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.
| 71 | 95 | 49 | 4 |
| 26 | 45 | 41 | 52 |
| 19 | 49 | 78 | 36 |
| 39 | 76 | 63 | 65 |
For each row, the minimum element is subtracted from all elements in that row.
| 67 | 91 | 45 | 0 | (-4) |
| 0 | 19 | 15 | 26 | (-26) |
| 0 | 30 | 59 | 17 | (-19) |
| 0 | 37 | 24 | 26 | (-39) |
For each column, the minimum element is subtracted from all elements in that column.
| 67 | 72 | 30 | 0 |
| 0 | 0 | 0 | 26 |
| 0 | 11 | 44 | 17 |
| 0 | 18 | 9 | 26 |
| (-19) | (-15) |
A total of 3 lines are required to cover all zeros.
| 67 | 72 | 30 | 0 | x |
| 0 | 0 | 0 | 26 | x |
| 0 | 11 | 44 | 17 | |
| 0 | 18 | 9 | 26 | |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 9. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 76 | 72 | 30 | 0 |
| 9 | 0 | 0 | 26 |
| 0 | 2 | 35 | 8 |
| 0 | 9 | 0 | 17 |
A total of 4 lines are required to cover all zeros.
| 76 | 72 | 30 | 0 | x |
| 9 | 0 | 0 | 26 | x |
| 0 | 2 | 35 | 8 | x |
| 0 | 9 | 0 | 17 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 76 | 72 | 30 | 0 |
| 9 | 0 | 0 | 26 |
| 0 | 2 | 35 | 8 |
| 0 | 9 | 0 | 17 |
This corresponds to the following optimal assignment in the original cost matrix.
| 71 | 95 | 49 | 4 |
| 26 | 45 | 41 | 52 |
| 19 | 49 | 78 | 36 |
| 39 | 76 | 63 | 65 |
The total minimum cost is 131.