The modified distribution method, is also known as MODI method or (u - v) method provides a minimum cost solution to the transportation problems. MODI method is an improvement over stepping stone method. This model studies the minimization of the cost of transporting a commodity from a number of sources to several destinations. The supply at each source and the demand at each destination are known. The objectives are to develop and review an integral transportation schedule that meets all demands from the inventory at a minimum total transportation cost.

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REVIEW OF MODIFIED DISTRIBUTION METHOD IN CURRENT BUSINESS

Limbore Nilesh Vitthal1

Saste Sachin Chandrakant2

Abstract

The modified distribution method, is also known as MODI method or (u - v) method provides a

minimum cost solution to the transportation problems. MODI method is an improvement over

stepping stone method. This model studies the minimization of the cost of transporting a

commodity from a number of sources to several destinations. The supply at each source and the

demand at each destination are known. The objectives are to develop and review an integral

transportation schedule that meets all demands from the inventory at a minimum total

transportation cost.

1. Introduction

The basic transportation problem was originally developed by F.L Hitchcock (1941) in his study

entitled "The distribution of product from several sources to numerous locations. In 1947 T.C

Koopmans independently published a study on "Optimum utilization of transportation system.

Subsequently the linear programming formulation and the associated systematic procedure for

solution were given by Gorge B. Dantzig(1951).

The Transportation problem is to transport various amounts of a single homogeneous commodity that are initially

stored at various origins, to different destinations in such a way that the total transportation cost is a minimum. It

can also be defined as to ship goods from various origins to various destinations in such a manner that the

transportation cost is a minimum. The availability as well as the requirements is finite. It is assumed that the cost of

shipping is linear. MODI method is an improvement over stepping stone method.

Here researcher study an important class of linear programs called the transportation model. This

model studies the minimization of the cost of transporting a commodity from a number of

sources to several destinations. The supply at each source and the demand at each destination are

known.

2. Purpose of the study

This research work represents transportation modeling approaches and forecasting techniques

addressing the transportation flow of cargo containers with semi-processed goods on the selected

routes from a certain number of suppliers with various production capacities to the certain points

of destination. The aim is to achieve the minimum cost of transportation flow and to forecast the

future for the company's activities. Since the cost minimization directly relates to the company's

profitability of which is representing operation efficiency that can be expressed as a fraction,

respective transportation modeling methods can be solved using modified distribution method.

The models were studied based on a real-life data and as example of transportation. Since the

forecast of future activities can be also related to the company's strategic planning. The

forecasting problem is solved by one of the most common forecasting techniques used in

business life, namely the trend adjusted forecast approach.

The paper is conducted in order to introduce the transportation problem solutions by applying

different methods of the transportation flow of a company, in order to find the points that could

be improved and minimize transportation costs of the company. The paper was also conducted in

order to show how basic figures of transportation flow can be transferred into a transportation

matrix which is the basis of any transportation problem. Understanding of transportation problem

methods can help to find an optimum solution for the transportation flow. Based on calculations

and results of different methods and approaches to the same transportation problem, using

different cases when demand was and wasn't equal to supply were also investigated. The

researcher is also looking into the forecasting problem to show how forecasting approaches can

help to predict transportation activities of the company in the future. The paper studied, with the

help of transportation modeling methods such as Northwest-corner, Lowest-Cost and Vogel's

Approximation, using real figures and data, destinations to the terminal, terminal expenses and

freight cost of transportation from the terminal of the final destination. The study investigates

possible ways of minimizing the cost of transportation by using handmade calculation. The

objective is to review an integral transportation schedule that meets all demands from the

inventory at a minimum total transportation cost.

3. Basic Assumption behind Transportation Problem

Let us consider a T.P involving m-origins and n-destinations. Since the sum of origin capacities

equals the sum of destination requirements, a feasible solution always exists. Any feasible

solution satisfying m+n-1 of the m + n constraints is a redundant one and hence can be deleted.

This also means that a feasible solution to a T.P can have at the most only m + n – 1 strictly

positive component, otherwise the solution will degenerate.

It is always possible to assign an initial feasible solution to a T. P. in such a manner that the rim

requirements are satisfied. This can be achieved either by inspection or by following some

simple rules. We begin by imagining that the transportation table is blank i.e. initially all Xij = 0.

The simplest procedures for initial allocation discussed in the following section.

