The Least Cost Method is a technique used to find an initial feasible solution to the Transportation Problem in operations research. The transportation problem involves determining the most cost-effective way to transport goods from multiple suppliers (or sources) to multiple consumers (or destinations), while meeting supply and demand constraints.

Goal:

The objective of the transportation problem is to minimize the total transportation cost while fulfilling the supply and demand requirements. The Least Cost Method is a straightforward heuristic that helps in finding an initial solution to this problem.

Key Components of the Transportation Problem:

1. Suppliers: These are the sources or origins of the goods.
2. Consumers: These are the destinations where the goods need to be delivered.
3. Cost Matrix: The transportation cost per unit between each supplier and consumer is given in a matrix form.
4. Supply and Demand: Each supplier has a certain amount of goods to supply, and each consumer has a specific demand.

Steps to Apply the Least Cost Method:

1. Create the Cost Matrix:

   • List the suppliers and consumers in a matrix where the cost of transporting goods from each supplier to each consumer is mentioned.
   • The supply for each supplier and the demand for each consumer is given alongside the matrix.

2. Identify the Cell with the Least Cost:

   • From the cost matrix, identify the cell (supplier-consumer pair) that has the least transportation cost.

3. Allocate the Maximum Possible Quantity:

   • Allocate as much as possible to the identified cell, considering the supply and demand constraints.
   • The amount allocated will be the minimum of the supply available for that supplier or the demand for that consumer.
   • Subtract the allocated amount from the corresponding supply and demand.

4. Cross Out the Row or Column:

   • Once a supplier’s entire supply or a consumer’s entire demand is fulfilled, cross out the respective row (if supply is exhausted) or column (if demand is fulfilled).

5. Repeat the Process:

   • Repeat steps 2-4 until all supplies and demands are satisfied.
   • Continue identifying the next cell with the least cost, allocating the maximum possible amount, and updating the supply and demand until all are met.

6. Obtain the Initial Feasible Solution:

   • Once all the supply and demand constraints are satisfied, the resulting allocations form the initial feasible solution to the transportation problem.

Example:

Consider a simple transportation problem with the following cost matrix, where the rows represent suppliers and the columns represent consumers:

Step 1: Identify the Cell with the Least Cost
The least cost in the cost matrix is 2, located at the intersection of Supplier 3 and Consumer 2.

Step 2: Allocate the Maximum Possible Quantity
The supply at Supplier 3 is 40, and the demand at Consumer 2 is 60. Allocate 40 units to (Supplier 3, Consumer 2). After the allocation:

• The supply for Supplier 3 becomes 0 (40 – 40 = 0).
• The demand for Consumer 2 becomes 20 (60 – 40 = 20).

Step 3: Cross Out the Row or Column
Since the supply at Supplier 3 is exhausted, cross out Supplier 3‘s row.

Step 4: Identify the Next Least Cost
Now, the least cost is 4, located at Supplier 1 and Consumer 1.

Step 5: Allocate the Maximum Possible Quantity
The supply for Supplier 1 is 30, and the demand for Consumer 1 is 40. Allocate 30 units to (Supplier 1, Consumer 1). After the allocation:

• The supply for Supplier 1 becomes 0 (30 – 30 = 0).
• The demand for Consumer 1 becomes 10 (40 – 30 = 10).

Step 6: Cross Out the Row or Column
Since the supply for Supplier 1 is exhausted, cross out Supplier 1‘s row.

Step 7: Identify the Next Least Cost
The next least cost is 5, located at Supplier 2 and Consumer 2.

Step 8: Allocate the Maximum Possible Quantity
The supply for Supplier 2 is 50, and the demand for Consumer 2 is 20. Allocate 20 units to (Supplier 2, Consumer 2). After the allocation:

• The supply for Supplier 2 becomes 30 (50 – 20 = 30).
• The demand for Consumer 2 becomes 0 (20 – 20 = 0).

Step 9: Cross Out the Column
Since the demand for Consumer 2 is fulfilled, cross out Consumer 2‘s column.

Step 10: Final Allocations
The remaining supplies and demands are as follows:

• Supplier 2 has 30 units of supply left, and Consumer 1 has a demand of 10.
• Allocate the remaining 10 units from Supplier 2 to Consumer 1.

At this point, all supplies and demands are satisfied.

Summary of Allocations:

Advantages of the Least Cost Method:

1. Simplicity: It is easy to implement and understand.
2. Quick Solution: It provides a fast way to get an initial feasible solution.
3. Effective: While it may not give the optimal solution, it generally provides a good approximation.

Disadvantages:

1. Not Guaranteed to be Optimal: The Least Cost Method only provides an initial feasible solution. It doesn’t guarantee the optimal solution.
2. Heuristic Nature: It is a heuristic method, meaning it doesn’t always lead to the best possible result.

Conclusion:

The Least Cost Method is a simple and effective approach to finding an initial feasible solution to the transportation problem. It focuses on allocating goods in the most cost-effective way by choosing the least expensive transportation routes first. While the solution may not be optimal, it provides a good starting point that can be further improved using optimization methods like the MODI Method or Stepping Stone Method.