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.
| 95 | 40 | 11 | 52 |
| 68 | 4 | 88 | 81 |
| 61 | 43 | 63 | 51 |
| 97 | 1 | 20 | 13 |
For each row, the minimum element is subtracted from all elements in that row.
| 84 | 29 | 0 | 41 | (-11) |
| 64 | 0 | 84 | 77 | (-4) |
| 18 | 0 | 20 | 8 | (-43) |
| 96 | 0 | 19 | 12 | (-1) |
For each column, the minimum element is subtracted from all elements in that column.
| 66 | 29 | 0 | 33 |
| 46 | 0 | 84 | 69 |
| 0 | 0 | 20 | 0 |
| 78 | 0 | 19 | 4 |
| (-18) | (-8) |
A total of 3 lines are required to cover all zeros.
| 66 | 29 | 0 | 33 | x |
| 46 | 0 | 84 | 69 | |
| 0 | 0 | 20 | 0 | x |
| 78 | 0 | 19 | 4 | |
| 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.
| 66 | 33 | 0 | 33 |
| 42 | 0 | 80 | 65 |
| 0 | 4 | 20 | 0 |
| 74 | 0 | 15 | 0 |
A total of 4 lines are required to cover all zeros.
| 66 | 33 | 0 | 33 | x |
| 42 | 0 | 80 | 65 | x |
| 0 | 4 | 20 | 0 | x |
| 74 | 0 | 15 | 0 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 66 | 33 | 0 | 33 |
| 42 | 0 | 80 | 65 |
| 0 | 4 | 20 | 0 |
| 74 | 0 | 15 | 0 |
This corresponds to the following optimal assignment in the original cost matrix.
| 95 | 40 | 11 | 52 |
| 68 | 4 | 88 | 81 |
| 61 | 43 | 63 | 51 |
| 97 | 1 | 20 | 13 |
The total minimum cost is 89.