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LP Mid I solns (2005)

# LP Mid I solns (2005) - Fall 2005 Solution to Midterm I...

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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 2 6 – 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.

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