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