The document outlines the presentation topic of Modified Distribution Method (MODI Method) for solving transportation problems. It first discusses the prerequisite methods of Least Cost Method, Vogel's Approximation Method and North-West Corner Method. It then explains the steps of MODI Method which involves setting up cost matrices for unallocated cells and introducing dual variables to find the implicit cost and evaluate unoccupied cells to determine if the initial solution can be improved. The document provides an example problem and solution to demonstrate the application of MODI Method.
This presentation covers the problems and solutions of North West Corner Method, Least cost Method and Vogels Approximation Method in Transportation Problem
Steps to solve Transportation models by North west corner method are given the presentation. North west corner method is one of the well known methods used to solve the transportation models.
This presentation covers the problems and solutions of North West Corner Method, Least cost Method and Vogels Approximation Method in Transportation Problem
Steps to solve Transportation models by North west corner method are given the presentation. North west corner method is one of the well known methods used to solve the transportation models.
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 or origins to a number of destinations.
Because of its special structure, the usual simplex method is not suitable for solving transportation problems. These problems require a special method of solution.
The Least Cost Method is another method used to obtain the initial feasible solution for the transportation problem. Here, the allocation begins with the cell which has the minimum cost. The lower cost cells are chosen over the higher-cost cell with the objective to have the least cost of transportation.
Transportation Problem In Linear ProgrammingMirza Tanzida
This work is an assignment on the course of 'Mathematics for Decision Making'. I think, it will provide some basic concept about transportation problem in linear programming.
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 or origins to a number of destinations.
Because of its special structure, the usual simplex method is not suitable for solving transportation problems. These problems require a special method of solution.
The Least Cost Method is another method used to obtain the initial feasible solution for the transportation problem. Here, the allocation begins with the cell which has the minimum cost. The lower cost cells are chosen over the higher-cost cell with the objective to have the least cost of transportation.
Transportation Problem In Linear ProgrammingMirza Tanzida
This work is an assignment on the course of 'Mathematics for Decision Making'. I think, it will provide some basic concept about transportation problem in linear programming.
This presentation is made to represent the basic transportation model. The aim of this presentation is to implement the transportation model in solving transportation problem.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Quality defects in TMT Bars, Possible causes and Potential Solutions.PrashantGoswami42
Maintaining high-quality standards in the production of TMT bars is crucial for ensuring structural integrity in construction. Addressing common defects through careful monitoring, standardized processes, and advanced technology can significantly improve the quality of TMT bars. Continuous training and adherence to quality control measures will also play a pivotal role in minimizing these defects.
Automobile Management System Project Report.pdfKamal Acharya
The proposed project is developed to manage the automobile in the automobile dealer company. The main module in this project is login, automobile management, customer management, sales, complaints and reports. The first module is the login. The automobile showroom owner should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
When a customer search for a automobile, if the automobile is available, they will be taken to a page that shows the details of the automobile including automobile name, automobile ID, quantity, price etc. “Automobile Management System” is useful for maintaining automobiles, customers effectively and hence helps for establishing good relation between customer and automobile organization. It contains various customized modules for effectively maintaining automobiles and stock information accurately and safely.
When the automobile is sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting automobiles for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item.
Also when the user tries to sale items which are not in stock, the system will prompt the user that the stock is not enough. Customers of this system can search for a automobile; can purchase a automobile easily by selecting fast. On the other hand the stock of automobiles can be maintained perfectly by the automobile shop manager overcoming the drawbacks of existing system.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Democratizing Fuzzing at Scale by Abhishek Aryaabh.arya
Presented at NUS: Fuzzing and Software Security Summer School 2024
This keynote talks about the democratization of fuzzing at scale, highlighting the collaboration between open source communities, academia, and industry to advance the field of fuzzing. It delves into the history of fuzzing, the development of scalable fuzzing platforms, and the empowerment of community-driven research. The talk will further discuss recent advancements leveraging AI/ML and offer insights into the future evolution of the fuzzing landscape.
