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:
49 | 65 | 50 | 78 |
40 | 31 | 60 | 79 |
77 | 35 | 74 | 33 |
88 | 52 | 29 | 56 |
Subtract row minima
We subtract the row minimum from each row:
0 | 16 | 1 | 29 | (-49) |
9 | 0 | 29 | 48 | (-31) |
44 | 2 | 41 | 0 | (-33) |
59 | 23 | 0 | 27 | (-29) |
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:
0 | 16 | 1 | 29 | x |
9 | 0 | 29 | 48 | x |
44 | 2 | 41 | 0 | x |
59 | 23 | 0 | 27 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 16 | 1 | 29 |
9 | 0 | 29 | 48 |
44 | 2 | 41 | 0 |
59 | 23 | 0 | 27 |
This corresponds to the following optimal assignment in the original cost matrix:
49 | 65 | 50 | 78 |
40 | 31 | 60 | 79 |
77 | 35 | 74 | 33 |
88 | 52 | 29 | 56 |
The optimal value equals 142.
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