Explanation:
In the assignment model, the following steps are involved to find the opportunity cost matrix
10
|
3
|
10
|
7
|
7
|
5
|
9
|
7
|
11
|
9
|
13
|
18
|
2
|
9
|
10
|
15
|
3
|
2
|
7
|
4
|
16
|
16
|
2
|
12
|
12
|
1.Subtract the smallest element of each row from every element of the corresponding row
7
|
0
|
7
|
4
|
4
|
0
|
4
|
2
|
6
|
4
|
11
|
16
|
0
|
7
|
8
|
13
|
1
|
0
|
5
|
2
|
14
|
4
|
0
|
10
|
10
|
Now, subtract the smallest element in each column from every element of the corresponding column.
7
|
0
|
7
|
0
|
2
|
0
|
4
|
2
|
2
|
2
|
11
|
16
|
0
|
3
|
6
|
13
|
1
|
0
|
1
|
0
|
14
|
4
|
0
|
6
|
8
|
Now, make an assignment in the opportunity cost matrix. For assigning initially check single zero rows and assign the value to it and then cut the remaining zero in the column, then find the column with single zero and then repeat this procedure till all the assignment is done

Now, if the no-of allocation is exactly equal to the size of matric, then the current solution is optimum otherwise perform optimality
In this case no-of allocations = 4, so the solution is not optimum
Now, for optimality
i) Mark all rows for which assignment have not been made (5th row)
ii) Mark all columns which have unassigned zero in the marked (column 3)
iii) Mark all row which have an assignment in the marked column (row 3)

Now, draw minimum no. of lines through the unmarked row and through the marked column to cover all zeros at least once.
Now, select the smallest element that does not have a line through then, subtract form all the elements that do not have a line through them, add it to every element that lies at the intersection of two lines and leave the remaining element matrix unchanged. Then make allocations in the new opportunity cost matrix.
Here smallest element is in 3rd row (3)
7
|
0
|
10
|
0
|
2
|
0
|
4
|
5
|
2
|
2
|
8
|
13
|
0
|
0
|
3
|
13
|
1
|
3
|
1
|
0
|
11
|
1
|
0
|
3
|
5
|
Now making assignment

The final allocation is
I - B, II → A, III → D, IV - E and V - C
Total cost = 3 + 5 + 9 + 4 + 2 = 23