Transportation Model and it's variants

Transportation Model and it's variants 

The transportation model deals with a special class of linear programming problem in which the objective is to transport a homogeneous commodity from various origins or factories to different destinations or markets at a total minimum cost.

To understand the problem more clearly, let us take an example and discuss the rationale of transportation problem. Three factories A, B and C manufactures sugar and are located in different regions. Factory A manufactures, b1 tons of sugar per year and B manufactures b2 tons of sugar per year and C manufactures b3 tons of sugar. The sugar is required by four markets W, X, Y and Z. The requirement of the four markets is as follows: Demand for sugar in Markets W, X, Yand Z is d1, d2, d3 and d4 tons respectively. The transportation cost of one ton of sugar from each factory to market is given in the matrix below. The objective is to transport sugar from factories to the markets at a minimum total transportation cost.

let us see how a general model for transportation problem appears. The general problem will have 'm' rows and 'n' columns i.e., m × n matrix.

Methods to find basic feasible solution of Transportation models 

1) North West Corner Method

2) Row Minima Method

3) Column Minima Method

4) Least Cost Method / Matrix Minima Method

5) Vogels Approximation Method / Opportunity Cost Method

MAXIMISATION PROBLEM: (PRODUCTION AND TRANSPORTATION

SCHEDULE FOR MAXIMIZATION)

This type of problems will arise when a company having many units manufacturing the same product and wants to satisfy the needs of various market centers. The production manager has to work out for transport of goods to various market centers to cater the needs. Depending on the production schedules and transportation costs, he can arrange for transport of goods from manufacturing units to the market centers, so that his costs will be kept at minimum. At the same time, this problem also helps him to prepare schedules to aim at maximizing his returns.

DEGENERACY IN TRANSPORTATION PROBLEM

Earlier, it is mentioned that the basic feasible solution of a transportation problem must have (m + n – 1) basis variables or allocations. This means to say that the number of occupied cells or loaded cells in a given transportation problem is 1 less than the sum of number of rows and columns in the transportation matrix. Whenever the number of occupied cells is less than (m + n – 1) , the transportation problem is said to be degenerate.

Degeneracy in transportation problem can develop in two ways. First, the problem becomes degenerate when the initial programme is designed by northwest corner or inspection or VAM, i.e. at the stage of initial allocation only.

To solve degeneracy at this stage, we can allocate extremely small amount of goods (very close to zero) to one or more of the empty cells depending on the shortage, so that the total occupied cells becomes m + n – 1. The cell to which small element (load) is allocated is considered to be an occupied cell. In transportation problems, Greek letter ‘∈’ represents the small amount. One must be careful enough to see that the smallest element epsilon is added to such an empty cell, which will enable us to write row number ‘ui ’ and column number ‘vj ’ without any difficulty while giving optimality test to the basic feasible solution by MODI method. That is care must be taken to see that the epsilon is added to such a cell, which will not make a closed loop, when we move horizontally and vertically from loaded cell to loaded cell.

(Note: Epsilon is so small so that if it is added or subtracted from any number, it does not change the numerical value of the number for which it added or from which it is subtracted.). Secondly, the transportation problem may become degenerate during the solution stages. This happens when the inclusion of a most favorable empty cell i.e. cell having highest opportunity cost results in simultaneous vacating of two or more of the currently occupied cells. Here also, to solve degeneracy, add epsilon to one or more of the empty cells to make the number of occupied cells equals to (m + n – 1).

TRANSSHIPMENT PROBLEM

We may come across a certain situation, that a company (or companies) may be producing the product to their capacity, but the demand arises to these products during certain period in the year or the demand may reach the peak point in a certain period of the year. This is particularly true that products like Cool drinks, Textbooks, Notebooks and Crackers, etc. The normal demand for such products will exist, throughout the year, but the demand may reach peak points during certain months in the year. It may not possible for all the companies put together to satisfy the demand during peak months. It is not possible to produce beyond the capacity of the plant. Hence many companies have their regular production throughout the year, and after satisfying the existing demand, they stock the excess production

in a warehouse and satisfy the peak demand during the peak period by releasing the stock from the warehouse. This is quite common in the business world. Only thing that we have to observe the inventory carrying charges of the goods for the months for which it is stocked is to be charged to the consumer. Take for example crackers; though their production cost is very much less, they are sold at very high prices, because of inventory carrying charges. When a company stocks its goods in warehouse and then sends the goods from warehouse to the market, the problem is known as Transshipment problem.

REDUNDANCY IN TRANSPORTATION PROBLEMS

Some times, it may very rarely happen or while writing the alternate solution it may happen or during modifying the basic feasible solution it may happen that the number of occupied cells of basic feasible solution or some times the optimal solution may be greater than m + n – 1. This is called redundancy in transportation problem. This type of situation is very helpful to the manager who is looking about shipping of available loads to various destinations. This is as good as having more number of independent simultaneous equations than the number of unknowns. It may fail to give unique values of unknowns as far as mathematical principles are concerned. But for a transportation manager, it enables him to plan for more than one orthogonal path for an or several cells to evaluate penalty costs, which obviously will be different for different paths.

(for details refer the book titled " Operations Research by P R Murthy)