Solve by Steps for Two-Phase Method
Max Z = 5x_{1} + 8x_{2}
Subject to
3x_{1} + 2x_{2 }≥ 3
x_{1 }+ 4x_{2} ≥ 4
x_{1 }+ x_{2 }≤ 5
& x_{1 }≥ 0, x_{2 }≥ 0
Answer
Standard LPP
Max Z = 5x_{1} + 8x_{2 }
Subject to
3x_{1} + 2x_{2 }- s_{1}+ a_{1 }= 3
x_{1 }+ 4x_{2} - s_{2}+ a_{2} = 4
x_{1 }+ x_{2 }+ s_{3} = 5
x_{1 }, x_{2 }, s_{1}, s_{2}, s_{3}, a_{1}, a_{2 }≥ 0
Auxiliary LPP
Max Z* = 0x_{1} + 0x_{2 }+ 0s_{1 }+ 0s_{2 }+ 0s_{3 }-1a_{1 }-1a_{2}
Subject to
3x_{1} + 2x_{2 }- s_{1}+ a_{1 }= 3
x_{1 }+ 4x_{2} - s_{2}+ a_{2} = 4
x_{1 }+ x_{2 }+ s_{3} = 5
x_{1 }, x_{2 }, s_{1}, s_{2}, s_{3}, a_{1}, a_{2 }≥ 0
As all Δ_{j }≥ 0, Max Z* = 0 and no artificial vector appears in the basis, we move to phase II.
Phase II
As all Δ_{j }≥ 0, optimal basic feasible solution is achieved. Thus the solution is Max Z = 40, x_{1 }= 0, x_{2} = 5