Page 59 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes
Example:
Maximise ‘Z’ = 5x + 3x [Subject to constraints]
1 2
= 3x + 5x +  15
1 2
= 5x + 2x  10
1 2
Where, x , x  0
1 2
Solution:

Step 1: Convert the inequalities into equalities by adding the slack variables.
3x + 5x + x = 15
1 2 3
5x + 2x + x = 10
1 2 4
Where, x and x are slack variables.
3 4
Step 2: Fit the data into a matrix form.
 Y 1 Y 2 S 1 S 2   x 1 
 x x x x   x   15 
A   1 2 3 4  X   2  B   
 3 5 1 0   x 3   10 
       
 5 2 0 1   x 4 
Step 3: First iteration of Simplex Method.
BV CB XB Y1 Y2 S1 S2 Min. Ratio
S1 0 15 3 5 1 0 15/3 = 5
S2 0 10 5 2 0 1 10/5 = 2(KR) ?
Zj 0 0
Cj 5 3
Zj – Cj -5 -3
( KC)

Therefore, Maximise Z = C X
B B
= (0 × 15) + (0 × 10)
= 0

Step 4: Second iteration of Simplex Method.
BV C B X B y 1 Y 2 S 1 S 2 Min. Ratio
S 1 0 15 – 2 (3) = 9 3 –1(3) = 0 5 – 0.4 (3) = 3.8 -- -- 9/3.8 = 2.37 (KR)?
S 2 5 10/5 = 2 5/5 = 1 2/5=0.4 -- -- 2/0.4 = 5
Z j 5 2
C j 5 3
Z j – C j 0 –1
( KC)
Therefore, Z = C X
B B
= (0 × 9) + (5 × 2)
= 10

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