This preview shows page 1. Sign up to view the full content.
55
Therefore, our problem is equivalent to
min
v
5(6 + 2
v
)
4
+(6+2
v
)(1 +
v
) + 6(1 +
v
)
2
subject to
6+2
v
≥
0
,
1+
v
≥
0
.
Using a log barrier function for these constraints, we obtain the unconstrained
problem
min
v
B
μ
(
v
)
where
B
μ
(
v
) = 5(6 + 2
v
)
4
v
)(1 +
v
) + 6(1 +
v
)
2
−
μ
log(6 + 2
v
)
−
μ
log(1 +
v
)
.
Notice that if 1 +
v
≥
0, then 6 + 2
v
≥
0. Therefore, the Frst log term can be
dropped from
B
μ
(
v
).
CHALLENGE 10.4.
(a) The central path is deFned by
Ax
=
b
,
A
∗
w
+
s
=
c
,
μF
±
(
x
)+
s
=
0
,
where
F
±
(
x
)=
X
−
1
e
,
w
is an
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

Click to edit the document details