T R A N S P O R T A T I O N P R O B L E M S

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.

Then the transportation problem is said to be BALANCED.

North-West Corner Method

Least-Cost methodVogel’s approximation method

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.

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

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.

2. Least-Cost method :

3 1

2 4 2

3 6

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

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