study pack on operations research

1

STUDY PACK

ON

OPERATIONS RESEARCH

PROFESSIONAL EXAMINATION I

2

STUDY PACK ON

OPERATIONS RESEARCH

PROFESSIONAL EXAMINATION I

@CIPM 2019

FOURTH EDITION

CHARTERED INSTITUTE OF PERSONNEL

MANAGEMENT OF NIGERIA

CIPM House, 1 CIPM Avenue, Off Obafemi Awolowo Way,

Opposite Lagos State Secretariat, Alausa, Ikeja, Lagos.

P.O.Box 5412, Marina, Lagos.Tel: 08105588421

E-mail: [email protected]

Website: www.cipmnigeria.org

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All rights reserved, no part of this publication may be reproduced, stored in

retrieval system, or transmitted in any form or by any means, electronically,

mechanical, photocopying or otherwise without permission of CIPM

NIGERIA.

mailto:[email protected]

http://www.cipmnigeria.org/

http://www.twitter.com/CIPMNIGERIA

http://www.youtube.com/cipmnigeria

3

FOREWORD

This third edition of our study pack has been made available for the

use of our professional students to assist them in effectively

accomplishing their HR professional goal as dictated by the Institute

from time to time.

The text is meant not only for Chartered Institute of Personnel

Management of Nigeria (CIPM) students, but also for researchers,

HR practitioners and organisations embarking on the promotion of

human capital development in its entirety. It has therefore been

written not only in a manner that users can pass CIPM professional

examinations without tears, but also to provide HR professional

practitioners further education, learning and development references.

Each chapter in the text has been logically arranged to sufficiently

cover all the various sections of this subject in the CIPM

examination syllabus in order to enhance systematic learning and

understanding of the students. The document, a product of in-depth

study and research is both practical and original. We have ensured

that topics and sub-topics are based on the syllabus and on

contemporary HR best practices.

Although concerted effort has been made to ensure that the text is up

to date in matters relating to theories and practice of contemporary

issues in HR, we still advise and encourage students to complement

the study text with other relevant literature materials because of the

elastic scope and dynamics of the HR profession.

Thank you and have a productive preparation as you navigate

through the process of becoming a professional in Human Resources

Management

Ajibola Ponnle.

REGISTRAR/CEO

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ACKNOWLEDGEMENT On behalf of the President/Chairman of the Governing Council and

the entire membership of the Chartered Institute of Personnel

Management of Nigeria (CIPM), we acknowledge the intellectual

prowess of Dr. Bolanle A. Oseni and Mr. Musa Sanni in writing this

well researched text for Operations Research. The meticulous work

of our reviewers, Prof. Nnamdi Mojekwu and Dr. Joshua Ajilore has

not gone unnoticed and is hereby acknowledged for the thorough

review of this publication.

We also commend and appreciate the efforts of members of the

Education Committee of the Institute for their unflinching support.

Finally, we appreciate the contributions of the National Secretariat

staff competently led by the Registrar/CEO, Mrs. Ajibola Ponnle and

the project team, Dr. Charles Ugwu, Mrs. Nkiru Ikwuegbuenyi, Miss

Charity Nwaigbo, Mrs. Livina Onukuba and Miss Opeoluwa Ojo.

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Table of Contents

Page

CHAPTER ONE:

MEANING OF OPERATIONS RESEARCH 1

1.0 Learning Objectives 1

1.1 Introduction 1

1.2 Origin and Development of Operation Research 1

1.3 Definition of Operation Research 3

1.4 Procedure of Operations Research 5

1.5 Types of Model Formulation 5

1.6 Principle and Steps involved in Modelling 7

1.7 Advantages of Model in Operations Research 9

1.8 Disadvantages of Model in Operations Research 9

1.9 Limitations of Operations Research 9

1.10 Summary 10

1.11 Review Questions 10

References and Further Readings 10

CHAPTER TWO:

ELEMENTS OF DECISION ANALYSIS 11


2.1 Introduction 11

2.2 Concepts of Decisions Process, Rationality in

Decision Making Process and Decision Analysis 11

2.3 Phases in Decision Making Process 15

2.4 Types of Decision Situation 16

2.5 Decision Making Under Risk 24

2.6 Decision Tree Approach 31

2.7 Summary 35

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CHAPTER THREE:

LINEAR PROGRAMMING MODEL 38


3.1 Introduction 38

3.2 Definition of Linear Programming 38

3.3 Characteristics of Linear Programming 40

3.4 Basic Assumptions of Linear Programming Problems 40

3.5 Application Areas of Linear Programming Problems 41

3.6 Some Important Concepts in Linear Programming 42

3.7 Linear Programming Formulation 44

3.8 Solution of Linear Programming 51

3.9 Primal and Duality in Linear Programming 62

3.10 Summary 68



CHAPTER FOUR:

TRANSPORTATION MODEL 70


4.1 Introduction 70

4.2 Methods of Solving Transportation Problem 72

4.3 Optimality Test 76

4.4 Formulation of Assignment Problem 80

4.5 Solution to Assignment Model ( Hungarian method) 81

4.6 The Branch and Bound Method 84

4.7 Summary 85


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CHAPTER FIVE:

PROJECT PLANNING AND SCHEDULING 88


5.1 Introduction 88

5.2 Some Basic Terms in Network 89

5.3 Methods of Project Planning and Scheduling Analysis 89

5.4 Project Evaluation Review Technique (PERT) 96

5.5 Summary 104



CHAPTER SIX:

INVENTORY CONTROL AND MANAGEMENT 108


6.1 Introduction 108

6.2 Inventory Control Terminology 108

6.3 Economic Order Quantity 109

6.4 EOQ Where Stockouts are Permitted 111

6.5 Summary 115



8

CHAPTER ONE

INTRODUCTION TO OPERATIONS RESEARCH

1.0 Learning Objectives

At the end of this chapter, readers should be able to:

• Explain the meaning of Operations Research

• Explain some concepts of Operations Research

• List and explain the principles and steps involved in

modeling

• Discuss the advantages and disadvantages of Operations

Research

1.1 Introduction

This chapter will introduce you to the basic terminologies in

operations research, including mathematical modelling, feasible

solutions, optimization and iterative computations. You will learn

that defining the problem correctly is the most important (and most

difficult) phase of practicing Operations Research (OR). The lecture

also emphasizes that while mathematical modelling is the

cornerstone of OR, unquantifiable factors (such as human behaviour)

must be accounted for in the final decision.

1.2 Origin and Development of Operation Research

During World War II, the military management in England looked at

tactical problems associated with air and land defence of the country.

Their objectives were to determine the most effective utilization of

limited resources, and how to use the newly invented radar. It was to

determine the effectiveness of the new type of bombers that had been

introduced. The establishment of this scientist team marked the first

formal OR activity. The name Operation Research (OR) came

apparently because the team was dealing with research on (military)

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operations. Since its birth, this new decision making field has been

characterised by the use of scientific knowledge through

interdisciplinary team efforts for the purpose of determining the best

utilization of limited resources.

The encouraging result achieved by the British operations research

team motivated the United State military management to start similar

activities. Areas covered by the U.S team included the study of

complex logistics problems; the invention of the new flight patterns,

the practice of sea mining and the effective utilization of electronic

equipment.

The development of operations research started at the pre-World War

era. During the pre-World War era, Taylor (1885) emphasized the

application of scientific analysis to methods of production. Taylor

(1885) conducted experiments in connection with a simple shovel.

His aim was to find the weight load of ore moved with shovel which

would result in a maximum of ore moved without fatigue. After

many experiments with varying weight, he obtained the optimum

weight load, which though much lighter than that commonly used,

provided maximum movement of ore in a day.

A Danish mathematician Erlang (1917), published his work on the

problem of congestion of telephone traffic. A few years after the

appearance of his work, the British Post Office accepted his work as

the basis for calculating circuit facilities. The formulae developed by

Erlang on waiting time are of fundamental importance to the theory

of telephone traffic. The work of Levinson in the 1930s was the first

industrial revolution that contributed mainly towards the

development of Operations research.

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In the era of post War II, the success of military teams attracted the

attention of industrial managers who were seeking solution to their

problems. Industrial Operations Research in UK and USA developed

along different lines. In the U.K., the potential of the field of

Operations research was further increased due to the nationalization

of a few key industries. Thus, operation research spread from

military to government, industrial, social and economic planning.

The progress of industrial operations research in USA was due to the

advent of second industrial revolution which resulted in the

replacement of man by machine as a source of control. This

revolution began around 1940s. Then in 1950, Operations Research

was introduced as a subject of academic study in American

Universities. Since then, Operations Research has been gaining ever

increasing importance for the students of Mathematics, Statistics,

Economics, Management and Engineering sciences. The Operations

Research Society of Nigeria was also formed in 1950 and today

almost all countries of the world now have one society or the other to

help government and private institutions in taking valuable decisions.

1.3 Definition (Operations Research)

Operations research is the attack of modern science on complex

problems arising in the direction and management of large systems

of men, machines, materials and money, industry, business,

government and defence. The distinctive approach is to develop a

scientific model of the system, incorporating measurements of

factors such as change and risk, with which to predict and compare

the alternative decisions, strategies or controls in order to help the

management determine its policy and action scientifically (British

Standard definition).

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Several other definitions exist for Operations Research, but the one

favoured by the author is given below.

Operation Research is the application of scientific methods to

problems arising from operations involving integrated systems of

men, machines, and materials. It normally utilizes the knowledge and

skill of an interdisciplinary research team to provide the managers of

such systems with optimum operations solution. From the above

definition, it is clear that OR has some essential features which

include:

i. Application of a model based scientific approach: The basis of

the Operations research is to construct models of problems in an

objective, factual manner and experimenting with these models

to show the results of various possible causes of action.

ii. System approach to organizations: An activity by any part of an

organization has some effect on the activity of every other part.

Thus, the optimum operation of some other part. Therefore, to

evaluate any decision, it will be pertinent to identify all the

possible interactions and determine their impact on the

organization as a whole.

iii. The recognition of risk and uncertainty: Operation Research

techniques do not remove the risks and uncertainties on their

own, but it can be used to highlight the effect of the risks and

uncertainties on the organization’s operations.

iv. Assistance to management decision making and control. The

role of an OR expert to his organisation is the provision of

information to assist the planners and decision makers. It should

be noted that the skill, experiences and sense of judgment of

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managers cannot be replaced by formal decision making

techniques.

1.4 Procedure for model building in Operations Research

An Operation Research (OR) study consists in building a model of

physical situation. An OR model is defined as an idealized

(simplified) representation of real-life system. Since OR deals with

building model of the physical situations, the managers makes use of

scientific methods in model building. In building the model the

following issues are firstly recognized and specifically defined.

i. Recognize that a problem exists and defining the problem.

ii. Identification of parameter, such as the variables that are under

the control of the decision maker. Here, the objectives of the

goal are defined e.g to maximize profit to minimize cost, to

maintain stable employment etc. The practical constraint to those

problems must be recognized e.g Budget, Labour, Requirement,

Machine Capacity, Time factor etc.

iii. Examinations of all alternative solutions to the problems, there

may be many solutions that may satisfy the stated objective.

iv. The solutions are put into practice and the “Optimal” solution is

chosen.

1.5 Types of Model Formulation

(i) Mathematical models: Mathematical models are symbolic

models but instead of words, equations or symbols are used to

express a simplified version of a complex problem. In mathematical

model, data can be manipulated by person in such a way that if

another person were to manipulate it, the same unique result would

be obtained. It should be noted that most of the models in operations

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research and decision analysis are mathematical models which are

approximate representation of reality and are used for optimization

purpose to get the best decision.

(ii) Descriptive models: The model explains the various operations

in non-mathematical language. It defines the functional relationships

and interactions between various operations.

(iii) Predictive models: The model explains or predicts the

behaviour of the system.

(iv) Prescriptive models: The model develops decision rules or

criteria for optimal solutions. The model is applicable to repetitive

problems, the solution process of which can be programmed without

managerial involvement.

(v) Physical models: The physical models have properties that

resemble the system they represent but differ in size. In this type of

model, properties of the real system are represented by the properties

themselves, frequently with a change of scale. They are either

prototypes of the real objects or have characteristics that reflect the

function of the real objects. The models can be divided into two

categories. These are the iconic models and analogue models. The

iconic models look exactly like the real objects but could be scaled

downward or upward or could employ a change in materials of the

real object. On the other hand, the analogue models may or may not

look exactly like the real objects. They explain specific few

characteristics of an idea and ignore other details in the object.

(vi) Deterministic models: This is a model where all the variables in

the model are completely defined and the outcomes are certain.