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
COLLEGE BUS MANAGEMENT SYSTEM PROJECT REPORT.pdfKamal Acharya
The College Bus Management system is completely developed by Visual Basic .NET Version. The application is connect with most secured database language MS SQL Server. The application is develop by using best combination of front-end and back-end languages. The application is totally design like flat user interface. This flat user interface is more attractive user interface in 2017. The application is gives more important to the system functionality. The application is to manage the student’s details, driver’s details, bus details, bus route details, bus fees details and more. The application has only one unit for admin. The admin can manage the entire application. The admin can login into the application by using username and password of the admin. The application is develop for big and small colleges. It is more user friendly for non-computer person. Even they can easily learn how to manage the application within hours. The application is more secure by the admin. The system will give an effective output for the VB.Net and SQL Server given as input to the system. The compiled java program given as input to the system, after scanning the program will generate different reports. The application generates the report for users. The admin can view and download the report of the data. The application deliver the excel format reports. Because, excel formatted reports is very easy to understand the income and expense of the college bus. This application is mainly develop for windows operating system users. In 2017, 73% of people enterprises are using windows operating system. So the application will easily install for all the windows operating system users. The application-developed size is very low. The application consumes very low space in disk. Therefore, the user can allocate very minimum local disk space for this application.
block diagram and signal flow graph representation
MODI Method (Operations Research)
1. Shri S'ad Vidya Mandal Institute Of Technology
BRANCH : MECHANICAL ENGINEERING
SEMESTER : 7TH SEM
YEAR : 2016-2017
SUBJECT : OPERATIONS RESEARCH (2171901)
PRESENTATION TOPIC : MODIFIED DISTRIBUTION METHOD (MODI METHOD)
ENROLLMENT NO : 130454119006
GUIDED BY : ASST. PROF. JAY SHAH
2. Outline of Presentation:-
1. Prerequisite :
Least cost Method
Vogel’s Approximation Method
North-West Corner Method
2. Modified Distribution Method (MODI) or u-v Method
3. References
3. Least cost method
Step 1: Balance the problem. If not add a dummy column or dummy row as the case may be and
balance the problem.
Step 2: Identify the lowest cost cell in the given matrix. Allocate minimum of either demand or
supply in that cell. If there is a same value of cost allocate to any cell as per your choice
according to demand and supply. Then eliminate that row or column in further procedure.
Step 3: Next search for lowest cost cell. Repeat this procedure till all allocation is not completed.
Step 4: Once all the allocations are over, i.e., availability and demand are satisfied, write
allocations and calculate the cost of transportation.
4. Problem : Use least cost method for the problem and find out basic feasible solution.
Solution :
Step 1 : Here, the matrix is unbalanced.
Step 2 : Using least cost method, the lowest cost cell in the given matrix is AF having cost 1. Allocating
minimum of either demand or supply in that cell i.e., 40 units to AF cell. Then eliminating that row from the
source but the demand of the market F is not fulfilled.
Units D E F Supply
0
B 3 4 3 60
C 6 2 8 70
Demand 40 40 2006
A 4 5 )04(1 04
5. Then next lowest cost is 2 in CE cell. Allocating 40 units in that cell and repeating this
procedure till all allocations are mode. Final allocation is as shown below
Units D E F Supply
A 4 5 1(40) 40
B 3(40) 4 3(20) 60
C 6(30) 2(40) 8 70
Demand 70 40 60 170
Units D E F Supply
B 3 4 3 60
C 6 2(40) 8 70 30
Demand 70 40 0 20
6. Calculating the total cost : 3*40 + 6*30 + 2*40 + 1*40 + 3*20 = Rs.480
The alternate least cost method solution as follow
Calculating the total cost : 3*60 + 6*10 + 2*40 1*40 + 8*20 =Rs. 480
Units D E F Supply
A 4 5 1(40) 40
B 3(60) 4 3 60
C 6(10) 2(40) 8(20) 70
70 40 60 170
7. Vogel’s Approximation Method (VAM)
VAM gives better initial solution than obtained by other methods. Here, the concept of
opportunity cost is considered.
