Example 1: (continued)
A
=
±

1
5
0
1
1
3

1
1
4
2
²
,
b
=
±
4

5
²
,
c
=

4
6

1

3

1
Optimality?
λ
*
=
solve(A’,c)
λ
*
=
±
1

1
²
,
which are unique Lagrange multipliers
⇒
x
*
is optimal, with
‘
(
x
*
) =
c
T
x
*
= 9.
UCSD Center for Computational Mathematics
Slide 5/37, Friday, January 28, 2011
Example 1: (continued)
A
=
±

1
5
0
1
1
3

1
1
4
2
²
,
b
=
±
4

5
²
,
c
=

4
6

1

3

1
Note that
b
x
=

3
2
1
2
0
0
0
is
another
basic solution of
Ax
=
b
, with
‘
(
b
x
) =
c
T
b
x
= 9.
UCSD Center for Computational Mathematics
Slide 6/37, Friday, January 28, 2011
Example 2:
A
=
±

1
5
0
1
1
3

1
1
4
2
²
,
b
=
±
4

5
²
,
c
=

8

2
6
3
3
rank(
A
) = 2
⇒
A
has
independent rows
.
Feasibility?
x
*
=
solve(A,b)
x
*
=
0
1
0

1
0
,
which is a basic solution of
Ax
=
b
UCSD Center for Computational Mathematics
Slide 7/37, Friday, January 28, 2011
Example 2: (continued)
A
=
±

1
5
0
1
1
3

1
1
4
2
²
,
b
=
±
4

5
²
,
c
=

8

2
6
3
3
Optimality?
λ
*
=
solve(A’,c)
gives an incompatibility message.
⇒
c
6∈
range(
A
T
) and the solution is unbounded.
UCSD Center for Computational Mathematics
Slide 8/37, Friday, January 28, 2011