Consider the following example:

Minimize $Z = C_1X_1 + C_2 X_2$ Subject to: $5X_1 + X_2 ≥ 10 \\ X_1 – X_2 ≥ 3 \\ X1, X2 ≥ 0$

Surplus variables and artificial variables have to be introduced in the constraints:

$5X_1 + X_2– S_1 + A_1 = 10 \\ X_1 – X_2– S_2 + A_2 = 3$

So the minimization function becomes:

$Z = C_1X_1 + C_2 X_2 – 0S_1 – 0S_2 + A_1 + A_2$

Phase 1:

Co-efficients of ‘0’ are assigned to all the variables (besides the artificial variables) in the objective function.

If it is a minimization problem, co-efficient of ‘1’ is assigned to each artificial variable.

If it is a maximization problem (or minimization converted to maximization), co-efficient of ‘-1’ is assigned to each artificial variable.

In our example, it is a minimization problem.

So the simplex table for initial calculation is as follows:

The simplex procedure is then carried on as usual, i.e. till all the ‘Cj – Zj’ values are zero or less than zero.

At the end of the iterations, the table obtained may look as follows:

Where all aij& bi are calculated as usual.

As can be seen, the artificial variables have been eliminated from the base.

So now we proceed with phase 2, by considering the above table as the initial table, but leaving out the columns of the artificial variables since the artificial variables have been removed from the base.

Phase 2:

In phase 2, the initial table to be considered is the last table obtained in phase 1, but without the columns of the artificial variables.

Also, the co-efficients of the variables of the objective function are now assigned their actual values.

The simplex procedure is then carried out as usual.

Phase 1 helps in finding an initial basic feasible solution, while Phase 2 takes the solution towards optimality.

If an artificial variable is still present in the base at the end of phase 1, then it indicates that the solution is infeasible.