CSC 445/545 Notes 1.4: Practice with the 2-phase method.

The problem is:

Maximize X1 + X2 + 2 * X3 + X4

subject to

 -X1          +     X3  +  2 X4  < = -3
 -X1   +   X2                    < = -7
 -X1   + 2 X2           +    X4  < = -5

X1, X2, X3, X4 > = 0

Question 1: Set up the phase 1 dictionary for this problem and make the first pivot.

Answer #1:

Phase 1: Before pivoting to make feasible.


X5 =  -3  +   X1            -   X3   - 2 X4   +   X0 
X6 =  -7  +   X1   -   X2                     +   X0 
X7 =  -5  +   X1   - 2 X2            -   X4   +   X0 
-----------------------------------------------
w  =   0                                    -     X0

Taking the first pivot:

X0  enters.   X6  leaves.  z = -0.000000 
The initial dictionary:
X5 =   4.00+   0.00 X1 +   1.00 X2 -   1.00 X3 -   2.00 X4 +   1.00 X6 
X0 =   7.00-   1.00 X1 +   1.00 X2 +   0.00 X3 +   0.00 X4 +   1.00 X6 
X7 =   2.00+   0.00 X1 -   1.00 X2 +   0.00 X3 -   1.00 X4 +   1.00 X6 
-----------------------------------------------
w  =  -7.00+   1.00 X1 -   1.00 X2 +   0.00 X3 +   0.00 X4 -   1.00 X6 

X1  enters.   X0  leaves.  z = -7.000000

Question 2: The optimal solution to the phase 1 problem is:

After   1 pivot:
X5 =   4.00+   1.00 X2 -   1.00 X3 -   2.00 X4 +   1.00 X6 +   0.00 X0 
X1 =   7.00+   1.00 X2 +   0.00 X3 +   0.00 X4 +   1.00 X6 -   1.00 X0 
X7 =   2.00-   1.00 X2 +   0.00 X3 -   1.00 X4 +   1.00 X6 +   0.00 X0 
-----------------------------------------------
w  =  -0.00+   0.00 X2 +   0.00 X3 +   0.00 X4 +   0.00 X6 -   1.00 X0 

The optimal solution: -0.000000
X1  =     7.0000   X2  =     0.0000   X3  =     0.0000   X4  =     0.0000   
X5  =     4.0000   X6  =     0.0000   X7  =     2.0000

Set up the initial dictionary for phase 2.

Recall, the objective function:

Maximize X1 + X2 + 2 * X3 + X4

Answer #2:

The initial dictionary is:

The initial dictionary:
X5 =   4.00+   1.00 X2 -   1.00 X3 -   2.00 X4 +   1.00 X6 
X1 =   7.00+   1.00 X2 +   0.00 X3 +   0.00 X4 +   1.00 X6 
X7 =   2.00-   1.00 X2 +   0.00 X3 -   1.00 X4 +   1.00 X6 
-----------------------------------------------
z  =   7.00+   2.00 X2 +   2.00 X3 +   1.00 X4 +   1.00 X6

Question 3: The final dictionary is:

X2  enters.   X7  leaves.  z = 7.000000 

After   1 pivot:
X5 =   6.00-   1.00 X3 -   3.00 X4 +   2.00 X6 -   1.00 X7 
X1 =   9.00+   0.00 X3 -   1.00 X4 +   2.00 X6 -   1.00 X7 
X2 =   2.00+   0.00 X3 -   1.00 X4 +   1.00 X6 -   1.00 X7 
-----------------------------------------------
z  =  11.00+   2.00 X3 -   1.00 X4 +   3.00 X6 -   2.00 X7

What can you conclude about this problem?

Answer #3:

The problem is UNBOUNDED.