We convert the given problem in standard form by introducing slack variable, surplus variable and artificial variable. Also assign the cost – 1 to artificial variable and the cost 0 to another variables.
We have
Phase-1
Max Z* = 0x1 + 0x2 + 0S1 + 0S2 – 1A1
S.T. 2x1 + x2 + S1 = 1
x1 + 4x2 – S2 + A1 = 6
x1, x2, S1, S2, A1 ≥ 0
The initial basic feasible solution is given by
x1 = x2 = 0, S1 = 1, A1 = 6
Now prepare the initial simplex table
Here, all values of Zj – Cj are not positive so choose the most negative value of Zj – Cj, i.e., Z2 – C2 is the most negative value. It will enter in the basis and treated as key column. Find key row by taking min \(\left\{\frac{X_B}{x_2}\right\}\).
\(= \left\{\frac11,\frac 64\right\}\)
= 1
1 is key element. Make other elements of key column zero by applying matrix row transformation. We have the first simplex table.
Here all values of Zj – Cj ≥ 0. Max Z* < 0 and an artificial variable A1 appears in the basis at positive level. In this case L.P.P. does not possess any feasible solution.