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Lecture 10
Feasible Directions and Vertices
UCSD Math 171A: Numerical Optimization
Philip E. Gill
(
pgill@ucsd.edu
)
Monday, February 2nd, 2009
The step to a constraint hyperplane
Let ¯
x
be any point (not necessarily feasible).
Let
p
be any nonzero direction.
Steps of
α
along
p
changes the value of the
i
th constraint as:
r
i
(¯
x
+
α
p
) =
r
i
(¯
x
) +
α
a
T
i
p
If
a
T
i
p
6
= 0, then constraint
a
T
i
x
≥
b
i
becomes active at
α
=
σ
i
,
where
r
i
(¯
x
+
σ
i
p
) = 0 =
r
i
(¯
x
) +
σ
i
a
T
i
p
UCSD Center for Computational Mathematics
Slide 2/44, Monday, February 2nd, 2009
From the previous slide, the step to constraint
i
satisﬁes
0 =
r
i
(¯
x
+
σ
i
p
) =
r
i
(¯
x
) +
σ
i
a
T
i
p
which means we can solve for
σ
i
, giving
σ
i
=

r
i
(¯
x
)
a
T
i
p
=
r
i
(¯
x
)
(

a
T
i
p
)
We associate the negative sign with the denominator for reasons
that will be explained later.
UCSD Center for Computational Mathematics
Slide 3/44, Monday, February 2nd, 2009
We have to be careful how we deﬁne
σ
i
when
a
T
i
p
= 0.
Deﬁnition
The step to the constraint
a
T
i
x
≥
b
i
from ¯
x
along a nonzero
p
is
given by:
σ
i
=
r
i
(¯
x
)

a
T
i
p
if
a
T
i
p
6
= 0
+
∞
if
a
T
i
p
= 0 and
r
i
(¯
x
)
>
0
 ∞
if
a
T
i
p
= 0 and
r
i
(¯
x
)
<
0
undeﬁned
if
a
T
i
p
= 0 and
r
i
(¯
x
) = 0
A
positive
σ
i
implies that the constraint is
“in front of us”
.
A
negative
σ
i
implies that the constraint is
“behind us”
.
UCSD Center for Computational Mathematics
Slide 4/44, Monday, February 2nd, 2009
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x
1
+
x
2
≥
4
x
1
≥ 
1
6
x
1

x
2
≤
18
x
1
+ 3
x
2
≥
6
3
≤
x
2
≤
6
(i.e.,
x
2
≥
3 and
x
2
≤
6)
constraint #1:
x
1
+
x
2
≥
4
,
a
T
1
=
(
1
1
)
,
b
1
=
1
constraint #2:
x
1
≥ 
1
,
a
T
2
=
(
1
0
)
,
b
2
=

1
constraint #3:

6
x
1
+
x
2
≥ 
18
,
a
T
3
=
(

6
1
)
,
b
3
=

18
constraint #4:
x
1
+ 3
x
2
≥
6
,
a
T
4
=
(
1
3
)
,
b
4
=
6
constraint #5:
x
2
≥
3
,
a
T
5
=
(
0
1
)
,
b
5
=
3
constraint #6:

x
2
≥ 
6
,
a
T
6
=
(
0

1
)
,
b
6
=

6
UCSD Center for Computational Mathematics
Slide 5/44, Monday, February 2nd, 2009
This is just
Ax
≥
b
, with
A
=
1
1
1
0

6
1
1
3
0
1
0

1
b
=
4

1

18
6
3

6
UCSD Center for Computational Mathematics
Slide 6/44, Monday, February 2nd, 2009
1
2
5
6
4
1
2
3
5
6
#5
#3
#1
#6
#4
#2
4
x
2
x
1
0
UCSD Center for Computational Mathematics
Slide 7/44, Monday, February 2nd, 2009
Consider the point ¯
x
and direction
p
such that
¯
x
=
3
3
!
and
p
=

1
0
!
r
=
A
¯
x

b
=
2
4
3
6
0
3
←
constraint #5 is active at ¯
x
⇒
¯
x
is feasible, with
A
(¯
x
) =
{
5
}
,
A
a
=
(
0 1
)
and
b
a
=
(
3
)
.
UCSD Center for Computational Mathematics
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