# hw6 - x 1 x 5 non basic at the upper bound s the slack...

This preview shows page 1. Sign up to view the full content.

AMS 540 / MBA 540 (Fall, 2010) Estie Arkin Linear Programming Homework Set # 6 Due in class on Tuesday, November 2, 2010. 1). [BJS 6.17] Show (prove) that if the problem { max cx | Ax = b, x 0 } has a ±nite optimal solution, then the new problem { max cx | Ax = b , x 0 } cannot be unbounded, no matter what value the vector b might take. 2). Use the dual simplex method to solve: max z = - 2 x 1 - x 3 s . t . x 1 + x 2 - x 3 5 x 1 - 2 x 2 + 4 x 3 8 x 1 , x 2 , x 3 0 3). Consider the following knapsack problem: max z = x 1 + x 2 + 8 x 3 + 2 x 4 + 5 x 5 3 x 1 + 3 x 2 + 4 x 3 + 4 x 4 + 2 x 5 7 0 x 1 , x 2 , x 3 , x 4 , x 5 1 (a). Consider the upper bound constraints as explicit (regular) constraints. What is the ±rst tableau for the standard simplex method? (No need to apply the simplex method.) (b). Set up the starting simplex tableau with x 2 , x 3 , x 4 non basic at the lower bound and
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x 1 , x 5 non basic at the upper bound, s the slack variable is basic. (c). Solve using the Simplex method with upper bounds, starting from the tableau in part (b). (d). Now consider the general (fractional) knapsack problem: max z = c 1 x 1 + . . . + c n x n a 1 x 1 + . . . + a n x n ≤ b ≤ x 1 , . . . , x n ≤ 1 where c i > 0, a i > 0 for i = 1 , . . . , n and b > 0. Show that for all such knapsack problems, there exists an optimal solution in which at most one of the variables is non integer. Hint: Consider BFS. (e). Suggest a simple way to solve the problem in part (d) (without using simplex)....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online