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(vii) Probabilistic models: In this model, the input/output variables

take the form of probability distributions. They represent the

likelihood of occurrence of an event.

(viii) Static models: They are one-time decision models. In these

models cause and effect occur almost simultaneously.

(ix) Dynamic models: These are models in which time often plays

an important role. The models are used for optimization of

multistage decision problems which require a series of decisions with

the outcome of each depending upon the results of the previous

decisions in the series.

1.6 Principle and Steps involved in Modelling

The following are the steps involved in modelling.

1. Definition of the problem

A description of the objective of the study must reflect on accurate

representation of the overall interest of the system. In similar

manners a study that does not account for all the decision alternatives

and limitation of the system is liable to held an inaccurate-solution.

2. Construction of the model

Once you have defined the problem, and then formulate a model. A

model is an abstract representation of the problem situation. The

main components of a mathematical model are as follows.

a. Decision variables and parameter.

The decision variables are the unknown to be determined from the

solution of the model.

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The parameter represents the controlled variables of the system

which may be deterministic or probabilistic. The decision variables

are under the control of the management.

b. Constraint or restrictions.

This refers to physical limitation of the system. These are practical

constraints that inhibit the decision maker in achieving his goals.

c. Objective function: This acts as an indicator for the achievement

of the optimum solution. This is the goal of the management. It must

be clearly defined as part of the model.

3.Solution of the problem

This aspect deals with solving the model. In mathematical models,

this is achieved by using well-defined optimization techniques and

the model is said to yield an “optimum” solution. In addition, to the

(optimum) solution of the model, one must also secure whenever

possible, additional information concerning the behaviour of the

solution due to changes in the system’s parameters. This is usually

referred to as “sensitivity analysis” which is needed when parameters

of the system cannot be estimated accurately. A situation where there

will be additional information to the system due to the behaviour of

the solution due to changes in the system in the parameters.

4.Validation of the problem

A model is valid if despite its inexactness in representing the system,

it can give a reliable prediction of the system performance. A

common method for testing the validity of a model is to compare its

performance with some past data available for the actual system. The

model will be valid if under similar conditions of the system. It must

be noted that such a validation method is not appropriation for none

existing system since there will be no available data for comparison.

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5. Implementation of the final results

This should be executed through the cooperation of both the

operators’ research team and those who will be responsible for

managing and operating the system.

1.7 Advantages of Model in Operations Research

i. An iconic model is concrete.

ii. It is easy to construct the model.

iii. It is easy to study model than the system itself.

iv. Better control.

v. Better decision.

vi. Better system

1.8 Disadvantages of Model in Operations Research

i. The model is not suitable for further implementation.

ii. It cannot be use to study the changes in operation of the system.

iii. It is not possible to make any modification in the model.

iv. Adjustment with changing situation cannot be done in the model.

1.9 Limitations of Operations Research

Below are the some of the limitations of operations research:

i. Model is only idealized representation of reality and should not be

regarded as absolute in any case.

ii. The validity of a model for a particular situation can be

ascertained only by conducting experiments on it.

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1.10Summary

In this chapter, we have explained some fundamental concepts such

as definition of Operation Research, basic principles of Operations

Research, OR objectives and methodology, modeling and models

characteristics in Operations Research, application in business,

Management, Banking and Finance.

1.11 Review Questions

1. List and explain any four principle of Modelling in Operation

Research.

2. Explain different types of model formulation in Operation

Research.

3. List and explain principles and steps involved in modelling.

References and Further Readings

Adekeye, K.S. (2008). Unpublished Lecture Note on Introduction to

Linear Programming.

Akinbade, F. (1996).Basic Operational research techniques.Panaf

Publishing Inc., Bariga, Lgaos, Nigeria

Banjoko, S.A. (2000). Production and operations management.Saban

publishers, Lagos.

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CHAPTER TWO

ELEMENTS OF DECISION ANALYSIS


At the end of this chapter the readers should be able to:

• Mention the importance of business decision

• Discuss the concept of Decision Analysis

• Mention different types of Decision Situation

• Describe decision tree

• Develop ability of using Decision Analysis in areas of practical

decision making.

2.1 Introduction

Decision analysis has been described as a scientific technique and is

useful for finding optimal course of actions and strategy when a

decision maker is confronted with many options. Decision making

can be complex or otherwise.

2.2 Concept of Decision Process and Decision Analysis,

Rationality in Decision Making Process

A normal human being is an experienced rational decision maker.

Decisions like what to eat for breakfast, lunch or dinner are

important decisions that can affect our life tremendously. There is a

saying, that “good decisions makers are made, they are not born”.

Decision problem plays a key role in economics, politics, social,

technological as well as personal life.

In business, various decisions have to be made. Management

decision unconsciously uses techniques similar to that in personal

decision making. However, it can be improved greatly with training

and this is necessary in view of the complex nature of business

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decision making. The life wire of an organization is management. If

management falters, the organization will find it difficult to survive.

Hence there is need for managers to be trained on related and

systematic ways of decision making.

Decision analysis has been described as a scientific technique that

consists of collection of principles and methods whose principal

objectives is to aid decision making by individuals, group of

individuals, management of organization and others who have to

make one decision or the other. In decision analysis, complex

decision problems are broken down into smaller elements which may

be probabilistic, differential or value oriented.

Decision analysis is useful for finding optimal course of actions and

strategy when a decision maker is confronted with many options and

an uncertain or risk filled pattern of future events. For example, a

frozen chicken supplier wishes to prepare large quantities of frozen

chicken provided consumer’s acceptance and demand for the product

will be high. Otherwise, he would prepare small quantities. Actual

acceptance can only be determined when he puts them out for sale.

The process of selecting the best consumer acceptance among

several alternatives where there is uncertainty of future demand can

be done best by decision analysis. Stages of decision making can be

categorized into two. These are the structural stage and the decisional

stage. The structural stage involves where the decision maker gets all

the information available, organizes them and identifies a specific

decision problem that needs to be solved. The decisional stage is the

stage where the decision maker develops methods and technologies

to assemble the information.

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2.2.1 Steps in Decision Analysis

Decision analysis involves the following steps:

(i) Definition of decision problem

(ii) Exploring available data and information

(iii) Stating alternative courses of action

(iv) Analysis of feasible alternatives

(v) Selection of best or optimal actions

(vii) Implementation of decision

(viii) Evaluation of results

2.2.2 Problems of Decision Analysis

Decision problem exist when individual or group of individuals’

organization known as decision makers have the need to select

among different options. The problems in decision analysis can be

classified into two categories. These are:

(i) Sequential decisions that involve uncertainty and probability

estimates. Sequential decisions are modeled using decision

trees and influence diagrams which allow sequential

organization of sequence of decisions and their probabilities

in diagrammatic form.

(ii) The multiple conflicting goals and objectives decisions

where groups of alternatives are to be compared on the basis

of goals. Multi-objective decisions are modeled using the

multi-attribute utility theory which enables one to compute

an alternative’s overall desirability based on its performance

on specified set of evaluation measures.

2.2.3 Structure of Decision Making Process

The basic structure of a decision problem especially under risk and

uncertainty are decision alternatives, states of nature and conditional

payoffs.

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Decision alternatives are alternatives course of actions that are

available and feasible. Practically, there are many alternative courses

of actions that can be considered in a decision problem and it may be

practically impossible to consider them exhaustively. We therefore,

identify and use the feasible ones in our analysis of the problem

before taking decision.

States of Nature are independent uncontrollable variables that can

occur with possibilities. They are outside the direct control of the

decision maker and they include all the possibilities of outcomes that

can occur. It should be noted that only one of them can occur in the

final decision making.

Conditional payoffs are the outcomes of the combination of events

and decision alternatives. It is associated with time and it is known as

planning or decision horizon.

Payoff Table

The payoff table is the table that is used to proffer solution to

decision problem. Information on decision alternatives, states of

nature and payoff are usually represented on a payoff table. The table

is a concise table containing summary of information available for

various alternatives and events in a decision making process. It is in

the form of a matrix in which the rows represent the decision

alternatives and the columns represent the states of nature which

could deal with high medium and low acceptance of a product.

For example, the payoff table below represents the amount gained

from three strategies (alternatives) and four conditions (state of

nature)

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Decision

alternative

States of nature

S1 S2 S3 S4

D1

D2

D3

4000

20000

20000

-100

5000

15000

600

400

-2000

18000

0

1000

The above table can also be represented in a matrix form, which is

called decision matrix. Therefore, the decision matrix for the above

payoff table is

4000 100 600 1800

20000 5000 400 0

2000 15000 2000 1000

− −

The rows of the matrix represent the decision alternatives and the

columns represent the states of nature.

2.3 Phases in Decision Making Process

The decision analysis cycle consists of four phases of operation. The

phases are:

(i) Decision framing phase: this is where the problems as well

as the possible available alternatives from which one can

select are clearly stated. The phase is also called the

structural phase. In this phase, the decision makers

collect and organize information relevant to the decision

problem. Tools used at this stage include performance

measures of consequences of each alternative as well as

diagrammatic representative of relations between

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decisions, uncertainties and consequences of each

alternative.

(ii) Deterministic analysis Phase: this is the phase which

accounts for certainties rather than uncertainties. Here,

graphical and diagrammatic models like influence

diagrams and flowcharts can be translated into

mathematical models. Necessary tools are used in this

phase for predicting consequences of alternatives and for

evaluating the decision alternatives.

(iii) Probabilistic analysis phase: this phase has been developed

to cater for uncertainties in the decision making

processes.

(iv) Evaluation phase: this is the phase where the decision tree is

evaluated to enable the identification of the decision

outcomes that correspond to sequence of decisions and

events represented in the decision trees.

2.4 Types of Decision Situation

There are four types of decision situations in decision analysis. These

are discussed below:

1. Decision making under certainty: In this situation, only one

state of nature exists for each alternatives. Since the decision

maker has perfect knowledge about the future outcomes, he

simply chooses the alternative with the optimum payoff.

2. Decision making under uncertainty: In this situation, more than

one states of nature exist but the decision maker lacks the

knowledge about the probabilities of their occurrence. Much

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of the business decision takes place under conditions of

uncertainty. Business decision may sometimes be taken with

incomplete knowledge or knowing that the outcomes of these

decisions are uncertain. In decision making under uncertainty,

it is assumed that the true state of nature belongs to the set of

all states in that decision environment and so the process does

not make use of probability estimates on these states.

The decision criteria used for judgment under this environment are:

(i) Maximax (Optimistic) Criterion

The decision maker finds the maximum possible payoff for

each alternatives and then chooses the alternative with the

maximum payoff within the group of the maximum.

(ii) Walds' Maximin (Pessimistic) Criterion

The decision maker finds the minimum possible payoff for

each alternative and then chooses the alternative with the

maximum payoff within the group of the minimum.

(iii) Minimax Regret Criterion

This criterion is also known as the savage criterion. Here, the

decision maker might experience regret after the decision has

been made and the states of nature (events) have occurred.

Thus, the decision maker tries to minimize this regret prior

to the selection of a particular alternative. Therefore, the

maximum amount of regret is determined for each

alternative and thereafter, the alternative with the minimum

of the above maximum regrets are chosen.

The following steps can be used to take decision in minimax

regret criterion:

• Identify the highest value in each column

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• Subtract the highest value from every value in its respective

column to obtain the regret

• For each decision alternative, obtain the maximum regret

• Find the minimum of this maximum regret to obtain the

minimax regret value

• Identify the decision alternatives corresponding to the

minimax regret as the optimum decision

(iv) Hurwicz Criterion

Hurwicz criterion is the criterion of realism. It is also called

the weighted average criterion because of the nature of its

determination. It is a compromise between the maximax and

the minmax decision criteria. The criterion is based on an

index of optimism or coefficient of optimism denoted by α,

where 0 < α < 1. The valuation function is

V(dj) = (α max dij) + ((1- α)min dij) , j = 1, 2, 3, …, m

The above function can be re-written to be

Weighted payoff = α(maximum payoff) + (1 – α)(minimum

payoff).

From the above expression, it is clear that if α = 0, then we

have the maximin criterion and if α = 1, then we have the

maximax criterion. Then, the best decision alternative will

be the one with the maximum weighted payoff.

(v) Laplace Criterion

This criterion is based on what is known as the principle of

insufficient reason. This criterion is also called equal

probabilities criterion. Under this criterion, equal

probabilities are assigned to each state of uncertainty when

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the probabilities are not known in the case of decision

environment under uncertainty.

The steps for determining the best decision under the

Laplace criterion are:

(a) For two states cases, we assign probabilities of 0.5 to each of

the states

(b) For three states cases, we assign probability 0.33 to each of

the states

(c) For n states we assign probability to each of the state.

