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© 2011 Pearson Education
TRANSPORTATION PROBLEM
Introduction:
In the application of linear programming techniques, the
transportation problem was probably one of the first
significant problems studied. When a large number of
variables (more than 2) are involved in a problem, the
solution by graphical method is not possible. The
simplex method provides an efficient technique which
can be applied for solving linear programming
problems of any magnitude, involving two or more
decision variables.
© 2011 Pearson Education
All linear programming problems can be solved by
simplex method, but certain special problems lend
themselves to easy solution by other methods. One of
such case is “Transportation Problem”.
Transportation problems are encountered in physical
distribution of goods. Source of supply, availability of
material or commodity for distribution, the requirement
of demand at particular place or destination or at
number of destinations are some of the parameters
involved in the problem.
© 2011 Pearson Education
The objective is to minimize the cost associated with such
transportation from place of supply to places of demand
within given constraints of availability and level of
demand. These distribution problems are amenable to
solution by a special type of linear programming model
known as “Transportation Model”. It can also be applied
to the maximization of some utility value such as financial
resources.
© 2011 Pearson Education
Transportation Modeling
 An interactive procedure that finds the least costly
means of moving products from a series of sources to
a series of destinations
 Can be used to
help resolve
distribution
and location
decisions
© 2011 Pearson Education
Transportation Modeling
 A special class of linear programming
 Need to know
1. The origin points and the capacity or supply per
period at each
2. The destination points and the demand per period
at each
3. The cost of shipping one unit from each origin to
each destination
Meaning and Definition
Transportation model is a special type of networks
problems dealing with shipping a commodity from
source (e.g., factories) to destinations (e.g., warehouse).
It deals with obtaining the minimum cost plan to
transport a commodity from a number of sources (m) to
number of destination (n). The objective of
transportation problem is to determine the amount to be
transported from each origin to each destination such
that the total transportation cost is minimized.
Structure of the Transportation Problem (TP)
Let, m = Origin n = Destinations
a i = Quantity of products available at origin i
b j = Quantity of products required at destination j
C ij = Cost of transporting one unit of product from origin i to
destination j
x ij = Quantity transported from origin i to destination j
The objective is to determine the quantity x ij to be transported
over all routes
(i, j) so as to minimize the total transportation cost.
The quantity of products available at the origins must satisfy the
quantity of products required at the destinations.
• If this is represented in matrix form, wherein each
cell contains a transportation
• variable, means the number of units shipped from
the row designated origin to the
• column designated destination. The amount of
supplies ai available at source i and the amount
demanded by bj at each destination j.
• Hence, ai and bj represent supply and demand
constraint.
Vogel’s Approximation Method
Vogel’s Approximation Method (VAM) is one of the
methods used to calculate the initial basic feasible
solution to a transportation problem. However, VAM
is an iterative procedure such that in each step, we
should find the penalties for each available row and
column by taking the least cost and second least cost.
In this presentation, you will learn how to find the
initial basic feasible solution to a transportation
problem such that the total cost is minimized.
Vogel’s Approximation Method Steps
Below are the steps involved in Vogel's approximation method of finding
the feasible solution to a transportation problem.
Step 1: Identify the two lowest costs in each row and column of the given
cost matrix and then write the absolute row and column difference. These
differences are called penalties.
Step 2: Identify the row or column with the maximum penalty and assign
the corresponding cell’s min(supply, demand). If two or more columns or
rows have the same maximum penalty, then we can choose one among
them as per our convenience.
Step 3: If the assignment in the previous satisfies the supply at the origin,
delete the corresponding row. If it satisfies the demand at that
destination, delete the corresponding column.
Step 4: Stop the procedure if supply at each origin is 0, i.e., every supply
is exhausted, and demand at each destination is 0, i.e., every demand is
satisfying. If not, repeat the above steps, i.e., from step 1.
Vogel’s Approximation Method Solved Example
Question:
Solve the given transportation problem using Vogel’s approximation method.
Solution:
For the given cost matrix,
Total supply = 50 + 60 + 25 = 135
Total demand = 60 + 40 + 20 + 25 = 135
Thus, the given problem is balanced transportation problem.
Now, we can apply the Vogel’s approximation method to
minimize the total cost of transportation.
Step 1: Identify the least and second least cost in each row and
column and then write the corresponding absolute differences
of these values. For example, in the first row, 2 and 3 are the
least and second least values, their absolute difference is 1.
These row and column differences are called penalties.
Step 2: Now, identify the maximum penalty and choose
the least value in that corresponding row or column.
Then, assign the min(supply, demand).
