hw5.sol[1]

hw5.sol[1] - IE 335 Operations Research - Optimization...

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Unformatted text preview: IE 335 Operations Research - Optimization Solutions to Homework 5 Fall 2011 Problem 20 5-5.(a) 1 2 3 4 5 6 7 8 0.5 1 1.5 2 2.5 3 3.5 4 y 1 y 2 Graphical Solution to Problem 20 y 1 + 2y 2 = 6 y 2 = 2 (2,2) (0,3) (0,2) 5-5.(b) y 1 + 2 y 2 + y 3 = 6 y 2 + y 4 = 2 y 1 ,y 2 ,y 3 ,y 4 A = 1 2 1 0 0 1 0 1 b = 6 2 T 5-5.(c) If the matrix formed by the columns has a non-zero determinant, this implies they are linearly independent and thus form a basis. Therefore, since det([ y 1 y 2 ]) = 1 , det([ y 2 y 3 ]) =- 1 , det([ y 2 y 4 ]) = 2, those sets form a basis. y 1 cannot form a basis in 2 dimensions, y 2 = 2 y 3 + y 4 and y 1 = y 3 which makes the 1 corresponding sets linearly dependent and thus cannot be used as a basis. 5-5.(d) We set the non-basic variables to 0 and solve for the remaining components of the solution. { y 1 ,y 2 } Feasible y = 2 2 0 0 T { y 2 ,y 3 } Feasible y = 0 2 2 0 T { y 2 ,y 4 } Infeasible y = 0 3 0- 1 T 5-5.(e) The first two components of each basic solution forms the equivalent solution in the original problem.The first two components of each basic solution forms the equivalent solution in the original problem....
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hw5.sol[1] - IE 335 Operations Research - Optimization...

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