# Transportation problem with Vogel’s approximation method.
Factory Warehouse Availability
A B C D
F1 3 2 5 2 15
F2 2 1 4 4 24
F3 2 3 4 3 21
Requirement 13 12 16 19 60
Vogel’s approximation method is more systematic and orderly than other two methods.
Following steps are given below
1. Identify the smallest unit cost and second smallest unit cost of each row or column for determining the penalties. To calculate penalty of each row or column we use the formula Penalty = (Second lowest – lowest) costs of that row/column
2. Identify the row or column with maximum of the penalties. Also find the cell with lowest cost in the selected row or column. Allocate the maximum possible units to this cell by comparing the demand and supply associated with that cell.
3. Delete the row or column whose demand or supply in exhausted
4. Continue the process till demand and supply is exhausted.
Transportation model
Factory Warehouse Availability
Penalty
A B C D
F1 3 2 5 2 15
3-2=1
F2 2 1
12 4 4 24-12=12
2-1=1
F3 2 3 4 3 21
3-2=1
Requirement 13 12-12=0 16 19 48
Penalty 3-2=1 2-1=1 5-4=1 3-2=1
1. The penalty of all column and rows are same. So we select x22 with lowest unit cost of tk 1. Here the maximum possible allocation of 12 units to x22 and remaining supply 24 – 12 = 12 units. Hence the column for warehouse B is deleted.
The shrunken matrix is
Factory Warehouse Availability
Penalty
A B C D
F1 3 2 5 2 15
3-2=1
F2 2
12 1
12 4 4 12-12=0
4-2=2
F3 2 3 4 3 21
3-2=1
Requirement 13-12=1 0 16 19 36
Penalty 3-2=1 2-1=1 5-4=1 3-2=1
2. The maximum penalty occurs of row 2 (F2) is 2. So we select with lowest unit cost of tk 2. Here the maximum possible allocation of 12 units to x21 and remaining demand 13 – 12 = 1 units. Hence the row for F2 is deleted.
The shrunken matrix is
Factory Warehouse Availability
Penalty
A B C D
F1 3 2 5 2
15 15-15=0
3-2=1
F2 2
12 1
12 4 4 0
4-2=2
F3 2 3 4 3 21
3-2=1
Requirement 1 0 16 19-15=4 21
Penalty 3-2=1 2-1=1 5-4=1 3-2=1
3. The penalty of all column and rows are same. So we select x14 and x31 with lowest unit cost of tk 2. Here the maximum possible allocation of 15 units to x14 and remaining demand 19 – 15 = 4 units. Hence the row for F1 is deleted.
Thus the final allocation matrix is
Factory Warehouse Availability
A B C D
F1 3 2 5 2
15 15
F2 2
12 1
12 4 4 12
F3 2
1 3 4
16 3
4 21
Requirement 13 12 16 19 60
So, it can be observed that the number of occupied cells in the solution is (m + n - 1) = (3 + 4 – 1) = 6. Hence the initial solution is feasible.
The total transportation cost is
z = x14 × c14 + x21 × c21 + x22 × c22 + x31 × c31 + x33 × c33 + x34 × c34
=15 × 2 + 12 × 2 + 12 ×1 + 1 × 2 + 16 × 4 + 4 × 3
= 144