In: Math
Maximize objective function P=3x+4y
subject to: x + y ≤ 7 x ≥ 0
x+4y ≤ 16 y ≥ 0
subject to
Iteration-1 |
Cj |
3 |
4 |
0 |
0 |
||
B |
CB |
XB |
x1 |
x2 |
S1 |
S2 |
MinRatio |
S1 |
0 |
7 |
1 |
1 |
1 |
0 |
7/1=7 |
S2 |
0 |
16 |
1 |
(4) |
0 |
1 |
16/4=4 |
Z=0 |
Zj |
0 |
0 |
0 |
0 |
||
Zj-Cj |
-3 |
-4 |
0 |
0 |
Negative minimum Zj-Cj is -4 and its column index is 2
Minimum ratio is 4 and its row index is 2
The pivot element is 4.
Entering =x2, Departing =S2,
.
Iteration-2 |
Cj |
3 |
4 |
0 |
0 |
||
B |
CB |
XB |
x1 |
x2 |
S1 |
S2 |
MinRatio |
S1 |
0 |
3 |
(3/4) |
0 |
1 |
-1/4 |
3/(3/4)=4 |
x2 |
4 |
4 |
1/4 |
1 |
0 |
1/4 |
4/(1/4)=16 |
Z=16 |
Zj |
1 |
4 |
0 |
1 |
||
Zj-Cj |
-2 |
0 |
0 |
1 |
Negative minimum Zj-Cj is -2 and its column index is 1.
Minimum ratio is 4 and its row index is 1
The pivot element is 3/4.
Entering =x1, Departing =S1
Iteration-3 |
Cj |
3 |
4 |
0 |
0 |
||
B |
CB |
XB |
x1 |
x2 |
S1 |
S2 |
MinRatio |
x1 |
3 |
4 |
1 |
0 |
4/3 |
-1/3 |
|
x2 |
4 |
3 |
0 |
1 |
-1/3 |
1/3 |
|
Z=24 |
Zj |
3 |
4 |
8/3 |
1/3 |
||
Zj-Cj |
0 |
0 |
8/3 |
1/3 |
all
optimal solution is arrived