optimal solution
Example: Product Planning
The second
simplex table was not optimal because there was positive value in the 
row under the x1
column. It mean that x1 will be the entering value and
according to the value 
x4 will be
the leaving value. The pivot row is the x4 row. To get 1 to the
pivot position we must divide whole x4 row by the pivot element - in
this case by 4/3
|
New pivot
line |
6 |
|
1 |
0 |
-1/4 |
3/4 |
6 |
|
Table 1: New
Pivot Line
The element
above the pivot is 2/3 in x2 row
and should be 0, so we have to multiply the new pivot row by 2/3 and this
new row subtract from the old x2
row. Note the changes in objective function coefficients in the very left
column – changes together with the changes of solution variables.
Pivot
|
(A) New pivot line multiplied by 2/3 |
2/3 |
0 |
-1/6 |
1/2 |
4 |
|
(B ) Old x2 row |
2/3 |
1 |
1/3 |
0 |
8 |
|
(B)-(A) |
0 |
1 |
1/2 |
-1/2 |
4 |
Table 2: Getting new x1 Row
The new
simplex tableau is optimal because the 
row contains no
positive variable see Table3:
|
|
|
6 |
7 |
0 |
0 |
b |
|
|
|
|
|
|
|
|
||
|
7 |
|
0 |
1 |
½ |
-1/2 |
4 |
|
|
6 |
|
1 |
0 |
-1/4 |
3/4 |
6 |
|
|
|
|
0 |
0 |
-2 |
-1 |
64 |
|
Table 3:
Optimal Simplex Tableau
Compare with
the optimal solution got by graphical method - optimum point.