Opportunity cost is the penalty occurring for not selecting right cell for the allocation. This
opportunity cost is found out by calculating the difference in each row or in each column for cost
coefficients As the value of difference is larger, higher will be the penalty for allocating in second
smallest cell instead of smallest cost cell. So, this shows the penalty for failing to make right
allocating to the smallest cell.
8. Steps involved in the (VAM) method :
1. Balance the problem. If not add a dummy column or dummy row to balance the problem.
2. Opportunity cost is found out by calculating the difference between the smallest and second
smallest element in each row and in each column. Enter the difference in respective column or in
respective row in brackets.
3. Select the row or column from the matrix which has maximum value. In that selected row or
column, select the cell which has minimum cost coefficient. Allocate smallest value of demand or
supply in the cell.
4. Eliminate that allocated row or column and prepare the new matrix.
5. Repeat this procedure until all allocation are not complete.
6. Once all the allocation are over, i.e. availability and demand are satisfied, write allocations and
calculate the cost of transportation.
Important Points to Remember:
1. If the coefficients in a row or column are same, than difference for that row or column is '0'.
2. In case of tie among the highest penalties, select the row or column having minimum cost. In case
of tie in minimum cost also, select the cell in which maximum allocation can be done. Again, if
there is a tie in maximum allocation value, select the cell arbitrarily for allocation.
9. Since, the matrix is balanced; we should start from step 2.
Here, finding out the difference between the smallest and second smallest for each row and
each column & entering in respective column and row. It is shown in brackets.
Example :- The paper manufacturing company has three warehouses located in
three different areas, says A, B & C. The company has to send from these
warehouses to three destinations, says D, E & F. The availability from warehouses A,
B & C is 40, 60, & 70 units. The demand at D, E and F is 70, 40 and 60 respectively.
The transportation cost is shown in matrix (in Rs.).
D E F Supply Row Difference
A 4 5 1 40 [3]
B 3 4 3 60 [0]
C 6 2 8 70 [4]
Demand 70 40 60 170
Column Difference [1] [2] [2]
10. Here, the maximum difference is 4 in the rows. So, selecting the column in that row which has
minimum cost i.e, ‘CE’ cell. Allocating 40 units to CE cell.
D E F Supply New supply
A 4 5 1 40 40
B 3 4 3 60 60
C 6 2(40) 8 70 30
Demand 70 40 60
New demand 70 0 60
Now, eliminating E column, repeating the above process, we will AF cell as the cell for allocating
the units. On repetitive process, we get finally allocated cells as shown in matrix.
11. D F Supply New supply Row diff.
A 4 1(40) 40 0 3
B 3 3 60 60 0
C 6 8 70 30 2
Demand 70 60
New demand 70 20
Column diff. 1 2
D F Supply New supply Row diff.
B 3 3(20) 60 40 0
C 6 8 30 30 2
Demand 70 20
New demand 70 0
Column diff. 3 5
12. Now allocating 40 units to BD cell and 30 units to CD cell, final matrix can be generated as
below:
D E F Supply
A 4 5 1(40) 40
B 3(40) 4 3(20) 60
C 6(30) 2(40) 8 70
Demand 70 40 60 170
Calculating the total cost: 3 40+ 6 30+ 2 40+ 1 40+3 20= Rs 480
VAM method provides most efficient allocation then any other method for obtaining the basic feasible solution.
Checking the above problem for degeneracy, the number of allocated cell must be equal to
R+C-1, i.e 3+3-1=5. Here the solution is non degenerated basic feasible solution.
13. North-West Corner method
•Following steps are involved in above method:
•Step 1: balance the problem . if not add a dummy column or dummy row as the case may be and
balance the problem.
•Step 2: start allocation from the left hand side top most corner cell and make allocations
depending on the availability and demand .if the availability is less than the requirement ,then
for that cell make allocation in unit which is equal to the availability .commonly speaking , verify
which is the smallest among the availability and requirement and allocate the smallest one to
the cell.
•Step 3: when availability or demand is fulfill for that row or column respectively , remove that
row or column from the matrix. Prepare new matrix.
