Here, the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The cost matrix: of the given assignment problem is

Column 1 contains no zero. Go to step 2.
Step 2 : Select the smallest element (1) in column 1 and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 : Examine the rows with only one zero.
Row P, Q and S contains exactly one zero, mark them by 0 and mark the other zeros in the column byX.
Thus, all the four assignments have been made.
∴ The optimal assignment schedule and total cost is

Here, the number of rows and columns are equal
∴ The given assignment problem is balanced.
Step I : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The given assignment problem is

Here column 2 has no zero. Go to Step 2.
Step 2 : Select the smallest element (10) and subtract it from all the elements in its column.

Step 3 : Examine the rows with only one zero mark that zero by Ԡ. Mark other zeros in its column by X.
Row 1 and Row 3 contains only one zero. Mark the other zeros by X
Column 2 contains exactly one zero. Mark it by Ԡ
Thus, all the 3 assignments have been made.
Hence, the optimal assignment schedule and total cost is

Total cost = Rs. 280
Thus, the optimal assignment (minimum) cost  =  Rs. 280

Here the number of rows and columns are equal.
\(\therefore\) The given assignment problem is balanced.
Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.
The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3: (Assignment):
Examine the rows with exactly one zero. Row B contains exactly one zero. Mark that zero by \(\square\) (i.e) Person B is assigned to Job 1. Mark other zeros in its column by ×.

Now, Row C contains exactly one zero. Mark that zero by \(\square\). Mark other zeros in its column by × .

Now, Row D contains exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its column by × .

Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

There is an assignment in Column 3 and column 4. Go to Column 5. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 9

Here \(\sum { { a }_{ i }=\sum { { b }_{ j } } =80 } \)
(i.e) Total Availability = Total Requirement
\(\therefore\) The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First Allocation:

Second Allocation:

Third Allocation:

Fourth Allocation:

Fifth Allocation:

Sixth Allocation:

Thus we have the following allocations:


Transportation schedule :
A⟶I, A⟶II, A⟶III, A⟶IV, B⟶I, C⟶IV, D⟶II
Total transportation cost:
= (6×5)+(6+1)+(17×3)+(5×3)+(15×3)+(12×3)+(19×1)
= 30+6+51+15+45+36+19
= Rs.202