In mathematics and economics, transportation theory or transport theory is a name given to the study of optimal transportation and allocation of resources.A transportation matrix is a way of understanding the maximum possibilities the shipment can be done. It is also known as decision variables because these are the variables of interest that we will change to achieve the objective, that is, minimizing the cost function.

1. Transportation Problems
Dr. Deepa Chauhan
Transportation Problems
Initial Basic Feasible Solutions
By
Dr. Deepa Chauhan
Associate Professor,
Applied Science & Humanities Department,
Axis Institute of Technology, Kanpur

2. Transportation Problems
Dr. Deepa Chauhan
Methods of Finding an Initial Basic Feasible Solution
There are mainly three methods to find initial basic feasible solution to a transportation problem
a) North West Corner Method (NWCM)
b) Least Cost Method (LCM) or Method of Matrix Minima
c) Vogel’s Approximation Method (VAM)
Solution by transportation method involves making a transportation model in the form of a matrix,
finding an initial basic feasible solution, performing optimality test and moving towards an optimal
solution. Initially, Initial Basic Feasible Solution (IBFS) is obtained then the result is used
to calculate the optimal solution. The popular methods to find IBFS of transportation problem are
North West Corner Method (NWCM), Least Cost Method (LCM), and Vogel's
Approximation Method (VAM).
The Vogel approximation (Unit penalty) method is an iterative procedure for computing a basic
feasible solution of a transportation problem. This method is preferred over the two methods
discussed in the previous sections, because the initial basic feasible solution obtained by this
method is either optimal or very close to the optimal solution.

3. Transportation Problems
Dr. Deepa Chauhan
North West Corner Method (NWCM)
Example 1: Find the initial feasible solution by North West Corner Method
Warehouse→
Factory↓
W1 W2 W3 W4 Factory
Capacity
F1
F2
F3
19
70
40
30
30
8
50
40
70
10
60
20
7
9
18
Warehouse
Requirement
5 8 7 14 Total=34
Solution: The above problem is a balanced problem because factory capacity is equal to the
warehouse requirement. We prepare 3X4 table with 3 factories and 4 warehouses.
Start giving allocations from North West Corner, shown as below:
North-West Corner
Available↓
5 (19) 2 (30) 7
6 (30) 3 (40) 9
4 (70) 14 (20) 18
Requirement→ 5 8 7 14
Given solution is initial basic feasible solution as m+n-1=No of allocation=6
By multiplying each individual allocation by its corresponding unit cost and adding, we get the
Total Transportation Cost=5X19+2X30+6X30+3X40+4X70+14X20
=₹ 1015

4. Transportation Problems
Dr. Deepa Chauhan
Example 2: Find the initial feasible solution by North West Corner Method
W1 W2 W3 W4 Supply
F1
F2
F3
F4
48
45
50
52
60
55
65
64
56
53
60
55
58
60
62
61
140
260
360
220
Demand 200 320 250 210
Solution: Since the total demand is equal to the total supply (980), it is a balanced transportation
problem. Now complete allocation is shown as below:
W1 W2 W3 W4 Supply
F1 48 (140) 60 56 58 140
F2 45 (60) 55 (200) 53 60 260
F3 50 65 (120) 60 (240) 62 360
F4 52 64 55 (10) 61 (210) 220
Demand 200 320 250 210
Given solution is initial basic feasible solution as m+n-1=No of allocation=7
Total transportation cost=140X48+60X45+200X55+120X65+240X60+10X55+210X61
=₹ 55,980

5. Transportation Problems
Dr. Deepa Chauhan
Least Cost Method (LCM)
Example 1: Find the initial feasible solution by Least Cost Method
Warehouse→
Factory↓
W1 W2 W3 W4 Supply
F1
F2
F3
19
70
40
30
30
8
50
40
70
10
60
20
7
9
18
Demand 5 8 7 14 Total=34
Solution: The above problem is a balanced problem because factory capacity is equal to the
warehouse requirement. Allocations are started with least cost cell, as follows
Least Cost
Warehouse→
Factory↓
W1 W2 W3 W4 Supply
F1 19 30 50 10 (7) 7
F2 70 (2) 30 40 (7) 60 9
F3 40 (3) 8 (8) 70 20 (7) 18
Demand 5 8 7 14 Total=34
Given solution is initial basic feasible solution as m+n-1 (3+4-1=6) =No of allocation=6
Total Transportation Cost= 2x70+40x3+8x8+40x7+10x7+20x7
=₹ 814

