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.
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
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.