This preview shows pages 1–2. Sign up to view the full content.
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.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
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.
 Fall '05
 BARD

Click to edit the document details