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:
35 | 19 | 30 | 67 |
53 | 23 | 61 | 21 |
2 | 49 | 6 | 47 |
5 | 80 | 61 | 88 |
Subtract row minima
We subtract the row minimum from each row:
16 | 0 | 11 | 48 | (-19) |
32 | 2 | 40 | 0 | (-21) |
0 | 47 | 4 | 45 | (-2) |
0 | 75 | 56 | 83 | (-5) |
Subtract column minima
We subtract the column minimum from each column:
16 | 0 | 7 | 48 |
32 | 2 | 36 | 0 |
0 | 47 | 0 | 45 |
0 | 75 | 52 | 83 |
(-4) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
16 | 0 | 7 | 48 | x |
32 | 2 | 36 | 0 | x |
0 | 47 | 0 | 45 | x |
0 | 75 | 52 | 83 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
16 | 0 | 7 | 48 |
32 | 2 | 36 | 0 |
0 | 47 | 0 | 45 |
0 | 75 | 52 | 83 |
This corresponds to the following optimal assignment in the original cost matrix:
35 | 19 | 30 | 67 |
53 | 23 | 61 | 21 |
2 | 49 | 6 | 47 |
5 | 80 | 61 | 88 |
The optimal value equals 51.
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