This preview shows pages 1–2. Sign up to view the full content.
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 BTILO
: 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
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.
 Spring '10
 Bertsekas
 Optimization

Click to edit the document details