3 = 2 or x 2 = 0 (Notice that the complementary slackness conditions involving equality constraints/free vari- ables: 4 x 2 + 10 x 3 7 x 4 = 5 or y 3 = 0 6 y 2 + 10 y 3 = 1 or x 3 = 0 7 y 3 = 8 or x 4 = 0 are always satisfied) (2pts for each complementary slackness for inequality constraints) Question 3 - Duality (25 points) Consider the linear programming problem max c T x s.t. Ax b (P) where A R m × n , b R m and c R n .
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