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:
82 | 71 | 93 | 31 |
55 | 92 | 72 | 17 |
31 | 29 | 99 | 76 |
5 | 82 | 61 | 81 |
Subtract row minima
We subtract the row minimum from each row:
51 | 40 | 62 | 0 | (-31) |
38 | 75 | 55 | 0 | (-17) |
2 | 0 | 70 | 47 | (-29) |
0 | 77 | 56 | 76 | (-5) |
Subtract column minima
We subtract the column minimum from each column:
51 | 40 | 7 | 0 |
38 | 75 | 0 | 0 |
2 | 0 | 15 | 47 |
0 | 77 | 1 | 76 |
(-55) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
51 | 40 | 7 | 0 | x |
38 | 75 | 0 | 0 | x |
2 | 0 | 15 | 47 | x |
0 | 77 | 1 | 76 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
51 | 40 | 7 | 0 |
38 | 75 | 0 | 0 |
2 | 0 | 15 | 47 |
0 | 77 | 1 | 76 |
This corresponds to the following optimal assignment in the original cost matrix:
82 | 71 | 93 | 31 |
55 | 92 | 72 | 17 |
31 | 29 | 99 | 76 |
5 | 82 | 61 | 81 |
The optimal value equals 137.
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