LP Mid I solns - Fall 2005 Solution to Midterm I Linear Programming 1(20(5 The point x is feasible(10 The corresponding columns of the A matrix are

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Fall 2005 Solution to Midterm I Linear Programming 1. (20%): (5%) The point x * is feasible. (10%) The corresponding columns of the A matrix are B = –4 5 –2 7– 2 1 26 2 . The determinant of B is: det( B ) = –16 + 10 – 84 – (8 – 24 – 70) = – 90 + 86 = – 4. Thus the matrix B is nonsingular so its columns are linearly independent. This implies that B is a basis so x * is a basic feasible solution (Theorem 2.4). (5%) Also, x * is an extreme point because all BFSs are extreme points (Theorem 2.3 2a. (5%) In this question, the first three constraints are linearly dependent; i.e., 2 × Constraint 1 = Constraint 2 – 4 × Constraint 3 This will not cause any difficulty in Phase I when a full set of artificial variables is used because the resulting equations will then be linearly independent. That is, the above relationship will no longer be true. 2b. (5%) The situation will be detected at the end of Phase I. The optimal basis will contain an artificial variable whose value is zero. Either the corresponding constraint can be removed or ignored.
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This note was uploaded on 11/06/2011 for the course ORI 391Q taught by Professor Bard during the Fall '05 term at University of Texas at Austin.

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LP Mid I solns - Fall 2005 Solution to Midterm I Linear Programming 1(20(5 The point x is feasible(10 The corresponding columns of the A matrix are

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