mthsc810-lecture07-1x2

mthsc810-lecture07-1x2 - MthSc 810: Mathematical...

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Unformatted text preview: MthSc 810: Mathematical Programming Lecture 7 Pietro Belotti Dept. of Mathematical Sciences Clemson University September 15, 2011 Reading for today: textbook, Sections 3.1 and 3.2 Reading for Sep. 20: Sections 3.3 and 3.4 Homework #4 out today due September 22. Recap Consider the Linear Optimization problem LP : min { c x : A x = b , x } . If P = { x R n + : A x = b } contains 1 extreme points, and LP admits an optimal solution, then there exists an optimal solution that is an extreme point. Even more: if P has at least one extreme point, then either there exists an optimal thats an extreme point, or the cost is (i.e., the problem is unbounded) The standard form is convenient min { c x : A x = b , x } . There exists an extreme point since P R n + . If A R m n , then rank ( A ) = m n . (If rank ( A ) < m , we can eliminate some equations) P is the intersection of a subspace ( A x = b ) with a cone (the first orthant R n + ) Basic solutions Select m linearly independent columns (variables) of A Call B this submatrix and denote A = ( B | N ) The variable vector x can be partitioned accordingly as x = x B x N ( x B R m , x N R n m ) So A x = b is equivalent to B x B + N x N = b Then x B = B 1 b B 1 N x N x = x B x N = B 1 b is a basic solution Why? there are n active constraints: A x = b and x N , or B N I x B x N = b If, in addition, B 1 b 0, we have a BFS. When is a BFS optimal?When is a BFS optimal?...
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mthsc810-lecture07-1x2 - MthSc 810: Mathematical...

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