14. •Step 4: then proceed allocating either sidewise or downward to satisfy the rim requirement
continue this until all the allocation are over.
•Step 5 : once all the allocations are over, i.e., availability and demand are satisfied, write
allocations and calculate the cost of transportation.
15. Solution:
Step 1: here the given matrix is balanced. I.e. the demand and supply is same. So the proceeding
towards step 2
step 2: starting the allocation from the left hand side top most corner cell AD. Here the demand
for the market is 70 units and supply available from the source A is 40 . So allocating the
minimum value of this two , we allocate 40 units to cell AD as follow and eliminating source A in
the next step.
D E F SUPPLY New supply
A 4(40) 5 1 40 × 0
B 3 4 3 60 60
C 6 2 8 70 70
demand 70 × 40 60
New 30 40 60
markets ( destinations )
Paper unit
(sources)
16. In the next matrix, since the demand for a market D has not been fulfilled , so
allocating 30 units to the BD cell, which is smallest among supply and demand
for the source B and demand from market D.
now ,since the demand for the market D has been fulfilled , and preparing the
next matrix removing market D . New N-w cell is BE ,which has demand of 4o
units , which can be satisfied by matrix B.
D E F SUPPLY New
supply
B 3(30) 4 3 60 × 30
C 6 2 8 70 70
demand 30 × 40 60
New 0 40 60
Paper unit
(sources)
markets ( destinations )
17. Allocating to the cell CE and CF , 10 and 60 units for fulfilling the demand
source C and proceeding towards step 4.
D E F SUPPLY
A 4 (40) 5 1 40
B 3 (30) 4 (30) 3 60
C 6 2 (10) 8 (60) 70
demand 70 40 60 170
markets ( destinations )
Paper unit
(sources)
E F SUPPLY New
supply
B 4(30) 3 30 × 0
C 2 8 70 70
demand 40 × 60
New 10 60
Paper unit
(sources)
markets ( destinations )
19. Modified Distribution Method ( MODI METHOD )
When a basic initial feasible solution is obtained, then we have to check for it’s optimality. An
optimal solution is one where there is no other set of roots that will further reduce the total
cost.
Further we have to evaluate each unoccupied cell in table to reduce total cost.
Evaluating the steps will result in the most optimal cost of transportation.
20. Modified Distribution Method ( MODI METHOD )
The steps to evaluate unoccupied cells are as follows :
1. Set up cost matrix for unallocated cells
2. Introduce dual variables corresponding to the supply and demand constraints. Let U
(i=1,2,3..m) and V (j=1,2,3…n) be the dual variables corresponding to supply and demand
constraints. Variables U and V are such that 𝑈𝑖 + 𝑉𝑗 = 𝑐𝑖𝑗. Now select any of the dual
variable as ‘0’ and find the other dual variables.
D E F Supply
A 4 5 1(40) 40
B 3 (40) 4 3(20) 60
C 6(30) 2(40) 8 70
Demand 70 40 60 170
Markets (Destinations)
Sources
21. Modified Distribution Method ( MODI METHOD )
Now the newly formed matrix will be:
Solving the filled cell allocation,
U1+V3= 1 U3+V1= 6 U2+V3= 3
U2+V1= 3 U3+V2= 2
Taking U3 (any) variable as ‘0’.
We get, V1=6; V2=2; V3=6, U1=-5; U2=-3
𝑈𝑖 𝑉𝑗 𝑉1 𝑉2 𝑉3
𝑈1 1
𝑈2 3 3
𝑈3 6 2
22. Modified Distribution Method ( MODI METHOD )
Rewriting the matrix with U and V values,
3. Find out implicit cost. Implicit cost is summation of dual variables of row and column for each
unoccupied cell. Then, vacant cell evaluation is carried out by taking the difference for each
unoccupied cell.
