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Unformatted text preview: CS 435 : Linear Optimization Fall 2008 Lecture 13: Introduction to Duality Lecturer: Sundar Vishwanathan Computer Science & Engineering Indian Institute of Technology, Bombay Let x be an extreme point. Suppose it is given by A x = b ; A 00 x < b 00 . We call A x = b as the de ning hyperplanes of x . We know that the neighbours of x are along the columns of A 1 . 1 Proof of correctness of Simplex algorithm Theorem 1 If the cost does not increase along any of the columns of A 1 then x is optimal. Proof: The columns of A 1 span R n . Let x opt be an optimal point. We need to show that c T x opt c T x . Since the columns of A 1 form a basis of R n (why?) the vector x opt x can be represented as a linear combination of them say as given below. x opt x = X j ( A 1 ) ( j ) (1) Premultiplying both sides with A yields A x opt A x = X j A ( A 1 ) ( j ) : (2) We know that A x opt b and A x = b hence A ( x opt x ) 0. That is, the lhs is a vector, all of whose components are non-positive. Also note that A ( A 1 ) ( j ) is an n ¢ 1 vector whose j th element is 1 and remaining elements are 0. Hence A x opt A x = B B B B B B @ 1 2 : : : n 1 C C C C C C A (3) This implies that j 0 for all j . Now, c T x opt c T x = X j c T ( A 1 ) ( j ) (4) Since the cost decreases along the columns of A 1 we have c T ( A 1 ) ( j ) 0 and since j 0 we conclude that P j c T ( A 1 ) j 0. Hence c T x opt c T x , as desired. £ Note: From the above theorem we infer that when the Simplex algorithm terminates it gives us an optimal solution....
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