1. Summary of Simplex Procedure
Add a slack variable for each less-than-or-equal constraint. Add an artificial variable for each equality constraint. Add both an artificial and a surplus variable for each greater-than-or-equal constraint.
For each artificial variable, assign a very large cost (negative profit) in the objective function.
The current solution is optimal if: For maximisation, all coefficients in the, row are zero or negative. For minimisation, all coefficients in row are zero or positive. If optimum has been reached, stop simplex procedure.
This is the variable associated with the largest positive (for maximisation) or largest negative (for minimisation) coefficient in the row.
For each row, calculate the ratio of the values in the "Solution Values" column divided by the coefficients in the entering variable column. Ignore any ratios that are negative. The leaving variable is the one associated with the row having the smallest ratio.
|New||=||Old||-||Coefficient in entering||*||Row obtained|
|Row||Row||Variable column of row||In step 6 b)|
For the j-th column: for each row, multiply the substitution coefficients by the cj value for that row and sum. The total is Zj. Repeat for all columns.
9. Return to step 3