(d) The expected payoffs are computed as in the expected value

criterion and the decision with the maximum payoff is

selected to be the best.

Example 2.1

The CIPM investment recently purchased a land at Lekki and the

management is trying to determine the types of building to be erected

on the land, namely shopping complex, high-rise building and blocks

of flat. The Institute knows that a large development that results in

low demand could be very disastrous. On the other hand, if CIPM

investment makes a conservative small development decision and

obtain a high demand, substantial profit will be obtained. Managers

of the company used three levels of demands and obtained the

following payoffs decision matrix:

Decision alternative

States of nature

Low Medium High

Build shopping complex

(d1)

200 100 400

Build high-rise houses (d2) 100 150 300

Build block of flats (d3) 300 400 200

Probability 0.25 0.3 0.45

27

Advice the institute investment on the decision to take using the

Laplace criterion assuming the probabilities values were not

provided.

Solution:

Since there are three states of nature in the problem, then we assign

probabilities P (S1) = P (S2) = P(S3) =1/3. Then the expected values

for each of the decision alternative are determined as follows:

For (d1) = (200*1/3) + (100*1/3) + (400*1/3) = 233.3

For (d2) = (100*1/3) + (150*1/3) + (300*1/3) = 183.3

For (d3) = (300*1/3) + (400*1/3) + (200*1/3) = 300

The best decision alternative = max {233.3, 183.3, 300} = 300 = d3.

Therefore, based on Laplace criterion, the CIPM investment is

advice to build blocks of flats.

Example 2.2

A steel manufacturing company in Nigeria is concerned with the

possibility of a strike. It will cost an extra N20,000.00 to acquire an

adequate stockpile. If there is a strike and the company has not

stockpiled, management estimates an additional expense of

N60,000.00 on account of lost sales. Should the company stockpile

or not under,

(a) Optimistic criterion (b) Wald criterion (c) Savage criterion

(d) Hurwicz criterion for 0.4 = (e) Laplace criterion

Solution: Using the given information, we construct the payoff table.


State of nature

Strike(S1) No Strike (S2)

Stockpile(A1) 20,000 20,000

No stockpile(A2) 60,000 0

28

(a) Using the optimistic criterion, we use the maximax

approach. The table below is constructed to determine the

maximum payoff for each decision alternative

The max {20000, 60000} = 60000. Therefore, using the

optimistic criterion, the company should choose the no

stockpile alternative.

(b) Using the pessimistic (Wald) criterion, we use the maximin


maximum payoff for each decision alternative.

The maximum (0 , 20000) is 20000. Therefore, using the Wald

criterion, the company should choose the stockpile alternative.

(c) Using the Savage criterion, we first construct a conditional

regret table. Following the step of the regret criterion, from

the pay off table, the highest value in column 1 is 60,000 and

the highest value in column 2 is 20,000.

The regret value = highest value – value.

Therefore, we have

D1S1 regret= 60,000 -20,000 = 40, 000

Decision

alternative

State of nature Maximum

value Strike(S1) No Strike

(S2)

Stockpile(A1) 20,000 20,000 20,000

No stockpile(A2) 60,000 0 60,000

Decision

alternative

State of nature Minimum

value Strike(S1) No Strike

(S2)

Stockpile(A1) 20,000 20,000 20,000

No stockpile(A2) 60,000 0 0

29

D2S1 regret = 60, 000 – 60, 000 = 0

D1S2 regret = 20,000 – 20000 = 0

D2S2 regret = 20,000 – 0 = 20,000.

The above regret values are summarized in the table below

The minimum of the maximum regret is the min {40000 , 20000} =

20000. Therefore, the minimax regret is N20,000. Hence, using the

Savage criterion, the company should choose the no stockpile

alternative.

(d) Using the Hurwicz criterion for 0.4 = , the formula for

weighted payoff is employed.

Recall that, weighted payoff = α(maximum payoff) + (1 –

α)(minimum payoff).

Therefore, for alternative 1, we have

Weighted payoff for A1 = 0.4(20000) + (0.6)(20000)= N

20,000

Similarly, for alternative 2, we have

Weighted payoff for A2 = 0.4(60000) + (0.6)(0)= N 24,000

The optimal decision is the alternative with the highest weighted

payoff. Therefore, the alternative with the maximum of the weighted

payoff is A2. Hence, the company should not stockpile.

(e) Using the Laplace criterion, we assign a probability value ½

to each of the state since there are two states in the problem

under consideration. Thus we have



regret Strike(S1) No Strike

(S2)

Stockpile (A1) 40,000 0 40,000

No stockpile (A2) 0 20,000 20,000

30

Expected payoff (Alternative 1) = ½ (20000 + 20000) = N

20,000

Expected payoff (Alternative 2) = ½ (60000 + 0) = N30,000

The optimal decision is the alternative with the highest expected

payoff. Therefore, alternative 2 is selected and hence the company

should not stockpile.

Examples 2.3

A major energy company offers a landowner N60,000 for the

exploration rights to natural gas on a certain site and the option for

future development. The option, if exercised, is worth an additional

N 660, 000 to the landowner, but this will occur only if natural gas is

discovered during the exploration phase. The landowner, believing

that the energy company’s interest is a good indication that gas is

present, is tempted to develop the field herself. To do so, she must

contracts with local outfits with expertise in exploration and

development. The initial cost is N100,000 which is lost if no gas is

found. If gas is discovered, however, the landowner estimates a net

profit of N2000,000. Determine the recommended decisions under

(a) Optimistic criterion (b) pessimistic(Wald) criterion

(c) Savage criterion (d) Hurwicz criterion for 0.1 = (e) Laplace criterion

Solution:

Let the decisions for the landowner be represented by Di, i = 1, 2,

where D1 = accept the offer and D2 = to explore and develop on her

own. The states of nature be represented by Si, i = 1, 2, where S1 =

There is no gas on the land and S2 = there is gas on the land. Then

the payoff table is

31

(a) Using the optimistic criterion, we use the maximax


maximum payoff for each decision alternative

Since max {660,2000} = 2000. Therefore, using the optimistic

criterion, D2 is the recommended decision. Hence, the landowner

should explore and develop on her own.

(b) Using the pessimistic (Wald) criterion, the maximin is

employed. The table below is constructed to determine the

minimum payoff for each decision alternative


State of nature

S1 S2

D1 60 660

D2 -100 2000



value S1 S2

D1 60 660 660

D2 -100 2000 2000


State of nature minimum

value S1 S2

D1 60 660 60

D2 -100 2000 -100

32

The max {60, -100} = 60. Therefore, using the Wald criterion, the

decision D1 is recommended. Hence, the landowner under this

criterion should accept the offer.

(c) Using the Savage criterion, we first construct a conditional regret

table. Following the step of the regret criterion, we have the table

below

Table of Regret payoffs

The minimum of the maximum regret is obtained by determining the

min {1340. 160} = 160. Therefore, the minimax regret is N160,

which is D2. Hence, using the Savage criterion, it is recommended

that the landowner should explore and develop on her own

(d) Using the Hurwicz criterion for 0.1 = , the formula for

weighted payoff is employed.

Weighted payoff = α (maximum payoff) + (1 – α) (minimum

payoff).

Therefore, weighted payoff for D1 = 0.1(60) + (0.9)(660)= N 610

Weighted payoff for D2 = 0.1(-100) + (0.9) (2000) =N 1,790

The optimal decision is the alternative with the highest weighted

payoff. Therefore, D2 is the alternative with the maximum of the

weighted payoff. Hence, it is recommended that the landowner

should explore and develop on her own.


regret Decision alternative S1 S2

D1 0 1340 1340

D2 160 0 160

33

(e) Using the Laplace criterion, we assign a probability value ½ to

each of the state since there are two states in the problem under

consideration. Thus we have

Expected payoff for D1 = ½ (60 + 660) = N 360

Expected payoff for D2 = ½ (-100 + 2000) = N1,900

The optimal decision is the alternative with the highest expected

payoff. From the calculation max {360, 1900} = 1900 which is the

expected payoff for D2. Therefore, decision D2 is recommended.

2.5 Decision Making under Risk

In decision making under risk, there are more than one states of

nature but the decision maker has sufficient information to allow him

assign probabilities to each of the states. The probabilities could be

obtained from past records or from the subjective judgment of the

decision maker. The payoffs associated with each alternative are

represented by probability distribution, and decision can be based on

the expected value criterion. Under conditions of risk, there exist a

number of decision criteria which can be used. These are:

(i) Expected value Criterion

The expected value criterion seeks the maximization of

expected profit or the minimization of expected cost.

The data of the problem assumes that the payoff (or cost)

associated with each decision alternative is probabilistic.

The criterion requires the calculation of the expected

value of each decision alternative which is the sum of

the weighted payoffs for that alternative, where the

weights are the probabilities assigned to the states of

nature that can happen. Thus, if dij is the decision

alternatives, pi is the probability and V(di) is the expected

value , then

34

1

( )n

i i i j

i

V d p d=

= , i = 1, 2, …, n is the number of

decision alternatives, and

j = 1,2, …, m is the number of states of nature.

It should be noted that the expected value criterion is also known as

expected monetary value (EMV) criterion. Thus V (di ) is also

known as the EMV of the problem. The alternative with the highest

EMV is selected to be the best alternative (maximization of expected

profit or minimization of the expected cost).

Example 2.4

Suppose that you want to invest N10,000 in the stock market by

buying shares in one of two banks: Guarantee trust bank and Zenith

bank. Shares in GT, though risk could yield a 50% return during the

next year. If the stock market conditions are not favourable, the stock

may lose 20% of its value. Zenith bank provides safe investments

with 15% return in a bull market and only 5% in a bear market. All

the publications you have consulted are predicting a 60% chance of a

bull market and 40% for a bear market. How should you invest your

money?

Solution: The problem is summarized in the table below


States of nature

Bull Market Bear market

GT stock 5000 -2000

Zenith stock 1500 500

Probability 0.6 0.4

35

The expected value can be obtained using the above formula for each

of the decision alternative. Therefore, for GT bank, we have,

V(d1) = (5000*0.6) + (-2000*0.4)= 2200

For Zenith bank, we have

V(d2) = (1500*0.6) + (500*0.4)= 1100

Since the GT bank stock has the higher EMV, and then the money

should be invested on GTB stock.

Example 2.5

Using the problem in Example 2.1 above: Advice the institute

investment on the decision to take using the expected value criterion

Solution:

Using1

( )n

i i i j

i

V d p d=

= , then we have

The expected monetary value for (d1) = (200*0.25) + (100*0.3) +

(400*0.45) = 260


(300*0.45) = 205


(200*0.45) = 285

The best decision alternative = max{260, 205, 285} = 285 = d3.

Therefore, the CIPM investment is advice to build blocks of flats.

(ii) Expected Opportunity Loss or Expected Regret Criterion

The expected opportunity loss (EOL) is an alternative approach to

maximizing EMV. The approach is to minimize expected

opportunity loss (EOL). The expected opportunity loss (or expected

value of regrets) represents the amount by which maximum possible

profit will be reduced under various possible stock actions. The

36

course of action that minimizes these losses or reductions is the

optimal decision alternative. The expected opportunity loss or

expected regret is determined using the following steps:

(i) Identify the maximum for each of the states of nature using the

table of payoffs.

(ii) Obtain the regret or opportunity loss table by using the

expression (regret = maximum payoff – payoff) for each of the states

of nature

(iii) Calculate the expected opportunity loss for each decision

alternative by multiplying the value of the regret in the table

in step b by the associated probabilities and then add the

values.

(iv) Select the alternative that yields the minimum EOL.

Example 2.6

Solve the problem in example 2.1 above using the Expected

opportunity loss criterion

Solution:

Table of payoff


States of nature

Low Medium High


(d1)

200 100 400



Probability 0.25 0.3 0.45

37

Steps a and b: From the table of payoff, the regret or opportunity

loss table is constructed and presented below

Regret or Opportunity Loss Table


States of nature

Low Medium High


(d1)

100 300 0



Step c: Determine the Expected loss for each decision alternative.

Thus, we have

Expected loss for d1 = (100*0.25) + (300*0.3) + (0 *0.45) = 115



Step d: The decision to be selected is the min{ 115, 170, 90} = 90 =

d3. Therefore, based on the Expected opportunity loss criterion, the

best decision is to advice the CIPM investment to build blocks of

flats.

Decision making under conflict

Situations exist in which two or more opponents with conflicting

objectives try to make decisions with each trying to win at chess,

candidates fighting an election, two enemies planning war tactics,

firms struggling to maintain their market shares etc. These situations

are different since the decision-maker is working against an

intelligent opponent. They are called decision making under conflict

38

or under partial uncertainty. The theory governing these types of

decision problems is called theory of games.

