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Operations Research

Notes = –40M – 50M
= –90M
Step 6: Second iteration of Simplex Method.

BV CB XB Y1 Y2 S1 S2 a1 a2 Min. Ratio
y2 –5 40/8 = 5 2/8 = 0.25 8/8 = 1 – – – – 5/0.25 = 20
a2 –M 50–5(4) = 30 3–0.25(4) = 2 4–1(4)= 0 – – – – 30/2 = 15(KR)
Zj –2M – 1.25 –5
Cj –3 –5
Zj – Cj –2M + 1.75 0
( KC)

Therefore, Z = (–5 × 5) + (–M × 30)
= –30M – 25
Step 7: Third iteration of Simplex Method.

7 C B X B Y 1 Y 2 S 1 S 2 a 1 a 2 Min. Ratio
y 2 –5 5–15 (0.25) = 1.25 0.25 – 1 (0.25) = 0 1–0 (0.25) = 1 – – – – –
y 1 –3 30/2 = 15 2/2 = 1 0/2 = 0 – – – – –
Z j –3 –5
C j –3 –5
Z j – C j 0 0

Z = C X
B B
= – 6.25 – 45
Maximise Z = – 51.25

Therefore, Minimise Z = 51.25

Example:
Minimise ‘Z’ = 12x + 20x [Subject to constraints]
1 2
6x + 8x  100
1 2
7x + 12x  120
1 2
Where, x , x  0
1 2
Solution:
Step 1: Conversion of the minimization case into maximisation case.
Maximise Z = – 12x – 20x [Subject to constraints]
1 2
Step 2: Convert the inequalities into equalities adding artificial and surplus variables.

6x + 8x – x + x = 100
1 2 3 5
7x + 12x – x + x = 120
1 2 4 6
Where, x and x are surplus variables and x and x are artificial variables.
3 4 5 6

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