# Operation Research

Edited by Jen Moreau

## OPERATION RESEARCH

Operations research can be explained to be an application of quantitative techniques in solving a variety of problems in many areas of life to better enhance decision making. Problem-solving in operation research is mostly towards optimizing performance, improvement or quality control system and usually involves mathematical and statistical models. Models formulated in operation Research unusually assign values to various variables and also shows the relationships between them. It then involves developing strategies of control by measuring, comparing and predicting problem situation by using the scientific model of the situation. The British call the technique operational research while the Americans call it management science and Europe and other English speaking countries refer to it as operation research[OR].

Hence the flexibility and dynamic nature of OR has made scientist, engineers, and managers consider it as a proper tool for decision making and productive management.

## HOW TO APPLY OPERATION RESEARCH

The analysis and study of OR are scientific in nature and hence the application can best be carried out provided certain steps are taken.

- 1The facts or principles that govern past operation could be adequately studied and understood.Advertisement
- 2Try to propose theories to explain or demonstrate these facts that govern the past operations.
- 3These theories are usually mathematically written in form of model formulas containing variables.
- 4Those facts or theories are then used to predict characteristic or future operations.

Below are some of the topics that are being studied in OR.

**LINEAR PROGRAMMING:** this has to do with how using limited available resources in maximizing profit and minimizing cost.

**TRANSPORTATION OR TRANSSHIPMENT PROBLEM:** this area of OR deals with a method of transporting goods and services from the source at a minimal cost but increase efficiency.

**ASSIGNMENT MODEL:** this area of LP deals with methods in assigning jobs to available people. It aims at helping to know how to assign N -" job in N -" sources to N -" people at N -" different location.

**QUEUE THEORY MODEL:** This help to tell the technique needed to serve customers on a queue.

**NETWORK ANALYSIS:** it helps to know how to carry out job schedule especially when large projects are involved eg road constructions, construction of shopping mall, building bridges provided a town planning exist. He does use this model of OR to schedule this jobs in order to meet up with the completion on or before the stipulated time and with the least cost.

**REPLACEMENT ANALYSIS:** this area of OR guides the management of an organization on when and how to carry out replacement exercise of equipment's at the minimum cost in order to improve efficiency in production.

### WHAT IS A MODEL

In OR the best way to solve complex problems of real life is creating a model for it, i.e. using simpler, less expensive, less cumbersome objects to represent the complexity of the problems. In other words, models show a clearer topic of the problem at hand. Then a model can be said to be a simplified representation of a complex reality. The main objective of any model is to use a simple inexpensive object to represent complex or uncertain situation an example of a model is mathematical or statistical models whereby symbols and equations are used to represent the complexity or uncertainty of real life problems.

Let's consider the expression bellow.

I = 12m if I represent annual income and m represent monthly income that means the annual income is 12 multiply by monthly income (12 * monthly income)

Mathematical models deal with mathematical language. Thus in building a model certain factors has to be considered. These are:

- 1The prevailing circumstance
- 2The expected result.

And in the optimization problem, one seeks or desire to either maximize or minimize a particular quantity called the objective function "f". this objective function depends on the finite numbers of input variables, these variables may be independent on another or be related to each other through one or more constraints. Eg

min = 2X1 -" X2 - - - - - - - - - - - => (1)

Subject to:-

X1 -" X2 ≥ 2

X2 ≥ 4

X1, X2 ≥ 0

From above Equation (1) is referred to as the objective function of the model. i.e to minimize possible cost whereas question (2) is referred to as the constraint of the model. We say X1 ≥ X2 by 2 and also X2 must not be less than 4.

==LINEAR PROGRAMMING==

LP is the method that helps in solving complex problems and also helps in decision making.it was mathematical tools thereby considering constraints application especially during the decision-making process. It is also a mathematical technique concerned with the allocation of scarce resources. it is a procedure that can be used to optimize an objective function bearing in mind certain constraint. These objectives may be to maximize profit or minimize cost. LP models are a simple mathematical representation of constraint optimization problems. I.e. the use of equations, is highly put in place for a problem to be solved using LP, it must be expressible using mathematical equations using equality or inequality signs.

For any problem to be solved using LP certain conditions must be fulfilled and these conditions are categorized in 2 section. They are:

- 1The component of the LP must be clearly stated
- 2Some assumptions of Lp must be undertaken.

The component of LP is further divided in four which must be clearly and adequately understood. these component are listed bellow.

**Objective function:**this is the objective or purpose of the problem eg minimization or maximization**Decision variable:**these are the unknown variables to be computed for in the models.**Structural constraints:**in problem-solving with limited resources using LP technique, the structural constraint are set of linear equation indicating the way the problem is to be solved eg ≥ symbol for minimizing problems ≤ for maximizing problems.**Parameters:**this is decision variables and numerical values used in the model.

