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.
| 23 | 6 | 69 | 49 |
| 49 | 29 | 94 | 85 |
| 50 | 83 | 73 | 62 |
| 39 | 85 | 65 | 58 |
For each row, the minimum element is subtracted from all elements in that row.
| 17 | 0 | 63 | 43 | (-6) |
| 20 | 0 | 65 | 56 | (-29) |
| 0 | 33 | 23 | 12 | (-50) |
| 0 | 46 | 26 | 19 | (-39) |
For each column, the minimum element is subtracted from all elements in that column.
| 17 | 0 | 40 | 31 |
| 20 | 0 | 42 | 44 |
| 0 | 33 | 0 | 0 |
| 0 | 46 | 3 | 7 |
| (-23) | (-12) |
A total of 3 lines are required to cover all zeros.
| 17 | 0 | 40 | 31 | |
| 20 | 0 | 42 | 44 | |
| 0 | 33 | 0 | 0 | x |
| 0 | 46 | 3 | 7 | x |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 17. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 0 | 0 | 23 | 14 |
| 3 | 0 | 25 | 27 |
| 0 | 50 | 0 | 0 |
| 0 | 63 | 3 | 7 |
A total of 3 lines are required to cover all zeros.
| 0 | 0 | 23 | 14 | |
| 3 | 0 | 25 | 27 | |
| 0 | 50 | 0 | 0 | x |
| 0 | 63 | 3 | 7 | |
| x | x |
The number of lines is smaller than 4. The smallest uncovered element is 3. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 0 | 0 | 20 | 11 |
| 3 | 0 | 22 | 24 |
| 3 | 53 | 0 | 0 |
| 0 | 63 | 0 | 4 |
A total of 4 lines are required to cover all zeros.
| 0 | 0 | 20 | 11 | x |
| 3 | 0 | 22 | 24 | x |
| 3 | 53 | 0 | 0 | x |
| 0 | 63 | 0 | 4 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 0 | 0 | 20 | 11 |
| 3 | 0 | 22 | 24 |
| 3 | 53 | 0 | 0 |
| 0 | 63 | 0 | 4 |
This corresponds to the following optimal assignment in the original cost matrix.
| 23 | 6 | 69 | 49 |
| 49 | 29 | 94 | 85 |
| 50 | 83 | 73 | 62 |
| 39 | 85 | 65 | 58 |
The total minimum cost is 179.