CS149-Notes-3-Equiv

CS149-Notes-3-Equiv - -b 0, also A x = ( A,-I ) p x Ax-b P...

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1 Equivalence of LP S and LP K Theorem: LP S and LP K are eqivalent problems. Proof: Let P 1 , c 1 and P 2 , c 2 be instances of any optimization problem. (Refer to Lecture 2, Slide 31 for a deFnition of optimization problems. Note that we are minimizing). Lemma: Assume that x 1 P 1 x 2 P 2 with c 2 T x 2 c 1 T x 1 and x 2 P 2 x 1 P 1 with c 1 T x 1 c 2 T x 2 Then if x 2 P 2 is optimal, then the analagous x 1 P 1 is optimal, and they have the same objective function value. Proof: Let y 1 be any element of P 1 . Let y 2 be an element of P 2 with c 2 T y 2 c 1 T y 1 ( as in the supposition). Then: c 1 T x 1 c 2 T x 2 c 2 T y 2 c 1 T y 1 . Therefore x 1 is optimal. a 1. Let LP K : min c T x s.t. x 0 , Ax b be given. Then with ˜ A := ( A, - I ) , ˜ b := b, ˜ c := p c 0 P , ˜ x := p x x s P we can describe an optimization problem in standard form ( LP S ) by: min ˜ c T ˜ x s.t. ˜ x 0 , ˜ A ˜ x = ˜ b Let ˜ x be a solution of LP S . Then x is a solution of LP K with ˜ c T ˜ x = ( c T , 0 T ) · p x x s P = c T x + 0 T · x s = c T x Now let x be a solution of LP K and let ˜ x := ( x Ax - b ) . Then, with ˜ x 0 since x 0 and Ax
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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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CS149-Notes-3-Equiv - -b 0, also A x = ( A,-I ) p x Ax-b P...

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