Lecture 10 - Feasible Directions and Vertices

# Lecture 10 - Feasible Directions and Vertices - The step to...

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Lecture 10 Feasible Directions and Vertices UCSD Math 171A: Numerical Optimization Philip E. Gill ( pgill@ucsd.edu ) Monday, February 2nd, 2009 The step to a constraint hyperplane Let ¯ x be any point (not necessarily feasible). Let p be any nonzero direction. Steps of α along p changes the value of the i th constraint as: r i x + α p ) = r i x ) + α a T i p If a T i p 6 = 0, then constraint a T i x b i becomes active at α = σ i , where r i x + σ i p ) = 0 = r i x ) + σ i a T i p UCSD Center for Computational Mathematics Slide 2/44, Monday, February 2nd, 2009 From the previous slide, the step to constraint i satisﬁes 0 = r i x + σ i p ) = r i x ) + σ i a T i p which means we can solve for σ i , giving σ i = - r i x ) a T i p = r i x ) ( - a T i p ) We associate the negative sign with the denominator for reasons that will be explained later. UCSD Center for Computational Mathematics Slide 3/44, Monday, February 2nd, 2009 We have to be careful how we deﬁne σ i when a T i p = 0. Deﬁnition The step to the constraint a T i x b i from ¯ x along a nonzero p is given by: 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 - ∞ if a T i p = 0 and r i x ) < 0 undeﬁned if a T i p = 0 and r i x ) = 0 A positive σ i implies that the constraint is “in front of us” . A negative σ i implies that the constraint is “behind us” . UCSD Center for Computational Mathematics Slide 4/44, Monday, February 2nd, 2009

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x 1 + x 2 4 x 1 ≥ - 1 6 x 1 - x 2 18 x 1 + 3 x 2 6 3 x 2 6 (i.e., x 2 3 and x 2 6) constraint #1: x 1 + x 2 4 , a T 1 = ( 1 1 ) , b 1 = 1 constraint #2: x 1 ≥ - 1 , a T 2 = ( 1 0 ) , b 2 = - 1 constraint #3: - 6 x 1 + x 2 ≥ - 18 , a T 3 = ( - 6 1 ) , b 3 = - 18 constraint #4: x 1 + 3 x 2 6 , a T 4 = ( 1 3 ) , b 4 = 6 constraint #5: x 2 3 , a T 5 = ( 0 1 ) , b 5 = 3 constraint #6: - x 2 ≥ - 6 , a T 6 = ( 0 - 1 ) , b 6 = - 6 UCSD Center for Computational Mathematics Slide 5/44, Monday, February 2nd, 2009 This is just Ax b , with A = 1 1 1 0 - 6 1 1 3 0 1 0 - 1 b = 4 - 1 - 18 6 3 - 6 UCSD Center for Computational Mathematics Slide 6/44, Monday, February 2nd, 2009 1 2 5 6 4 1 2 3 5 6 #5 #3 #1 #6 #4 #2 4 x 2 x 1 0 UCSD Center for Computational Mathematics Slide 7/44, Monday, February 2nd, 2009 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
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## Lecture 10 - Feasible Directions and Vertices - The step to...

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