handout11

# handout11 - Recap computing the step to a constraint Math...

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Math 171A: Linear Programming Lecture 11 Vertices Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Monday, January 31, 2011 Recap: computing the step to a constraint The step to the constraint a T i x b i from a feasible point ¯ x along a nonzero p is: σ i = r i x ) - a T i p if a T i p 6 = 0 + if a T i p = 0 and r i x ) > 0 undefined if a T i p = 0 and r i x ) = 0 UCSD Center for Computational Mathematics Slide 2/41, Monday, January 31, 2011 Example: Consider the constraints Ax b , with A = 1 1 1 0 - 6 1 1 3 0 1 0 - 1 , and b = 4 - 1 - 18 6 3 - 6 UCSD Center for Computational Mathematics Slide 3/41, Monday, January 31, 2011 1 2 5 6 4 1 2 3 5 6 #5 #3 #1 #6 #4 #2 4 x 2 x 1 0

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Consider the point ¯ x and direction p such that ¯ x = 3 3 ! and p = - 1 0 ! r = A ¯ x - b = 2 4 3 6 0 3 constraint #5 is active at ¯ x ¯ x is feasible, with A x ) = { 5 } , A a = ( 0 1 ) and b a = ( 3 ) . UCSD Center for Computational Mathematics Slide 5/41, Monday, January 31, 2011 Ap = - 1 - 1 6 - 1 0 0 constraint #5 A a p = ( 0 1 ) - 1 0 = 0 0 p is a feasible direction. UCSD Center for Computational Mathematics Slide 6/41, Monday, January 31, 2011 1 2 5 6 4 1 2 3 5 6 #5 #3 #1 #6 #4 #2 p = - 1 0 ¯ x = 3 3 4 x 2 x 1 0 At ¯ x we compute the step to each of the constraint hyperplane. r x ) = 2 4 3 6 0 3 and Ap = - 1 - 1 6 - 1 0 0 Step to constraint #1: σ 1 = r 1 x ) ( - a T 1 p ) = 2 ( - ( - 1)) = 2 UCSD Center for Computational Mathematics Slide 8/41, Monday, January 31, 2011
1 2 5 6 4 1 2 3 5 6 #5 #3 #1 #6 #4 #2 p = - 1 0 ¯ x = 3 3 4 x 2 x 1 0 r x ) = 2 4 3 6 0 3 and Ap = - 1 - 1 6 - 1 0 0 Step to constraint #2: σ 2 = r 2 x ) ( - a T 2 p ) = 4 ( - ( - 1)) = 4 UCSD Center for Computational Mathematics Slide 10/41, Monday, January 31, 2011 1 2 5 6 4 1 2 3 5 6 #5 #3 #1 #6 #4 #2 p = - 1 0 ¯ x = 3 3 4 x 2 x 1 0 r x ) = 2 4 3 6 0 3 and Ap = - 1 - 1 6 - 1 0 0

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