Sometimes, it is possible to cross out all the zeros in the reduced matrix in two or more ways.
If you can choose a zero cell arbitrarily, then there will be multiple optimal solutions with the same total pay-off for assignments made. In such a case, the management may select that set of optimal assignments, which is more suited to their requirement.
Example: Multiple Optimal SolutionsConsider the following assignment problem: The Spicy Spoon restaurant has four payment counters. There are four persons available for service. The cost of assigning each person to each counter is given in the following table.
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 1 | 8 | 15 | 22 |
| B | 13 | 18 | 23 | 28 |
| C | 13 | 18 | 23 | 28 |
| D | 19 | 23 | 27 | 31 |
Assign one person to one counter to minimize the total cost.
Solution.
After applying steps 1 to 3 of the Hungarian Method, we obtain the following matrix.
Table
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 3 | 6 | 9 | |
| B | 1 | 2 | 3 | |
| C | 1 | 2 | 3 | |
| D | ||||
Now by applying the usual procedure, we get the following matrix.
Table
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 2 | 5 | 8 | |
| B | 1 | 2 | ||
| C | 1 | 2 | ||
| D | 1 | |||
The resulting matrix suggest the alternative optimal solutions as shown in the following tables.
Table
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 2 | 4 | 7 | |
| B | 1 | |||
| C | 1 | |||
| D | 2 | 1 | ||
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 2 | 4 | 7 | |
| B | 1 | |||
| C | 1 | |||
| D | 2 | 1 | ||
The persons B & C may be assigned either to job 2 or 3.
The two alternative assignments are:
| A1 + B2 + C3 + D4 1 + 18 + 23 + 31 = 73 |
A1 + B3 + C2+ D4 1 + 23 + 18 + 31 = 73 |