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.