This document discusses three methods for finding a basic feasible solution to transportation problems: 1. The North-West Corner Method which starts assigning supply from the upper left corner, comparing supply and demand and moving horizontally or vertically as needed. 2. The Least-Cost Method which assigns supply based on the lowest cost cells, striking out columns and rows as demand is met. 3. Vogel's Approximation Method which iteratively assigns supply in a round-robin fashion. The document provides an example applying the North-West Corner Method, explaining the step-by-step process of assigning supply and tracking remaining amounts.

1. TRANSPORTATION PROBLEMS
 Finding Basic Feasible Solution :
There are three methods
1. North-West Corner Method :
Consider an example,
P Q R S T SUPPLY
A 2 11 10 3 7 4
B 1 4 7 2 1 8
C 3 9 4 8 12 9
DEMAND 3 3 4 5 6 21
 Make sure that total number of units of supply = total
number of units of demand. In this example it is 21.
North-West Corner
Method
Least-Cost method
Vogel’s
approximation
method

2. Then the transportation problem is said to be BALANCED.
Source A :
 Starting from the upper left (North-
West) corner, compare the supply
and demand of A and P, and assign
the maximum possible demand to
cell AP. Make sure the demand
doesn’t exceed the supply. If it
exceeds, then move on to the next row.
In this ex, demand = 3 supply = 4. Therefore D<S. 3 units can
be assigned to AP. The remaining supply is 4-3=1.
P Q R S T SUPPLY
A 2 11 10 3 7 4/1
B 1 4 7 2 1 8
C 3 9 4 8 12 9
DEMAND 3/0 3 4 5 6 21
Now the demand of P is 0. Move onto the next column to
satisfy the demand of Q. Again compare demand and supply.
Here D>S ( 3 > 1) . Hence only 1 Unit can be assigned to AQ.
3
If D<S assign the demand to
the respective destination
and proceed horizontally.
If D>S assign the demand
to the respective
destination without
exceeding the supply and
proceed vertically.

3. P Q R S T SUPPLY
A 2 11 10 3 7 4/1/0
B 1 4 7 2 1 8
C 3 9 4 8 12 9
DEMAND 3/0 3/2 4 5 6 21
 Now the demand is 2 and supply is 0. Therefore move onto
the next row. Source A has completed delivering all the
units. We next move onto Source B.
Source B :
Comparing demand and supply of B and Q we find that D<S .
Therefore we can assign 2 units to BQ
P Q R S T SUPPLY
A 2 11 10 3 7 4/1/0
B 1 4 7 2 1 8/6
C 3 9 4 8 12 9
DEMAND 3/0 3/2/0 4 5 6 21
3 1
3 1
2

4.  Since demand is 0 we move on to next column to satisfy the
demand of R.
Continuing this way we obtain the following table.
P Q R S T SUPPLY
A 2 11 10 3 7 4/1/0
B 1 4 7 2 1 8/6/2/0
C 3 9 4 8 12 9/6/0
DEMAND 3/0 3/2/0 4/0 5/3/0 6 21
 Minimum transportation cost is given by
Z=(3×2)+(11×1)+(4×2)+(7×4)+(2×2)+(8×3)+(12×6)= Rs 153.
3 1
2 4 2
3 6

5. 2. Least-Cost method :
Consider an example,
P Q R S T SUPPLY
A 2 11 10 3 7 4
B 1 4 7 2 1 8
C 3 9 4 8 12 9
DEMAND 3 3 4 5 6 21
 Check for the minimum cost in the entire table. In this
example it is BP(1) and BT(1). Chose anyone arbitrarily.
Let us chose BP. Now compare the values of demand and
supply and assign the maximum possible demand to BP.
Strike out the 1st column since the demand becomes 0.
P Q R S T SUPPLY
A 2 11 10 3 7 4
B 1 4 7 2 1 8/5
C 3 9 4 8 12 9
DEMAND 3/0 3 4 5 6 21
3

6. Again check for the minimum cost in the remaining columns
and rows. In this ex, it is BT(1).Assign the maximum possible
demand for BT.
Now the supply has become 0. Strike out the entire 2nd row.
P Q R S T SUPPLY
A 2 11 10 3 7 4
B 1 4 7 2 1 8/5/0
C 3 9 4 8 12 9
DEMAND 3/0 3 4 5 6/1 21
Continuing this way we obtain the following table.
P Q R S T SUPPLY
A 2 11 10 3 7 4/0
B 1 4 7 2 1 8/5/0
C 3 9 4 8 12 9/6/2/1/0
DEMAND 3/0 3/0 4/0 5/1/0 6/1/0 21
3 5
3 5
4
3 4 1 1