Page 73 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes = (–12 × 15.04) + (–20 × 1.28)
= –180.48 – 25.6
Therefore, Maximise Z = – 206.08
Therefore, Minimise Z = 206.08

Example: Solve the following LPP using Simplex Method.
Maximise ‘Z’ = x + 1.5x + 2x + 5x [Subject to constraints]
1 2 3 4
3x + 2x + 4x + x  6
1 2 3 4
2x + x + x + 5x  4
1 2 3 4
2x + 6x – 8x + 4x = 0
1 2 3 4
x + 3x – 4x + 3x = 0
1 2 3 4
Where, x , x , x , x  0
1 2 3 4
Solution:
Step 1: Convert the inequalities into equalities by adding slack variables and surplus variables.
3x + 2x + 4x + x + x = 6
1 2 3 4 5
2x + x + x + 5x + x = 4
1 2 3 4 6
2x + 6x – 8x + 4x + x = 0
1 2 3 4 7
x + 3x – 4x + 3x + x = 0
1 2 3 4 8
Where, x and x are slack variables and x and x are artificial variables.
5 6 7 8
Step 2: The standard form of objective function.

Maximise ‘Z’ = x + 1.5x + 2x + 5x ± 0x ± 0x – Mx – Mx
1 2 3 4 5 6 7 8
Step 3: Fit the data into a matrix form.
 x 1 
 
 Y 1 Y 2 Y 3 y 4 S 1 S 2 a 1 a 2   x 2 
   x 
6
 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8   3   
4
 3 2 4 1 1 0 0 0  x  
A    X    4    B   
0
 2 1 1 5 0 1 0 0   x 5   
 
 2 6 8 4 0 0 1 0   x   
0
  6  
   
 1 3 4 3 0 0 0 1   x 7 
 x 
 8 
Step 4: First iteration of Simplex Method.
BV CB XB Y1 Y2 Y3 Y4 S1 S2 a1 a2 Min. Ratio
S1 0 6 3 2 4 1 1 0 0 0 6/2 = 3
S2 0 4 2 1 1 5 0 1 0 0 4/1 = 4
a1 –M 0 2 6 –8 4 0 0 1 0 0/6 = 0(KR)?
a2 –M 0 1 3 –4 3 0 0 0 1 0/3 = 0
Zj –3M –9M 12M –7M
Cj 1 1.5 2 5
Zj – Cj –3M–1 –9M–1.5 +12M–2 –7M–5
( KC)

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