subject to
.
.
After introducing surplus,artificial
variables
subject to
Iteration-1 | Cj | 0 | -5 | -3 | -2 | 0 | -M | -M | ||
B | CB | XB | x1 | x2 | x3 | x4 | S1 | A1 | A2 | MinRatio XB/x3 |
A1 | -M | 2 | -1 | 1 | (1) | 0 | 0 | 1 | 0 | 2/1=2→ |
A2 | -M | 5 | -1 | 0 | 2 | 1 | -1 | 0 | 1 | 5/2=2.5 |
Z=-7M | Zj | 2M | -M | -3M | -M | M | -M | -M | ||
Zj-Cj | 2M | -M+5 | -3M+3↑ | -M+2 | M | 0 | 0 |
Negative minimum Zj-Cj is
-3M+3 and its column index is 3.
Minimum ratio is 2 and its row index is 1
The pivot element is 1.
Entering =x3, Departing =A1,
Iteration-2 | Cj | 0 | -5 | -3 | -2 | 0 | -M | ||
B | CB | XB | x1 | x2 | x3 | x4 | S1 | A2 | MinRatio XB/x4 |
x3 | -3 | 2 | -1 | 1 | 1 | 0 | 0 | 0 | --- |
A2 | -M | 1 | 1 | -2 | 0 | (1) | -1 | 1 | 1/1=1→ |
Z=-M-6 | Zj | -M+3 | 2M-3 | -3 | -M | M | -M | ||
Zj-Cj | -M+3 | 2M+2 | 0 | -M+2↑ | M | 0 |
Negative minimum Zj-Cj is
-M+2 and its column index is 4
Minimum ratio is 1 and its row index is 2
The pivot element is 1.
Entering =x4, Departing =A2,
Iteration-3 | Cj | 0 | -5 | -3 | -2 | 0 | ||
B | CB | XB | x1 | x2 | x3 | x4 | S1 | MinRatio |
x3 | -3 | 2 | -1 | 1 | 1 | 0 | 0 | |
x4 | -2 | 1 | 1 | -2 | 0 | 1 | -1 | |
Z=-8 | Zj | 1 | 1 | -3 | -2 | 2 | ||
Zj-Cj | 1 | 6 | 0 | 0 | 2 |
Since all
Hence, optimal solution is arrived
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