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 | 85 | 99 | 63 |
96 | 61 | 65 | 98 |
15 | 15 | 15 | 92 |
91 | 60 | 16 | 91 |
Subtract row minima
We subtract the row minimum from each row:
0 | 63 | 77 | 41 | (-22) |
35 | 0 | 4 | 37 | (-61) |
0 | 0 | 0 | 77 | (-15) |
75 | 44 | 0 | 75 | (-16) |
Subtract column minima
We subtract the column minimum from each column:
0 | 63 | 77 | 4 |
35 | 0 | 4 | 0 |
0 | 0 | 0 | 40 |
75 | 44 | 0 | 38 |
(-37) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
0 | 63 | 77 | 4 | x |
35 | 0 | 4 | 0 | x |
0 | 0 | 0 | 40 | x |
75 | 44 | 0 | 38 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 63 | 77 | 4 |
35 | 0 | 4 | 0 |
0 | 0 | 0 | 40 |
75 | 44 | 0 | 38 |
This corresponds to the following optimal assignment in the original cost matrix:
22 | 85 | 99 | 63 |
96 | 61 | 65 | 98 |
15 | 15 | 15 | 92 |
91 | 60 | 16 | 91 |
The optimal value equals 151.
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