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:
12 | 42 | 51 | 68 |
64 | 6 | 63 | 2 |
26 | 6 | 65 | 29 |
19 | 20 | 9 | 27 |
Subtract row minima
We subtract the row minimum from each row:
0 | 30 | 39 | 56 | (-12) |
62 | 4 | 61 | 0 | (-2) |
20 | 0 | 59 | 23 | (-6) |
10 | 11 | 0 | 18 | (-9) |
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:
0 | 30 | 39 | 56 | x |
62 | 4 | 61 | 0 | x |
20 | 0 | 59 | 23 | x |
10 | 11 | 0 | 18 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 30 | 39 | 56 |
62 | 4 | 61 | 0 |
20 | 0 | 59 | 23 |
10 | 11 | 0 | 18 |
This corresponds to the following optimal assignment in the original cost matrix:
12 | 42 | 51 | 68 |
64 | 6 | 63 | 2 |
26 | 6 | 65 | 29 |
19 | 20 | 9 | 27 |
The optimal value equals 29.
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