sol2 - AMS 540 / MBA 540 (Fall, 2009) Estie Arkin Homework...

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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....
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