Fill in the cost matrix of an assignment problem and click on 'Solve'. The optimal assignment will be determined and a step by step explanation of the hungarian algorithm will be given.
Fill in the cost matrix (random cost matrix):
This is the cost matrix.
| 75 | 49 | 30 | 41 |
| 50 | 57 | 44 | 87 |
| 39 | 98 | 73 | 23 |
| 29 | 34 | 27 | 45 |
For each row, the minimum element is subtracted from all elements in that row.
| 45 | 19 | 0 | 11 | (-30) |
| 6 | 13 | 0 | 43 | (-44) |
| 16 | 75 | 50 | 0 | (-23) |
| 2 | 7 | 0 | 18 | (-27) |
For each column, the minimum element is subtracted from all elements in that column.
| 43 | 12 | 0 | 11 |
| 4 | 6 | 0 | 43 |
| 14 | 68 | 50 | 0 |
| 0 | 0 | 0 | 18 |
| (-2) | (-7) |
A total of 3 lines are required to cover all zeros.
| 43 | 12 | 0 | 11 | |
| 4 | 6 | 0 | 43 | |
| 14 | 68 | 50 | 0 | x |
| 0 | 0 | 0 | 18 | x |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 4. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 39 | 8 | 0 | 7 |
| 0 | 2 | 0 | 39 |
| 14 | 68 | 54 | 0 |
| 0 | 0 | 4 | 18 |
A total of 4 lines are required to cover all zeros.
| 39 | 8 | 0 | 7 | x |
| 0 | 2 | 0 | 39 | x |
| 14 | 68 | 54 | 0 | x |
| 0 | 0 | 4 | 18 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 39 | 8 | 0 | 7 |
| 0 | 2 | 0 | 39 |
| 14 | 68 | 54 | 0 |
| 0 | 0 | 4 | 18 |
This corresponds to the following optimal assignment in the original cost matrix.
| 75 | 49 | 30 | 41 |
| 50 | 57 | 44 | 87 |
| 39 | 98 | 73 | 23 |
| 29 | 34 | 27 | 45 |
The total minimum cost is 137.