Hw4_Sol - Math464 - HW 4 Due on Friday, Feb 12 1 Linear...

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Math464 - HW 4 Due on Friday, Feb 12 1 Linear Optimization (Spring 2010) Brief solutions to Homework 4 1. Procedure for finding basic solutions: As in the original case, we still start by choosing the indices for m linearly independent columns, B (1) ,...,B ( m ). The indices of the non- basic x j ’s ( j 6 = B (1) ,...,B ( m )) are divided into two disjoint sets, say, L and U . We set x j = 0 j L and x j = u j j U . As before, we then solve the system A x = b for the basic variables x B (1) ,...,x B ( m ) . Theorem equivalent to Theorem 2.4 in BT-ILO : Just replace the statement (b) in the original theorem by the following: (b’) If i 6 = B (1) ,...,B ( m ) , then either x i = 0 or x i = u i . The proof is essentially similar to the proof of the original theorem. Assume that x R n is a feasible point, and there exists indices B (1) ,...,B ( m ) that satisfy statements (a) (as in the original theorem) and statement (b’) above. The active constraints are A x = b ; x j = 0 , j L ; and x j = u j , j U . We get A x = n X i =1 A i x i = m X i =1 A B ( i ) x B ( i ) + X j U A j u j = b . We can write the above system as follows: m X i =1 A B ( i ) x B ( i ) = b - X j U A j u j . Since the basic columns of A , i.e., A B (1) ,...,A B ( m ) are linearly independent, the above system of active constraints has a unique solution, and hence x is a bfs. Conversely, assume that
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Hw4_Sol - Math464 - HW 4 Due on Friday, Feb 12 1 Linear...

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