2.6 Decision Tree Approach

A decision tree is a graphical representation of the decision process

indicating decision alternatives, states of nature, probabilities

attached to the states of nature and conditional benefits and losses. It

consists of a network of nodes and branches. These are decision

nodes represented by a square and state of nature node represented

by a circle. Alternative courses of action (strategies) originate from

the decision node as main branches (decision branches). At the end

of each decision branch there is a state of nature node from which

emanates chance events in the form of sub-branches (chance

branches). The respective payoffs and the probabilities associated

with alternative courses and the chance events are shown alongside

these branches. At the terminal of the chance branches are shown the

expected values of the outcome. The general approach used in

decision tree analysis is to work backward through the tree from

right to left, computing the expected value of each chance node.

Thus, the expected value (EV) can be computed using, EV(d) = G(d)

+ (P * L)

Where G is the gain outcome, L is the loss outcome and P is the

probability. The decision node which leads to the chance node with

the highest expected value is chosen to be the best decision

alternative.

Example 2.7

A company wishing to try the possibilities of marketing three

products A, B, C under three market growths S1, S2 and S3 has

obtained the following data to aid its decision making.

39

Decision Alternatives

States of nature

S1 S2 S3

A 20 100 300

B -60 200 1000

C -40 400 700

If P(S1) = 0.5, P(S2) = 0.3, and P(S3) = 0.2 and a consultant predicts

that if condition S1 exists then product A will be the optimal

decision; if S2 exists, product B will be the optimal decision and if S3

exists , product C will be the optimal decision. Draw a decision tree

for the problem.

Solution: 0.5 20

0.3 100

0.2 300

0.5 - 60

0.3 200

0.2 1000

0.5 -40

0.3

400

0.2 700

1

A

aaaaa

B

aaaaa

C

aaaaa

40

The expected value for product A = (20*0.5) + (100*0.3) +

(300*0.2) = 100

The expected value for product B = (-60*0.5) + (200*0.3) +

(1000*0.2) = 230

The expected value for product C = (-40*0.5) + (400*0.3) +

(700*0.2) = 240

The decision alternative with the highest expected value is product

C. Therefore, decision C is the optimal decision.

Example 2.8

A company dealing with newly invented telephonic device is faced

with the problem of selecting the following strategies:

(i) Manufacture the device itself

(ii) To be paid on a royalty basis by another manufacturer

(iii) Sell the rights for its invention for a lump sum

The profit in thousands of naira that can be expected in each case and

the probabilities associated with the sales volumes are shown in the

following table:

Event probability Manufacture Royalties Sell the

right

High demand

Medium

demand

Low demand

0.2

0.3

0.5

100

30

-10

40

25

15

20

20

20

(a) Represent the company’s problem in the form of a decision tree

(b) Determine the optimal decision for the company

41

Solution:

Let A represent the manufacture strategy, B represent royalty

strategy and C represent the sell the right strategy. Therefore, the

decision tree for the problem is

0.2 100

0.3 30

0.5 -10

0.2 40

0.3 25

0.5 15

0.2 20

0.3 20

0.5 20

(b)

The expected value for strategy A = (100*0.2) + (30*0.3) +

(-10*0.5) = 24

The expected value for strategy B = (40*0.2) + (25*0.3) + (15*0.5)

= 23

The expected value for strategy C = 20(0.2 + 0.3+ 0.5) = 20

The decision alternative with the highest expected value is strategy

A. Therefore, the optimal decision for the company is to manufacture

the telephonic device itself to get the maximum expected profit of

N24,000.

2.7 Summary

• Decision analysis as a scientific course which entails collection

of principles and

• technique and which aim at helping organisation and individuals

to always make the

1

A

aaaaa

B

aaaaa

C

aaaaa

42

• best decision has been extensively discussed.

• The decision maker uses a specific technique to make an optimal

choice.

• Decision alternatives are the possible courses of actions.

• States of nature are the possible events that can occur in a

decision problem.

• Payoff, is the outcome resulting from specific decision

alternative.

• The maximax criterion is a criterion in which we select the

maximum of all the

• maximum payoffs.

• The maximin criterion is a criterion in which we select the

maximum of the

• minimum payoffs.

• The minimax criterion or regret involves selecting the minimum

of the maximum

• regret or opportunity loss.

• The four states of decision environment are

(i) Decision making under certainty

(ii) Decision making under uncertainty

(iii) Decision making under risk

(iv) Decision making under conflict.


1. A dress buyer for a large department store must place orders

with a dress manufacturer nine months before the dresses are

needed. One decision is as to the number of knee-length

dresses to stock. The ultimate gain to the department store

depends both on this decision and on the fashion prevailing

43

nine months later. The buyer’s estimates of the gain (in

thousands of naira) are given in the table below

States of nature

Alternative decision Knee

lengths are

high

fashion (S1)

Knee lengths

are

acceptable

(S2)

Knee lengths

are not

acceptable (S3)

Oder none (D1)

Order a little (D2)

Order moderately (D3)

Order a lot (D4)

-30

-10

60

80

0

30

45

40

80

35

-30

-45

Determine the recommended decisions under

(a) Optimistic criterion (b) pessimistic(Wald) criterion (c) Savage

criterion (d) Hurwicz criterion for 0.5 = (e) Laplace

criterion

2. A large steel manufacturing company has three options with

regards to production: (i) produce commercially (ii) build pilot plant

(iii) stop producing steel. The management has estimated that their

pilot plant, if built, has 0.8 chance of high yield and 0.2 chance of

low yield. If the pilot plant does show a high yield, management

assigns a probability of 0.75 that the commercial plant will also have

a high yield. If the pilot plant shows a low yield there is only a 0.1

chance that the commercial plant will show a high yield. Finally,

management’s best assessment of the yield on a commercial –size

plant without building a pilot plant first has a 0.6 chance of high

yield. A pilot plant will cost N300,000. The profits earned under

44

high and low yield conditions are N12,000,000 and N1,200,000

respectively. Find the optimum decision for the company.


Adedayo, O.O.Ojo, O. &Obamiro, J.K. (2006). Operations Research

in decision analysis and production management. Pumark Nigeria

Limited, Lagos

Adekeye, K.S. (2008). Unpublished Lecture Note on Introduction To

Linear Programming.

Akinbade, F. (1996).Basic Operational research techniques.Panaf


Banjoko, S.A. (2000). Production and Operations

Management.SabanPublishers, Lagos.

45

CHAPTER THREE

LINEAR PROGRAMMING


At the end of this chapter, the readers should be able to:

• Define Linear Programming as a tool in Operation Research

• Discuss the main characteristics of Linear Programming

• Formulate correctly a Linear Programming Problem

• Explain simplex (algorithm) solution of a Linear Programming

• Solve two-variable Linear Programming Problem by simplex

algorithm and graphical methods.

3.1 Introduction

This chapter aims to provide decision makers good foundation of the

knowledge and skills that managers and other decision body requires

in formulation of policies and making good decision among various

decision alternatives.

3.2 Definition of Linear programming

Linear programming can be defined in any of the following way:

i. Linear programming is a mathematical method for solving a large

class of special problems where one is attempting either to maximize

benefit while using limited resources or to minimize cost while

meeting certain requirement.

ii. Linear programming is a technique for allocating limited

resources optimally (maximize profit or minimize costs) among a

number of competing demands while operating within the specified

constraint.

46

iii. Linear programming is a technique used to determine the best

utilization of limited resources to reach desired objectives of either

maximizing the benefit (profit) or minimizing the costs.

Linear Programming is one of the most commonly used techniques

in Operations Research. It applies to optimization models in which

the objectives and constraint are strictly linear. The technique is

applicable in a wide range of applications, including, agriculture,

Industry, transportation, economics, behavioral science, social

sciences and military

Managers use a range of mathematical techniques in performance

management, risk analysis, financial management, decision-making,

policy-making and formulation among others. In most businesses,

there are insufficient resources available to do as many things as

management would wish, and the problem is to decide how the

resources should be allocated to obtain best results. To achieve this,

linear programming technique can be used.

Linear programming (LP) is a tool for solving optimization

problems. It is a mathematical modeling “system” which has found

widespread uses in providing decision making with an efficient

means for resolving complex operational alternatives. Linear

programming is applicable to the general category of problems that

require the optimization of a linear objective function subject to

linear constraints. This constraint exists because the activities under

consideration usually compete for scarce resources. Optimization

may require minimization of cost, time, resources, etc, or it may

require the maximization of profit, yield or other desirable output

measure(s) depending on the situation under study.

47

3.3 Characteristics of Linear Programming Problem

i. The objective function must be well defined. The objective

function is a function of the decision variables subject to

satisfying the constraints. There must be alternatives courses of

action.

ii. Alternative courses of action: There must be alternatives courses

of action, one of which achieves the objectives.

iii. Additivity of resources and activities: The sum of the resources

used by different activities must be equal to the total quantity of

resources used by each activity for all the resources individually

and collectively.

iv. Linearity of the objective function and constraints: The objective

function and its constraints must be expressed as a mathematical

equation or inequalities and these must be linear equation or

inequalities.

v. Non-negativity of decision variables: All decision variables

should be non-negative.

vi. Single Valued Expectation: Resources and activities are known

with certainty, thus a deterministic programming model is

obtained.

3.4 Basic Assumptions of Linear Programming Problem (LPP)

i. Certainty: The co-efficient of the decision variables in linear

programming are assumed with certainty. Such as profit,

contribution and the amount of resources require per unit of are

known constraint.

48

ii. Divisibility: it should be noted that the decision variables can take

fractional values. It is feasible to have fractional value in the

resources and production activities in linear programming models.

iii. Non-negativity: The decision variable cannot have negative

values. It is not possible for a company or an establishment to

produce negative items.

iv. Additivity: This means that the total of all activities equal the

sum of each individual activity.

v. Proportionality: We are sure that proportionality exists in the

objectives and the constraints. This means that, if production of one

unit of product used two of a particular scare resource, then making

five units of that product uses ten resources.

3.5 Application Areas of Linear Programming

Linear programming is widely used in different areas such as:

1. Product mix problem

2. The diet problem

3. Advertising/promotion planning i.e. product promotion

4. Investment planning

5. Production/Inventory planning

6. Optimal cargo shipment

7. Multi period problem with sub-contracting option

8. Strategic/military effectiveness analysis

49

9. Manufacturing

10. Agricultural Application; Farm Economic and Management

11. Military Application

12. Production Management; Production planning, Blending

Problem.

13. Financial Management; Port folio selection and Profit n

planning.

14. Marketing Management; Media Selection, Physical Distribution

and others.

15. Personal management; Starting Problem, Job evaluation and

others.

16. Education

17. Administration

18. Awarding Contracts.

19. Capital Budgeting and others.

3.6 Some Important Concepts in Linear Programming

The following are some of the concepts that are commonly used in

the application of linear programming.

1. Feasible solution

This is defined as a set of values of decision variables which satisfy

all the constraints and non-negativity conditions of a linear

programming problem simultaneously.

50

2. Infeasible solution

This is defined as a set of values of decision variables which do not

satisfy all the constraints and non-negativity conditions of a linear

programming problem simultaneously.

3. Basic solution

For a set of m simultaneous equation in n variables ( ),n m a

solution obtained by setting ( )n m− variables equal to zero and

solving for remaining m equations in m variables is called a basic

solution.

4. Basic feasible solution

A basic feasible solution is a solution in which all basic variables

have non-negative values. Basic feasible solution is of two types:

(a) Degenerate: A basic feasible solution is called degenerate if value

of at least one basic variable is zero.

(b) Non-degenerate: A basic feasible solution is called non-

degenerate if values m − basic variables are non-zero and positive.

5. Optimum Basic feasible solution

A basic feasible solution which optimizes (minimum/maximum) the

objective function value of the given linear programming problem

(LPP) is called an optimum basic feasible solution.

6. Unbounded solution

A solution which can increase or decrease the value of objective

function of the linear programming problem (LPP) indefinitely is

called unbounded solution.

51

3.7 Linear Programming Problem Formulation

To formulate a linear programming problem means to transform a

real life problem into a standardized format consisting of

mathematical equations and inequalities. Linear programming

problem formulation comprises of four components. These are (i)

Objective function (ii) Decision variables (iii) Constraints (iv)

Parameters and (v) Non-negativity constraints

(i) Objective function

This simply refers to a mathematical equation that relates the

decision variables in a linear form. After the decision variables have

been identified, the next thing is to identify the objective of the

decision maker and thereafter represent it in a mathematical equation

that relates the decision variables in a linear form. In linear

programming, there are two types of objective function. These are

maximization and minimization. Maximization problem might

involve profits, revenues, efficiency or rate of return. On the other

hand, minimization problem involves cost time, distances travelled

and so on. The profit, cost, etc. per unit of output or input is

summarized by an objective function. For example, objective

function can be Maximize 6 9P x y= + , where P is the profit, x is

the quantity of product 1 and y is the quantity of product 2

OR

Minimize 12.5 7C x y= + , where C is the cost, x is the quantity of

product 1 and y is the quantity of product 2.

