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 | 79 | 22 | 16 |
21 | 21 | 47 | 52 |
11 | 18 | 74 | 47 |
62 | 72 | 9 | 52 |
Subtract row minima
We subtract the row minimum from each row:
6 | 63 | 6 | 0 | (-16) |
0 | 0 | 26 | 31 | (-21) |
0 | 7 | 63 | 36 | (-11) |
53 | 63 | 0 | 43 | (-9) |
Subtract column minima
Because each column contains a zero, subtracting column minima has no effect.
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
6 | 63 | 6 | 0 | x |
0 | 0 | 26 | 31 | x |
0 | 7 | 63 | 36 | x |
53 | 63 | 0 | 43 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
6 | 63 | 6 | 0 |
0 | 0 | 26 | 31 |
0 | 7 | 63 | 36 |
53 | 63 | 0 | 43 |
This corresponds to the following optimal assignment in the original cost matrix:
22 | 79 | 22 | 16 |
21 | 21 | 47 | 52 |
11 | 18 | 74 | 47 |
62 | 72 | 9 | 52 |
The optimal value equals 57.
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