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# handout12 - Recap Vertices(aka corner points Math 171A...

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Math 171A: Linear Programming Lecture 12 Finding a Vertex Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Wednesday, February 2nd, 2011 Recap: Vertices (aka corner points) A vertex is a feasible point at which there are at least n linearly independent constraints active. i.e., the active-constraint matrix A a has rank n at a vertex. UCSD Center for Computational Mathematics Slide 2/30, Wednesday, February 2nd, 2011 x 2 x 1 #2 #1 #4 #3 #5 A a = 1 1 1 2 0 1 A a = 1 1 1 0 (nondegenerate vertex) (degenerate vertex) A = 1 1 1 0 - 1 0 0 1 1 2 When can we guarantee that Ax b has a vertex? How do we compute a vertex? Theorem (Existence of a vertex) Suppose that F = { x : Ax b } has at least one point. If rank( A ) = n then F has at least one vertex. UCSD Center for Computational Mathematics Slide 4/30, Wednesday, February 2nd, 2011

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Proof: (Constructive) The assumptions are: There is at least one point x 0 (say) such that Ax 0 b . There is at least one subset of n independent rows of A . Consider any feasible point x 0 . Define the active set at x 0 , i.e., A 0 x 0 = b 0 Assume that rank( A 0 ) < n (otherwise there is nothing to prove!) A 0 may be any shape (because it may have dependent rows). UCSD Center for Computational Mathematics Slide 5/30, Wednesday, February 2nd, 2011 We construct a sequence of feasible points, labeled as x 0 , x 1 , . . . , x k , x k +1 , . . . with active-set matrices A 0 , A 1 , . . . , A k , A k +1 , . . . such that rank( A k +1 ) > rank( A k ) Next we describe how to construct x k +1 from x k (i.e., the k th iteration) UCSD Center for Computational Mathematics Slide 6/30, Wednesday, February 2nd, 2011 k th iteration: Step 1 The previous step gives x k , A k
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