Vogel's Approximation Method-Operation Research-Handouts, Lecture notes of Operational Research

Operations Research (OR) refers to the science of decision making. This course elaborate like linear, nonlinear and discrete optimization. This lecture handout was provided by Sir Avikshit Gupte. It includes: Vogal, Matrix, Minimum, Provides, Associated, Column. Transportation, Penalty, Chooding, Exhausted

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Vogel's Approximation Method (VAM)
This method is based on the 'difference' associated with each row and column in the matrix giving unit cost
of transportation cij. This 'difference' is defined as the arithmetic difference between the smallest and next to the
smallest element in that row or column. This difference in a row or column indicates the minimum unit penalty
incurred in failing to make an allocation to the smallest cost cell in that row or column. This difference also provides
a measure of proper priorities for making allocations to the respective rows and column. In other words, if we take a
row, we have to allocate to the cell having the least cost and if we fail to do so, extra cost will be incurred for a
wrong choice, which is called penalty. The minimum penalty is given by this difference. So, the procedure
repeatedly makes the maximum feasible allocation in the smallest cost cell of the remaining row or column, with the
largest penalty. Once an allocation is fully made in a row or column, the particular row or column is eliminated.
Hence and allocation already made cannot be changed. Then we have a reduced matrix. Repeat the same procedure
of finding penalty of all rows and columns in the reduced matrix, choosing the highest penalty in a row or column
and allotting as much as possible in the least cost cell in that row or column. Thus we eliminate another fully
allocated row or column, resulting in further reducing the size of the matrix. We repeat till all supply and demand
are exhausted.
A summary of the steps involved in Vogel's Approximation Method is given below:
STEP 1: Represent the transportation problem in the standard tabular form.
STEP 2: Select the smallest element in each row and the next to the smallest element in that row. Find the
difference. This is the penalty written on the right hand side of each row. Repeat the same for each column. The
penalty is written below each column.
STEP 3: Select the row or column with largest penalty. If there is a tie, the same can be broken arbitrarily.
STEP 4: Allocate the maximum feasible amount to the smallest cost cell in that row or column.
STEP 5: Allocate zero else where in the row or column where the supply or demand is exhausted.
STEP 6: Remove all fully allocated rows or columns from further consideration. Then proceed with the remaining
reduced matrix till no rows or columns remain.
Let us apply Vogel's Approximation Method to the above example as given below in table12
Table 12
Origin
Destination
Supply Row
difference
P
Q
R
A
5
7
8
70
(2)
B
4
30 4
6
30
0
(0)
C
6
7
7
50
(1)
65
42
43
12
Column difference (1) (3) (1)
Largest Penalty
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Vogel's Approximation Method (VAM)

This method is based on the 'difference' associated with each row and column in the matrix giving unit cost of transportation cij. This 'difference' is defined as the arithmetic difference between the smallest and next to the smallest element in that row or column. This difference in a row or column indicates the minimum unit penalty incurred in failing to make an allocation to the smallest cost cell in that row or column. This difference also provides a measure of proper priorities for making allocations to the respective rows and column. In other words, if we take a row, we have to allocate to the cell having the least cost and if we fail to do so, extra cost will be incurred for a wrong choice, which is called penalty. The minimum penalty is given by this difference. So, the procedure repeatedly makes the maximum feasible allocation in the smallest cost cell of the remaining row or column, with the largest penalty. Once an allocation is fully made in a row or column, the particular row or column is eliminated. Hence and allocation already made cannot be changed. Then we have a reduced matrix. Repeat the same procedure of finding penalty of all rows and columns in the reduced matrix, choosing the highest penalty in a row or column and allotting as much as possible in the least cost cell in that row or column. Thus we eliminate another fully allocated row or column, resulting in further reducing the size of the matrix. We repeat till all supply and demand are exhausted.

A summary of the steps involved in Vogel's Approximation Method is given below:

STEP 1: Represent the transportation problem in the standard tabular form.

STEP 2: Select the smallest element in each row and the next to the smallest element in that row. Find the difference. This is the penalty written on the right hand side of each row. Repeat the same for each column. The penalty is written below each column.

STEP 3 : Select the row or column with largest penalty. If there is a tie, the same can be broken arbitrarily.

STEP 4 : Allocate the maximum feasible amount to the smallest cost cell in that row or column.

STEP 5 : Allocate zero else where in the row or column where the supply or demand is exhausted.

STEP 6 : Remove all fully allocated rows or columns from further consideration. Then proceed with the remaining reduced matrix till no rows or columns remain.

Let us apply Vogel's Approximation Method to the above example as given below in table

Table 12 Origin Destination Supply Row P Q R difference A 5 7 8 70 (2) B 4 30 4

C 6 7 7 50 (1)

Demand 65 42 43 12 Column difference (1) (3) (1) Largest Penalty

The difference between the smallest and next to the smallest element in each row and in each column is calculated. This is indicated in the parenthesis. We choose the maximum from among the differences. The first individual allocation will be to the smallest cost of a row or column with the largest difference. So we select the column Q (penalty = 3) for the first individual allocation, and allocate to ( B, Q ) as much as we can, since this cell has the least cost location. Thus 30 units from B are allocated to Q. This exhausts the supply from B. However, there is still a demand of 12 units from Q. The allocations to other cells in that column are 0, as indicated. The next step is to write down the reduced matrix (as in table 13) eliminating row B (as it is exhausted).

Table 13

Origin Destination Supply Row P Q R^ difference A 65 5

C 6 7 7 50 (1)

Demand 65 12 43 0 Column difference (1) (0) (1)

From the table 13, (2) is the largest unit difference corresponding to the row A. This leads to an allocation in the corresponding minimum cost location in row A , namely cell ( A, P ). The maximum possible allocation is only 65 as required by P from A and allocation of 0 to others in the row A. Column P is thus deleted and the reduced matrix is given in table 14.

Maximum difference is 1 in row A and in column C. Select arbitrarily A and allot the least cost cell ( A, Q ) 5 units. Delete row A.

Now, we have only one row C and two columns Q and R (Table 15) indicating that all the available amount from C has to be moved to Q and R as per their requirements. Hence we have the table 15

Table 14

Origin Destination Supply Row difference P R A 5 7

C 7 7 50

Demand 12 43 7 Column difference (0) (1)