Transportation model makes the following basic assumptions:
Availability of commodity : The supply available at different sources is equal to or more than the total demand of different destinations when it is equal it is called a balanced problem
 Transportation of commodity/terms: The model assumes that all items can be conveniently transported from sources to destinations.
 Certainty of per unit transportation cost: There is a definite cost of transportation of items from sources to destinations.
 Independent cost per unit: The per unit cost is independent of the quantitytransported from sources to destinations.
 Transportation cost on any given route is proportional to the number of units transported.
 Objective function: The objective is to minimize the total transportation cost for the entire organization.
Example 5.1
A car manufacturing company has plants at cities A, B and C. Its destination centers are located at cities X and Y. The capacity of three plants during the next quarter is 1000, 1500 and 1200 cars. The quarterly demand of the two destination centers is 2300 and 1400 cars. The train transportation cost per car per km is Rs. 8. The chart below shows the distance in km between the plants and the distribution centers
X 
Y 
1000 
2690 
1250 
1350 
1275 
850 
How many cars should be transported from which plant to which destination center to minimize cost?
Sol. Step I. In the above problem the market distances and the cost of transportation per km is given. This must be converted into the costs.

X 
Y 
A 
2000 
5380 
B 
2500 
2700 
C 
2550 
1770 
Let X_{ij} be the number of cars transported from plants to destinations centers (X_{ij} 0) since the total supply (1000 + 1500 + 1200 = 3700) happens to equal the total demand (2300 + 1400 = 3700), the transportation model is balanced.


X 
Y 
Supply 

A 
X_{11} 
X_{12} 
1000 
Source 
B 
X_{21} 
X_{22} 
1500 

C 
X_{31} 
X_{32} 
1200 

Demand 
2300 
1400 
3700 
Step II. Objective is to minimize the cost of Iran sport at ion
i.e. Z = 2000 X_{11} + 5380X_{12} + 2500 X_{21} + 2700X_{22} + 2550X_{31} + 1700 X_{32}
Step III. Constraints
(a) Availability or supply of cars
X_{11} + X_{12} = 1000 (Source A)
X_{21} + X_{22} = 1500 (Source B)
X_{31} + X_{32} = 1200 (Source C)
It may be seen that the constraints are equal to number of plants, i.e. there are three constraints.
(b) Requirement of demand of cars at destination centers
X_{11} + X_{21} + X_{31} = 2300 (Destination center X)
X_{12} + X_{22} + X_{32} = 1400 (Destination center Y)
It may be seen that in this problem there are (3 × 2 = 6) variables and (3 + 2 = 5) constraints. In general, such a solution will involve (m × n) variables and (m + n) constraints.
It is also clear that the objective function and the constraint equations are linear in nature, hence this problem can be solved by simplex method of linear programming. However, since 6 variables are involved (in real life situations there will be much more) the calculations will be very long and timeconsuming requiring the help of computers. Also, the simplex method is more suitable for maximization problems whereas the transportation problem require the minimization of the objective function.
SOLUTION OF THE TRANSPORTATION MODEL
Step I: Make a transportation model
Distribution centers


X 
Y 
Supply 
Plants 
A 
Rs.2000 X_{11}=1000 
Rs.=5380 
1000 

B 
Rs.2500 X_{21}=1300 
Rs. 2700 X_{22}=1400 
+1500 

C 
Rs.2550 
Rs.1700 
+1200 


2300 
+1400 
=3700 

Demand 



The above problem is balanced or self contained wherever it is not so, a dummy source or destination, as the case may be, is created to balance the supply and demand.
Step II. Finding a basic feasible solution
Basic feasible solution can be found out by using different methods. One technique developed by Dantzig is called the ‘North West corner rule’.
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