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Unformatted text preview: CO 350 Linear Optimization (Spring 2009) Solution to Assignment 2 [10] 1. We begin by eliminating the free variable x 2 using the equality constraint. That is, since x 2 = x 1 1, we can replace all instances of x 2 by x 1 1. This gives the following LP: maximize 2 x 1 + ( x 1 1) + 3 x 3 subject to x 1 + ( x 1 1) + x 3 2 x 1 x 3 2 x 1 , x 3 . We now convert the problem to standard inequality form by multiplying 1 to the first constraint. Note that the set of feasible solutions and optimal solutions do not change when we drop the constant from the objective function. maximize 3 x 1 + 3 x 3 subject to 2 x 1 x 3  3 x 1 x 3 2 x 1 , x 3 . Adding slack variables s 1 and s 2 gives the standard equality form: maximize 3 x 1 + 3 x 3 subject to 2 x 1 x 3 + s 1 = 3 x 1 x 3 + s 2 = 2 x 1 , x 3 , s 1 , s 2 . [10] 2. First Solution: The LP problem is equivalent to maximize 3 u 1 3 v 1 x 2 + 2 x 3 (+2) subject to u 1 v 1 + 2 x 3 2 u 1 + v 1 2 x 3  2 u 1 + v 1 x 2  1 x 3 1 u 1 , v 1 , x 2 , x 3 , where we used the substitutions x 1 = u 1 v 1 and x 3 = x 3 + 1 . Alternative Solution: Isolate the free variable x 1 in the first constraint: x 1 = 4 2 x 3 , and substitute x 3 = x 3 + 1 . We obtain the equivalent LP problem: maximize x 2 4 x 3 (+8) subject to x 2 + 2 x 3 1 x 3 1 x 2 , x 3 . 1 [10] 3. Denote the given problem by (P). Then the dual of (P) is given by (D) min 3 y 1 + 2 y 2 + 4 y 3 + 2 y 4 subject to y 1 y 2 + 2 y 3 + 2 y 4 5 3 y 1 y 3 + 3 y 4 5 y 1 + 3 y 2 + 2 y 3 y 4 3 . [10] 4. (Exercise 1 in pages 5455 of the course notes.) (a) The original primal and dual are ( P ) maximize c T x subject to Ax = b x and ( D ) minimize b T y subject to A T y c, where A = ( a ij ) R m n , b = ( b i ) R m , and c = ( c j ) R n . Then the modified primal and dual are ( P ) maximize c T x subject to A x = b x and ( D ) minimize ( b ) T y subject to ( A ) T y c, where A = ( a ij ) R m n and b = ( b i ) R m . The vector c in ( P ) and ( D ) is the same as in ( P ) and ( D )....
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 Fall '07
 S.Furino,B.Guenin

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