Evaluation of cell = 𝑐𝑖𝑗 − (𝑈𝑖+𝑉𝑗)
𝑈𝑖 𝑉𝑗 6 2 6
-5 1
-3 3 3
0 6 2
23. Examine the sign of each dij
1. If dij > 0 For i and j then solution is optimal.
2. If dij = 0 For all i and j then solution will remains unaffected but an alternative solution exists.
3. If one or more dij < 0 then the initial solution can be improved by entering unoccupied cells in basis
with the largest negative value of dij
Construct a closed loop For the unoccupied cell with largest negative value of dij . Start the closed loop
with the selection of unoccupied cells & mark (+) sign in the cell and trace a path along the rows (columns)
to unoccupied cell and mark (-) sign and continue to column (row) to an occupied cell with (+) sign and (-)
sign alternatively & back to the selected unoccupied cell.
Select the smallest value among the cells with (-) sign and allocate this value to the selected unoccupied
cell and add it to other occupied cells marked with (+) sign and subtract it from the occupied cells marked
with (-) sign.
Obtain the initial solution and calculate a new total cost.
The procedure terminate when all dij ≥ 0 For all unoccupied cells.
24. Find the initial basic feasible solution of the following transportation problem by
northwest corner method and then optimize the solution using U-V method
(MODI)
Destination
Source
D1 D2 D3 D4 Supply
S1 3 1 7 4 250
S2 2 6 5 9 350
S3 8 3 3 2 400
Demand 200 300 350 150
26. Phase 2 : Application of u-v method to optimize the solution.
v1 = v2 = v3 = v4 =
u1 = 3 1 7 4
u2 = 2 6 5 9
u3 = 8 3 3 2
200 50
100250
250 150
Equation for finding out the values of u & v is, ui + vi = cij (allocated cells)
Then find the Penalties by eq :- pij = ui + vi – cij (unallocated cells)
C13 = 0+0-7 = -7
C14 = 0-1-4 = -5
C21 = 5+3-2 = 6
C24 = 5-1-9 = -5
C31 = 3+3-8 = -2
C32 = 3+1-3 = 1
If all the values are ≤ 0 𝑂𝑝𝑡𝑖𝑚𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑
But here 6 is the maximum positive value at C21, so new basic cell is C21.
27. 3 1 7 4
2 6 5 9
8 3 3 2
200 50
250
250
100
150
Take the smallest (-ve) value and add it to both (+ve) values & subtract it from both (-
ve) values
Now, find out the new values of u & v, by eq ui + vi = cij (allocated cells) for next table
v1 = v2 = v3 = v4 =
u1 = 3 1 7 4
u2 = 2 6 5 9
u3 = 8 3 3 2
200
250
50
250
100
150
28. Then find the Penalties by eq :- pij = ui + vi – cij (unallocated cells)
C11 = 0-3-3 = -6
C13 = 0+0-7 = -7
C14 = 0-1-4 = -5
C24 = 5-1-9 = -5
C31 = 3-3-8 = -8
C32 = 3+1-3 = 1
3 1 7 4
2 6 5 9
8 3 3 2
If all the values are ≤ 0 𝑂𝑝𝑡𝑖𝑚𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑
But here 1 is the maximum positive value at C32, so new basic cell is C32.
200
250
50 100
250 150
29. Take the smallest (-ve) value and add it to both (+ve) values & subtract it from both (-
ve) values
Now, find out the new values of u & v, by eq ui + vi = cij (allocated cells) for next table
v1 = v2 = v3 = v4 =
u1 = 3 1 7 4
u2 = 2 6 5 9
u3 = 8 3 3 2
200
250
50 200
150
150
Then find the Penalties by eq :- pij = ui + vi – cij (unallocated cells)
C11 = 0-2-3 = -5
C13 = 0+1-7 = -6
C14 = 0-0-4 = -4
C22 = 4+1-6 = -1
C24 = 4+0-9 = -5
C31 = 2-2-8 = -8
Calculating the total transportation cost : 1*250 + 2*200 + 5*150 + 3*50 + 3*200 + 2*150 =
Rs.2450
30. Reference:-
Content from Operation Research by Dr.Akshay A. Pujara & Dr. Ravi Kant.
Tables (Self Prepared by group members)