North-west corner method (NWCM) Assignment Help

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North-west corner method (NWCM)

In this method, the allocation of products starts at the north-west corner (or the top left corner) of the transportation table. The method is explained below.

Table: Transportation Problem in Linear Programming

1470_transportation problem in linear programming.png

Step 1: Assign the maximum possible quantity of products to the top left corner cell of the transportation problem.

Step 2: After the allocation, adjust the supply and demand numbers.

Step 3: If the supply in the first row is exhausted, move down to the corresponding cell in second row and assign the possible quantity of products to that cell. If the demand in the column is first satisfied, move horizontally to the next cell in the second column and assign the quantity of products.

Step 4: Continue the same procedure till the entire requirements are met.

Step 5: Check for feasibility of the solution.

Problem

Distances between factory and its warehouses and demand at each warehouse are given in the table.

Table: Transportation Table

Factory/Warehouse                 

W1

W2

W3

Supply

F1

16

22

14

200

F2

18

14

18

150

F3

8

14

16

100

Demand

175

125

150

 

Find out the solution for transporting the goods at a minimum cost for the given transportation problems using North-West corner method.

Solution

Step 1: Now we assign the maximum number of goods that can be transported from 'F1' to 'W1', in the cell (F1, W1); i.e. 175.

Step 2: Then we move to the cell (F1, W2) and assign the possible number of goods; i.e. 25.

Step 3: Now, we move to the cell (F2, W2) and assign the possible number of goods; i.e. 100.

Step 4: Now, we move to the cell (F2, W3) and assign the remaining goods being supplied by F2 to W3; i.e. 50.

Step 5: The remaining goods are assigned in the cell (F3, W3); i.e. 100.

Thus, the solution of the transportation problem is  

2488_north west corner method.png

The number of occupied cells is 5 that is equal to (m + n - 1) cells, where m, n are the number of origins and destinations respectively. So, the solution obtained is a feasible solution.

Therefore, the cost associated with the solution is (175×16) + (25 × 22) + (100 × 14) + (50 × 18) + (100 × 16) = Rs. 7250.   Email based Operations Management assignment help - homework help at Expertsmind

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