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.
| 86 | 27 | 81 | 84 |
| 73 | 19 | 20 | 82 |
| 58 | 51 | 55 | 84 |
| 47 | 83 | 8 | 92 |
For each row, the minimum element is subtracted from all elements in that row.
| 59 | 0 | 54 | 57 | (-27) |
| 54 | 0 | 1 | 63 | (-19) |
| 7 | 0 | 4 | 33 | (-51) |
| 39 | 75 | 0 | 84 | (-8) |
For each column, the minimum element is subtracted from all elements in that column.
| 52 | 0 | 54 | 24 |
| 47 | 0 | 1 | 30 |
| 0 | 0 | 4 | 0 |
| 32 | 75 | 0 | 51 |
| (-7) | (-33) |
A total of 3 lines are required to cover all zeros.
| 52 | 0 | 54 | 24 | |
| 47 | 0 | 1 | 30 | |
| 0 | 0 | 4 | 0 | x |
| 32 | 75 | 0 | 51 | x |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 1. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 51 | 0 | 53 | 23 |
| 46 | 0 | 0 | 29 |
| 0 | 1 | 4 | 0 |
| 32 | 76 | 0 | 51 |
A total of 3 lines are required to cover all zeros.
| 51 | 0 | 53 | 23 | |
| 46 | 0 | 0 | 29 | |
| 0 | 1 | 4 | 0 | x |
| 32 | 76 | 0 | 51 | |
| x | x |
The number of lines is smaller than 4. The smallest uncovered element is 23. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 28 | 0 | 53 | 0 |
| 23 | 0 | 0 | 6 |
| 0 | 24 | 27 | 0 |
| 9 | 76 | 0 | 28 |
A total of 4 lines are required to cover all zeros.
| 28 | 0 | 53 | 0 | x |
| 23 | 0 | 0 | 6 | x |
| 0 | 24 | 27 | 0 | x |
| 9 | 76 | 0 | 28 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 28 | 0 | 53 | 0 |
| 23 | 0 | 0 | 6 |
| 0 | 24 | 27 | 0 |
| 9 | 76 | 0 | 28 |
This corresponds to the following optimal assignment in the original cost matrix.
| 86 | 27 | 81 | 84 |
| 73 | 19 | 20 | 82 |
| 58 | 51 | 55 | 84 |
| 47 | 83 | 8 | 92 |
The total minimum cost is 169.