Page 115 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes
Example: Solve example 5 by least cost method.
Solution:

To make our understanding better, let us go through the Table 5.15.
Table 5.15

Retail shops
Factories 1 2 3 4 Supply
1 3 5 7 6 50
2 2 5 8 2 75
3 3 6 9 2 25
Demand 20 20 50 60

Applying the least cost method-

Table 5.16

Retail shops
Factories 1 2 3 4 Supply
1 3 5 7 6 50
30
20
2 2 5 8 2 75
55
20
20
3 3 6 9 2 5 25
Demand 20 20 50 60

We observe that c = 2, which is the minimum transportation cost. So, x = 20.
21 21
Proceeding in this way, we observe that x = 55, x = 5, x = 20, x = 30, x = 20.
24 34 12 13 33
Number of basic variables = m + n –1 = 3 + 4 – 1 = 6.
The initial basic feasible solution:
= 20 × 2 + 55 × 2 + 5 × 2 + 20 × 5 + 30 × 7 + 20 × 9
= 650.

5.7.3 Algorithm for Vogel’s Approximation Method (VAM)

1. Calculate penalties for each row and column by taking the difference between the smallest
cost and next highest cost available in that row/column. If there are two smallest costs,
then the penalty is zero.
2. Select the row/column, which has the largest penalty and make allocation in the cell
having the least cost in the selected row/column. If two or more equal penalties exist,
select one where a row/column contains minimum unit cost. If there is again a tie, select
one where maximum allocation can be made.

3. Delete the row/column, which has satisfied the supply and demand.
4. Repeat steps (1) and (2) until the entire supply and demands are satisfied.
5. Obtain the initial basic feasible solution.

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