# sol2 - AMS 540 MBA 540(Fall 2009 Estie Arkin Homework Set 2...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AMS 540 / MBA 540 (Fall, 2009) Estie Arkin Homework Set # 2: Solution notes (1). (a) Let S be a feasible set, x 1 , x 2 S , 0 1 then: A ( x 1 + (1 ) x 2 ) = Ax 1 + (1 ) Ax 2 = b + (1 ) b = b Also: 0 and x 1 0 imply x 1 0, and 1 0 and x 2 0 imply (1 ) x 2 0, so x 1 +(1 ) x 2 0. Thus x 1 + (1 ) x 2 S . (b) Let S be the set of all optimal solutions, x 1 , x 2 S , 0 1 then by part (a) x 1 + (1 ) x 2 S . Also c ( x 1 + (1 ) x 2 ) = cx 1 + (1 ) cx 2 = z + (1 ) z = z , where z = min { cx | x S } , therefore x 1 + (1 ) x 2 S . 2). Proof: Any point x that is feasible to the new LP is also feasible to the old LP. Since x is optimal for the old LP we have that cx cx for any x feasible to the old LP. Therefore cx cx also for any x feasible to the new LP....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online