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:
40 | 58 | 92 | 11 |
44 | 9 | 23 | 86 |
5 | 18 | 86 | 67 |
67 | 75 | 45 | 76 |
Subtract row minima
We subtract the row minimum from each row:
29 | 47 | 81 | 0 | (-11) |
35 | 0 | 14 | 77 | (-9) |
0 | 13 | 81 | 62 | (-5) |
22 | 30 | 0 | 31 | (-45) |
Subtract column minima
Because each column contains a zero, subtracting column minima has no effect.
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
29 | 47 | 81 | 0 | x |
35 | 0 | 14 | 77 | x |
0 | 13 | 81 | 62 | x |
22 | 30 | 0 | 31 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
29 | 47 | 81 | 0 |
35 | 0 | 14 | 77 |
0 | 13 | 81 | 62 |
22 | 30 | 0 | 31 |
This corresponds to the following optimal assignment in the original cost matrix:
40 | 58 | 92 | 11 |
44 | 9 | 23 | 86 |
5 | 18 | 86 | 67 |
67 | 75 | 45 | 76 |
The optimal value equals 70.
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