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 |