4. Modified Distribution Method (MODI) or (u - v) method basic steps for solving

The modified distribution method, also known as MODI method or (u - v) method provides a

minimum cost solution to the transportation problem. In the stepping stone method, we have to

draw as many closed paths as equal to the unoccupied cells for their evaluation. To the contrary,

in MODI method, only closed path for the unoccupied cell with highest opportunity cost is

drawn.

1. Determine an initial basic feasible solution using any one of the three methods given below:

•North West Corner Rule

•Matrix Minimum Method

•Vogel Approximation Method

2. Determine the values of dual variables, ui and vj, using cij= ui + vj

3. Compute the opportunity cost using Δij= cij – ( ui + vj ).

4. Check the sign of each opportunity cost. If the opportunity costs of all the unoccupied cells are

either positive or zero, the given solution is the optimal solution. On the other hand, if one or

more unoccupied cell has negative opportunity cost, the given solution is not an optimal solution

and further savings in transportation cost are possible.

5. Select the unoccupied cell with the smallest negative opportunity cost as the cell to be

included in the next solution.

6. Draw a closed path or loop for the unoccupied cell selected in the previous step. Please note

that the right angle turn in this path is permitted only at occupied cells and at the original

unoccupied cell.

7. Assign alternate plus and minus signs at the unoccupied cells on the corner points of the

closed path with a plus sign at the cell being evaluated.

8. Determine the maximum number of units that should be shipped to this unoccupied cell. The

smallest value with a negative position on the closed path indicates the number of units that can

be shipped to the entering cell. Now, add this quantity to all the cells on the corner points of the

closed path marked with plus signs, and subtract it from those cells marked with minus signs. In

this way, an unoccupied cell becomes an occupied cell.

9. Repeat the whole procedure until an optimal solution is obtained.

5. Example

Consider the transportation problem presented in the following table.

Distribution centre

D1 D2 D3 D4 Supply

Plant P1 19 30 50 12 7

P2 70 30 40 60 10

P3 40 10 60 20 18

Requirement 58715

Determine the optimal solution of the above problem.

Solution-

An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table

1.

Table 1

Distribution centre

D1 D2 D3 D4 Supply

Plant

P1 19 30 50 7

P2 30 60 10

P3 60 18

Requirement 5 8 7 15 35

Initial basic feasible solution

Total Cost= 12 X 7 + 70 X 3 + 40 X7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.

Calculating ui and vj using cij= ui + vj

Substituting u1 = 0, we get

u1 + v4 = c14 0 + v4 = 12 or v4 = 12

u3 + v4 = c34 u3 + 12 = 20 or u3 = 8

u3 + v2 = c32 8 + v2 = 10 or v 2 = 2

u3 + v1 = c31 8 + v1 = 40 or v1 = 32

u2 + v1 = c21 u2 + 32 = 70 or u2 = 38

u2 + v3 = c23 38 + v3 = 40 or v3 = 2

Table 2

Distribution centre

D1 D2 D3 D4 Supply ui

Plant

P1 19 30 50 7 0

P2 30 60 10 38

P3 60 18 8

Requirement 5 8 7 15

vj 32 2 2 12

Calculating opportunity cost using Δij= cij – ( ui + vj ).

Unoccupied cells Opportunity cost

(P1 , D1) c11 – ( u1 + v1 ) = 19 – (0 + 32) = –13

(P1 , D2) c12 – ( u1 + v2 ) = 30 – (0 + 2) = 28

(P1 , D3) c13 – ( u1 + v3 ) = 50 – (0 + 2) = 48

(P2 , D2) c22 – ( u2 + v2 ) = 30 – (38 + 2) = –10

(P2 , D4) c14 – ( u2 + v4 ) = 60 – (38 + 12) = 10

(P3 , D3) c33 – ( u3 + v3 ) = 60 – (8 + 2) = 50

Table 3

Distribution centre

D1 D2 D3 D4 Supply ui

Plant

P1 7 0

P2 10 38

P3 18 8

Requirement 5 8 7 15

vj 32 2 2 12

Now choose the smallest (most) negative value from opportunity cost (i.e., –13) and draw a

closed path from P1D1. The following table shows the closed path.

Table 4

Choose the smallest value with a negative position on the closed path(i.e., 2), it indicates the

number of units that can be shipped to the entering cell. Now add this quantity to all the cells on

the corner points of the closed path marked with plus signs and subtract it from those cells

marked with minus signs. In this way, an unoccupied cell becomes an occupied cell.

