mthsc810-lecture07-2x2

mthsc810-lecture07-2x2 - MthSc 810 Mathematical Programming...

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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 that’s 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.0, we have a BFS....
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mthsc810-lecture07-2x2 - MthSc 810 Mathematical Programming...

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