Here, the maximum penalty is 3 and the least value in
the corresponding column is 2. For this cell, min(supply,
demand) = min(50, 40) = 40
Allocate 40 in that cell and strike the corresponding
column since in this case demand will be satisfied, i.e.,
40 – 40 = 0.
Step 3: Now, find the absolute row and column differences for
the remaining rows and columns. Then repeat step 2.
Here, the maximum penalty is 3 and the least cost in that
corresponding row is 3. Also, the min(supply, demand) =
min(10, 60) = 10
Thus, allocate 10 for that cell and write down the new supply
and demand for the corresponding row and column.
Supply = 10 – 10 = 0
Demand = 60 – 10 = 50
As supply is 0, strike the corresponding row.
Step 4: Repeat the above step, i.e., step 3. This will give the below
result.
In this step, the second column vanishes and the min(supply, demand) =
min(25, 50) = 25 is assigned for the cell with value 2.
Step 5: Again repeat step 3, as we did for the previous step.
In this case, we got 7 as the maximum penalty and 7 as the least cost of the
corresponding column.
Step 6: Now, again repeat step 3 by calculating the absolute differences
for the remaining rows and columns.
Step 7: In the previous step, except for one cell, every row and column
vanishes. Now, allocate the remaining supply or demand value for that
corresponding cell.
Total cost = (10 × 3) + (25 × 7) + (25 × 2) + (40 × 2) +
(20 × 2) + (15 × 3)
= 30 + 175 + 50 + 80 + 40 + 45
= 420
Vogel’s Approximation Method Problems
1. Consider the transportation problem given below. Solve this problem
by Vogel’s approximation method.
Origin
Destination
Supply
D1 D2 D3 D4
O1 3 1 7 4 300
O2 2 6 5 9 400
O3 8 3 3 2 500
Demand 250 350 400 200
From
To
Supply
D1 D2 D3
A1 6 8 10 150
A2 7 11 11 175
A3 4 5 12 275
Demand 200 100 350
2. Find the solution for the following transportation
problem using VAM.
3. Use Vogel’s Approximation Method to find a basic
feasible solution for the following.
Sources
Destinations
Supply
D1 D2 D3
S1 4 5 1 40
S2 3 4 3 60
S3 6 2 8 70
Demand 70 40 60

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Linear Programming.pptx

  • 1. © 2011 Pearson Education TRANSPORTATION PROBLEM Introduction: In the application of linear programming techniques, the transportation problem was probably one of the first significant problems studied. When a large number of variables (more than 2) are involved in a problem, the solution by graphical method is not possible. The simplex method provides an efficient technique which can be applied for solving linear programming problems of any magnitude, involving two or more decision variables.
  • 2. © 2011 Pearson Education All linear programming problems can be solved by simplex method, but certain special problems lend themselves to easy solution by other methods. One of such case is “Transportation Problem”. Transportation problems are encountered in physical distribution of goods. Source of supply, availability of material or commodity for distribution, the requirement of demand at particular place or destination or at number of destinations are some of the parameters involved in the problem.
  • 3. © 2011 Pearson Education The objective is to minimize the cost associated with such transportation from place of supply to places of demand within given constraints of availability and level of demand. These distribution problems are amenable to solution by a special type of linear programming model known as “Transportation Model”. It can also be applied to the maximization of some utility value such as financial resources.
  • 4. © 2011 Pearson Education Transportation Modeling  An interactive procedure that finds the least costly means of moving products from a series of sources to a series of destinations  Can be used to help resolve distribution and location decisions
  • 5. © 2011 Pearson Education Transportation Modeling  A special class of linear programming  Need to know 1. The origin points and the capacity or supply per period at each 2. The destination points and the demand per period at each 3. The cost of shipping one unit from each origin to each destination
  • 6. Meaning and Definition Transportation model is a special type of networks problems dealing with shipping a commodity from source (e.g., factories) to destinations (e.g., warehouse). It deals with obtaining the minimum cost plan to transport a commodity from a number of sources (m) to number of destination (n). The objective of transportation problem is to determine the amount to be transported from each origin to each destination such that the total transportation cost is minimized.
  • 7. Structure of the Transportation Problem (TP) Let, m = Origin n = Destinations a i = Quantity of products available at origin i b j = Quantity of products required at destination j C ij = Cost of transporting one unit of product from origin i to destination j x ij = Quantity transported from origin i to destination j The objective is to determine the quantity x ij to be transported over all routes (i, j) so as to minimize the total transportation cost. The quantity of products available at the origins must satisfy the quantity of products required at the destinations.
