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.
| 3 | 79 | 85 | 60 |
| 41 | 32 | 27 | 22 |
| 8 | 38 | 46 | 48 |
| 8 | 4 | 56 | 8 |
For each row, the minimum element is subtracted from all elements in that row.
| 0 | 76 | 82 | 57 | (-3) |
| 19 | 10 | 5 | 0 | (-22) |
| 0 | 30 | 38 | 40 | (-8) |
| 4 | 0 | 52 | 4 | (-4) |
For each column, the minimum element is subtracted from all elements in that column.
| 0 | 76 | 77 | 57 |
| 19 | 10 | 0 | 0 |
| 0 | 30 | 33 | 40 |
| 4 | 0 | 47 | 4 |
| (-5) |
A total of 3 lines are required to cover all zeros.
| 0 | 76 | 77 | 57 | |
| 19 | 10 | 0 | 0 | x |
| 0 | 30 | 33 | 40 | |
| 4 | 0 | 47 | 4 | x |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 30. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 0 | 46 | 47 | 27 |
| 49 | 10 | 0 | 0 |
| 0 | 0 | 3 | 10 |
| 34 | 0 | 47 | 4 |
A total of 3 lines are required to cover all zeros.
| 0 | 46 | 47 | 27 | |
| 49 | 10 | 0 | 0 | x |
| 0 | 0 | 3 | 10 | |
| 34 | 0 | 47 | 4 | |
| 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 | 46 | 44 | 24 |
| 52 | 13 | 0 | 0 |
| 0 | 0 | 0 | 7 |
| 34 | 0 | 44 | 1 |
A total of 4 lines are required to cover all zeros.
| 0 | 46 | 44 | 24 | x |
| 52 | 13 | 0 | 0 | x |
| 0 | 0 | 0 | 7 | x |
| 34 | 0 | 44 | 1 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 0 | 46 | 44 | 24 |
| 52 | 13 | 0 | 0 |
| 0 | 0 | 0 | 7 |
| 34 | 0 | 44 | 1 |
This corresponds to the following optimal assignment in the original cost matrix.
| 3 | 79 | 85 | 60 |
| 41 | 32 | 27 | 22 |
| 8 | 38 | 46 | 48 |
| 8 | 4 | 56 | 8 |
The total minimum cost is 75.