Solution
This is the cost matrix.
| 17 | 1 | 10 | 64 |
| 56 | 71 | 81 | 4 |
| 36 | 34 | 2 | 61 |
| 82 | 10 | 90 | 13 |
Subtract row minima
For each row, the minimum element is subtracted from all elements in that row.
| 16 | 0 | 9 | 63 | (-1) |
| 52 | 67 | 77 | 0 | (-4) |
| 34 | 32 | 0 | 59 | (-2) |
| 72 | 0 | 80 | 3 | (-10) |
Subtract column minima
For each column, the minimum element is subtracted from all elements in that column.
| 0 | 0 | 9 | 63 |
| 36 | 67 | 77 | 0 |
| 18 | 32 | 0 | 59 |
| 56 | 0 | 80 | 3 |
| (-16) | | | |
Cover all zeros with a minimum number of lines
A total of 4 lines are required to cover all zeros.
| 0 | 0 | 9 | 63 | x |
| 36 | 67 | 77 | 0 | x |
| 18 | 32 | 0 | 59 | x |
| 56 | 0 | 80 | 3 | x |
The optimal assignment
Because there are 4 lines required, an optimal assignment exists among the zeros.
| 0 | 0 | 9 | 63 |
| 36 | 67 | 77 | 0 |
| 18 | 32 | 0 | 59 |
| 56 | 0 | 80 | 3 |
This corresponds to the following optimal assignment in the original cost matrix.
| 17 | 1 | 10 | 64 |
| 56 | 71 | 81 | 4 |
| 36 | 34 | 2 | 61 |
| 82 | 10 | 90 | 13 |
The total minimum cost is 33.