In the previous section, we used vogel approximation method (Unit cost penalty method) to solve a transportation problem. In this section, we provide another example to enhance your knowledge.
Let's concentrate on the following example and find the optimal solution:
Example
2 : Vogel Approximation MethodConsider the transportation problem presented in the following table:
| Destination | ||||
|---|---|---|---|---|
| Origin | 1 | 2 | 3 | Supply |
| 1 | 2 | 7 | 4 | 5 |
| 2 | 3 | 3 | 1 | 8 |
| 3 | 5 | 4 | 7 | 7 |
| 4 | 1 | 6 | 2 | 14 |
| Demand | 7 | 9 | 18 | 34 |
Solution.
Table 1
| Destination | |||||
|---|---|---|---|---|---|
| Origin | 1 | 2 | 3 | Supply | Penalty |
| 1 | 7 | 4 | 2 | ||
| 2 | 3 | 3 | 1 | 8 | 2 |
| 3 | 5 | 4 | 7 | 7 | 1 |
| 4 | 1 | 6 | 2 | 14 | 1 |
| Demand | 9 | 18 | 34 | ||
| Penalty | 1 | 1 | 1 | ||
The highest penalty occurs in the first row. The minimum cij in this row is c11 (i.e., 2). Hence, x11 = 5 and the first row is eliminated.
Now again calculate the penalty. The following table shows the computation of penalty for various rows and columns.
Final table
| Destination | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Origin | 1 | 2 | 3 | Supply | Penalty | |||||
| 1 | 7 | 4 | 2 | - | - | - | - | - | ||
| 2 | 3 | 2 | 2 | 2 | 2 | 3 | 3 | |||
| 3 | 5 | 7 | 1 | 1 | 3 | 3 | 4 | - | ||
| 4 | 6 | 1 | 1 | 4 | - | - | - | |||
| Demand | |
|
|
34 | ||||||
| Penalty | 1 | 1 | 1 | |||||||
| 2 | 1 | 1 | ||||||||
| - | 1 | 1 | ||||||||
| - | 1 | 6 | ||||||||
| - | 1 | - | ||||||||
| - | 3 | - | ||||||||
5 X 2 + 2 X 3 + 6 X 1 + 7 X 4 + 2 X 1 + 12 X 2 = 76.
Now, you must take a break because you really deserve it. We will see you at the next section when you are ready again.