Page 81 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes
Figure 3.3: Graphical Representation of Alternative Non-basic Feasible Solutions

Example:
Maximise ‘Z’ = 3x + 2x
1 2
Sub. to, x  4
1
x  6
2
3x + 2x  18
1 2
x , x  0
1 2
Solution:
Maximise ‘Z’ = 3x + 2x
1 2
Sub. to x + S = 4
1 1
x + S = 6
2 2
3x + 2x + S = 18
1 2 3
Where, S , S and S are slack variables.
1 2 3
First Iteration

BV CB XB X1 X2 S1 S2 S3 Min. Ratio
S1 0 4 1 0 1 0 0 4/1 = 4 (KR) ?
S2 0 6 0 1 0 1 0 –
S3 0 18 3 2 0 0 1 18/3 = 6
Zj 0 0
Cj 3 2
Zj – Cj –3 –2

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