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:
27 | 42 | 46 | 78 |
94 | 4 | 1 | 99 |
75 | 79 | 38 | 3 |
34 | 27 | 3 | 59 |
Subtract row minima
We subtract the row minimum from each row:
0 | 15 | 19 | 51 | (-27) |
93 | 3 | 0 | 98 | (-1) |
72 | 76 | 35 | 0 | (-3) |
31 | 24 | 0 | 56 | (-3) |
Subtract column minima
We subtract the column minimum from each column:
0 | 12 | 19 | 51 |
93 | 0 | 0 | 98 |
72 | 73 | 35 | 0 |
31 | 21 | 0 | 56 |
(-3) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
0 | 12 | 19 | 51 | x |
93 | 0 | 0 | 98 | x |
72 | 73 | 35 | 0 | x |
31 | 21 | 0 | 56 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
0 | 12 | 19 | 51 |
93 | 0 | 0 | 98 |
72 | 73 | 35 | 0 |
31 | 21 | 0 | 56 |
This corresponds to the following optimal assignment in the original cost matrix:
27 | 42 | 46 | 78 |
94 | 4 | 1 | 99 |
75 | 79 | 38 | 3 |
34 | 27 | 3 | 59 |
The optimal value equals 37.
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