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Problem Set 12

1. Convert the following LP to standard form:

min  

s. t.

≥ |1 − 2 + 33| 

≥ |−1 + 32 − 3| 

1, 2, 3 ≥ 0 

Equivalently the problem can be restated as

min  

s. t.

≥ (1 − 2 + 33) 

≥ −(1 − 2 + 33) 

≥ (−1 + 32 − 3) 

≥ −(−1 + 32 − 3) 

1,

2,

3 ≥0. 

Now we can put this LP in standard form. Note that, since z is greater than or equal to an absolute

value, it must be non-negative.

max − 

s. t.

1 − +(

1 − 2 + 33)

= 0 

2 − − (1 − 2 + 33) = 0 

3 − + (−1 + 32 − 3) = 0 

4 − − (−1 + 32 − 3) = 0 

1, 2, 3, , 1, 2, 3, 4 ≥ 0 

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2.  Consider the following partial simplex tableau for a max model (we have used the notation to

denote the coefficients of the basic variables in the profit function):

Profit→  20 30 25 0 0 0

  Basic 1  2  3  1  2  3  RHS

3 0 1 1 -2 0 1001 1 0 0 1 0 200

-5 0 0 -2 4 1 400

Indirect Effect  

Relative Profit  −  

(a)  Write down the LP which gave rise to the above tableau.

(b)  What are the basic-variables?

(c) What are the non-basic variables?

(d)  What are the values of all the variables and the value of the objective function at the corner

point corresponding to the tableau?

(e) Complete the tableau.

(f)  Is the current solution optimal? If not, which variable should enter and which should exit?

(g) Use the simplex method to find the optimal solution and the optimal profit.

(a) max 201 + 302 + 253 

s. t.31 + 3 + 1 − 22 = 100 

1 + 2 + 2 = 200 

−51 − 21 + 42 + 3 = 400 

1, 2, 3, 1, 2, 3 ≥ 0 

(b)  Basic: 2, 3, 3 

(c) Non-Basic: 1, 1, 2 

(d) 1 = 0, 2 = 200, 3 = 100 

1 = 0 , 2 = 0 , 3 = 400 

= 8,500 

(e) 

20 30 25 0 0 0

  Basic 1  2  3  1  2  3  RHS

25 3  3 0 1 1 -2 0 100 -500

30 2  1 1 0 0 1 0 200 +200

0 3  -5 0 0 -2 4 1 400 +100←exit

Indirect Effect   105 30 25 25 -20 0

Relative Profit  −   -85 0 0 -25 20 0

(f) No. The relative profit of 2 is 20. 2: enter 3: exit

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  (g)

All relative profits ≤0 ⇒ optimal solution:

1∗ = 1∗ = 3∗ = 0,   2∗ = 100,   3∗ = 200,   2∗ = 100,   ∗ = 10,500. 

3. Fill in the missing elements in the following optimal simplex tableau for a maximization problem:

20 30 0 0

  Basic 1  2  1  2  RHS

1 0 1/4 3

4 0 1 3/8 3/2

 

 −   0 0 -3/4 -1/4  ∗ = 15 

Explain how you filled in the missing elements. (Note: ∗ denotes the optimal profit.)

Let the profit function be 11 + 42 since from the  column, we know that the coefficient of 2 

is 4. We also know 1∗ = 3, 2∗ = 3/2 since they are the basic variables. Thus,

31 +3

2× 4 = 15 ⇒ 1 =

15 − 6

3= 3

20 30 0 0

 

Basic 1 

2 

1 

2 

RHS3 1  1 0 1/12 1/4 3

4 2  0 1 3/8 -1/8 3/2

  3 4 3/4 1/4

 −   0 0 -3/4 -1/4  ∗ = 15 

20 30 25 0 0 0

  Basic 1  2  3  1  2  3  RHS

25 3  1/2 0 1 0 0 1/2 300

30 2  9/14 1 0 1/2 0 -1/4 100

0 2  -5/14 0 0 -1/2 1 1/4 100

Indirect Effect   80 30 25 15 0 5Relative Profit  −   -20 0 0 -15 0 -5

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  Now the top column can be filled. From the fact that  corresponding to 1 = 3/4, we get

3 + 4 × 3 8⁄ = 3 4⁄   ⇒ =1

3�3

4− 3

2 = − 1

4 

where

 is the coefficient of

1 in the first constraint. Similarly, denoting the coefficient of

2 in the

second constraint by , we get

3

4+ 4 =

1

4 

⇒ = − 1

8 

This completes the tableau.

4.  Convert the following LP to standard form:

min –3 + 4 − 2 + 5 

s. t.

4 − + 2 −   = −2 

+ + 3 − ≤  14 

−2 + 3 − + 2 ≥  2 

≥ 0,  ≥  0,  ≥  0,   unrestricted in sign

min −3 + 4 − 2 + 5 − 5  

s. t.

−4 + − 2 + − = 2 

+ + 3 − +  + 1 = 14 

−2 + 3 − + 2 − 2  − 2 = 2 

, , , , , 1, 2 ≥ 0