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CO 350 Linear Optimization (Spring 2009)
Solution to Assignment 3
[20]
1.
(a)
B
=
{
1
,
4
,
6
}
,
x
B
= (2
,
0
,
0
,
5
,
0
,
4)
T
.
(b)
z
= 13
(c)
x
3
enters.
t
= min
{
5
/
2
,
2
}
= 2, so
x
1
leaves. The new tableau, for basis
{
3
,
4
,
6
}
is
z
+ 2
x
1
+
x
2
+ 5
x
5
= 17
2
x
1

x
2

2
x
5
+
x
6
=
8

2
x
1
+ 9
x
2
+
x
4

5
x
5
=
1
x
1

2
x
2
+
x
3
+ 2
x
5
=
2
(d)
The new tableau is optimal, as all ¯
c
j
≤
0. (So the objective value cannot increase if 1
,
2
,
or
5 enters the basis.) The original LP problem has an optimal solution. [0
,
0
,
2
,
1
,
0
,
8]
T
is an optimal
solution with objective value 17.
[10]
2.
The LP problem is unbounded: ¯
c
4
>
0 and ¯
a
i
4
≤
0 for all
i
. That is, if we choose
x
4
to enter,
x
4
can become arbitrarily large.
From the tableau we can see that we can get feasible solutions
x
(
t
),
t
≥
0, as follows:
z
(
t
) = 42 + 4
t
x
1
(
t
) = 0
x
2
(
t
) = 7 +
t
x
3
(
t
) = 3 + 2
t
x
4
(
t
) =
t
x
5
(
t
) = 0
x
6
(
t
) = 8 + 3
t.
[10]
3.
(a)
The tableau corresponding to
B
=
{
1
,
2
,
3
}
, which is obtained after elementary row operations,
is
z

9
2
x
4
+
5
2
x
5

29
x
6
=

77
2
x
1
+
x
4
+
x
6
=
9
x
2

5
x
4
+
2
x
5

24
x
6
=

31
x
3

1
2
x
4
+
1
2
x
5

2
x
6
=

3
2
It is not feasible.
The tableau corresponding to
B
=
{
1
,
4
,
5
}
is
z

2
3
x
2

7
3
x
3

25
3
x
6
=

43
3
x
1
+
1
3
x
2

4
3
x
3
+
2
3
x
6
=
2
3

1
3
x
2
+
4
3
x
3
+
x
4
+
16
3
x
6
=
25
3

1
3
x
2
+
10
3
x
3
+
x
5
+
4
3
x
6
=
16
3
1
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View Full Document This tableau is feasible.
(b)
Consider the feasible tableau (
T
1
) from (a) corresponding to the basis
B
1
=
{
1
,
4
,
5
}
. We shall
use the largest coeﬃcient rule when choosing entering variables.
Since
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This note was uploaded on 04/11/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Fall '07 term at Waterloo.
 Fall '07
 S.Furino,B.Guenin

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