CO 350 Linear Optimization (Spring 2009)
Solutions to Assignment 5
[10]
1.
(a)
The variables that may enter are
x
4
and
x
5
. For
x
4
all coeﬃcients in
A
4
are positive. Since in
each case
b
i
= 0 any basis element may be leaving. Thus we have (
x
4
,x
2
),(
x
4
,x
3
) and (
x
4
,x
6
). If
x
5
is entering then
x
2
is leaving, so we have (
x
5
,x
2
).
(b)
The current matrix is
2
1
0
2

4 0
β

1 1
x
4
enters so we compute (after substituting
β
= 1) the lexicographic minimum of
(0
,
1
,

2
,
0) = min
±
(0
,
2
,
1
,
0)
2
,
(0
,
2
,

4
,
0)
2
,
(0
,
1
,

1
,
1)
1
²
therefore
x
3
is leaving.
(c)
We have the same matrix as in part (b). Note that (0
,
1
,
0
.
5
,
0)
>
L
(0
,
1
,

2
,
0), so we need to
ensure that
(0
,
1
,

2
,
0)
>
L
±
0
,
1
,

1
β
,
1
β
²
⇒ 
2
>

1
β
⇒
0
.
5
> β.
To have
x
6
leaving the basis, we need 0
.
5
> β >
0.
[20]
2.
A
=
³
1 1 2 0 2
2 3 3 1 2
´
,
b
=
³
3
7
´
,
c
= [5
,
8
,
4
,
2
,
3]
T
.
We use
x
4
as basic variable. Phase 1 is the LP
A
=
³
1 1 2 0 2 1
2 3 3 1 2 0
´
,
b
=
³
3
7
´
,
c
= [0
,
0
,
0
,
0
,
0
,

1]
T
.
Iteration 1,
B
=
{
6
,
4
}
and
x
*
B
= [3
,
7]
T
1. Solve
A
T
B
y
=
c
B
³
1 0
0 1
´³
y
1
y
2
´
=
³

1
0
´
⇒
y
=
³

1
0
´
2. Choose entering variable
c
1
=
c
1

A
T
1
y
= 0

[1
,
2][

1
,
0]
T
= 1
>
0
So
x
1
enters the basis.
1