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Unformatted text preview: EE236A (Fall 200708) Lecture 11 The simplex method extreme points adjacent extreme points one iteration of the simplex method degeneracy initialization numerical implementation 111 Idea of the simplex method move from one extreme point to an adjacent extreme point with lower cost until an optimal extreme point is reached invented in 1947 (George Dantzig) usually developed and implemented for LPs in standard form questions 1. how do we characterize extreme points? (answered in lecture 3) 2. how do we move from an extreme point to an adjacent one? 3. how do we select an adjacent extreme point with a lower cost? 4. how do we find an initial extreme point? The simplex method 112 Extreme points to check whether x is an extreme point of a polyhedron defined by a T i x b i , i = 1 , . . . , m check that Ax b define A I = a T i 1 a T i 2 . . . a T i K , I = { i 1 , . . . , i K } where I is the set of active constraints at x : a T k x = b k , k I, a T k x &lt; b k , k negationslash I x is an extreme point if and only if rank ( A I ) = n The simplex method 113 Degeneracy an extreme point is nondegenerate if exactly n constraints are active at x A I is square and nonsingular ( K = n ) x = A 1 I b I , where b I = ( b i 1 , b i 2 , . . . , b i n ) an extreme point is degenerate if more than n constraints are active at x extremality is a geometric property (depends on P ) degeneracy/nondegeneracy depend on the representation of P ( i.e. , A and b ) The simplex method 114 Assumptions we will develop the simplex algorithm for an LP in inequality form minimize c T x subject to Ax b with A R m n we assume throughout the lecture that rank ( A ) = n if rank ( A ) &lt; n , we can reduce the number of varables implies that the polyhedron has at least one extreme point (page 324) implies that if the LP is solvable, it has an optimal extreme point (page 327) until page 1120 we assume that all the extreme points are nondegenerate The simplex method 115 Adjacent extreme points extreme points are adjacent if they have n 1 common active constraints moving to an adjacent extreme point given extreme point x with active index set I and an index k I , find an extreme point x that has the active constraints I \ { k } in common with x 1. solve the n equations a T i x = 0 , i I \ { k } , a T k x = 1 2. if A x , then { x + x  } is a feasible halfline: A ( x + x ) b 3. otherwise, x = x + x , where = min i : a T i x&gt; b i a T i x a T i x The simplex method 116 comments step 1: equations are solvable because A I is nonsingular step 3: &gt; because a T i x &gt; means i negationslash I , hence a T i x &lt; b i (for nondegenerate x ) new active set is I = I \ { k } { j } where j = argmin i : a T i x&gt; b i a T i x a T i x...
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 Spring '07
 Dr.Vandenber

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