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Unformatted text preview: -b 0, also A x = ( A,-I ) p x Ax-b P = Ax-Ax + b = b Therefore x is a solution of LP S . or these it follows that: c T x = c T x + 0 T ( Ax-b ) = ( c T , T ) p x Ax-b P = c T x 2 2. Let LP S be given as min c T x s.t. x , Ax = b . Then with A := p A-A P , b := p b-b P , c := c, x := x we can describe an optimization problem in canonical form ( LP K ) by: min c T x s.t. x , A x b From these it follows: x is exactly the feasible solution for LP S when x is a feasible solution for LP K . Furthermore, it follows that c T x = c T x . The proof concerning this is similar to the rst case....
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- Spring '09