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):
Size: 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10
This is the original cost matrix:
22 | 4 | 81 | 66 |
81 | 5 | 66 | 7 |
54 | 45 | 21 | 51 |
1 | 89 | 24 | 54 |
Subtract row minima
We subtract the row minimum from each row:
18 | 0 | 77 | 62 | (-4) |
76 | 0 | 61 | 2 | (-5) |
33 | 24 | 0 | 30 | (-21) |
0 | 88 | 23 | 53 | (-1) |
Subtract column minima
We subtract the column minimum from each column:
18 | 0 | 77 | 60 |
76 | 0 | 61 | 0 |
33 | 24 | 0 | 28 |
0 | 88 | 23 | 51 |
(-2) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
18 | 0 | 77 | 60 | x |
76 | 0 | 61 | 0 | x |
33 | 24 | 0 | 28 | x |
0 | 88 | 23 | 51 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
18 | 0 | 77 | 60 |
76 | 0 | 61 | 0 |
33 | 24 | 0 | 28 |
0 | 88 | 23 | 51 |
This corresponds to the following optimal assignment in the original cost matrix:
22 | 4 | 81 | 66 |
81 | 5 | 66 | 7 |
54 | 45 | 21 | 51 |
1 | 89 | 24 | 54 |
The optimal value equals 33.
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