Transportation problem is a type of linear programming problem in Operational Research. It is used to find a minimum cost of transporting. It uses certain methods to minimize the cost and time of transportation. The methods that are used are:
What is Vogelâ€™s Approximation Method?
Vogelâ€™s Approximation Method (VAM) is a technique used to find a good initial feasible solution to transportation problems. It takes into account the costs associated with each route alternative. It is preferred over the Least Cost Method and the North-West Corner Approach because the initial feasible solution obtained through this approach is an â€œoptimalâ€ or â€œvery close to optimal solution.â€
Feasible Solution by VAM homework answers allows you to find a possible solution to transportation problems. VAM is the most popular of all the transportation techniques that are used widely. It involves lengthy calculations to arrive at a minimum cost. As it is a difficult topic to understand, you first need to understand the steps that are taken to reach an optimal solution.
Steps involved in Vogelâ€™s Approximation Method
In order to understand VAM better, it is essential to understand the steps involved.
What is the most Feasible Solution by VAM?
When all the above steps are followed, you are left with a cost which is known as the most feasible solution. The cost is mostly known as opportunity cost as we chose one price from available alternatives.
Let us understand with the help of an illustration.
The below is a transportation table that shows the distance between a factory and its warehouse. It also includes the demand at each warehouse.
Step 1: Calculate the penalty for each row and column from the two lowest shipping costs.
The penalties for the row X is (4-3) =1. Similarly, for rows, Y and Z are 1 and 2 respectively
The penalties for the column A is (8-5) =3. Similarly, for B and C are 1 and 2 respectively.
Step 2: Identify the row and column with maximum penalty. Let us take column A for example with a penalty of 8.
Step 3: Select the cell with the least cost (X, A), i.e. 5 and assign the maximum possible unit, i.e. 100.
Step 4: Eliminate the row or column if the quantity is exhausted. Therefore, now the transportation problem will be reduced as below.
Step 5: The process is repeated until an optimal solution is found.
The next biggest difference is 3, so we assign maximum units 200 to (B, Y), i.e. 4. So we are left with the below.
Once again, we find the biggest difference which is 2 and assign 100 units to the cell (Y, C), i.e. 3, which eliminates the row Y.
So now, we are left with the below transportation table.
Now, we see that an assignment of 200 units to (Z, A), i.e. 9 and 100 units to (Z, C), i.e. 5completes the table.
Therefore, the cost associated with this solution can be determined as (100 units Ã— $5) + (200 units Ã— $4) + (100 units Ã— $3) + (200 units Ã— $9) + (100 units Ã— $5) = $3,900.
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