(ii) Decision variables

These represent choices available to the decision maker in terms of

amount of either inputs or outputs. For example, some problems

52

involve a combination of inputs that will minimize total costs while

others involve selecting a combination of outputs that will maximize

profits or revenues. In the above example, x and y are the decision

variables.

(iii) The Constraints

The constraints are classified as either structural or technical and

non-negative. A structural constraint relates the amount of a

particular resources used to the amount of resources available. The

structural constraints of LPPs are mathematical statements of the

limitations placed on the decision maker in the problem setting.

There are three types of constraints:

Less or equal to (≤), Greater or equal to ( ≥) and Equal to (=) where,

Constraint implies an upper limit on the amount of scarce

resources (e.g. machine hours, labour hours, raw materials)

Constraint specifies a lower bound that must be achieved in the

final solution.

= Constraint specifies exactly what a decision variable should equal.

(iv) Parameters

The coefficients in the objective function or the coefficients in the

functional constraints and the right-hand sides of the functional

constraints are known as the parameters.

Hence, the general linear programming problem is in the form:

Optimize

53

1 1 2 2 (1)Z C X C X C= + + +

Subject to:

11 1 12 2 1 1

2 1 22 2 2 2

1 1 2 2

(2)

n n

i n n

m m mn n m

a X a X a X b

a X a X a X b

a X a X a X b

+ + +

+ + +

+ + +

1 2, , , 0 (3)nX X X

It should be noted in equation (2) that, can be , and = . Also

m n

In the above expression, equation (1) is the objective function which

could be either maximize or minimize, equation (2) are the

constraints equation while equation (3) is the non-negative

constraints. The non-negativity requirements serve to guarantee

economically meaningful solutions.

Equation (2) and (3) above can be written in matrix form as:

,AX b

where,

11 12 1 1 1

21 22 2 2 2

1 2

, ,

n

n

m m mn n m

a a a X b

a a a X bA X

a a a X b

= =

iia and ib are unknown coefficients and X’s are unknown decision

variables.

54

Example 3.1 (Maximization Problem)

The NDE Plastics Company has a government contract to produce

three different plastic valves. These valves must be highly heated and

pressure resistant and the company has developed a three-stage

production process that will provide the valves with the necessary

properties involving work in three different chambers. Chamber A

provides the necessary pressure resistance and can process valves for

1200 minutes each week. Chamber B provides heat resistance and

can process valves for 900 minutes a week. Chamber C tests the

valves and can work for 1300 minutes a week. The three valves and

the time in minutes required in each Chamber are tabulated in the

table below:

Types of

Valves

Chamber A Chamber B Chamber C

Exhaust 5 7 4

Intake 3 2 10

By-pass 2 4 5

The government will buy all the valves that can be produced and the

company will receive the following profit margin on each valve:

Exhaust N1.50, Intake N1.35 and By-pass N1.00. Formulate a linear

programming model to determine how many valves of each type the

company should produce each week in order to maximize profit.

Solution:

1. The decision variables are: 1X for the number of exhaust valve

2X for the number of intake valve

55

3X for the number of by-pass valve.

2. The constraints and parameters are

1 2 3

1 2 3

1 2 3

5 3 2 1200

7 2 4 900

4 10 5 1300

X X X

X X X

X X X

+ +

+ +

+ +

3. The objective function is

Max Z 1 2 31.50 1.35 1.00X X X= + +

4. Non-negativity constraints are:

1 20, 0X X and 3 0X

The complete formulation is:

Max Z 1 2 31.50 1.35 1.00X X X= + +

Subject to:

1 2 3

1 2 3

1 2 3

5 3 2 1200

7 2 4 900

4 10 5 1300

X X X

X X X

X X X

+ +

+ +

+ +

1 2 3, 0X X X

₦Example 3.2 (Maximization Problem)

A resourceful home decorator manufactures two types of lamps, say,

A and B. Both lamps go through two techniques, cutter and finisher.

Lamp A requires two hours of cutter’s time and one hour of

56

finisher’s time. Lamp B requires one hour of cutter’s time and two

hours of the finisher’s time. The cutter has 104 hours and finishers

76 hours of available time each month. The profit on one lamp A is

₦6 and the profit on one lamp B is ₦11. Assuming that the

manufacturer sells all the lamps produced and wishes to maximize

his profit, formulate the problem as a linear programming problem.

Solution:

There are two decision variables in the example. These are lamps A

and B. Let 1X represent the number of lamp A produced and 2X

represent the number of lamp B. Since the profit on lamb A is N6

and the profit on lamp B is ₦11, then the objective function is:

1 26 11Z X X= +

The constraints are formulated as follows:

Lamb A requires 2 hours of cutter’s time while lamp B requires 1

hour of cutter’s time of finisher. Thus, the total cutter’s time for lamp

A and B is:

1 22 104 ( ' )X X Cutter time+

Similarly, the total finisher’s time for both lamps is

1 22 76 ( ' )X X Finisher s time+

Thus the standard LPP is

57

1 26 11Max Z X X= +

Subject to:

1 2

1 2

1 2

2 104

2 76

, 0

X X

X X

X X

+

+

Example 3.3 (Minimization Problem)

A person requires 10, 12 and 12 units of chemicals A, B and C

respectively for his garden. A liquid product contains 5, 2 and 1 unit

of A, B and C respectively per jar; and a dry product contains 1, 2

and 4 unit of A, B and C respectively, per carton. If the product sells

for ₦300 per jar and the dry products sells for ₦200 per carton.

Formulate the LP model to minimize the cost.

Chemical Unit per jar Unit per

carton

Units needed

Chemical A 5 1 10

Chemical B 2 2 12

Chemical C 1 4 12

Cost ₦300 ₦200

Solution:

i. Let 1X denote the number of jars bought

2X the number of cartons

ii. The constraints and the parameters are

58

1 2

1 2

1 2

5 10

2 2 12

4 12

X X

X X

X X

+

+

+

non-negativity constraints 1 20, 0X X

The complete formulation is:

Min 1 2300 200P X X= +

Subject to:

1 2

1 2

1 2

1 2

5 10

2 2 12

4 12

, 0

X X

X X

X X

X X

+

+

+

3.8 Solution of Linear Programming Problem (LPP)

There are various methods that can be used in solving LPP. Some of

which are:

(i) Graphical method

(ii) Simplex method

(iii) Revised Simplex method or Artificial variable technique (Big-M

methods and Two-Phase method).

In this text, we shall discuss and apply the Graphical and Simplex

methods only.

59

3.8.1 Graphical Solution of Linear Programming

In linear programming, the objective is to maximize or minimize

several conditions or constraints. If there are only two decision

variables, we use a graphical method. We start by setting the

inequalities to linear equations. The following are the steps in

solving LPP graphically.

Step 1: Define variables

Step 2: Construct inequalities to represent the constraints

Step 3: Plot the constraints on a graph

Step 4: Identify the feasible region. This is the area that represents

the combinations of 1X and 2X that are possible in the light of the

constraints

Step 5: Construct an objective function

Step 6: Identify the values of 1X and 2X that lead to the optimum

value of the objective function. This might be a maximization

(highest) value or minimization (minimum) value depending on the

objective.

*Note that in graphical method 1X is plotted along X- axis while

2X is plotted along Y-axis.

Example 3.4

A tailor has the following material available: 16m2 cotton, 11m2 silk

and 15m2 wool. A suit requires the following: 22m cotton,

21m silk

and 21m wool. A gown requires the following:

21m cotton,

60

22m silk and 23m wool. If a suit sells for ₦3000.00 and a gown for

₦5000.00, Formulate LP model for the above problem. How many

of each garment should the tailor make to obtain the maximum

amount of money?

Solution:

Let Suit = X and Y = Gown

Suit ( )X Gown ( )Y Material

2 X Y 16

X 2Y 11

X 3Y 15

Profit Z 3000 5000

LP Model:

Max. Z = 3000 X + 5000Y

Subject to:

2 X + Y ≤ 16 … (1)

X + 2Y ≤ 11 … (2)

X + 3Y ≤ 15 … (3)

X , Y ≥ 0 (non-negativity)

From equation 1, when X = 0, 2(0) + Y = 16

Y =16 (0, 16)

whenY = 0, 2 X + 0 = 16,

X = 8 (8, 0)

From equation 2, when, X = 0, 0 +2 Y = 11

61

Y = 11

2= 5.5 (0, 5.5)

whenY = 0, X + 2(0) = 11

X = 11 (11, 0)

From equation 3, when X = 0, 0 + 3 Y = 15

Y = 5 (0, 5)

whenY = 0, X + 3(0) = 15

X = 15 (15, 0)

Y LP Graph:

15 16 X Y

10

Eq. (1) 8 16

(11, 5.5) 2x+y<16 Eq. (2) 11 5.5

5 5.5

x+2y<11 Eq. (3) 5 15

FEASIBLE x+3<15

0 REGION

2 4 6 8 10 12 14 16 X

3.8.2 Simplex Method

The simplex method is a more general systematic and efficient

approach that works on all LPP’s regardless of the number of

decision variables. The iterative method can be used where the

graphical approach can no longer produce results or fail or otherwise

62

stated. The simplex method is a procedure for solving a set of such

equations simultaneously. To use this method, the problem has to be

stated in standardized manner. That is, the inequality is converted to

an equation by adding an extra variable called slack variable.

In general, simplex method is a step by step arithmetic method of

solving LP problems whereby one moves progressively from a

position of, say zero production, and therefore zero contribution,

until no further contribution can be made. Each step produces a

feasible solution and each step produces an answer better than the

one before i. e either greater contribution in maximizing problems, or

less cost in minimizing problems.

Definition of basic terms in Simplex Method

(i) Slack Variable: The variable added to the left hand side of the

constraint to make the inequality constraint to be equality. It should

be noted that whenever the constraint has the sign less than or equal

to ( ) , it is mandatory to introduce slack variable to it so as to get

starting solution.

(ii) Surplus Variable: The variable subtracted from the left hand

side of the constraint to make the inequality to become equality. The

surplus variable is introduced whenever the constraint has the sign

greater than or equal to ( ).

3.8.2 Simplex Method Procedure

The steps to follow in this method are as stated below:

63

Step 1: The problem must be expressed in the standard form by

introducing slack variable.

Step 2: Select the original variables to be non-basic variables (equal

to zero) and the slack variables to be the basic variables.

Step 3: To determine the entry basic variable, this is done by

selecting the non-basic variables whose coefficient is largest

negative for maximizing and positive for minimizing on the

objective function.

Step 4: To determine the leaving form variable, it is done by select

the basic variable which have the minimum ratio by dividing the

right hand side by the pivot number.

Step 5: Perform the pivot operation to get a new tableau and the

basic feasible solution.

Step 6: Determine whether the solution is optimal. In a maximize

(optimize) problem, a basic variable is spherical it the co-efficient of

the non-basic variable are all negative or zero (positive or zero for

minimize).

The current basic feasible solution is optimal if and only if every

coefficient in the new objective function is non-negative. If it is then

stop, if otherwise go to step 3.

Example 3.4

Using simplex method, solve the following linear programming

problem,

64

5 6 (1)

. : 3 2 120 (2)

4 6 260 (3)

, 0

Max C x y

s t x y

x y

x y

= +

+

+

Solution:

Objective function max 5 6C x y= + will become

1

2

1 2

max 5 6 0

. 3 2 120

4 6 260

, , , 0

C x y

s t x y s

x y s

x y s s non negative

− − =

+ + =

+ + =

−

Simplex Table

C x y

1s 2s Solution

L.H.S

1

1

2

C

s

s

1

0

0

+

5

3

4

−

6

2

(6)

−

0

1

0

0

0

1

0

120

260

. .1

. .2

. .3

coeff of eq

coeff of eq

coeff of eq

To determine on entry basic variable, we shall enter through y with

the largest negative on objective function.

65

We can now determine the leaving variable by considering the least

ratio, i.e. the least out of 120

2 and

260

6which is of course

260,

6so 6

will serve as our pivot element for table 1.

