North West Corner Rule
The North West corner rule is a method for computing
a basic feasible solution of a transportation problem, where the basic
variables are selected from the North – West corner ( i.e., top left
corner ).
The standard instructions for a transportation model are paraphrased below. Please read them carefully.
Steps in North West Corner Rule
 Select the upper lefthand corner cell of the transportation
table and allocate as many units as possible equal to the minimum
between available supply and demand, i.e., min(s1, d1).
 Adjust the supply and demand numbers in the respective rows and
columns.
 If the demand for the first cell is satisfied, then move horizontally
to the next cell in the second column.
 If the supply for the first row is exhausted, then move down to
the first cell in the second row.
 If for any cell, supply equals demand, then the next allocation
can be made in cell either in the next row or column.
 Continue the process until all supply and demand values are exhausted.

This trial routing method is often far from optimal. 
Example 1: North West Corner Rule  Transportation Problem
The Amulya Milk Company has three plants located throughout a state
with production capacity 50, 75 and 25 gallons. Each day the firm must
furnish its four retail shops R_{1}, R_{2}, R_{3},
& R_{4} with at least 20, 20 , 50, and 60 gallons respectively.
The transportation costs (in Rs.) are given below.
Plant 
Retail Shop 
Supply 
R_{1} 
R_{2} 
R_{3} 
R_{4} 
P_{1} 
3 
5 
7 
6 
50 
P_{2} 
2 
5 
8 
2 
75 
P_{3} 
3 
6 
9 
2 
25 
Demand 
20 
20 
50 
60 

The economic problem is to distribute the available product to different
retail shops in such a way so that the total transportation cost is
minimum
Solution.
Starting from the North west corner, we allocate min (50, 20) to P_{1}R_{1},
i.e., 20 units to cell P_{1}R_{1}. The demand for the
first column is satisfied. The allocation is shown in the following
table.
Table 1
Plant 
Retail Shop 
Supply 
R_{1} 
R_{2} 
R_{3} 
R_{4} 
P_{1} 

5 
7 
6 
50 30 
P_{2} 
2 
5 
8 
2 
75 
P_{3} 
3 
6 
9 
2 
25 
Demand 
20 
20 
50 
60 

Now we move horizontally to the second column in the first row and
allocate 20 units to cell P_{1}R_{2}. The demand for
the second column is also satisfied.
Table 2
Plant 
Retail Shop 
Supply 
R_{1} 
R_{2} 
R_{3} 
R_{4} 
P_{1} 


7 
6 
50 30 10 
P_{2} 
2 
5 
8 
2 
75 
P_{3} 
3 
6 
9 
2 
25 
Demand 
20 
20 
50 
60 

Proceeding in this way, we observe that P_{1}R_{3}
= 10, P_{2}R_{3} = 40, P_{2}R_{4} =
35, P_{3}R_{4} = 25. The resulting feasible solution
is shown in the following table.
Final Table
Plant 
Retail Shop 
Supply 
R_{1} 
R_{2} 
R_{3} 
R_{4} 
P_{1} 



6 
50 
P_{2} 
2 
5 


75 
P_{3} 
3 
6 
9 

25 
Demand 
20 
20 
50 
60 

Here, number of retail shops(n) = 4, and
Number of plants (m) = 3
Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6.
Initial basic feasible solution
The total transportation cost is calculated by multiplying each x_{ij}
in an occupied cell with the corresponding c_{ij} and adding
as follows:
20 X 3 + 20 X 5 + 10 X 7 + 40 X 8 + 35 X 2 + 25 X 2 = 670

