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
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 Spring '10
 Bertsekas
 Optimization, Trigraph, BMW Sports Activity Series, 10 m, Det, 1 j

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