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Q.1 Maximise z = 3×1 + 4×2 Subject to constrains 5×1 + 4×2ï‚£ 200; 3×1 + 5×2ï‚£ 150; 5×1 + 4×2 100; 8×1 + 4×2 80, x1 0, x2 0

January 08, 2013 By: Meliza Category: 1st SEM

Solution
Find Solution Using Simplex Method

Cj

3

4

0

0

0

0

B

CB

XB

X1

X2

S1

S2

S3

S4

MinRatio

S1

0

200

5

4

1

0

0

0

50

S2

0

150

3

5

0

1

0

0

30

S3

0

100

5

4

0

0

1

0

25

S4

0

80

8

4

0

0

0

1

20

Zj

0

0

0

0

0

0

Cj – Zj

3

4

0

0

0

0

Entering = X2, Departing = S4, Key Element = 4
R4(new) = R4(old) / 4 = R4(old) × 1/4
R1(new) = R1(old) – 4 R4(new)
R2(new) = R2(old) – 5 R4(new)
R3(new) = R3(old) – 4 R4(new)

Cj

3

4

0

0

0

0

B

CB

XB

X1

X2

S1

S2

S3

S4

MinRatio

S1

0

120

-3

0

1

0

0

-1

S2

0

50

-7

0

0

1

0

-5/4

S3

0

20

-3

0

0

0

1

-1

X2

4

20

2

1

0

0

0

1/4

Zj

8

4

0

0

0

1

Cj – Zj

-5

0

0

0

0

-1

Optimum Solution is arrived at with value of variables as :
X1 = 0
X2 = 20
Maximise Z = 80

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