Revised Simplex Method We use Revised Simplex Method

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Revised Simplex Method: We use Revised Simplex Method for large scale Linear programming problem

Revised Simplex Method: We use Revised Simplex Method for large scale Linear programming problem (LPP). Revised simplex Method uses only small portion of the Simplex Table and saves lot of unnecessary computations. We consider the following example for the sake of illustrations as the given problem is not a large LPP.

Example 1: Max. Z = 40 x 1 + 30 x 2 Subject to

Example 1: Max. Z = 40 x 1 + 30 x 2 Subject to constraints: 5 x 1 + 4 x 2 ≤ 400 x 1 < 60 x 2 < 75 x 1, x 2 > 0

Step I. Standarizing the given LPP: Max. Z = 40 x 1 + 30

Step I. Standarizing the given LPP: Max. Z = 40 x 1 + 30 x 2 +0∙s 1+ 0∙s 2+ 0∙s 3 Subject to constraints: 5 x 1 + 4 x 2 + s 1 =400 x 1 + s + 2 = 60 = x 2 + s 3 = 75 = x 1, x 2 s 1, s 2 s 3 ≥ 0 ≥

Step I: We express our set of constraints in the following form x 1

Step I: We express our set of constraints in the following form x 1 P 1+x 2 P 2+…+xn. Pn=b where P 1, P 2, …, Pn are the columns of our coefficients of decision and slack variables and b is a column vector of right hand side constants. In our example, we have

We prepare the following Simplex table: Cb X B-1 b b 0 s 1

We prepare the following Simplex table: Cb X B-1 b b 0 s 1 1 0 0 400 0 s 2 0 1 0 60 0 s 3 0 0 1 75 Step II: Calculate Simplex Multipliers

We prepare the following table. We choose the variable with minimum ratio as the

We prepare the following table. We choose the variable with minimum ratio as the leaving variable Cb Xb B-1 b 0 s 1 1 0 0 400 5 400/5=80 0 0 s 2 0 1 0 60 1 60/1=60 1 0 s 3 0 0 1 75 0 - 0 Pivot Column P 1

Thus, s 2 is the leaving variable and the new updated table is produced

Thus, s 2 is the leaving variable and the new updated table is produced below;

Cb Xb B-1 0 s 1 1 -5 0 100 40 x 1 0

Cb Xb B-1 0 s 1 1 -5 0 100 40 x 1 0 60 0 s 3 0 0 1 75 b

Cb Xb B-1 b 0 s 1 1 -5 0 100 -40 x 1

Cb Xb B-1 b 0 s 1 1 -5 0 100 -40 x 1 0 60 0 1 75 s 3

And C 3=���� 3 -c 3=(-15/2, -5/2, 0) [█(1@0@0)]-0=-15/2

And C 3=���� 3 -c 3=(-15/2, -5/2, 0) [█(1@0@0)]-0=-15/2

Cb Xb B-1 b Pivot Column P 1 Ratio 0 s 1 1 -5

Cb Xb B-1 b Pivot Column P 1 Ratio 0 s 1 1 -5 0 100 4 100/4=25 1 -40 x 1 0 60 0 - 0 0 1 75/1=1 0 s 3

Thus s 1 will leave and x 2 will enter

Thus s 1 will leave and x 2 will enter

Cb Xb B-1 30 x 2 1/4 -5/4 0 25 40 x 1 0

Cb Xb B-1 30 x 2 1/4 -5/4 0 25 40 x 1 0 60 0 s 3 -1/4 5/4 1 50 b

Now, we have Cb=B-1

Now, we have Cb=B-1