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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 Fall '05
 BARD
 Linear Programming, Optimization, basic feasible solution, artificial variables, xi – di, standard simplex form

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