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:
32 | 29 | 33 | 17 |
2 | 81 | 68 | 51 |
9 | 46 | 9 | 36 |
32 | 29 | 61 | 72 |
Subtract row minima
We subtract the row minimum from each row:
15 | 12 | 16 | 0 | (-17) |
0 | 79 | 66 | 49 | (-2) |
0 | 37 | 0 | 27 | (-9) |
3 | 0 | 32 | 43 | (-29) |
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:
15 | 12 | 16 | 0 | x |
0 | 79 | 66 | 49 | x |
0 | 37 | 0 | 27 | x |
3 | 0 | 32 | 43 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
15 | 12 | 16 | 0 |
0 | 79 | 66 | 49 |
0 | 37 | 0 | 27 |
3 | 0 | 32 | 43 |
This corresponds to the following optimal assignment in the original cost matrix:
32 | 29 | 33 | 17 |
2 | 81 | 68 | 51 |
9 | 46 | 9 | 36 |
32 | 29 | 61 | 72 |
The optimal value equals 57.
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