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:
84 | 13 | 65 | 1 |
71 | 2 | 6 | 39 |
67 | 35 | 31 | 80 |
30 | 81 | 45 | 52 |
Subtract row minima
We subtract the row minimum from each row:
83 | 12 | 64 | 0 | (-1) |
69 | 0 | 4 | 37 | (-2) |
36 | 4 | 0 | 49 | (-31) |
0 | 51 | 15 | 22 | (-30) |
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:
83 | 12 | 64 | 0 | x |
69 | 0 | 4 | 37 | x |
36 | 4 | 0 | 49 | x |
0 | 51 | 15 | 22 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
83 | 12 | 64 | 0 |
69 | 0 | 4 | 37 |
36 | 4 | 0 | 49 |
0 | 51 | 15 | 22 |
This corresponds to the following optimal assignment in the original cost matrix:
84 | 13 | 65 | 1 |
71 | 2 | 6 | 39 |
67 | 35 | 31 | 80 |
30 | 81 | 45 | 52 |
The optimal value equals 64.
HungarianAlgorithm.com © 2013-2024