Graphical solution procedure, Operation Research

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Question: Max Z = 80x1 + 55x2

Subject to

4x1+ 2x2 ≤ 40

            2x1 + 4x2 ≤ 32

 x1 ≥ 0 , x2 ≥ 0

 

Answer

The first constraint 4x1+ 2 x2 ≤ 40, can be written in a form of equation

4x1+ 2 x2 = 40

Place x1 =0, then x2 = 20

Place x2 =0, then x1 = 10

Therefore, the coordinates are (0, 20) and (10, 0)

 

The second constraint 2x1 + 4x2 ≤ 32, can be written in a form of equation

2x1 + 4x2 =32

Place x1 =0, then x2 = 8

Place x2 =0, then x1 = 16

Therefore, the coordinates are (0, 8) and (16, 0)

 

The graphical presentation is

84_Graphical_Solution_Procedure.png

 

The corner positions of feasible region are A, B and C. Thus the coordinates for the corner points are

A (0, 8)

B (8, 4) (Crack the two equations 4x1+ 2 x2 = 40 and 2x1 + 4x2 =32 to obtain the coordinates)

C (10, 0)

 

We are given that Max Z = 80x1 + 55x2

At A (0, 8)

Z = 80(0) + 55(8) = 440

 

At B (8, 4)

Z = 80(8) + 55(4) = 860

 

At C (10, 0)

Z = 80(10) + 55(0) = 800

 

The maximum value is achieved at the point B. Thus Max Z = 860 and x1 = 8, x2 = 4

Posted Date: 7/4/2012 3:36:51 AM | Location : United States







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