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Chapter: 12th Business Maths and Statistics : Chapter 10 : Operations Research

Methods of finding initial Basic Feasible Solutions: Least Cost Method (LCM)

Operations Research: Transportation Problem: Methods of finding initial Basic Feasible Solutions: Least Cost Method (LCM)

Methods of finding initial Basic Feasible Solutions

There are several methods available to obtain an initial basic feasible solution of a transportation problem. We discuss here only the following three. For finding the initial basic feasible solution total supply must be equal to total demand.



Method: Least Cost Method (LCM)

The least cost method is more economical than north-west corner rule,since it starts with a lower beginning cost. Various steps involved in this method are summarized as under.

Step 1: Find the cell with the least(minimum) cost in the transportation table.

Step 2: Allocate the maximum feasible quantity to this cell.

Step:3: Eliminate the row or column where an allocation is made.

Step:4: Repeat the above steps for the reduced transportation table until all the allocations are made.

 

Example 10.3

Obtain an initial basic feasible solution to the following transportation problem using least cost method.


Here Oi and Dj denote ith origin and jth destination respectively.

Solution:

Total Supply = Total Demand = 24

 The given problem is a balanced transportation problem.

Hence there exists a feasible solution to the given problem.

Given Transportation Problem is:


The least cost is 1 corresponds to the cells (O1, D1) and (O3, D4)

Take the Cell (O1, D1) arbitrarily.

Allocatemin (6,4) = 4 units to this cell.


The reduced table is


The least cost corresponds to the cell (O3, D4). Allocate min (10,6) = 6 units to this cell.


The reduced table is


The least costis 2 corresponds to the cells (O1, D2), (O2, D3), (O3, D2), (O3, D3)

Allocate min (2,6) = 2 units to this cell.


The reduced table is


The least cost is 2 corresponds to the cells (O2, D3), (O3, D2), (O3, D3)

Allocate min ( 8,8) = 8 units to this cell.


The reduced table is


Here allocate 4 units in the cell (O3, D2)


Thus we have the following allocations:


Transportation schedule :

O1→ D1, O1→D2, O2→D3, O3→D2, O3→D4

Total transportation cost

=  (4×1)+ (2×2)+(8×2)+(4×2)+(6×1)

=  4+4+16+8+6

=Rs. 38.


 

Example 10.4

Determine how much quantity should be stepped from factory to various destinations for the following transportation problem using the least cost method


Cost are expressed in terms of rupees per unit shipped.

Solution:

Total Capacity = Total Demand

 The given problem is balanced transportation problem.

Hence there exists a feasible solution to the given problem.

Given Transportation Problem is


First Allocation:


Second Allocation:


Third Allocation:


Fourth Allocation:


 Fifth Allocation:


Sixth Allocation:


Transportation schedule :

T→ H, T→P, B→C, B→H, M→H, M→K

The total Transportation cost = ( 5×8) + (25×5)+ (35×5) + (5×11)+ (18×9) + (32×7)

=  40+125+175+55+162+224

=  ₹ 781


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