Least Cost Method for Transportation Problems¶

The Least Cost Method (LCM) is a technique used to find an initial feasible solution for transportation problems in linear programming. The goal of the LCM is to minimize the initial transportation cost by selecting the cell with the lowest transportation cost and making allocations there first. This method often results in a better starting solution compared to the North-West Corner Rule.

Steps to Apply the Least Cost Method¶

Identify the cell with the lowest transportation cost in the entire cost matrix.
Allocate as much as possible to that cell without exceeding the supply or demand.
Adjust the supply and demand based on the allocation:
Subtract the allocated amount from both the corresponding row's supply and the column's demand.
Cross out the row or column if the supply or demand is exhausted.
If both are exhausted simultaneously, cross out one and reduce the remaining demand/supply of the other to zero.
Repeat steps 1-4 until all supplies and demands are met.

Example Problem¶

Consider three factories (A, B, C) supplying goods to three warehouses (X, Y, Z). The transportation costs per unit, supplies, and demands are given below:

	X	Y	Z	Supply
A	2	3	1	30
B	5	4	8	40
C	5	6	8	20
Demand	20	50	20

Step 1: Find the cell with the lowest cost.¶

The lowest cost is 1 (A-Z).

Step 2: Allocate as much as possible to A-Z.¶

The supply at A is 30, and the demand at Z is 20.
Allocate 20 units to A-Z (minimum of 30 and 20).
Update the remaining supply at A to 30 - 20 = 10, and the remaining demand at Z to 20 - 20 = 0.
Cross out column Z because the demand is met.

Step 3: Find the next lowest cost in the remaining matrix.¶

The next lowest cost is 2 (A-X).

Step 4: Allocate to A-X.¶

The supply at A is now 10, and the demand at X is 20.
Allocate 10 units to A-X (minimum of 10 and 20).
Update the remaining supply at A to 10 - 10 = 0, and the remaining demand at X to 20 - 10 = 10.
Cross out row A because the supply is exhausted.

Step 5: Find the next lowest cost in the remaining matrix.¶

The remaining costs are:

	X	Y
B	5	4
C	5	6

The next lowest cost is 4 (B-Y).

Step 6: Allocate to B-Y.¶

The supply at B is 40, and the demand at Y is 50.
Allocate 40 units to B-Y (minimum of 40 and 50).
Update the remaining supply at B to 40 - 40 = 0, and the remaining demand at Y to 50 - 40 = 10.
Cross out row B because the supply is exhausted.

Step 7: Allocate the remaining units.¶

The only remaining cell is C-Y with a demand of 10 and a supply of 20.
Allocate 10 units to C-Y.
Update the remaining supply at C to 20 - 10 = 10, and the remaining demand at Y to 10 - 10 = 0.

Final Allocation¶

The allocations are as follows:

	X	Y	Z
A	10	0	20
B	0	40	0
C	0	10	0

Summary¶

The Least Cost Method aims to minimize the initial transportation cost by prioritizing cells with the lowest cost.
This method provides a better starting solution compared to the North-West Corner Rule, but further optimization may be required using techniques like the stepping-stone method or the MODI (Modified Distribution) method to achieve the optimal solution.

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