In the previous section, we provided an example of Goal Programming. Another example is given below to enhance your knowledge.
We will use the simplex method to solve this problem.
Minimize z = P1d1− + 2P2d2− + P2d3− + P3d1+
subject to
x1 + x2 + d1− - d1+ = 350
x1 + d2− = 200
x2 + d3− = 300
x1, x2, d1−, d2−, d3−, d1+ ≥ 0
Solution.
Substituting x1 = 0, x2 = 0, d1+ = 0
Therefore, d1− = 350, d2− = 200,
d3− = 300
Table 1
cj | 0 | 0 | P1 | 2P2 | P2 | P3 | ||
---|---|---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | d1− | d2− | d3− | d1+ | Solution values b (=XB) |
P1 | d1− | 1 | 1 | 1 | 0 | 0 | -1 | 350 |
2P2 | d2− | 1 | 0 | 0 | 1 | 0 | 0 | 200 |
P2 | d3− | 0 | 1 | 0 | 0 | 1 | 0 | 300 |
zj-cj | P3 | 0 | 0 | 0 | 0 | 0 | -1 | 0 |
P2 | 2 | 1 | 0 | 0 | 0 | 0 | 700 | |
P1 | 1 | 1 | 0 | 0 | 0 | -1 | 350 |
Key column = x1 column
Minimum positive value = Min(350/1, 200/1) = 200
So, d2− row is
the key row.
Therefore, d2− departs & x1 enters
Table 2
cj | 0 | 0 | P1 | 2P2 | P2 | P3 | ||
---|---|---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | d1− | d2− | d3− | d1+ | Solution values b (=XB) |
P1 | d1− | 0 | 1 | 1 | -1 | 0 | -1 | 150 |
0 | x1 | 1 | 0 | 0 | 1 | 0 | 0 | 200 |
P2 | d3− | 0 | 1 | 0 | 0 | 1 | 0 | 300 |
zj-cj | P3 | 0 | 0 | 0 | 0 | 0 | -1 | 0 |
P2 | 0 | 1 | 0 | -2 | 0 | 0 | 300 | |
P1 | 0 | 1 | 0 | -1 | 0 | -1 | 150 |
Table 3
cj | 0 | 0 | P1 | 2P2 | P2 | P3 | ||
---|---|---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | d1− | d2− | d3− | d1+ | Solution values b (=XB) |
0 | x2 | 0 | 1 | 1 | -1 | 0 | -1 | 150 |
0 | x1 | 1 | 0 | 0 | 1 | 0 | 0 | 200 |
P2 | d3− | 0 | 0 | -1 | 1 | 1 | 1 | 150 |
zj-cj | P3 | 0 | 0 | 0 | 0 | 0 | -1 | 0 |
P2 | 0 | 0 | -1 | -1 | 0 | 1 | 150 | |
P1 | 0 | 0 | -1 | 0 | 0 | 0 | 0 |
cj | 0 | 0 | P1 | 2P2 | P2 | P3 | ||
---|---|---|---|---|---|---|---|---|
cB | Basic variables B |
x1 | x2 | d1− | d2− | d3− | d1+ | Solution values b (=XB) |
0 | x2 | 0 | 1 | 0 | 0 | 1 | 0 | 300 |
0 | x1 | 1 | 0 | 0 | 1 | 0 | 0 | 200 |
P3 | d1+ | 0 | 0 | -1 | 1 | 1 | 1 | 150 |
zj-cj | P3 | 0 | 0 | -1 | 1 | 1 | 0 | 150 |
P2 | 0 | 0 | 0 | -2 | -1 | 0 | 0 | |
P1 | 0 | 0 | -1 | 0 | 0 | 0 | 0 |
The optimal solution is:
x1 = 200, x2 = 300, d1− = 0, d2− = 0, d3− = 0,
d1+ = 150.