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

Step 2: Select the minimum 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. Mark the zero by ׈ and draw a vertical line.
After examining the rows, examine the columns with exactly one zero. Mark the zero by ׈ and draw a horizontal line.

Only 4 assignments are made
Number not lying on the line are

and minimum of these numbers are 1.
This number 1 should be subtracted from the above number and 1 should be added to the numbers which are on the intersecting lines (ie. 52,43,59,30,21)and the other numbers remains the same.
A new cost matrix is as follows and repeat step 3.
New cost matrix is

Thus, all the six 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 1: Select a minimum element in each row and subtract this from all the elements in its row.
∴ The cost matrix of the given assignment problem is

Here column c, d and e have no zeros. Go to step 2.
Step 2 : Select the minimum element in each column and subtract this from all the elements in its column.

Since each row and column contains exactly atleast one zero, assignments can be made.
Step 3: (Assignment)
Examine the rows with exactly one zero. Mark that zero by ׈ and draw a vertical line.

Also examine the columns with exactly one zero. Mark that zero by ׈ and draw a horizontal line.
Here, numbers not lying on the line are and minimum of these number is 15.


Now, subtract 15 from all these numbers and numbers lying on the line remains the same.
Hence, the new cost matrix is as follows.
Again repeat step 3 (Assignment)

Thus, all the 5 assignments have been made.
The optimal assignment schedule and total cost is

Here total supply = 6 + 1 + 10 = 7
total demand = 7 + 5 + 3 + 2 = 17
total supply = total demand
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution for the given transportation problem.
NWC:
I. allocation:

[∵ highest penalty = 6. 1 & min (1, 2) = 1]
II. allocation:

[∵ highestpenalty= 5. In D₂, leastcost= 3&min (5,6) = 5]
III. allocation:

[∵ highest penalty = 5. In O₁, & least cost = 2 &min (7, 1) = 1]
IV. allocation:

[∵ highest penalty = 4. In O₃, least cost is 5 &min (6, 10) = 6]
V. allocation:

[In O₃ least cost = 9 & min (1,4) = 1]
VI. allocation:


[∵ min (3, 3) = 3]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₁ → D₂, O₂ → D₄, O₃ → D₃  O₃ → D₁ and O₃ → D₄
Hence, the total transportation cost
= 1(2) + 5(3) + 1(1) + 6(5) + 3(15) + 1(9)
= 2 + 15 + 1 + 30 + 45 + 9 = Rs. 102

Here total supply = 25 + 35 + 40 = 100
total requirement = 30 + 25 + 45 100
total supply = total requirement
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution for the given transportation problem.
North West Corner Rule:
I - allocation:

[∵ min (25, 30) = 25]
II - allocation:

[∵ min (5, 35) = 5]
III - allocation:

[∵ min (25, 30) = 25]
IV - allocation:

[∵ min (5, 45) = 5]
V - allocation:

[∵ min (40, 40) = 40]
Thus, the allocations are

∴ The transportation schedule is
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₂ → D₃, S₃ → D₃
Hence, the total transportation cost
= 25(9) + 5(6) + 25(8) + 5(4) + 40(9)
= 225 + 30 + 200 + 20 + 360
= Rs. 835

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one.
To balance it, introduce dummy columns with all the entries zero.
∴ The revised assignment problem is

Step 1 : Select the smallest element in each row and subtract this from all the elements in its row.

Since each row and column has atleast one zero, assignments can be made.
Step 2 : Examine the rows with atleast one row.
Row A & B have exactly one row. Mark them by ׈ and mark X by other zeros in its column.
Row C & F also has only one zero.
Here only 4 vacant space can be assigned to 4 trucks.
∴ The optimal assignments schedule and total cost is

∴ The optimal assignment (minimum) cost = Rs. 12