6. Transportation Problems
Dr. Deepa Chauhan
Example 2: Using Least-cost method to solve initial solution of the following problem:
Destination
Source D1 D2 D3 Capacity
S1 10 13 6 10
S2 16 7 13 12
S3 8 22 2 8
Demand 6 11 13 30
Solution: The above problem is a balanced problem because demand is equal to supply. By least
cost method, allocations are as follows
Destination
Source D1 D2 D3 Capacity
S1 10 (5) 13 6 (5) 10
S2 16 (1) 7 (11) 13 12
S3 8 22 2 (8) 8
Demand 6 11 13 30
Given solution is initial basic feasible solution as m+n-1=No of allocation=5
Total Transportation Cost by LCM= 10x5+6x5+16x1+7x11+2x8= ₹ 189
Example 3: Find the initial feasible solution by Least Cost Method
Origin
Destination Supply
2 7 4 5
3 3 1 8
5 4 7 7
1 6 2 14
Demand 7 9 18

7. Transportation Problems
Dr. Deepa Chauhan
Solution:
Origin
Destination Supply
2 7 4 5
3 3 1 8
5 4 7 7
1 6 2 14
Demand 7 9 18 Total=34
The above problem is a balanced problem because demand is equal to supply. By least cost method,
allocations are as follows
Origin
Destination Supply
2 7 (2) 4 (3) 5
3 3 1 (8) 8
5 4 (7) 7 7
1 (7) 6 2 (7) 14
Demand 7 9 18 Total=34
Given solution is initial basic feasible solution as m+n-1=No of allocation=6
Total Transportation Cost by LCM= 7x2+4x3+1x8+4x7+1x7+2x7= ₹ 83

8. Transportation Problems
Dr. Deepa Chauhan
Vogel’s Approximation Method (VAM) or Unit Cost Penalty Method
Vogel’s approximation method is an analytic method and is preferred to the method described
earlier. Vogel’s approximation method makes effective use of the cost information and yields a
better initial solution than obtained by the other methods.
Example 1: Find the initial feasible solution by Vogel’s method
Warehouse→
Factory↓
W1 W2 W3 W4 Supply
F1
F2
F3
19
70
40
30
30
8
50
40
70
10
60
20
7
9
18
Demand 5 8 7 14 Total=34
Solution: The above problem is a balanced problem because supply is equal to the demand.
Step I: We first construct the cost and requirement table:
Table-1
Warehouse→
Factory↓
W1 W2 W3 W4 Supply
F1 19 30 50 10 7
F2 70 30 40 60 9
F3 40 8 70 20 18
Demand 5 8 7 14 Total=34

9. Transportation Problems
Dr. Deepa Chauhan
Step II: We add in above table one row and one column. The additional column is called row penalty
and additional row is called column penalty. In these added rows and columns, we find the
difference between the lowest and second lowest cost entries.
Table-2
Warehouse→
Factory↓
W1 W2 W3 W4 Supply Row
Penalties
F1 19 30 50 10 7
F2 70 30 40 60 9
F3 40 8 (8) 70 20 18
Demand 5 8 7 14 Total=34
Column
Penalties
Step III:
 We select the row or column from table for which penalty is maximum and accordingly allocate
maximum possible amount to cell with lowest cost.
 If there are more than one largest penalty rows or columns, select the row or column having
minimum cost.
 In case of tie in the minimum cost, select the cell which can have maximum allocation. If there is
tie among maximum allocation cell also, select the cell arbitrarily for allocation.
Table-3
Warehouse→
Factory↓
W1 W2 W3 W4 Supply Row
Penalties
F1 19 30 50 10 7 19-10=9
F2 70 30 40 60 9 40-30=10
F3 40 8 (8) 70 20 18 20-8=12
Demand 5 8 7 14 Total=34
Column
Penalties
40-19=21 30-8=22 50-40=10 20-10=10