Simplex Table 2

C x

Y

1s 2s Solution

2

1

C

s

y

1

0

0

1

5

3

2

3

−

0

0

1

0

1

0

1

2

1

6

−

260

100

3

130

3

1

1

2

6

2

1

6

y c

y s

s

+

− +

Since we are left with only ,x we need to look for the minimum

ratio, i.e. the least out of

100 5

3 3 and

130 2

3 3 which of course,

100 5

3 3 = 20.

So,5

3 will serve as our pivot element.

66

c x y 1s 2s solution

3C

x

y

1

0

0

0

1

0

0

0

1

3

5

3

5

2

5−

1

5

6

5

29

30

−

−

−

280

20

30

2

1

3

5

2

3

x c

s

x y

+

− +

Therefore, for the firm to maximize the contribution, 20 of items x

should be produced and 30 of items y should be produced.

Example 3.6

Max Z = 16x1 + 17x2 + 10x3

Subject to:

x1 + x2 + 4x3 ≤ 2000

2x1 + x2 + x3 ≤ 3600

x1+ 2x2 + 2x3 ≤ 2400

x1 ≤ 30

Solution

Z – 16x1 – 17x2 – 10x3 + 0s1 + 0s2 + 0s3 + 0s4 = 0

Subject to

x1+ x2 + 4x3 + s1 = 2000

2x1 + x2 + x3 +s2 = 3600

67

x1+ 2x2 + 2x3 + s3 = 2400

x1 + s4 = 30

x1, x2, x3, s1, s2, s3, s4 ≥ 0

Step 1

Table 1 ( Non-Basic)

Basic Z x1 x2 x3 s1 s2 s3 s4 Solution

Z 1 -16 -17 -10 0 0 0 0 0

s1 0 1 1 4 1 0 0 0 2000

s2 0 2 1 1 0 1 0 0 3600

s3 0 1 2 [2] 0 0 1 0 2400

s4 0 1 O 0 0 0 0 1 30

To determine an entering basic variable, we shall enter through with

the largest negative objective function. We can determine the leaving

variable by considering the least value out of 2000

1 = 2000,

3600

1

= 3600, 2400

2 = 1200, and

30

0 = ∞, which is

2400

2, so (2) will

serve as pivot element for table 1 and S3 is the pivot row.

Step 2

Reduce the coefficient of [2] to [1] by dividing through by 2,

then eliminate s3 from the equation and then use suitable multiple of

x2 equation to get the new table given below:

68

Table 2 Non-Basic

Algorithm Basic Z x1 x2 x3 s1 s2 s3 s4 Solution

17x2 +Z Z 1 15

2

0 7 0 0 17

2

0 0

s1 s1 0 1

2

0 3 1 0 1

2

0 800

s1 – x2 s2 0 3

2

0 0 0 1 1

2

−

0 2400

3

2

s

x3 0 1

2

1 1 0 0 1

2

0 1200

s4 0 1 0 0 0 0 0 1 30

Step 3

x1contain the most negative coefficient so x1 is the entering variable,

the ratio are

8001600

1

2

= ,2400

16003

2

= , 1200

24001

2

= 2400 and 30

301= .

Therefore, s4 is the leaving variable [1] and is the pivot element.

69

(Basic) Z x1 x2 x3 s1 s2 s3 S4 Solution

Z 1 0 0 7 0 0 17

2

15

2

20625

s1 0 0 0 3 1 0 1

2−

1

2−

785

s2 0 0 0 0 0 1 1

2

−

3

2−

2355

x2 0 0 1 1 0 0 1

2

1

2

−

1185

x1 0 1 0 0 0 0 0 1 30

Since all the coefficients of Z equation are positive variables, the

solution is optimal. The optimal solution is x1 = 30, x2 = 1185, x3 =

0, s1 = 785, s2 = 2355, s3 = 0.

Therefore, Z =16(30) + 17(1185) + 10(0) = 20625

3.9 Primal and Duality in Linear Programming

Every linear programming problem is associated with another linear

programming problem called the DUAL of the problem. The original

problem is called the Primal, while the inverse of it is called the

DUAL. The optimal solution of either problem reveals information

concerning the optimal solution of the other. If the optimal solution

of either primal or dual is known, the optimal solution of the other is

also available.

The DUAL problem of the LPP is obtained by:

(i) Changing the variable from say X, to any convenient variable

say W.

70

(ii) Transposing the coefficients matrix.

(iii) Interchanging the role of constant terms and the coefficients of

the objective function.

(iv) Reverting the inequality.

(v) Minimizing the objective function instead of maximizing it or

vice-versa.

Consider a PRIMAL general LPP:

1 1 2 2p n nMax Z C X C X C X= + + +

Subject to:

11 1 12 2 1 1

2 1 22 2 2 2

1 1 2 2

n n

i n n

m m mn n m

a X a X a X b

a X a X a X b

a X a X a X b

+ + +

+ + +

+ + +

1 2, , , 0nX X X

71

The DUAL problem of the PRIMAL is:

1 1 2 2p n nMin Z bW b W b W= + + +

Subject to:

11 1 12 2 1 1

2 1 22 2 2 2

1 1 2 2

n n

i n n

m m mn n m

a W a W a W C

a W a W a W C

a W a W a W C

+ + +

+ + +

+ + +

1 2, , , 0nW W W

The solution of the DUAL can be read from the final simplex tableau

of the PRIMAL and vice-versa using the following rules

(i) The optimal value of the primal objective function is equal to

the optimal value of the dual objective function.

(ii) The value of the objective function row (Z) that corresponds to

the slack variables in the final simplex tableau for the primal are

the values of the corresponding optimal dual variables.

Example 3.7

Write the DUAL of the LPP:

1 23Min Z X X= +

subject to :

1 2

1 2

1 2

2 3 2

1

, 0

X X

X X

X X

+

+

72

Solution:

The DUAL to the primal LPP is:

1 22pMax Z W W= +

subject to:

1 2

1 2

1 2

2 3

3 1

, 0

W W

W W

W W

+

+

---------- non negativity

It should be noted that,

(i) the optimal value of the PRIMAL objective function is equal to

the optimal value of the DUAL objective function i.e.

.p DMax Z Min Z=

(ii) the value of the objective function row that correspond to the

slack variables in the final simplex tableau for the primal are the

values of the corresponding optimal dual variables in the final

simplex tableau.

It should be noted that in formulating the DUAL from the PRIMAL,

a column is made for each limitation and a constraint row for each

element in the objective function.

Example 3.8

A chemical manufacturer processes two chemicals, NAOH and

H2SO4 in varying proportion to produce three products, A, B and C.

He wishes to produce at least 150 units of product A, 200 units of

product B, and 60 units of product C. Each ton of NAOH yields 3

73

units of A, 5 units of B, and 3 unit of C. Similarly, each ton of H2SO4

yields 5 units of A, 5 units of B, and 1 unit of C. If NAOH cost ₦40

per ton and H2SO4 cost ₦50 per ton, advise the manufacturer how to

minimize his cost.

Solution

Let 1X represent number of tons of NAOH, and 2X represent the

number of tons of H2SO4. Since the manufacturer is interested in

minimizing his cost, then the LPP is of minimization. The objective

function is 1 240 50 .Z X X= + thus, the LPP is

1 240 50Min Z X X= + .

Subject to,

1 23 5 150X X+ Product A constraint

25 5 200X X+ Product B constraint

1 23 60X X+ Product C constraint

1 2, 0X X

Since the problem is minimization, we can formulate the inverse or

DUAL of the problem. This is:

150 200 60pMax Z A B C= + +

Subject to,

74

3 5 3 40

5 5 50

, , 0

A B C

A B C

A B C

+ +

+ +

Solve the maximization problem in a standard form as follows.

Initial simplex tableau

Basic variable Main variable

A B C

Slack variable

1S 2S

Current

solution

Ratio

S1

S2

3 5 3

5 5 1

1 0

0 1

40

50

8

10

ZD 150 200 60 0 0 0

Tableau 2


A B C

Slack variable

1S 2S

Current

solution

Ratio

B

S2

3

5 5

3

5

2 0 -2

1

5 0

-1 1

408

5=

5010

5=

13.3

5

ZD 30 0 60 0 0 -1600

75

Tableau 3


A B C

Slack variable

1S 2S

Current solution

B

A

0 1 6

5

1 0 -1

1

2

3

10−

1

2

−

1

2

5

5

ZD 0 0 30 -25 -15 -1750

Tableau3 is the optimal solution to the LPP. Thus, we have:

*S1 = -25, which implies purchase of 25 tons of NAOH

S2 = -15, implies purchase of 15 tons of H2SO4.

These two variables have a total cost of ₦1750.

Valuation of constraints: Rows B and A indicate the cost of changing

the limitations of 200B and 150A. If either of these limitations are

changed by one unit, then the total cost will change by ₦5. The -30

under column C indicates an over production of 30 units.

3.10 Summary

The chapter has discussed how to measure of objective function and

constraints in linear programming, formulate linear programming

problems, solving linear programming problems using graphical

problem. Furthermore, the chapter discussed the meaning of the dual

price and the slack, the calculations involved and setting up initial

76

tableau for the simplex method. Also discussed are transforming the

tableau to a situation and Interpretation of the final tableau.


1. A firm manufactures two products each of which must be

processed through departments I and II. Department I produces 3 of

the first product and 2 of the second and it can only produce 120 in a

week. Also department II can produce 300 in a week with 4 of the

first product and 8 of the second. I f the first product cost ₦120 per

carton and the second cost ₦N200 per carton. Express the above

statement in linear programming to maximize the contribution.

2. Kajewole Sawmill has been requested to supply the following

minimum units of planks to GAG Construction Company: 20 units

of grade I planks, 7 units of grade II and 10 units of grade III. The

Saw miller’s vehicle operates from two different locations A and B.

Three number of units delivered on each trip from A and B are as

follows:

Grade I Grade II Grade III

From A

From B

4

2

1

1

1

2

If A and B daily trips are made, from A cost ₦300 and each from B

cost ₦400, find the numbers daily trips from A and B which

minimizes total cost and hence the maximum cost of each Sawmiller

using graphical method.

77


Akingbade, Funso (1996). Based Operational Research Techniques,

Panaf Publishing Inc.Bariga, Lagos, Nigeria.


Management.Saban Publishers, Lagos.

Bronson, R. and Naadimuthu, G. (2004).Schaum’s Outline of Theory

and Problems of Operations research, 2nd Edition, Tata McGraw-Hill

Publishing Company Limited, New Delhi

Carter, M.W. and Price, C.C. (2001). Operations Research, A

Practical Introduction.

Curwin Jon & Slater Roger (2001).Quantitative methods for business

decisions.5th ed., Thomson, Gray publishing Singapore.

Divivedi, D.W. (2002). Management economics 6th Ed. Vikis

Publishing House PVT, newDelhi

Kalavathu, S. (2000).Operational research.vikas publishing House.

PVT Ltd, New Delhi, India.

Lucey, T. (2002). Quantitative Techniques, 6th Edition, BookPower,

London

78

CHAPTER FOUR

TRANSPORTATION MODEL


At the end of this chapter, the readers should be able to:

• Describe the nature of a transportation problem

• Know the technique of solving unbalanced transportation

problem

• Understand the concept of degeneracy

• Find the initial feasible solution using the North West Corner

Rule

• Find the initial feasible solution using the Least Cost Method

4.1 Introduction

Transportation model is a special class of Linear Programming

problem in which the objective is to transport a single commodity or

goods from various sources or origin to different destinations at a

minimum total cost.

The objective is to determine the quantity shipped from each source

to each destination that minimize the total shipping cost while

satisfying both the supply units and the demand requirements.

In the statement of transportation problem, the total supply available

at the origin (ai), the total quantity demanded by the destination (bij)

and the cost (Cij) of transporting or shipping a unit of the commodity

from a known origin to a known destination are given.

Table 4.1: General Pattern of a Transportation Problem

79

Destination

1 2 3 … N Supply

Origin 1 C11 C12 . … . a1

2 C21 .. . … . a2

M . . . … Cmn Am

Demand b1 b2 b3 … bn

An example of a transportation problem is given below:

Balanced Transportation Problem

Destination

1 2 3 4 Supply

Origin 1 23 17 26 23 90

2 28 27 23 29 70

3 23 24 37 24 60

Demand 60 80 40 50 210

From above Mathematical model, it could be seen that a

transportation problem is a special class of a linear programming

problem. Again, a transportation problem is said to be balanced if

1 1

m n

i j

i j

a b= =

= ; otherwise

1 1

m n

i j

i j

a b= =

is said to be unbalance

It is a necessary condition that a transportation problem must be

balanced before it could be solvable. That is total supply must be

equal to total demand. This condition is also known as the “Rim

Condition”. But when the problem is an unbalanced one, there is the

80

need to create a dummy origin (row) or destination (column) for the

difference between total supply and demand with zero cost in order

to create the balance.

