# An Example of Two Phase Simplex Method

**Topics:**Optimization, BMW, BMW Sports Activity Series

**Pages:**3 (671 words)

**Published:**October 18, 2010

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 = 10

0 0...

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