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.
| 92 | 89 | 42 | 75 |
| 14 | 40 | 60 | 42 |
| 99 | 65 | 67 | 90 |
| 14 | 21 | 16 | 75 |
For each row, the minimum element is subtracted from all elements in that row.
| 50 | 47 | 0 | 33 | (-42) |
| 0 | 26 | 46 | 28 | (-14) |
| 34 | 0 | 2 | 25 | (-65) |
| 0 | 7 | 2 | 61 | (-14) |
For each column, the minimum element is subtracted from all elements in that column.
| 50 | 47 | 0 | 8 |
| 0 | 26 | 46 | 3 |
| 34 | 0 | 2 | 0 |
| 0 | 7 | 2 | 36 |
| (-25) |
A total of 3 lines are required to cover all zeros.
| 50 | 47 | 0 | 8 | x |
| 0 | 26 | 46 | 3 | |
| 34 | 0 | 2 | 0 | x |
| 0 | 7 | 2 | 36 | |
| 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.
| 52 | 47 | 0 | 8 |
| 0 | 24 | 44 | 1 |
| 36 | 0 | 2 | 0 |
| 0 | 5 | 0 | 34 |
A total of 3 lines are required to cover all zeros.
| 52 | 47 | 0 | 8 | |
| 0 | 24 | 44 | 1 | |
| 36 | 0 | 2 | 0 | x |
| 0 | 5 | 0 | 34 | |
| 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.
| 52 | 46 | 0 | 7 |
| 0 | 23 | 44 | 0 |
| 37 | 0 | 3 | 0 |
| 0 | 4 | 0 | 33 |
A total of 4 lines are required to cover all zeros.
| 52 | 46 | 0 | 7 | x |
| 0 | 23 | 44 | 0 | x |
| 37 | 0 | 3 | 0 | x |
| 0 | 4 | 0 | 33 | x |
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 52 | 46 | 0 | 7 |
| 0 | 23 | 44 | 0 |
| 37 | 0 | 3 | 0 |
| 0 | 4 | 0 | 33 |
This corresponds to the following optimal assignment in the original cost matrix.
| 92 | 89 | 42 | 75 |
| 14 | 40 | 60 | 42 |
| 99 | 65 | 67 | 90 |
| 14 | 21 | 16 | 75 |
The total minimum cost is 163.