Vogel's approximation method (VAM)

Vogel's Approximation Method is the most preferred method over the above two methods as it usually results in an optimal or a near optimal solution. The method is explained below.

Step 1: Calculate a penalty for each row and column of the transportation table. The penalty for a row/column is the difference between the least cost and the next least cost of that row/column.

Step 2: Identify the row or column with the largest penalty value and assign the possible quantity of products to that cell having the least unit cost in that row or column. In case of a tie, select the row or column that has minimum cost.

Step 3: Adjust the supply and requirement values after the allocation is made.

Step 4: Delete that row or column where the supply or requirement is zero.

Step 5: Calculate the values of penalty to all the rows and columns for the reduced transportation problem and repeat the same procedure till the entire requirement has been met.

Problem 

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

Table: Transportation Table

Solve Problem using Vogel's Approximation method.

Solution

Step 1: Compute the penalty for each row and column of the transportation problems.  The penalty for the first row is, (16 - 14) = 2. Similarly the values of penalty for the second and the third row are 4 and 6.

Similarly, the values of penalty for the first, second and the third columns are 8, 0 and 2 respectively.

Step 2: Identify the row or column with the largest penalty value. (Here the first column with a penalty value of 8).

Step 3: The cell with the least cost is chosen and the possible number of goods is assigned to that cell. Therefore, assign 100 to the cell (F₃, W₁).

Step 4: If the remaining row supply or column demand is zero, remove that row/column.

Now, the transportation problem can be reduced to (refer table below):

Step 5: The process is repeated for the reduced transportation problem till the entire supply at the factories is assigned to satisfy the demand at different warehouses.

Now, the 'W₂' column has the highest penalty, i.e. 8. Therefore, assign 125 units to the cell (F₂, W₂) since the cell has the least cost in the 'W₂' column.

Then the transportation problem can further be reduced (Refer table below).

Now, the 'W₃' column has the highest penalty, i.e. 4. Assign 150 units to the cell (F₁, W₃) since the cell has the least cost. Then remove the 'W₃' column and the remaining units are assigned to the cells (F₁, W₁) and (F₂, W₁). Thus, 50 units are assigned to the cell (F₁, W₁) and 25 units to the cell (F₂, W₁).

Since the number of cells occupied is 5, i.e., (3+3-1), the solution obtained is a feasible solution. Thus, the cost associated with the solution is, (50 × 16) + (25 × 18) + (100 × 8) + (125 × 14) + (150 × 14) = Rs.5,900.

The solution obtained by this method is the same as that obtained by the least cost method.                   

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