Math
484.1 November 8, 2007 Midterm 2
5 problems, 15 pts each. By: Nathan Pish
1. Solve linear program where all xi >= 0
x1 |
x2 |
1 |
-x5 |
|
1 |
2 |
3 |
4 |
=
x3 |
1 |
2 |
0 |
-1 |
=
-x2 |
0 |
2 |
-3 |
0 |
àmin |
First, we make the tableaux standard.
x1 |
x2 |
x5 |
1 |
|
1 |
2 |
-4 |
3 |
=
x3 |
-1 |
-2 |
-1
* |
0 |
=
x2 |
0 |
2 |
0 |
-3 |
àmin |
Then, we pivot on the -1 in the x5 column.
x1 |
x2 |
x2 |
1 |
|
-3 |
-6 |
4 |
3 |
=
x3 |
-1 |
-2 |
-1 |
0 |
=
x5 |
0 |
2 |
0 |
-3 |
àmin |
Finally, we combine the x2
columns.
x1 |
x2 |
1 |
|
-3 |
-2 |
3 |
=
x3 |
-1 |
-3 |
0 |
=
x5 |
0 |
2 |
-3 |
àmin |
We get an optimal tableaux with min = -3
at x1 = x2 = 0, x5 = 0, x3 = 3.
2. Solve linear program where all xi >= 0
2x1 |
-x2 |
1 |
x5 |
|
1 |
2 |
3 |
4 |
=
x3 |
1 |
2 |
0 |
-1 |
=
-x4 |
0 |
2 |
-3 |
0 |
àmax |
First, we need to make a
standard tableaux.
x1 |
x2 |
x5 |
1 |
|
2 |
-2 |
4 |
3 |
=
x3 |
-2 |
2 |
1 |
0 |
=
x4 |
0 |
2 |
0 |
3 |
àmin |
This is an optimal tableaux with min
= 3 à max = -3 at x1 = x2
= x5 = 0, x4 = 0, x3 = 3.
3. Solve transportation problem.
This can be solved in phase 1.
potentials |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
-1 |
1 |
|
0 |
2
(1) |
4
(3) |
3
(3) |
2
(1) |
1
(1) |
2
(2) |
1
6 |
2
(3) |
1
3 |
9 |
0 |
1
6 |
2
(1) |
3
(3) |
1
1 |
1
(1) |
0
1 |
1
(0) |
3
(4) |
1
0 |
8 |
-1 |
2
(2) |
2
(2) |
1
(0) |
3
(1) |
1
6 |
1
(0) |
3
(1) |
0
1 |
2
2 |
9 |
0 |
2
(1) |
1
1 |
0
6 |
1
(0) |
2
(2) |
1
(1) |
2
(1) |
3
(4) |
1
1 |
8 |
|
6 |
1 |
6 |
1 |
6 |
1 |
6 |
1 |
6 |
|
Min = 28
4. Solve transportation problem.
This can also be solved in phase 1.
potentials |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
|
0 |
3
(3) |
1
1 |
3
(2) |
2
(2) |
1
4 |
2
(2) |
4
(3) |
2
(1) |
1
4 |
9 |
0 |
0
6 |
1
(0) |
3
(2) |
1
(1) |
1
(0) |
5
(5) |
1
2 |
3
(2) |
1
(0) |
8 |
0 |
2
(2) |
2
(1) |
1
6 |
3
(3) |
1
2 |
1
(0) |
3
(1) |
1
1 |
2
(1) |
9 |
-1 |
2
(1) |
3
(1) |
2
(0) |
1
1 |
2
(0) |
1
1 |
2
4 |
3
(1) |
2
2 |
8 |
|
6 |
1 |
6 |
1 |
6 |
1 |
6 |
1 |
6 |
|
Min = 34
5. Solve job assignment problem.
1* |
2 |
3 |
4 |
2 |
2 |
3 |
4 |
1* |
2 |
3 |
1* |
5 |
3 |
1 |
1 |
3 |
4 |
3 |
1* |
5 |
5 |
2* |
1 |
3 |
We start by subtracting 1 from every row.
0 |
1 |
2 |
3 |
1 |
1 |
2 |
3 |
0 |
1 |
2 |
0 |
4 |
2 |
0 |
0 |
2 |
3 |
2 |
0 |
4 |
4 |
1 |
0 |
2 |
Now, we subtract 1 from the third column
0* |
1 |
1 |
3 |
1 |
1 |
2 |
2 |
0* |
1 |
2 |
0* |
3 |
2 |
0 |
0 |
2 |
2 |
2 |
0* |
4 |
4 |
0* |
0 |
2 |
The zeros are marked in this last table, and
we get the min by adding the original values together. Min = 6