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Unformatted text preview: x cannot be an optimal solution to this LP. 3). Consider an LP: ( p 1 and p 2 are given constants.) max z =p 1 x 1 + p 2 x 2 x 1x 2 = 0 ≤ x 1 , x 2 ≤ 1 (a). What are the extreme points and extreme directions of this LP? (b). Write the equivalent LP in terms of the extreme points and extreme directions. (c). Now suppose that p 1 > p 2 . What is the optimal solution to the LP? (Make sure to state z as well as x 1 , x 2 .)...
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This note was uploaded on 01/31/2011 for the course AMS 540 taught by Professor Arkin,e during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Arkin,E

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