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:
17 | 1 | 10 | 64 |
56 | 71 | 81 | 4 |
36 | 34 | 2 | 61 |
82 | 10 | 90 | 13 |
Subtract row minima
We subtract the row minimum from each row:
16 | 0 | 9 | 63 | (-1) |
52 | 67 | 77 | 0 | (-4) |
34 | 32 | 0 | 59 | (-2) |
72 | 0 | 80 | 3 | (-10) |
Subtract column minima
We subtract the column minimum from each column:
0 | 0 | 9 | 63 |
36 | 67 | 77 | 0 |
18 | 32 | 0 | 59 |
56 | 0 | 80 | 3 |
(-16) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
0 | 0 | 9 | 63 | x |
36 | 67 | 77 | 0 | x |
18 | 32 | 0 | 59 | x |
56 | 0 | 80 | 3 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 0 | 9 | 63 |
36 | 67 | 77 | 0 |
18 | 32 | 0 | 59 |
56 | 0 | 80 | 3 |
This corresponds to the following optimal assignment in the original cost matrix:
17 | 1 | 10 | 64 |
56 | 71 | 81 | 4 |
36 | 34 | 2 | 61 |
82 | 10 | 90 | 13 |
The optimal value equals 33.
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