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.
| 17 | 66 | 45 | 93 |
| 61 | 77 | 55 | 48 |
| 56 | 2 | 88 | 26 |
| 4 | 5 | 53 | 95 |
For each row, the minimum element is subtracted from all elements in that row.
| 0 | 49 | 28 | 76 | (-17) |
| 13 | 29 | 7 | 0 | (-48) |
| 54 | 0 | 86 | 24 | (-2) |
| 0 | 1 | 49 | 91 | (-4) |
For each column, the minimum element is subtracted from all elements in that column.
| 0 | 49 | 21 | 76 |
| 13 | 29 | 0 | 0 |
| 54 | 0 | 79 | 24 |
| 0 | 1 | 42 | 91 |
| (-7) |
A total of 3 lines are required to cover all zeros.
| 0 | 49 | 21 | 76 | |
| 13 | 29 | 0 | 0 | x |
| 54 | 0 | 79 | 24 | x |
| 0 | 1 | 42 | 91 | |
| 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.
| 0 | 48 | 20 | 75 |
| 14 | 29 | 0 | 0 |
| 55 | 0 | 79 | 24 |
| 0 | 0 | 41 | 90 |
A total of 3 lines are required to cover all zeros.
| 0 | 48 | 20 | 75 | |
| 14 | 29 | 0 | 0 | x |
| 55 | 0 | 79 | 24 | |
| 0 | 0 | 41 | 90 | |
| x | x |
The number of lines is smaller than 4. The smallest uncovered element is 20. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 0 | 48 | 0 | 55 |
| 34 | 49 | 0 | 0 |
| 55 | 0 | 59 | 4 |
| 0 | 0 | 21 | 70 |
A total of 4 lines are required to cover all zeros.
| 0 | 48 | 0 | 55 | x |
| 34 | 49 | 0 | 0 | x |
| 55 | 0 | 59 | 4 | x |
| 0 | 0 | 21 | 70 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 0 | 48 | 0 | 55 |
| 34 | 49 | 0 | 0 |
| 55 | 0 | 59 | 4 |
| 0 | 0 | 21 | 70 |
This corresponds to the following optimal assignment in the original cost matrix.
| 17 | 66 | 45 | 93 |
| 61 | 77 | 55 | 48 |
| 56 | 2 | 88 | 26 |
| 4 | 5 | 53 | 95 |
The total minimum cost is 99.