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:
42 | 98 | 80 | 29 |
81 | 22 | 97 | 5 |
38 | 70 | 70 | 51 |
31 | 52 | 49 | 47 |
Subtract row minima
We subtract the row minimum from each row:
13 | 69 | 51 | 0 | (-29) |
76 | 17 | 92 | 0 | (-5) |
0 | 32 | 32 | 13 | (-38) |
0 | 21 | 18 | 16 | (-31) |
Subtract column minima
We subtract the column minimum from each column:
13 | 52 | 33 | 0 |
76 | 0 | 74 | 0 |
0 | 15 | 14 | 13 |
0 | 4 | 0 | 16 |
(-17) | (-18) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
13 | 52 | 33 | 0 | x |
76 | 0 | 74 | 0 | x |
0 | 15 | 14 | 13 | x |
0 | 4 | 0 | 16 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
13 | 52 | 33 | 0 |
76 | 0 | 74 | 0 |
0 | 15 | 14 | 13 |
0 | 4 | 0 | 16 |
This corresponds to the following optimal assignment in the original cost matrix:
42 | 98 | 80 | 29 |
81 | 22 | 97 | 5 |
38 | 70 | 70 | 51 |
31 | 52 | 49 | 47 |
The optimal value equals 138.
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