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:
8 | 34 | 24 | 83 |
75 | 63 | 50 | 77 |
57 | 34 | 93 | 15 |
67 | 8 | 82 | 79 |
Subtract row minima
We subtract the row minimum from each row:
0 | 26 | 16 | 75 | (-8) |
25 | 13 | 0 | 27 | (-50) |
42 | 19 | 78 | 0 | (-15) |
59 | 0 | 74 | 71 | (-8) |
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 | 26 | 16 | 75 | x |
25 | 13 | 0 | 27 | x |
42 | 19 | 78 | 0 | x |
59 | 0 | 74 | 71 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 26 | 16 | 75 |
25 | 13 | 0 | 27 |
42 | 19 | 78 | 0 |
59 | 0 | 74 | 71 |
This corresponds to the following optimal assignment in the original cost matrix:
8 | 34 | 24 | 83 |
75 | 63 | 50 | 77 |
57 | 34 | 93 | 15 |
67 | 8 | 82 | 79 |
The optimal value equals 81.
HungarianAlgorithm.com © 2013-2025