10. Transportation Problems
Dr. Deepa Chauhan
Step IV: Now we delete that column or row on which the requirement is fulfilled and then we
construct reduced table
Table-4
Warehouse→
Factory↓
W1 W3 W4 Supply Row
Penalties
F1 19 (5) 50 10 7 19-10=9
F2 70 40 60 9 60-40=20
F3 40 70 20 10 40-20=20
Demand 5 7 14 Total=34
Column
Penalties
40-19=21 50-40=10 20-10=10
Step V: We repeat the step 3 and 4 till all the allocations are completed. The successive reduced
tables are shown below:
Table-5
Warehouse→
Factory↓
W3 W4 Supply Row
Penalties
F1 50 10 2 40
F2 40 60 9 20
F3 70 20 (10) 10 30
Demand 7 14 Total=34
Column
Penalties
50-40=10 20-10=10

11. Transportation Problems
Dr. Deepa Chauhan
Table-6
Warehouse→
Factory↓
W3 W4 Supply Row
Penalties
F1 50 10 (2) 2 40
F2 40 60 9 20
Demand 7 4 Total=34
Column
Penalties
50-40=10 60-10=50
Table-7
Warehouse→
Factory↓
W3 W4 Supply Row
Penalties
F2 40 (7) 60 (2) 9 20
Demand 7 2 Total=34
Column
Penalties
Step V: Finally we construct a table for giving required feasible solution
Given solution is initial basic feasible solution as m+n-1=No of allocation=6
Transportation cost by Vogel’s method
Z=5*19+10*2+40*7+60*2+8*8+20*10 = ₹ 779
Warehouse→
Factory↓
W1 W2 W3 W4 Supply
F1 19 (5) 30 50 10 (2) 7
F2 70 30 40 (7) 60 (2) 9
F3 40 8 (8) 70 20 (10) 18
Demand 5 8 7 14 Total=34

12. Transportation Problems
Dr. Deepa Chauhan
Example 2: Find the initial feasible solution by Vogel’s method
W X Y Z Availability
A 7 2 5 5 30
B 4 4 6 5 15
C 5 3 3 2 10
D 4 -1 4 2 20
Demand 20 25 15 15
Solution:
Table-1
W X Y Z Availability
A 7 2 5 5 30
B 4 4 6 5 15
C 5 3 3 2 10
D 4 -1 4 2 20
Demand 20 25 15 15 Total=75
The above problem is a balanced problem because availability is equal to the demand.
Table-2
W X Y Z Availability Row
Penalty
A 7 2 5 5 30 3
B 4 4 6 5 15 1
C 5 3 3 2 10 1
D 4 -1 (20) 4 2 20 3
Demand 20 25 15 15 Total=75
Column
Penalty
1 3 1 3

13. Transportation Problems
Dr. Deepa Chauhan
Table-3
W X Y Z Availability Row
Penalty
A 7 2 5 5 30 3
B 4 4 6 5 15 1
C 5 3 3 2 (10) 10 1
Demand 20 5 15 15 Total=75
Column
Penalty
1 1 2 3
Table-4
W X Y Z Availability Row
Penalty
A 7 2 (5) 5 5 30 3
B 4 4 6 5 15 1
Demand 20 5 15 5 Total=75
Column
Penalty
3 2 1 0
Table-5
W Y Z Availability Row
Penalty
A 7 5 5 25 2
B 4 (15) 6 5 15 1
Demand 20 15 5 Total=75
Column
Penalty
3 1 0

14. Transportation Problems
Dr. Deepa Chauhan
Table-6
Final table
W X Y Z Availability
A 7 (5) 2 (5) 5 (15) 5 (5) 30
B 4 (15) 4 6 5 15
C 5 3 3 2 (10) 10
D 4 -1 (20) 4 2 20
Demand 20 25 15 15 Total=75
Given solution is initial basic feasible solution as m+n-1=No of allocation=7
Total transportation cost=35+10+75+25+60+20-20= ₹ 205
W Y Z Availability
A 7 (5) 5 (15) 5 (5) 25
Demand 5 15 5 Total=75