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OPERATION RESEARCH
TRANSPORTATION PROBLEM
The transportation problem is a special type of linear programming
problem, where the objective is to minimize the cost of distributing a
product from a number of sources to a number of destinations.
TRANSPORTATION PROBLEM
THE GENERAL MATHEMATICAL MODEL
If xij (≥0) is the number of units shipped from ith source to jth destination, then equivalent LPP model will be
For a feasible solution to exist, it is necessary that total capacity equals total to the
n m
ai bj
Minimize Z
m n
i 1 j
cij xij
j
Subject to
n
xij ai
j 1
m
For i = 1,2, ..... , m (supply)
xij bj
i 1
For j = 1,2, ..... , n (demand)
requirements. If i.e. If total supply = total demand then it is a balanced
xij 0 .
transportation problem otherwise it is called unbalanced Transportation
problem. There will be (m + n - 1) basic independent variables out of (m x n)
variables
What are the understanding assumptions?
m
ai bj , ai (supply) and bj (demand) are all positive integers.
i 1 j 1
1. Only a single type of commodity is being shipped from an origin to a
destination.
2. Total supply is equal to the total demand.
n
3. The unit transportation cost of the item from all sources to destinations is
certainly and preciously known.
4. The objective is to minimize the total cost.
• 2
North West Corner Rule
Example 1: The ICARE Company has three plants located throughout a state with
production capacity 50, 75 and 25 gallons. Each day the firm must furnish its four retail
shops R1, R2, R3,
& R4 with at least 20, 20 , 50, and 60 gallons respectively. The transportation costs
(in Rs.) are given below.
The economic problem is to distribute the available product to
different retail shops in such a way so that the total transportation
cost is minimum?
Solution: Starting from the North West corner, we allocate min (50, 20) to P1R1, i.e., 20
units to cell P1R1. The demand for the first column is satisfied. The allocation is shown in
the following table.
Table 1
Company
Retail
Supply
R1 R2 R3 R4
P1 6 50
P2 2 5 75
P3 3 6 9 25
Demand 20 20 50 60
Now we move horizontally to the second column in the first row and allocate 20 units to cell P1R2. The demand for the second column is also satisfied.
Proceeding in this way, we observe that P1R3 = 10, P2R3 = 40, P2R4 = 35, P3R4 = 25. The resulting feasible solution is shown in the following table.
Here, number of retail shops (n) = 4, and
Number of plants (m) = 3.Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6.
Initial basic feasible solution
The initial basic feasible solution is x11=20, x12=5, x13=20, x23=40, x24=35, x34=25 and minimum cost of transportation=20 X 3 + 20 X 5 + 10 X 7 + 40 X 8 + 35 X 2 + 25
X 2 = 670
Matrix Minimum Method
Example 2: Consider the transportation problem presented in the following table:
Solution:
Factory
Retail shop
Supply
1 2 3 4
1 3 6 50
2 5 8 75
3 3 6 25
Demand 20 20 50 60
Initial basic feasible solution
The initial basic feasible solution is x12=20, x13=30, x21=20, x24=55, x233=20,
x34=5 and minimum cost of transportation=20 X 2 + 20 X 5 + 30 X 7 + 55 X 2
+ 20 X 9 + 5 X 2 = 650.
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OR presentation.pptx

  • 2. The transportation problem is a special type of linear programming problem, where the objective is to minimize the cost of distributing a product from a number of sources to a number of destinations. TRANSPORTATION PROBLEM
  • 3. THE GENERAL MATHEMATICAL MODEL If xij (≥0) is the number of units shipped from ith source to jth destination, then equivalent LPP model will be For a feasible solution to exist, it is necessary that total capacity equals total to the n m ai bj Minimize Z m n i 1 j cij xij j Subject to n xij ai j 1 m For i = 1,2, ..... , m (supply) xij bj i 1 For j = 1,2, ..... , n (demand) requirements. If i.e. If total supply = total demand then it is a balanced xij 0 . transportation problem otherwise it is called unbalanced Transportation problem. There will be (m + n - 1) basic independent variables out of (m x n) variables
  • 4. What are the understanding assumptions? m ai bj , ai (supply) and bj (demand) are all positive integers. i 1 j 1 1. Only a single type of commodity is being shipped from an origin to a destination. 2. Total supply is equal to the total demand. n 3. The unit transportation cost of the item from all sources to destinations is certainly and preciously known. 4. The objective is to minimize the total cost.
  • 5. • 2 North West Corner Rule Example 1: The ICARE Company has three plants located throughout a state with production capacity 50, 75 and 25 gallons. Each day the firm must furnish its four retail shops R1, R2, R3, & R4 with at least 20, 20 , 50, and 60 gallons respectively. The transportation costs (in Rs.) are given below. The economic problem is to distribute the available product to different retail shops in such a way so that the total transportation cost is minimum?
  • 6. Solution: Starting from the North West corner, we allocate min (50, 20) to P1R1, i.e., 20 units to cell P1R1. The demand for the first column is satisfied. The allocation is shown in the following table. Table 1 Company Retail Supply R1 R2 R3 R4 P1 6 50 P2 2 5 75 P3 3 6 9 25 Demand 20 20 50 60
  • 7. Now we move horizontally to the second column in the first row and allocate 20 units to cell P1R2. The demand for the second column is also satisfied. Proceeding in this way, we observe that P1R3 = 10, P2R3 = 40, P2R4 = 35, P3R4 = 25. The resulting feasible solution is shown in the following table. Here, number of retail shops (n) = 4, and Number of plants (m) = 3.Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6. Initial basic feasible solution The initial basic feasible solution is x11=20, x12=5, x13=20, x23=40, x24=35, x34=25 and minimum cost of transportation=20 X 3 + 20 X 5 + 10 X 7 + 40 X 8 + 35 X 2 + 25 X 2 = 670
  • 8. Matrix Minimum Method Example 2: Consider the transportation problem presented in the following table: Solution: Factory Retail shop Supply 1 2 3 4 1 3 6 50 2 5 8 75 3 3 6 25 Demand 20 20 50 60 Initial basic feasible solution The initial basic feasible solution is x12=20, x13=30, x21=20, x24=55, x233=20, x34=5 and minimum cost of transportation=20 X 2 + 20 X 5 + 30 X 7 + 55 X 2 + 20 X 9 + 5 X 2 = 650.