The assumption made in an LP model

**Linearity:**this tells that all decision variables are raised to the power of unity.**Divisibility:**decision variables are also allowed to have a real number such as 2.2X1, 3.2X2 etc.**Non-negativity****Proportionality****Additivity****Independence of variables**

**Steps to be taken in formulating an LP model.**

- 1Identification of the decision variables because they go along away to help us to decipher the nature of the problem.
- 2the purpose or plan.State the objective function, i.e.
- 3Identification of constraints and then representing them using mathematical statements of equality or inequality signs.

We can now write a generalized LP model for a maximum problem.

Pmax = A1 X1 + A2X2 + A3X3 + --------- AnXn

Subject to

b11 X1 + b12X2 + b13X3 + --------- b1nXn ≤ d1

b21 X1 + b22X2 + b23X3 + --------- b2nXn ≤ d2

b31 X1 + A32X2 + b33X3 + --------- b3nXn≤ d3

bm1 X1 + bm2X2 + bm3X3 + --------- bmnXn≤ dm X1, X2, Xn,≥0

Matrix Representation of LP model

b11 b12 b13--------b1n Xn d1

b21 b22 b23--------b2n Xn d2 =

b31 b32 b33--------b3n Xn d3

bm1 bm2 bm3--------bmn Xn dm

bij X1 + bijX2 + bijX3 + --------- bijXn≤ di

Example of Linear programming
**EXAMPLE 1;**

*A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.*

*If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?*

*The question asks for the optimal number of calculators, so my variables will stand for that:*
*x: number of scientific calculators produced*
*y: number of graphing calculators produced*

*Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > -"x + 200. The profit relation will be my optimization equation: P = -"2x + 5y. So the entire system is:*

*P = -"2x + 5y, subject to: *

*100 < x < 200 *

*80 < y < 170*

*y > �"x + 200*

*The feasibility region graphs as:*

*When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators".*

**Duality LP problems**

Minimization problems involving duality in LP duality occurs in LP problems in such a way the every LP model can be formulated and analyzed using the concept of primal and dual. When considering any LP problem, they're usually two associated factor, these are

- 1The primal
- 2The dual.

The primal is the original LP problem while the dual is the formulated problem that is derived from the primal. Now, if the original problem is a profit maximization problem, the other will be a cost minimization problem.

Below are some occurrence in primal to dual conversion

- 1The number of decision variable in the primal objective function constraints is equal to the number of constraints in the derived dual LP.
- 2The number of constraint in the primal LP is equal to the number decision variables in dual LP objection.
- 3The coefficient of the decision variables of the objective function of the primal LP becomes the limit of the constraints in the dual LP.
- 4The limit of the constraints in the primal LP becomes the coefficient of the decision variable in the dual objective function.

## TRANSPORTATION MODEL

We're now going into another special programming problem using mathematical techniques. Unlike the ones treated earlier, it deals with the actual minimization overhead cost of production of items or the maximization of total profit.

In this LP (model) problem we're studying the techniques which involve the possibility of transporting or transshipping goods and services either by land, air or sea from source or origin to destination or site.

A real life situation to illustrate this idea is the transshipping of petroleum products from source (refinery) to destination (filling station or gas station)

### Assumption made in transport model.

- 1This is the principle of homogeneity.The materials to be transported must be same regardless of their specific source or destination.
- 2The cost of transporting each material or unit is assumed to be equal
- 3The route for transporting goods from a particular destination is unique.

Examples 2;

**The Least-Cost Method**

This method usually provides a better initial basic feasible solution than the North-West

Corner method since it takes into account the cost variables in the problem.

- 1If there is a tie then choose arbitrarily.Assign as much as possible to the cell with the smallest unit cost in the entire tableau.
- 2If a row and column are both satisfied then cross out only one of them.Cross out the row or column which has satisfied supply or demand.
- 3Adjust the supply and demand for those rows and columns which are not crossed out.
- 4When exactly one row or column is left, all the remaining variables are basic and are assigned the only feasible allocation.Advertisement

For the above example:

_ Cells (1; 2) and (3; 1) both have zero cost so we arbitrarily choose the first and assign x12 = 15. Cross out column 2. The amount left in row 1 is 5.

_ x31 = 10. Cross out column 1. The amount left in row 3 is 5.

_ x23 = 15. Cross out column 3. The amount left in row 2 is 10.

_ Only column 4 is now left and so all the variables in this column will be basic.

The only feasible allocation is x14 = 5, x24 = 10 and x34 = 5.

This provides the initial basic feasible solution x12 = 15, x31 = 10, x23 = 15, x14 = 5, x24 = 10, x34 = 5. All the other variables are non-basic and are therefore equal to zero.

Again, we have 6 basic variables as required.

## Referencing this Article

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