PPT on Vogel's Approximation Method

1. Vogel’s Approximation Method (VAM)
Step-I: Compute the penalty values for each row
and each column. The penalty will be equal
to the difference between the two smallest
shipping costs in the row or column.
Step-II: Identify the row or column with the largest
penalty. Find the first basic variable which
has the smallest shipping cost in that row or
column. Then assign the highest possible
value to that variable, and cross-out the row
or column which is exhausted.
Step-III: Compute new penalties and repeat the
same procedure until all the rim
requirements are satisfied.

2. An example for Vogel’s Method
Find the IBFS of the following transportation problem by using Penalty
Method.
Supply
6 7 8
15 80 78
D1 D2 D3
15 5 5
10
15
O1
O2
Demand

3. Step 1: Compute the penalties in each row and
each column .
Supply Row Penalty
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15

4. Step 2: Identify the largest penalty and choose
least cost cell to corresponding this penalty
Supply Row Penalty
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15

5. Step-3: Allocate the amount 5 which is minimum of
corresponding row supply and column demand and
then cross out column2
Supply Row Penalty
5
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15

6. Step-4: Recalculate the penalties
Supply Row Penalty
5
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 78-8=70
8-6=2
78-15=63
15 X 5
5
15

7. Step-5: Identify the largest penalty and choose
least cost cell to corresponding this penalty
Supply Row Penalty
5
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 78-8=70
8-6=2
78-15=63
15 X 5
5
15

8. Step-6: Allocate the amount 5 which is minimum of
corresponding row supply and column demand, then
cross out column3
Supply Row Penalty
5 5
6 7 8
15 80 78
Demand
Column Penalty 15-6=9
8-6=2
78-15=63
15 X X
5
15

9. Step-7: Finally allocate the values 0 and 15 to
corresponding cells and cross out column 1
Supply
0 5 5
6 7 8
15
15 80 78
Demand X X X
X
X
D3
O1
O2
D1 D2

10. Solution of the problem
Now the Initial Basic Feasible Solution of the
transportation problem is
X11=0, X12=5, X13=5, and X21=15 and
Total transportation cost = (0x6)+(5x7)+(5x8)+(15x15)
= 0+35+40+225
= 300.