### What Is the Most Feasible Solution by VAM? How to Do Homework Answers?

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:

- Least Cost Method,
- Northwest Corner Method,
- Modi/ UV Method, and
- Vogel’s Approximation Method

**What is Vogel’s Approximation Method?**

Vogel’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 “optimal” or “very close to optimal solution.”

**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 Vogel’s Approximation Method**

In order to understand VAM better, it is essential to understand the steps involved.

- Identify the minimum and next to minimize transportation costs in each row and column and calculate the difference. This difference is known as a penalty.
- Identify the row and column with the maximum penalty.
- In the designated row or column, select the cell with the least cost and assign maximum possible units.
- Eliminate the row or column where all the quantities are exhausted and recalculate the penalties.
- Return to step 2 and repeat the steps until you obtain an initial feasible solution.

**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.

Factory/Warehouse | A | B | C | Supply |

X | 5 | 4 | 3 | 100 |

Y | 8 | 4 | 3 | 300 |

Z | 9 | 7 | 5 | 300 |

Demand | 300 | 200 | 200 |

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.

Factory/Warehouse | A | B | C | Supply | Penalty |

X | 5 | 4 | 3 | 100 | 1 |

Y | 8 | 4 | 3 | 300 | 1 |

Z | 9 | 7 | 5 | 300 | 2 |

Demand | 300 | 200 | 200 | ||

Penalty | 3 | 3 | 2 |

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.

Factory/Warehouse | A | B | C | Supply | Penalty |

Y | 8 | 4 | 3 | 300 | 1 |

Z | 9 | 7 | 5 | 300 | 2 |

Demand | 300 | 200 | 200 | ||

Penalty | 1 | 3 | 1 |

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.

Factory/Warehouse | A | C | Supply | Penalty |

Y | 8 | 3 | 300 | 5 |

Z | 9 | 5 | 300 | 4 |

Demand | 300 | 200 | ||

Penalty | 1 | 2 |

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.

Factory/Warehouse | A | C | Supply |

Z | 9 | 5 | 300 |

Demand | 300 |

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 × $5) + (200 units × $4) + (100 units × $3) + (200 units × $9) + (100 units × $5) = $3,900.

**Why do students need the help of ****Feasible Solution by VAM homework answers?**

As we have seen from the above illustration, the VAM method is a lengthy one and takes a lot of time to arrive at a good initial solution. If the concepts are clear to you, you do not need much help, but incase everything goes above your head; you need expert advice. There are a number of websites that can assist you with your assignments and projects. These homework help sites employ experts with academic and industrial knowledge. With **Feasible Solution by VAM homework answers, **you will be able to

- Understand the different steps in the VAM method,
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- Follow the methods to arrive at an optimal feasible solution.

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