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Unformatted text preview: 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 ) > 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 6 1 1 3 1 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 Consider the point ¯ x and direction p such that ¯ x = 3 3 ! and p = 1 ! r = A ¯ x b = 2 4 3 6 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 ← constraint #5 A a p = ( 0 1 ) 1 = 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 ¯ x = 3 3 4 x 2 x 1 At ¯ x we compute the step to each of the constraint hyperplane. r (¯ x ) = 2 4 3 6 3 and Ap =  1 1 6 1 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 ¯ x = 3 3 4 x 2 x 1 r (¯ x ) = 2 4 3 6 3 and Ap =  1 1 6 1 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...
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This note was uploaded on 03/19/2012 for the course MATH 171a taught by Professor Staff during the Spring '08 term at UCSD.
 Spring '08
 staff
 Linear Programming

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