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:
28 | 49 | 85 | 70 |
93 | 59 | 96 | 34 |
28 | 15 | 77 | 81 |
12 | 23 | 95 | 64 |
Negate all values
Because the objective is to maximize the total cost we negate all elements:
-28 | -49 | -85 | -70 |
-93 | -59 | -96 | -34 |
-28 | -15 | -77 | -81 |
-12 | -23 | -95 | -64 |
Make the matrix nonnegative
The cost matrix contains negative elements, we add 96 to each entry to make the cost matrix nonnegative:
68 | 47 | 11 | 26 |
3 | 37 | 0 | 62 |
68 | 81 | 19 | 15 |
84 | 73 | 1 | 32 |
Subtract row minima
We subtract the row minimum from each row:
57 | 36 | 0 | 15 | (-11) |
3 | 37 | 0 | 62 | |
53 | 66 | 4 | 0 | (-15) |
83 | 72 | 0 | 31 | (-1) |
Subtract column minima
We subtract the column minimum from each column:
54 | 0 | 0 | 15 |
0 | 1 | 0 | 62 |
50 | 30 | 4 | 0 |
80 | 36 | 0 | 31 |
(-3) | (-36) |
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
54 | 0 | 0 | 15 | x |
0 | 1 | 0 | 62 | x |
50 | 30 | 4 | 0 | x |
80 | 36 | 0 | 31 | x |
The optimal assignment
Because there are 4 lines required, the zeros cover an optimal assignment:
54 | 0 | 0 | 15 |
0 | 1 | 0 | 62 |
50 | 30 | 4 | 0 |
80 | 36 | 0 | 31 |
This corresponds to the following optimal assignment in the original cost matrix:
28 | 49 | 85 | 70 |
93 | 59 | 96 | 34 |
28 | 15 | 77 | 81 |
12 | 23 | 95 | 64 |
The optimal value equals 318.
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