Step 1:
Observe that the given problem is a restricted assignment problem. So we assign high cost ‘$\infty$’ to the prohibited cells.
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 6 | 6 | $\infty$ | 3 | 7 |
| B | 8 | 5 | 3 | 4 | 5 |
| C | 10 | 4 | 6 | $\infty$ | 4 |
| D | 8 | 3 | 7 | 8 | 3 |
| E | 7 | 6 | 8 | 10 | 2 |
Step 2: Row minimum
Subtract the smallest element in each row from every element in its row.
The matrix obtained is given below:
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 3 | 3 | $\infty$ | 0 | 4 |
| B | 5 | 2 | 0 | 1 | 2 |
| C | 6 | 0 | 2 | $\infty$ | 0 |
| D | 5 | 0 | 4 | 5 | 0 |
| E | 5 | 4 | 6 | 8 | 0 |
Step 3: Column minimum
Subtract the smallest element in each column of assignment matrix obtained in step 2 from every element in its column.
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 3 | $\infty$ | 0 | 4 |
| B | 2 | 2 | 0 | 1 | 2 |
| C | 3 | 0 | 2 | $\infty$ | 0 |
| D | 2 | 0 | 4 | 5 | 0 |
| E | 2 | 4 | 6 | 8 | 0 |
Step 4:
Draw minimum number of vertical and horizontal lines to cover all zeros.
First cover all rows and columns which have maximum number of zeros.
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 3 | $\infty$ | 0 | 4 |
| B | 2 | 2 | 0 | 1 | 2 |
| C | 3 | 0 | 2 | $\infty$ | 0 |
| D | 2 | 0 | 4 | 5 | 0 |
| E | 2 | 4 | 6 | 8 | 0 |
Step 5:
From step 4, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.
∴ Select smallest element from all the uncovered elements, i.e., 2 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 5 | $\infty$ | 0 | 6 |
| B | 2 | 4 | 0 | 1 | 4 |
| C | 1 | 0 | 0 | $\infty$ | 0 |
| D | 0 | 0 | 2 | 3 | 0 |
| E | 0 | 4 | 4 | 6 | 0 |
Step 6:
Draw minimum number of vertical and horizontal lines to cover all zeros.
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 5 | $\infty$ | 0 | 6 |
| B | 2 | 4 | 0 |
| D | 0 | 0 | 2 | 3 | 0 |
| E | 0 | 4 | 4 | 6 | 0 |
Step 7:
From step 6, minimum number of lines covering all the zeros are 5, which is equal to order of the matrix, i.e., 5.
$\therefore$ Select a row with exactly one zero, enclose that zero in $(\square)$ and cross out all zeros in its respective column.
Similarly, examine each row and column and mark the assignment ( $\square$ ).
$\therefore$ The matrix obtained is as follows:
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 5 | $\infty$ | 0 |
center;">14C100$\infty$0D00230E04460
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 5 | $\infty$ | 0 | 6 |
| B | 2 | 4 | 0 | 1 | 4 |
| C | 1 | 0 | 0 | $\infty$ | 0 |
| D | 0 | 0 | 2 | 3 | 0 |
| E | 0 | 4 | 4 | 6 | 0 |
| Operator | Machine |
| 1 | 2 | 3 | 4 | 5 |
| A | 0 | 5 | $\infty$ | 0 | 6 |
| B | 2 | 4 | 0 | 1 | 4 |
| C | 1 | 0 | 0 | $\infty$ | 0 |
| D | 0 | 0 | 2 | 3 | 0 |
| E | 0 | 4 | 4 | 6 | 0 |
Step 8:
The matrix obtained in step 7 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
| Operator | Machine | Cost |
| A | 4 | 3 |
| B | 3 | 3 |
| C | 2 | 4 |
| D | 5 | 3 |
| E | 1 | 7 |
| Total | 20 |
| Operator | Machine | Cost |
| A | 4 | 3 |
| B | 3 | 3 |
| C | 2 | 4 |
| D | 1 | 8 |
| E | 5 | 2 |
| Total | 20 |
| Operator | Machine | Cost |
| A | 4 | 3 |
| B | 3 | 3 |
| C | 5 | 4 |
| D | 2 | 3 |
| E | 1 | 7 |
| Total | 20 |
∴ Minimum cost = 20 units.