An Example of Two Phase Simplex Method
An Example of Two Phase Simplex Method
Consider the following LP problem. max z = 2x1 + 3x2 + x3
s.t. x1 + x2 + x3 · 40
2x1 + x2 ¡ x3 ¸ 10
¡x2 + x3 ¸ 10 x1; x2; x3 ¸ 0
It can be transformed into the standard form by introducing 3 slack variables x4, x5 and x6. max z = 2x1 + 3x2 + x3
s.t. x1 + x2 + x3 + x4 = 40
2x1 + x2 ¡ x3 ¡ x5 = 10
¡x2 + x3 ¡ x6 = 10 x1; x2; x3; x4; x5; x6 ¸ 0
There is no obvious initial basic feasible solution, and it is not even known whether there exists one. We can use Phase I method to ¯nd out. Consider the following LP problem derived from the original one by relaxing the second and third constraints and introducing a new objective function. min x7 + x8, (or max w = ¡x7 ¡ x8)
s.t. x1 + x2 + x3 + x4 = 40
2x1 + x2 ¡ x3 ¡ x5 + x7 = 10
¡x2 + x3 ¡ x6 + x8 = 10 x1; x2; x3; x4; x5; x6; x7; x8 ¸ 0
This problem (Phase I) has an initial basic feasible solution with basic variables being x4, x7 and x8. If the minimum value of x7 + x8 is 0, then both x7 and x8 are 0. As the result, the optimal solution of the Phase I problem is an basic feasible solution of the original problem.
If the minimum value of x7 +x8 is bigger than 0, then the original problem is not feasible. We construct tableaus to solve the Phase I problem. The objective value w should be written in terms of non-basic variables: w = ¡x7 ¡ x8 = ¡20 + 2x1 ¡ x5 ¡ x6:
The initial tableau is shown below (the basic variables are shown in bold font). w x1 x2 x3 x4 x5 x6 x7 x8
1 -2 0 0 0 1 1 0 0 = -20
0 1 1 1 1 0 0 0 0 = 40
0 2 1 -1 0 -1 0 1 0 = 10
0 0 -1 1 0 0 -1 0 1 = 10
The entering and leaving variables would be x1 and x7 respectively: w x1 x2 x3 x4 x5 x6 x7 x8
1 0 1 -1 0 0 1 1 0 = -10
0 0 0.5 1.5 1 0.5 0 -0.5 0 = 35
0 1 0.5 -0.5 0 -0.5 0 0.5 0 = 5
0 0 -1 1 0 0 -1 0 1 = 10
The entering and leaving variables would be x3 and x8 respectively: w x1 x2 x3 x4 x5 x6 x7 x8
1 0 1 0 0 0 0 1 1 = 0
0 0 2 0 1 0.5 1.5 -0.5 -1.5 = 20
0 1 0 0 0 -0.5 -0.5 0.5 0.5
Consider the following LP problem. max z = 2x1 + 3x2 + x3
s.t. x1 + x2 + x3 · 40
2x1 + x2 ¡ x3 ¸ 10
¡x2 + x3 ¸ 10 x1; x2; x3 ¸ 0
It can be transformed into the standard form by introducing 3 slack variables x4, x5 and x6. max z = 2x1 + 3x2 + x3
s.t. x1 + x2 + x3 + x4 = 40
2x1 + x2 ¡ x3 ¡ x5 = 10
¡x2 + x3 ¡ x6 = 10 x1; x2; x3; x4; x5; x6 ¸ 0
There is no obvious initial basic feasible solution, and it is not even known whether there exists one. We can use Phase I method to ¯nd out. Consider the following LP problem derived from the original one by relaxing the second and third constraints and introducing a new objective function. min x7 + x8, (or max w = ¡x7 ¡ x8)
s.t. x1 + x2 + x3 + x4 = 40
2x1 + x2 ¡ x3 ¡ x5 + x7 = 10
¡x2 + x3 ¡ x6 + x8 = 10 x1; x2; x3; x4; x5; x6; x7; x8 ¸ 0
This problem (Phase I) has an initial basic feasible solution with basic variables being x4, x7 and x8. If the minimum value of x7 + x8 is 0, then both x7 and x8 are 0. As the result, the optimal solution of the Phase I problem is an basic feasible solution of the original problem.
If the minimum value of x7 +x8 is bigger than 0, then the original problem is not feasible. We construct tableaus to solve the Phase I problem. The objective value w should be written in terms of non-basic variables: w = ¡x7 ¡ x8 = ¡20 + 2x1 ¡ x5 ¡ x6:
The initial tableau is shown below (the basic variables are shown in bold font). w x1 x2 x3 x4 x5 x6 x7 x8
1 -2 0 0 0 1 1 0 0 = -20
0 1 1 1 1 0 0 0 0 = 40
0 2 1 -1 0 -1 0 1 0 = 10
0 0 -1 1 0 0 -1 0 1 = 10
The entering and leaving variables would be x1 and x7 respectively: w x1 x2 x3 x4 x5 x6 x7 x8
1 0 1 -1 0 0 1 1 0 = -10
0 0 0.5 1.5 1 0.5 0 -0.5 0 = 35
0 1 0.5 -0.5 0 -0.5 0 0.5 0 = 5
0 0 -1 1 0 0 -1 0 1 = 10
The entering and leaving variables would be x3 and x8 respectively: w x1 x2 x3 x4 x5 x6 x7 x8
1 0 1 0 0 0 0 1 1 = 0
0 0 2 0 1 0.5 1.5 -0.5 -1.5 = 20
0 1 0 0 0 -0.5 -0.5 0.5 0.5