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.
| 74 | 46 | 85 | 81 |
| 6 | 7 | 21 | 11 |
| 69 | 94 | 49 | 13 |
| 44 | 42 | 62 | 88 |
For each row, the minimum element is subtracted from all elements in that row.
| 28 | 0 | 39 | 35 | (-46) |
| 0 | 1 | 15 | 5 | (-6) |
| 56 | 81 | 36 | 0 | (-13) |
| 2 | 0 | 20 | 46 | (-42) |
For each column, the minimum element is subtracted from all elements in that column.
| 28 | 0 | 24 | 35 |
| 0 | 1 | 0 | 5 |
| 56 | 81 | 21 | 0 |
| 2 | 0 | 5 | 46 |
| (-15) |
A total of 3 lines are required to cover all zeros.
| 28 | 0 | 24 | 35 | |
| 0 | 1 | 0 | 5 | x |
| 56 | 81 | 21 | 0 | x |
| 2 | 0 | 5 | 46 | |
| x |
The number of lines is smaller than 4. The smallest uncovered element is 2. We subtract this value from all uncovered elements and add it to all elements covered twice.
| 26 | 0 | 22 | 33 |
| 0 | 3 | 0 | 5 |
| 56 | 83 | 21 | 0 |
| 0 | 0 | 3 | 44 |
A total of 4 lines are required to cover all zeros.
| 26 | 0 | 22 | 33 | x |
| 0 | 3 | 0 | 5 | x |
| 56 | 83 | 21 | 0 | x |
| 0 | 0 | 3 | 44 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 26 | 0 | 22 | 33 |
| 0 | 3 | 0 | 5 |
| 56 | 83 | 21 | 0 |
| 0 | 0 | 3 | 44 |
This corresponds to the following optimal assignment in the original cost matrix.
| 74 | 46 | 85 | 81 |
| 6 | 7 | 21 | 11 |
| 69 | 94 | 49 | 13 |
| 44 | 42 | 62 | 88 |
The total minimum cost is 124.