Now again calculate the values for u i & vj and opportunity cost. The resulting matrix is shown

below.

Table 5

Distribution centre

D1 D2 D3 D4 Supply ui

Plant

P1 7 0

P2 10 51

P3 18 8

Requirement 5 8 7 15

vj 19 2 –11 12

Choose the smallest (most) negative value from opportunity cost (i.e., –23). Now draw a closed

path from P2D2 .

Table 6

Now again calculate the values for ui & vj and opportunity cost

Table 7

Distribution centre

D1 D2 D3 D4 Supply ui

Plant

P1 7 0

P2 10 28

P3 18 8

Requirement 5 8 7 15

vj 19 2 12 12

Since all the current opportunity costs are non–negative, this is the optimal solution. The

minimum transportation cost is: 19 X 5 + 12 X 2 + 30 X 3 + 40 X 7 + 10 X 5 + 20 X 13 = Rs.

799

6. Different Cases in Transportation Problems

6.1 Degeneracy

If the basic feasible solution of a transportation problem with m origins and n destinations has

fewer than m + n – 1 positive xij (occupied cells), the problem is said to be a degenerate

transportation problem.

Degeneracy can occur at two stages:

1. At the initial solution

2. During the testing of the optimal solution

To resolve degeneracy, we make use of an artificial quantity (ε). The quantity ε is assigned to

that unoccupied cell, which has the minimum transportation cost. For calculation purposes, the

value of ε is assumed to be zero. The quantity ε is so small that it does not affect the supply and

demand constraints.

6.2 Maximization in a Transportation Problem

There are certain types of transportation problems where the objective function is to be

maximized instead of being minimized. These problems can be solved by converting the

maximization problem into a minimization problem. The conversion can be done by subtracting

each of the profit elements associated with the transportation routes from the largest profit

element. The resulting values so obtained represents opportunity cost because they corresponds

to the difference in profit earned by that routs and the largest profit that could be earned by any

of the routs.

6.3 Prohibited Routes

Sometimes there may be situations, where it is not possible to use certain routes in a

transportation problem. For example, road construction, bad road conditions, strike, unexpected

floods, local traffic rules, weight or size condition etc. We can handle such type of problems in

different ways:

•A very large cost represented by M or ∞ is assigned to each of such routes, which are not

available.

•To block the allocation to a cell with a prohibited route, we can cross out that cell.

The problem can then be solved in its usual way.

7. Conclusion

Understanding of transportation problem methods are help to find an optimum solution for the

transportation flow. Based on calculations and results of different methods and approaches to the

same transportation problem, using different cases when demand was and wasn't equal to

supply. The researcher is also looking into the forecasting problem to how forecasting

approaches are help to predict transportation activities of the company in the future.

8. Reference

1. V.K. Kapoor, Operations Research, Sultan Chand & Sons, Fifth Revised Edition,(1994)

9. Bibliography

1. Panneerselvam, Operations Research, PHI Learning Pvt. Ltd., Second Edition,(2006)

2. Hamdy A. Taha, Operations Research-An Introduction, Prentice Hall, Fifth Revised

Edition, (1997)

3. Ronald L. Rardin, Optimization in operations research, Prentice Hall, (1998)

4. Render, Quantitative Analysis for Management, Pearson Education India, (2008)

5. Jay H. Heizer, Barry Render, Principles of operations management, Prentice Hall, (2001)

10. Webliography

1. http://mbaassignment.blogspot.in/2010/05/transportation-problem-and-modi-method.html

2. http://in.answers.yahoo.com/question/index?qid=20110220220648AAyzVBM

3. http://www.universalteacherpublications.com/univ/ebooks/or/Ch5/stepst.htm

4. https://publications.theseus.fi/bitstream/handle/10024/13384/Farnalskiy_Denys.pdf

This paper introduces an environment-driven, artificial intelligence model for sustainable policymaking in the European countries, focusing on Ukraine. It develops regional clusters using artificial neural networking and then it dynamically optimises budgeting allocations. It is a hybrid, environment-driven model which: i) clusters regionalised-data, using Kohonen's self-organising map; and ii) optimises budget allocations, using simplex modified distribution method (U–V-MODI). Model benefits focus on: i) regional public policies; ii) environmental development; and iii) core-periphery balanced growth. Results reveal an innovative plan that: i) activates participation of the environmental stakeholders in public policymaking ii) reforms regions based on sustainability criteria set; and iii) optimises regional funding.

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