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. Step 1: Compute the penalties in each row and each column . Step 2: Identify the largest penalty and choose least cost cell to corresponding this penalty. Step-3: Allocate the amount 5 which is minimum of corresponding row supply and column demand and then cross out column2. Step-4: Recalculate the penalties. Step-5: Identify the largest penalty and choose least cost cell to corresponding this penalty. Step-6: Allocate the amount 5 which is minimum of corresponding row supply and column demand, then cross out column3. Step-7: Finally allocate the values 0 and 15 to corresponding cells and cross out column 1\ 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.

1. Submitted to
Ms. Tarika Nandedkar ma’am
Submitted By
1. Jai Kumar Pandit
2. Rahul prajapati
3. Shivam yadav
4. Aditya Pal
CLASS- MBA 2nd SEM (Sec-A)
Session- 2018
ASSIGNMENT ON
Vogel’s Approximation Method

2. 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.

3. Step-III: Compute new penalties and
repeat the same procedure until all the rim
requirements are satisfied.

4. D1 D2 D3 SUPPLY
A 6 7 8 10
B 15 80 78 15
Demand 15 5 5
An example for Vogel’s Method

5. Step 1: Compute the penalties in each
row and each column .
D1 D2 D3 Supply Row
penalt
y
A 6 7 8 10 7-6=1
B 15 80 78 15 78-
15=63
Deman
d
15 5 5
Colum
penalty
15-6=9 80-
7=73
78-
8=70

6. Step 2: Identify the largest
penalty and choose least cost cell
to corresponding this penalty.
D1 D2 D3 Supply Row
penalty
A 6 7 8 10 7-6=1
B 15 80 78 15 78-
15=63
Deman
d
15 5 5
Colum
penalty
15-6=9 80-
7=73
78-
8=70

7. Step-3: Allocate the amount 5 which is minimum of
corresponding row supply and column demand and
then cross out column2.
D1 D2 D3 Supply Row penalty
A 6 7 8 10 7-6=1
B 15 80 78 15 78-
15=63
Deman
d
15 5 5
Colum
penalty
15-6=9 80-
7=73
78-
8=70
5

8. Step-4: Recalculate the penalties.
D1 D2 D3 Supply Row penalty
A 6 7 8 10 8-6=2
B 15 80 78 15 78-
15=63
Dema
nd
15 5
Colum
penalt
y
15-6=9 78-
8=70
5

9. Step-5: Identify the largest penalty and choose least
cost cell to corresponding this penalty
D1 D2 D3 Supply Row penalty
A 6 7 8 10 8-6=2
B 15 80 78 15 78-
15=63
Deman
d
15 5
Colum
penalty
15-6=9 78-8=70
5

10. Step-6: Allocate the amount 5 which is
minimum of corresponding row supply and
column demand, then cross out column3
D1 D2 D3 Supply Row penalty
A 6 7 8 10 8-6=2
B 15 80 78 15 78-
15=63
Dema
nd
15
Colum 15-
5 5

11. Step-7: Finally allocate the values
0 and 15 to corresponding cells
and cross out column 1
D1 D2 D3 Supply
A 6 7 8
B 15 80 78
Deman
d
5 5
15

12. 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.

13. Thank you