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 | 29 | 25 | 25 |
88 | 22 | 60 | 23 |
86 | 10 | 23 | 17 |
82 | 20 | 98 | 64 |
Subtract row minima
We subtract the row minimum from each row:
0 | 17 | 13 | 13 | (-12) |
66 | 0 | 38 | 1 | (-22) |
76 | 0 | 13 | 7 | (-10) |
62 | 0 | 78 | 44 | (-20) |
Subtract column minima
We subtract the column minimum from each column:
0 | 17 | 0 | 12 |
66 | 0 | 25 | 0 |
76 | 0 | 0 | 6 |
62 | 0 | 65 | 43 |
(-13) | (-1) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
0 | 17 | 0 | 12 | x |
66 | 0 | 25 | 0 | x |
76 | 0 | 0 | 6 | x |
62 | 0 | 65 | 43 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 17 | 0 | 12 |
66 | 0 | 25 | 0 |
76 | 0 | 0 | 6 |
62 | 0 | 65 | 43 |
This corresponds to the following optimal assignment in the original cost matrix:
12 | 29 | 25 | 25 |
88 | 22 | 60 | 23 |
86 | 10 | 23 | 17 |
82 | 20 | 98 | 64 |
The optimal value equals 78.
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