|
6 |
7 |
0 |
0 |
b |
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|
|
|
|
||||
|
7 |
|
0 |
1 |
1/2 |
-1/2 |
4 |
|
|
6 |
|
1 |
0 |
-1/4 |
3/4 |
6 |
|
|
|
0 |
0 |
-2 |
-1 |
64 |
Reading the optimal tableau
Example: Product Planning
The optimal solution of the example Example: Product planning is in Table 1.
|
6 |
7 |
0 |
0 |
b |
|||
|
|
|
|
|
||||
|
7 |
|
0 |
1 |
1/2 |
-1/2 |
4 |
|
|
6 |
|
1 |
0 |
-1/4 |
3/4 |
6 |
|
|
|
0 |
0 |
-2 |
-1 |
64 |
Table 1 : Optimal Solution
The outside variables are equal to zero:
X3 = 0
X4 = 0
The vector of optimal basic solution is than:
The values of solution variables are to found in the b column. The first value in B column is the value of x2 because the first row is labelled x2 and the column vector under the variable x2 has +1 in first row. Similarly, the value in column b in second line is the value of x1.
X1 =6
X2 = 4
If at least one of solution variables is equal to zero, the solution is degenerated. This solution is not degenerated.
If at least one of the values in 
under outside variables is equal to zero, there is another basic alternative solution ( the variable with zero can be added to the solution with no changes in the objective function value) and that means the model has undefined number of optimal solution. This is more understandable in the graphical solution – Alternative solution. This example has only one optimal solution.
