Question

Three fertilizers factories X, Y and Z located at different places of the country produce 6,4 and 5 lakh tones of urea respectively. Under the directive of the central government, they are to be distributed to 3 States A, B and C as 5, 3 and 7 lakh respectively. The transportation cost per tones in rupees is given below:

FACTORIES	State A	State B	State C
X	11	17	16
Y	15	12	14
Z	20	12	15

Find out suitable transportation pattern at minimum cost by North West Corner method and Least Cost method.

Answer 1

Answer a) Solution Using North-West Corner Rule:

Step 1: First, prepare the following table:

Step 2:

The rim values for X=6 and A=5 are compared.
The smaller of the two i.e. min(6,5) = 5 is assigned to X A
This meets the complete demand of A and leaves 6 - 5 = 1 units with X

Step 3: The rim values for X=1 and B=3 are compared.
The smaller of the two i.e. min(1,3) = 1 is assigned to X B
This exhausts the capacity of X and leaves 3 - 1 = 2 units with B

Step 4:

The rim values for Y=4 and B=2 are compared.
The smaller of the two i.e. min(4,2) = 2 is assigned to Y B
This meets the complete demand of B and leaves 4 - 2 = 2 units with Y

Step 5:

The rim values for Y=2 and C=7 are compared.
The smaller of the two i.e. min(2,7) = 2 is assigned to Y C
This exhausts the capacity of Y and leaves 7 - 2 = 5 units with C

Step 6:

The rim values for Z=5 and C=5 are compared.
The smaller of the two i.e. min(5,5) = 5 is assigned to Z C

The above table shows the feasible solution. Hence, we get the minimum total cost as mentioned below:

(Note: Cost In Lakhs = Urea (Lakh Tonnes) X Cost Per Tones)

Answer b) Solution Using Least Cost Rule:

Step 1: First, prepare the following table:

Step 2: Pick the cell with the least cost and allocate as many units as possible without exceeding the demand or supply:

The smallest transportation cost is 11 in cell XA
The allocation of this cell is min(6,5) = 5.
This satisfies the entire demand of A and leaves 6 - 5 = 1 units with X

Step 2:

The smallest transportation cost is 12 in cell YB
The allocation of this cell is min(4,3) = 3.
This satisfies the entire demand of B and leaves 4 - 3 = 1 units with Y

Step 3:

The smallest transportation cost is 14 in cell YC
The allocation of this cell is min(1,7) = 1.
This exhausts the capacity of Y and leaves 7 - 1 = 6 units with C

Step 4:

The smallest transportation cost is 15 in cell ZC
The allocation of this cell is min(5,6) = 5.
This exhausts the capacity of Z and leaves 6 - 5 = 1 units with C

Step 5:

The smallest transportation cost is 16 in cell XC
The allocation to this cell is min(1,1) = 1.

The above table shows the feasible solution. Hence, we get the minimum total cost as mentioned below:

(Note: Cost In Lakhs = Urea (Lakh Tonnes) X Cost Per Tones)