  • 8. • If this is represented in matrix form, wherein each cell contains a transportation • variable, means the number of units shipped from the row designated origin to the • column designated destination. The amount of supplies ai available at source i and the amount demanded by bj at each destination j. • Hence, ai and bj represent supply and demand constraint.
  • 9. Vogel’s Approximation Method Vogel’s Approximation Method (VAM) is one of the methods used to calculate the initial basic feasible solution to a transportation problem. However, VAM is an iterative procedure such that in each step, we should find the penalties for each available row and column by taking the least cost and second least cost. In this presentation, you will learn how to find the initial basic feasible solution to a transportation problem such that the total cost is minimized.
  • 10. Vogel’s Approximation Method Steps Below are the steps involved in Vogel's approximation method of finding the feasible solution to a transportation problem. Step 1: Identify the two lowest costs in each row and column of the given cost matrix and then write the absolute row and column difference. These differences are called penalties. Step 2: Identify the row or column with the maximum penalty and assign the corresponding cell’s min(supply, demand). If two or more columns or rows have the same maximum penalty, then we can choose one among them as per our convenience. Step 3: If the assignment in the previous satisfies the supply at the origin, delete the corresponding row. If it satisfies the demand at that destination, delete the corresponding column. Step 4: Stop the procedure if supply at each origin is 0, i.e., every supply is exhausted, and demand at each destination is 0, i.e., every demand is satisfying. If not, repeat the above steps, i.e., from step 1.
  • 11. Vogel’s Approximation Method Solved Example Question: Solve the given transportation problem using Vogel’s approximation method.
  • 12. Solution: For the given cost matrix, Total supply = 50 + 60 + 25 = 135 Total demand = 60 + 40 + 20 + 25 = 135 Thus, the given problem is balanced transportation problem. Now, we can apply the Vogel’s approximation method to minimize the total cost of transportation. Step 1: Identify the least and second least cost in each row and column and then write the corresponding absolute differences of these values. For example, in the first row, 2 and 3 are the least and second least values, their absolute difference is 1.
  • 13. These row and column differences are called penalties.
  • 14. Step 2: Now, identify the maximum penalty and choose the least value in that corresponding row or column. Then, assign the min(supply, demand). Here, the maximum penalty is 3 and the least value in the corresponding column is 2. For this cell, min(supply, demand) = min(50, 40) = 40 Allocate 40 in that cell and strike the corresponding column since in this case demand will be satisfied, i.e., 40 – 40 = 0.
  • 15.
  • 16. Step 3: Now, find the absolute row and column differences for the remaining rows and columns. Then repeat step 2. Here, the maximum penalty is 3 and the least cost in that corresponding row is 3. Also, the min(supply, demand) = min(10, 60) = 10 Thus, allocate 10 for that cell and write down the new supply and demand for the corresponding row and column. Supply = 10 – 10 = 0 Demand = 60 – 10 = 50 As supply is 0, strike the corresponding row.
  • 17.
  • 18. Step 4: Repeat the above step, i.e., step 3. This will give the below result. In this step, the second column vanishes and the min(supply, demand) = min(25, 50) = 25 is assigned for the cell with value 2.
  • 19. Step 5: Again repeat step 3, as we did for the previous step. In this case, we got 7 as the maximum penalty and 7 as the least cost of the corresponding column.
  • 20. Step 6: Now, again repeat step 3 by calculating the absolute differences for the remaining rows and columns.
  • 21. Step 7: In the previous step, except for one cell, every row and column vanishes. Now, allocate the remaining supply or demand value for that corresponding cell.
  • 22. Total cost = (10 × 3) + (25 × 7) + (25 × 2) + (40 × 2) + (20 × 2) + (15 × 3) = 30 + 175 + 50 + 80 + 40 + 45 = 420
  • 23. Vogel’s Approximation Method Problems 1. Consider the transportation problem given below. Solve this problem by Vogel’s approximation method. Origin Destination Supply D1 D2 D3 D4 O1 3 1 7 4 300 O2 2 6 5 9 400 O3 8 3 3 2 500 Demand 250 350 400 200
  • 24. From To Supply D1 D2 D3 A1 6 8 10 150 A2 7 11 11 175 A3 4 5 12 275 Demand 200 100 350 2. Find the solution for the following transportation problem using VAM.
  • 25. 3. Use Vogel’s Approximation Method to find a basic feasible solution for the following. Sources Destinations Supply D1 D2 D3 S1 4 5 1 40 S2 3 4 3 60 S3 6 2 8 70 Demand 70 40 60

Editor's Notes

  1. This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.
  2. This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.