A manufacturing company has 3 factories and 4 warehouses. It needs to transport products from factories to warehouses to meet demand. The document provides transportation costs between each factory-warehouse pair and asks to construct a transportation table and solve for initial feasible solutions using Northwest Corner Method, Least Cost Method, and Vogel's Approximation Method. It presents a sample problem with costs and asks to apply the three methods to find the lowest cost solution.

2. INTRODUCTION
Transportation plays a vital role in
any economy.
Transportation is a key
element for the success of a
business enterprise which is a
manufacturing organisation.
Products or goods
manufactured by any firm
need to be distributed
efficiently and effectively to
its distributors, dealers, sub-
dealers etc.

3. STRUCTURE OF A TRANSPORT PROBLEM
IN A TRANSPORTATION PROBLEM, WE MAKE FOLLOWING ASSUMPTIONS:
 The number of supply centre are finites and known
 The number of demand centers are finite and known
 Supply from each supply centre is constant for the given problem.
 Demand at each Demand centre is constant for the given problem.
 Cost of transporting one unit from each supply centre to each demand centre
is constant for the given problem.

4. INITIAL FEASIBLE SOLUTION
NWCR
(NORTH – WEST CORNER
METHOD)
LCM
(LEAST COST METHOD)
VAM
(VOGEL’S APPROXIMATION
METHOD)

5. Problem No 01
A company has 3 factories F1, F2 and F3 with production capacities of
11, 13 and 19 unit (in thousand). It has four warehouses W1, W2, W3
and W4 with demands of 6,10,12 and 15 respectively. Unit cost of
transportation is given from each warehouse
W1 W2 W3 W4
F1 42 32 50 26
F2 34 36 28 46
F3 64 54 36 82
Cost in Rs
Based on the above information
1) Construct a Transportation table
2) Find Initial Feasible Solution By
a) Northwest Corner Method
(NWCM)
b) Least Cost Method (LCM)
c) Vogel’s Approximation Method
(VAM)

6. STEP 1: - PROBLEM IS
BALANCED/UNBALANCED
Supply Demand
F1  11 W1 = 6
F2  13 W2 = 10
F3  19 W3 = 12
W4 = 15
Total = 43 Total = 43
A company has 3 factories F1, F2 and F3
with production capacities of 11, 13 and
19 unit (in thousand). It has four
warehouses W1, W2, W3 and W4 with
demands of 6,10,12 and 15 respectively.
Unit cost of transportation is given from
each warehouse
The problem is balanced as supply units are equal to Demand Units i.e.
43 = 43

7. STEP 2:- MINIMIZATION/ MAXIMISATION
The problem is of Minimization as this
problem is related to cost.
STEP 3:- CONSTRUCTION OF TRANSPORTATION
TABLE
W1 W2 W3 W4
F1 42 32 50 26
F2 34 36 28 46
F3 64 54 36 82
Cost in Rs

8. NWCM (North West Corner Method]

9. NWCR (North West Corner Method]
Factory Warehouse Units Cost Price Total Cost
F1 W1 6 42 6 x 42 = 252
F1 W2 5 32 5 x 32 = 160
F2 W2 5 36 5 x 36 = 180
F2 W3 8 28 8 x 28 = 224
F3 W3 4 36 4 x 36 = 144
F3 W4 15 82 15 x 82 = 1230
TOTAL COST BY NWCR = RS. 21,90,000

10.  In LCM , we start giving allocation from the lowest cost cell in the Transportation table
 Then we proceed to the next available lowest cost and so on until all supply and
demand centres are exhausted.
LEAST COST METHOD
Find the lowest
cost from the
matrix
i.e. 26

11. Find next
lowest
value
Is 28

12. LEAST COST METHOD

15. From Warehouse Units Cost Total
F1 W4 11 26 11 x 26 = 286
F2 W1 1 34 1 x 34 = 34
F2 W3 12 28 12 x 28 = 336
F3 W1 5 64 5 X 64 = 320
F3 W2 10 54 10 x 54 = 540
F3 W4 4 82 4 x 82 = 328
1844 x 1000 =
1844000
Total Transportation Cost () = Rs. 18,44,000

16. Vogel’s Approximation Method
(VAM)
 Vogel’s Approximation method is based on the concept of penalty
 Penalty = Difference between two lowest costs for Row or Column
 We calculate penalty for each row and column
 Rule of allocation is to compare all penalties (of all rows and columns)
 Then we start with that row or column which has maximum penalty and in that
row or column we give allocation in minimum cost.
 Rule of Allocation : Maximum Penalty and Minimum cost.

18. P1 with Row penalty
F1 = 32 – 26 = 6
F2 = 34 – 28 = 6
F3 = 54- 36 = 18

19. W1 = 42 – 34 = 8
W2 = 36 – 32 = 4
W3 = 36 – 28 = 8
W4 = 46 – 26 = 20

20. Choose the highest of the penalty
Choose the 20 as it is highest and W4 is the
column and from that column choose the least
cost 26

22. As this row i.e. F1 is exhausted then the Cross that row penalty

23. P2 with Row penalty
F2 = 34 – 28 = 6
F3 = 54- 36 = 18
P2 For Column Penalty
W1 = 64 – 34 = 30
W2 = 54 – 36 = 18
W3 = 36 – 28 = 8
W4 = 82 – 46 = 36*

24. Choose the smallest cost i.e. 46 and W4 requirement is 4 and that
demand is over then close the W4

25. W4 is completed then close the
W4 Column Penalties
Calculation of P3 Row Penalty
F2 = 34 – 28 = 6
F3 = 54 – 36 = 18
Calculation of P3 column penalty
W1 = 64 – 34 = 30*
W2 = 54 – 36 = 18
W3 = 36 – 28 = 8

26. The highest penalty is 30 then choose W1
Choose the smallest cost i.e. 34
Allocate the 6 units of W1

27. W1 is exhausted and column penalty is closed

28. With same process we will solve the P4 penalty of row and column
As the penalty of F3 and W2 penalty = 18
If this case then select that penalty where we allocate
maximum allocation
We will choose F3

30. Now least of the column from W2 i.e. 36 and allocate the units

32. From Warehouse Units Cost Total
F1 W4 11 26 11 x 26 = 286
F2 W1 6 34 6 x 34 = 204
F2 W2 3 36 3 x 36 = 108
F2 W4 4 46 4 x 46 = 184
F3 W2 7 54 7 x 54 = 378
F3 W3 12 36 12 x 36 = 432
Total cost by VAM 1592 *1000 =
1592000

33. IFS Method IFS Cost
1) NWCR Rs. 21,90,000
2) LCM Rs. 18,44,000
3) VAM Rs. 15,92,000
SUMMARY

34. Problem No 02
A company has three plants A, B, C for which capacities are 7, 10 and 18
units. It has four warehouses P,Q,R, S for which demands are 5, 8, 7 and
15 units.
Unit cost of transportation is given from each warehouse
Cost in Rs
Based on the above information
1) Construct a Transportation table
2) Find Initial Feasible Solution By
a) Northwest Corner Method
(NWCR)
b) Least Cost Method (LCM)
c) Vogel’s Approximation Method
(VAM)
Plant P Q R S
A 38 60 100 24
B 140 60 80 120
C 80 20 120 40