Vam

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 .
SupplyRow 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
SupplyRow 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
SupplyRow 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
SupplyRow 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
SupplyRow 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
SupplyRow 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
X
11
=0, X
12
=5, X
13
=5, and X
21
=15 and
Total transportation cost = (0x6)+(5x7)+(5x8)+(15x15)
= 0+35+40+225
= 300.