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# handout19 - Recap The phase-1 Linear Program Math 171A...

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Math 171A: Linear Programming Lecture 19 Finding a Feasible Point II Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Friday, February 18th, 2011 Recap: The phase-1 Linear Program To find a feasible x for minimize x R n c T x subject to Ax b | {z } general constraints , x 0 | {z } simple bounds Solve the linear program : minimize θ R , x R n θ subject to a T 1 x + θ b 1 a T 2 x + θ b 2 . . . . . . a T m x + θ b m x 0 , θ 0 UCSD Center for Computational Mathematics Slide 2/27, Friday, February 18th, 2011 Phase-1 LP example Example: min c T x subject to Ax b , x 0 , for c = 2 1 ! , A = 1 1 1 2 - 1 1 , b = 1 1 - 2 UCSD Center for Computational Mathematics Slide 3/27, Friday, February 18th, 2011 x 2 x 1 #1 #2 #3 #5 #4

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The point x 0 = 0 is infeasible, with r ( x 0 ) = Ax 0 - b = 1 1 1 2 - 1 1 x 0 - 1 1 - 2 = - 1 - 1 2 UCSD Center for Computational Mathematics Slide 5/27, Friday, February 18th, 2011 θ 0 = max { b i } = max { 1 , 1 , - 2 } = 1 The phase-1 problem is: minimize θ R , x R n θ subject to x 1 + x 2 + θ 1 x 1 + 2 x 2 + θ 1 - x 1 + x 2 + θ ≥ - 2 x 1 , x 2 , θ 0 with initial vertex x 0 = 0 , θ 0 = 1 UCSD Center for Computational Mathematics Slide 6/27, Friday, February 18th, 2011 x 2 x 1 #1 #2 #3 #5 #4 x 2 x 1 #1 #2 #3 #5 #4
It is not necessary to shift feasible general constraints. e.g., constraint #3 ( - x 1 + x 2 ≥ - 2) is feasible at x 0 = 0. x 2 x 1 #1 #2 #3 #5 #4 If constraint #3 is unshifted, the phase-1 LP is: minimize θ R , x R n θ subject to x 1 + x 2 + θ 1 x 1 + 2 x 2 + θ 1 - x 1 + x 2 ≥ - 2 x 1 , x 2 , θ 0 The definition of x 0 and θ 0 does not change. UCSD Center for Computational Mathematics Slide 10/27, Friday, February 18th, 2011 In this case, the phase-1 LP is: minimize ¯ x R n +1 ¯ c T ¯ x subject to ¯ A ¯ x ¯ b , ¯ x

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