Page 55 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes Step 4: Fit the data into second iteration of Simplex Method.
BV CB XB Y1 Y2 S1 S2 S3 Min. Ratio
Y1 5 2/1 = 2 1/1 = 1 1/1 = 1 – – – –
S2 0 10 – 2(5) = 0 5 – 1(5) = 0 2–1(5) = –3 – – – –
S3 0 12 – 2(3) = 6 3 – 1(3) = 0 8–1(3) = 5 – – – –
Zj 5 5
Cj 5 3
Zj – Cj 0 2

Therefore, Z = C X
B B
= (5 × 2) + (0 × 0) + (0 × 6)
Therefore, Z = 10
Therefore, Maximum value of ‘Z’ = 10

Example:
Maximise ‘Z’ = 2x + 3x [Subject to constraints]
1 2
x + x  1
1 2
3x + x  4
1 2
Where, x , x  0
1 2
Solution:
Step 1: Conversion of inequalities into equalities by adding slack variables.

x + x + x = 1
1 2 3
3x + x + x = 4
1 2 4
Where x and x are slack variables.
3 4
Step 2: Identify the coefficients.

 Y 1 Y 2 S 1 S 2   x 1 
 x x x x   x   1
A   1 2 3 4  X   2   B   
 1 1 1 0   x 3  4
   
 3 1 0 1   x 4 

Step 3: First iteration of Simplex Method.

BV CB XB Y1 Y2 S1 S2 Min. Ratio
?
S1 0 1 1 1 1 0 1/1 = 1 (KR)?
S2 0 4 3 1 0 1 4/1 = 4
Zj 0 0
Cj 2 3
Zj – Cj -2 -3
( KC)

50 LOVELY PROFESSIONAL UNIVERSITY