4.2 Methods of Solving Transportation Problem

The solution to transportation problem involves two phases. The

first phase is to obtain the initial basic feasible solution and the

second phase is obtaining the optimal solution.

There are various methods of obtaining the initial basic feasible

solution. However,

we shall look at,

(i) North West Corner Route

(ii) Least Cost Method

Note: that any of the two methods must satisfy the following

conditions in order to obtain the initial solution:

(a) The problem must be balanced

(b) The number of Cell’s allocation must be equal to m + n – 1,

where m and n are numbers of rows and columns respectively. Any

solution not satisfying the two conditions is called “Degenerate

Solution”.

(i) North West Corner Route (NWCR)

The method is the simplest but does not take into account the

cost of transportation for all the possible alternative routes, hence

it is the most inefficient as it has the highest total transportation

cost compared to other methods.

81

The algorithm needed to solve a transportation problem by

NWCR is:

Step 1: Begin by allocating to the North West Cell of

transportation matrix the allowable minimum of the supply and

demand capacities of that cell (min (a1, b1)).

Step 2: Check if allocation made in the first step is equal to the

supply (demand) available at the first row (column), then cross-

out the exhausted row (column) so that no further assignment

can be made to the said row (column). Move horizontally

(vertically) to the next cell and apply step 1.

In case a1 = b1, move to cell (2, 2) and apply step 1

Step 3: step 2 should be continued until exactly one row or

column is left uncrossed in the transportation matrix. Then,

make allowable allocation to that row or column and stop.

Otherwise, return to step 1.

Example 4.1

A multinational company located in Abuja has 3 plants (A, B, C)

where its goods can be produced with production capacity of 70,

60, 50 per month respectively for a particular product. These

units are to be distributed to 4 points (X, Y, W, Z) of

consumption with the demand of 70, 70, 30 and 10 per month

respectively.

The following table gives the transportation cost (in Nigeria)

from various plants to the various points of consumption.

82

Obtain the initial basic feasible solution by NWCR

Solution:

The table given below resulted from the NCWR algorithm

applied to this problem.

DESTINATION

Plant X Y W Z Supply

A

31

70

28 37 32 70 0

B

29 28

60

34 30 60 0

C 34 35

10

38

30

35

10

50 40 10 0

Demand 70

0

70

10

0

30

0

10

0

180

Total cost = N{70 (31) + 60 (28) + 10 (35) + 30 (38) + 10 (35)}

= N5,690

DESTINATION

X Y W Z

Plant A 31 28 37 32

B 29 28 34 30

C 34 35 38 35

83

Explanation to the above allocation beginning from Cell X11

• Allocate 70 to cell X11 in order to satisfy the minimum

of demand and supply capacities, zero balance is left for

both demand and supply.

Hence, row 1 and column 1 are crossed out

• Move to cell X22 and allocate 60. The supply balance is

zero while the demand balance is 10. Hence, row 2 is

crossed out

• Next, move to cell X32 and allocate 10 giving the balance

of zero for demand and 40 for the supply. Column 2 is

therefore crossed out.

• Then move to cell X33 and allocate 30 to exhaust the

demand and having 10 balance for supply. Hence,

column 3 is crossed out.

• Finally, 10 is allocated to cell X34.

(ii) Least Cost Method (LCM)

In this method the least route is always the focus for

allocation. It is a better method to NWCR because costs are

taken into consideration for allocation. The algorithm is

given below:

Step 1: Assign as much as possible to the smallest unit cost

(ties are broken arbitrarily). Also bear in mind the idea of

allowable minimum of supply and demand capacities as

done in NWCR.

Step 2: Cross out the exhausted row or column accordingly.

If both row and column are exhausted simultaneously, only

one is crossed out (in order to avoid degenerated case).

84

Step 3: Check for the smallest cost in the uncrossed row or

column and assign the allowable quantity. Repeat this

process until left with exactly one uncrossed row or column.

Example 4.2

Use the data in example 4.1 to determine the initial basic

feasible solution using the least cost method.

Solution:

Using the above steps gives the following table below

DESTINATION

Plant X Y W Z Supply

A

31 28

70

37 32

70 0

B

29

40

28

20

34 30 60 40 0

C 34

10

35 38

30

35

10

50 40 10 0

Demand 70

10

0

70

20

0

30

0

10

0

180

Total cost = N{70 (28) + 20 (28) + 40 (29) + 10 (34) + 30

(38) + 10 (35)} = N5,510

85

4.3 Optimality Test

The second phase of solving transportation problem is the

optimality test. In this phase it is also required that obtained

initial solution should be balanced and non–degenerated.

There are two popular methods usually used for

optimization. The stepping stone method and the modified

distribution (MOD 11) method. In this study pack, we shall

look at the stepping stone method.

Stepping Stone Method

The principle of simplex method where the given solution

minimizes the objective function only if the relative cost

coefficients of the non–basic variables are greater than or

equal to zero, is applied in carrying out the optimality test.

The method consists of computing relative cost coefficients

for each non–basic variable. The relative cost coefficient is

obtained by increasing a non-basic variable by one unit and

then computes the resulting change in the total transportation

cost. The most negative of the (m – 1) (n – 1) relative cost

coefficients will be taken as new basic variable.

Example 4.3

Given the following transportation problem:

Source Destination

X Y Z Supply

A 4 6 8 40

B 5 8 12 50

Demand 30 25 35 90

(a) Allocate by NWCR

(b) Obtain the optimal solution from the allocation result in

86

(a) by stepping stone method.

Solution:

(a) Destination

Source X Y Z Supply

A 4

30

6

10

8 40 10 0

B 5 8

15

12

35

50 35 0

Demand 30 0 25 15 0 35 0 90

Total cost = N{ (30 x 4) + (10 x 6) + (15 x 8) + (35 x 12)}

= N720

(b) Let us increase the non–basic variable X21 from 0 to

1. Then we need to adjust some allocations in order to

satisfy the demand and supply constraints as shown in the

following diagram below:

X Y

A –1 +1

B +1 –1

The relative cost coefficient (Cij) is computed as

C21 = 5 (+1) + 4 (–1) + 6 (+1) + 8 (–1)

= 5 – 4 + 6 – 8 = – 1

Similarly for the non –basic variable X13, we have

C13 = 8 (+1) + 12 (–1) + 8 (+1) + 6 (–1)

= 8 – 12 + 8 – 6 = – 2

Since C13 is more negatives than C21, X13 becomes new basic

variable by adjusting the quantities in cells X13, X23, X22 and

X12 so that one of the old basic variables becomes non –

basic. The adjustment is as follows:

87

6

10 –

8

+

8

15 +

12

35 –

= 10 will make X12 non basic, hence X13 = 10, X22 = 25

and X23 = 25, X12 = 0

These new values in the overall table will yield the following

table below:

Source X Y Z Supply

A

4

30

6

8

10

40

B

5 8

25

12

25

50

Demand 30 25 35 90

New total cost = (30 x 4) + (10 x 8) + (25 x 8) + (25 x 12)

= N700

Again, let us increase non –basic variable X21 by 1 in the

new table, we have

X Y Z

A –1 +1

B +1 –1

C21 = 5 (+1) + 4 (–1) + 8 (+1) + 12 (–1)

= 5 – 4 + 8 – 12 = – 3

88

Similarly, C12 = 6 (+1) + 8 (–1) + 12 (+1) + 8 (–1) = 6 –

8 + 12 – 8 = 2

hence, it is X21 that can be made basic variable as follows:

4

30 –

8

10 +

5

+

12

25 –

If = 30, X11 will become a non–basic variable, hence X21 = 30, X13

= 40, X23 = – 5 and X11 = 0

These new values in the overall table will yield the following

Source X Y Z Supply

A

4

6

8

40

40

B

5 8

25

12

-5

50

Demand 30 25 35 90

The total cost here is equal to N(40 x 8) + (30 x 5) + (25 x 8)

+ (12 x -5) = N610

Since the non–basic variable X11 and X12 give positive

relative cost coefficients, the optimal level has been reached

and therefore, iteration stops.

Hence, the optimal solution is N610.

30

89

4.4 Formulation of Assignment Problem

Assignment problem can be formally defined as: Having n jobs, n

facilities and the effectiveness of each facility for each job, the

optimized measure of effectiveness is based on assigning each

facility to one and only one job.

The mathematical expression of an assignment problem is given by

Minimize Z = 1 1

m n

ij ij

i j

C X= =

Subject to 1

1n

ij

j

X=

= (i=1,2,........n)

1

1n

ij

i

X=

= (j=1,2,........n)

The application areas of this problem are in

i. Assigning various jobs to various heads of department in an

organization

ii. Assigning jobs to various machines

iii. Assigning tractors to trailers in order to pick them up to

centralized depot etc.

4.5 Solution to Assignment Model

The Hungarian method or reduced matrix method is the commonly

used technique in solving assignment problem.

The method consists of the following algorithm

Step 1: Develop a cost table from a given balanced problem. If the

problem is unbalanced, a dummy row or column with zero cost is

used to balance up.

Step 2: Determine the opportunity cost table as follows

90

a. Subtract the lowest entry in each column of the cost table from

all entries in that column.

b. Subtract the lowest entry in each row of the table obtained in

2(a) from all the entries in that row.

Step 3: Determine whether an optimal assignment has been made.

The procedure is to draw straight line vertically or horizontally

through the total opportunity cost in each such a manner as to

minimize the number of lines necessary to cover all zero cells.

An optimal assignment is reached when the number of lines is equal

to the number of rows or columns. If an optimal assignment is not

feasible, we move to next step.

Step 4: (a) Select the smallest number in the table that is not covered

by a straight line and subtract this number from all numbers not

covered by a straight line.

(b) Add the same lowest number selected in 4(a) to the number lying

in the intersection of any two lines.

Step 5: Go to step 3

Example 4.4

A multinational company has just opened four new branches located

at four different locations (A, B, C, D) in Nigeria and wants to assign

branch managers (X, Y, W, Z) who will head these branches. Based

on the following table of cost (in thousand Naira) implications on

each manager, determine the optimal assignment.

91

Branch

Manager A B C D

X 51 55 64 44

Y 90 46 115 64

W 56 46 96 72

Z 70 75 86 71

Solution:

Applying step 2(a) to the problem we have

Table 1

Applying step 2(a) to table 1, we have

Table 2

A B C D

X 0 9 0 0

Y 39 0 51 20

W 5 0 32 28

Z 0 10 3 8

Also, applying step 3 led to straight lines drawn on zero cells in table

2.

A B C D

X 0 9 0 0

Y 39 0 51 20

W 5 0 32 28

Z 19 29 22 27

92

Here, the lines in table 2 are not equal to the number of rows or

columns. Hence, we proceed to step 4.

Table 3

A B C D

X 3 12 0 0

Y 39 0 48 17

W 5 0 29 25

Z 0 10 0 5

Application of step 3 to table 3 resulted to the straight lines. Hence,

we move again to step 4.

Table 4

A B C D

X 3 17 0 0

Y 34 0 43 12

W 0 0 24 20

Z 0 15 0 5

Here, since the number of lines is equal to 4, the problem is optimal.

We can then assign:

Manager X to Branch D

Manager Y to Branch B

Manager W to Branch A

Manager Z to Branch C

Therefore,

Total cost = 44 + 46 + 56 + 86=232

In summary, transportation problem is a special class of linear

programming problem whose structure allows the development of an

93

efficient computational technique. The technique is based on duality

theory. The following methods

(i) The North–West Corner Rule (NWCR)

(ii) Least Cost Method (LCM)

(iii) Vogel’s approximation method (VAM): etc. are the

common methods used to obtained the initial basic feasible

solution for any transportation problem.

The stepping stone method is used to obtain the optimal allocation,

while the Hungarian method is used to solve the assignment

problem.

4.6 The Branch and Bound method

The Branch and Bound method of assignment model uses an

iterative approach to find an optimal assignment of demand to

destinations. This method used an approach similar to the decision

tree method. The optimum solution is reached when the objective

function value is a minimum value.

4.7 Summary

• The transportation problem is a special type of linear

programming problem

• Two methods have been used to find the initial feasible

solution

• (i) the North West Corner Rule (ii) the Least Cost Method

• The objective of the transportation problem is to minimise

total transportation cost

• The number of cells having items allocated to them in the

initial feasible solution must be equal to m + n - 1, where m

and n are the number of rows and columns respectively. If

this criterion is not met then degeneracy occurs.

94


1. The table below shows the supply of tones of Groundnut

from Zaria, Kaduna and Kano and the demand for

Groundnut at Ibadan, Osogbo and Lagos. The figures in the

cells are unit cost (₦) of transporting a tone of groundnut

from source to destination.

Destinations

Sources Ibadan Osogbo Lagos Supply

Zaria 90 25 120 1500

Kaduna 80 60 120 2400

Kano 130 110 150 1100

Demand 800 2900 1300

(a) Determine the initial transportation cost using

(i) North west corner rule (ii) Least cost method

(b) Comment on your results in (i) and (ii) above

2. A civil engineering company in Lagos has just secured a contract

from the federal Government of Nigeria to rehabilitate four

major roads in the Federal Capital city Abuja. The company

wants to assign four site engineers who will supervise the jobs.

Based on the following table of cost in millions of Naira,

determine, using the Hungarian method, the optimal assignment.

Road

Site Engineers A B C D

1 51 55 64 44

2 90 46 115 64

3 56 46 96 72

4 70 75 86 71

95


Adedayo, O.O., Ojo, O. &Obamiro, J.K. (2006). Operations

Research in decision analysis and Production Management. Pumark

Nigeria Limited, Lagos


Linear Programming.

Akinbade, F. (1996).Basic Operational Research Techniques.Panaf




Kambo N.S. (1991). Mathematical programming Techniques.

Affiliated East-west press private limited, New Delhi, India.

96

CHAPTER FIVE

PROJECT PLANNING AND SCHEDULING



• define some concepts in a Network Analysis

• highlight rules in drawing a Network graph

• discuss extensively maxima flows in Network Analysis

• explain with illustrative examples, the shortest routes in Network

analysis

5.1 Introduction

A network is a graph which consists of a number of nodes or

junction points each joined to some or all of the others by arcs or

links or lines. A network is a graph such that a flow can take place in

the branches of the graph. A network may or may not be oriented

(orientation information, profit etc.) examples of network include

road networks, liquid networks.

On completion of this chapter, the readers should be able to state,

explain and apply network analysis to business-oriented situations as

a way of mapping out the series of tasks that are necessary to

complete a complex project.

Project planning and scheduling techniques, is a set of operations

research techniques needed and useful for planning, scheduling and

controlling large and complex project. The diagrammatical

representation of these techniques, known as network diagram,

consists of arrows (activities) and nodes (events).

97

Some possible areas of application of network analysis include:

i. Preparation of proposals and bids for large project.

ii. Projects such as moving to new house, office/house

renovation etc.

iii. Construction project such as house, bridge, road, highways

and many others.

5.2 Some Basic Terms in Network

a. Activities: an activity represents an action, project operation

which is a time and resource required effort to complete a

particular path of overall project. It has a starting point and a

point where it ends i.e. ( ).

b. Dummy activity: this is an activity which carried zero

duration because it does not consume either time or resource

i.e. (….. ).

c. Event: an event represents the start or completion of an

activity and as such does not consume time and resource. It

is usually represented by a circle (0) as node in the network

diagram.

5.3 Methods of Project Planning and Scheduling Analysis

i. Critical path method (CPM) analysis

ii. Program evaluation review technique (PERT) analysis

5.3.1 Network Calculations

The objective of network calculations is to determine the overall

duration of the project so that either a delivery date can be given to

the customer or so that we can consider what are the alterations

98

necessary for the project to be completed on or before a date to

which we are already committed.

To perform the network calculations, two things are required:

(a) An activity network representing the project

(b) The durations of all the activities in that network.

(c) Earliest Start Time (EST): The earliest start time for each

activity is calculated from the beginning of the network by

totaling all preceding activity durations (d). Where two or

more activities lead into one event, the following activity

cannot begin until the preceding activities are completed.

Consequently, the last of these activities to finish

determines the start time for the subsequent activity.

Therefore, when calculating earliest start time, work from

the beginning of the network and use the largest numbers at

junctions.

(d) Latest Finish Time (LFT): This is calculated from the end

of the project by successively subtracting activities

durations from the project finish time. Where two or more

activities stem from one event, the earliest of the times will

determine the last finish time for previous activities.

Therefore, when calculating latest finish times, begin from

the end of the network and use the smallest numbers at

junctions.

(e) Earliest Finish Time (EFT): The earliest finish time for any

activity is determined by that activity’s earliest start time

and its duration i.e. for any activity: EFT= EST+d

99

(f) Where d = duration.

(g) Latest Start Time (LST): The latest start time for any

activity is determined by that activity’s latest finish time

and its duration, i.e. for any activity: LST= LFT- d

(h) Total Float: The total float is the difference between the

times available for any activity and the time required.

(i) Note: Time available= LFT- EST

(j) Time required= d

(k) Total float= LFT – EST – d

(l) The Critical Path: The critical path is the largest path

through the network. Any delay in the activities on the

critical path will delay the completion of the project,

whereas delay in activities not on the critical path will

initially use up some of the total float on that path and not

affect the project completion time.

5.3.2 Critical Part Review Method (CPM) Analysis

The analysis of identifying the critical path CPM involves three

phases. These phases are:-

i. Forward pass computations,

ii. Backward pass computations, and

iii. Float (slack) computation

5.3.3 Forward Pass Computation

This is also known as computation of earliest time. The

cumulative durations of jobs from the start and to the end of a

100

project are obtained by this computation. Computation starts at

node 1 and advances recursively to the end of the node.

The following steps are required for forward pass.

Step 1: Set to indicate that the project starts at time 0

Step 2: obtain the next by the general formula:

= Maximum of all for all I, j leading to the events

Is earliest expected time at time zero

is earliest expected time of the successor event j

is the expected time of activity ij (i.e. i )

I = 1, 2, 3,… n-1

J = 2, 3, 4,… n

5.3.4 Background Pass Computation

This is known as latest time computation. The computation starts

from the last event (n) and proceed in descending order of the

node to the initial event (i). The following steps are required:

Steps 1: set

Step 2: obtain the next by the general formula

101

= minimum of all for all I, j coming from i

Where is the latest time for event n

is the latest allowable time for event i

is the latest allowable time for event j, and is the expected

time for activity

5.3.5 Float (Slack) computations

Float or slack refers to the amount of time by which a particular

activity or event can be delayed without the time schedule of the

network being affected. The term ‘float’ is used in CPM when

activities are considered, while slack is used in PERT when

events are involved. For the computation, float or slack can be

obtained by taking away the earliest expected latest allowable

time i.e. float (slack) = . The events (activities) with

zero float (slack) will give the expected critical path.

Example 5.1

A renovation project in Lagos has to pass through six tasks. The

duration and the precedence order of these tasks are given below:

Tasks (activity) Predecessor activity Duration in days

A -- 24

B A 50

C A 78

D B 103

E C 130

F D, E 155

102

ii. Construct the network diagram for the project

iii. Determine: (a) the forward pass (b) the backward pass

(c) the critical path

Solution:

b. (i) the forward passes, using the formula

50

130 1 2

3

4

5 6

A

24 B

78 C E

D 103

F 155

103

(ii) The backward passes are also obtained by formula:

(iii) The float=

Event Float(slack)

1 0 0 0

2 24 24 0

3 52 50 2

4 54 52 2

5 79 79 0

6 104 104 0

The critical path is 1-2-3-5-6 with activities A-C-E-F

104

The critical length is 24+28+27+25=104 days.

5.4 Project Evaluation Review Technique (PERT)

The duration of activities in PERT are based on three-time

estimate of performance which is probabilistic in nature. The

three-time estimates are:

i. The most optimistic time

ii. The most pessimistic time

iii. The most likely time

Where, ,

Hence, the expected activity time

The variance of time required to complete an activity is given by

( )2

2 1

6p ot t

= −

Example 5.2

A project consists of seven activities whose three-time estimates

are given below.

105

Activity Activity Name Time estimates (days)

1-2 A 3 3 12

1-3 B 3 6 9

1-4 C 6 6 15

2-5 D 3 3 3

2-5 E 6 15 24

4-6 F 6 15 21

5-6 G 9 15 27

a. Find the expected duration and variance for each

activity.

b. Draw a network diagram.

c. What is the expected project length (duration)?

a. Calculate the variance and standard deviation of the project

length.

Solution:

Activity’s Time

estimates

= ( )2

2 1

6p ot t

= −

A 3 12 3 4.5

B 3 9 6 6 1

C 6 15 6 7.5

D 3 3 3 3 0

E 6 24 15 15 9

F 6 21 15 14.5

G 9 27 15 16 9

106

(b) The Network diagram

D=3

A=4.5 G=16

E=15

B=6

C=7.5

F=14.5

(c) The critical path: to find the critical path,

The possible routes are:

1-2-5-6, total duration = 4.5+3+16=23.5 days

1-3-5-6, total duration = 6+15+16 = 37 days

1-4-6, total duration = 7.5+14.5 = 22 days

Hence, 1-3-5-6 is the critical path with 37 days.

Therefore, the expected project length is 37 days.

(d) The variance of the project length is 1+9+9 = 19 days

Hence, standard deviation = = 4.36 days

1

2 5

6 3

4

107

As a result, project planning and scheduling is an important concept

that deals with project management and it helps in having an

efficient way of reducing or having low cost of its completion time.

The two well-known methods of project analysis are CPM and

PERT. These are respectively used for activities, which are

deterministic and probabilistic in nature.

Example 5.3

Draw the network diagram for the following problem

Activity Preceding

Activity

1 -

2,3,4 1

5 2

6 3

7 5

8 6

9 7,8

10 3

11 4

12 9,10,11

108

5

2 7

3 6 8

9

10

4

11

Example 5.4: (a) Draw the network diagram for the following

problem:

Activity Immediate Predecessor Activity

Durations (Days)

A - 2

B A 3

C A 4

D A 5

E B 6

F C,D 3

G D 4

H B 7

I E,F,G 2

J G 3

1

12

109

(b) identify the available paths and determine the critical path.

Solution:

B E H

3 6 7

A C F I

2 4 4 3 3 2

D J

5 dummy 0 dummy 0 3

G 3

3

(b) The available paths are:

A B H A,B,H

A C F I OR A,C,F,I

A B E I A,B,E,I

A C F I A,D,G,J

1

00

2

20

4

70

6

110

8

140

5

70

7

110

3

50

110

Hence, the critical path is A D G J, because it is the

longest path i.e

2 + 5 +9 +3 = 19days.

Now that you have drawn the correct logical structure of the

activities, the minimum completion time for the project and have

identified which activities are on the critical path.

Example 5.5: Calculate the Earliest Event Time (EETs) of

example5.2 network diagram drawn above.

B E H

3 6 7

A C F I

2 4 4 3 3 2

D J

5 dummy 0 dummy 0 3

G 3

3

Explanations

Take your time to understand the following explanations very well

and understand the logic involved in calculating the EETs.

Event 1: The earliest event time is 0(starting time)

Event 2: The earliest event of Activity A and it is the point at which

Activities B, C and D could start. These Activities must take place

1 00 2 20 4 70 6 110 8 14

0

5 70 7 11

110

3 50

111

after activity A. Activity A take 2 days. The earliest that Activities

B,C and D could start is after 2 days.

Event 3: Activity H cannot start until Activity B has been

completed. Activity B can not start until Activity A has been

completed at event 2. Therefore, the earliest that Activity H could

start is the EET at event 2(2 days) plus the activity duration of B (3

days) which is

2+3 =5 days

Similarly, for event 4, EET is 2+5 = 7days, for Event 5, EET is 2+5

= 7 Days and so on.

Example 5.6

Calculate the latest event time (LET) of example 2 network diagram

drawn above.

The latest event time is enter into the segment at the bottom right

hand side of each event circle. This latest event time as previously

defined as the latest time by which the previous activity (or

activities) must finish if the entire project is to be completed in the

minimum possible time. It should be noted here that, the latest time

is the same as the earliest event time for any event on the critical

path. Therefore, the LETs are then entered into the l-network

diagram given below:

112

Explanations:

Event 8: This is the end of the project. The LET = EET = 14days.

Event 7: LET (11 days) is the LET for event 8(14 days) minus the

duration of Activity J (3 days).

Event 6: LET (12 days) is the LET for event 8(14 adys) minus the

duration od Activity I (2 days).


duration od Activity H (7 days).


duration od Activity F (3 days).

Similarly, LETs for event 3,2,1 follows the same procedures.

5.5 Summary

In this chapter, we have learnt the following:

- Explanation and definition of network terminologies: activity,

event, dummy and float.

- Drawing a Network diagram based on Arrow-on-node diagram.

- State and explain the origin of PERT and CPM techniques as aids

to efficient project management.

- List some application of PERT and CPM in project management.

-Identifying all the paths, critical path and critical activities in a

network diagram and calculation.

113

-Explain and evaluate the earliest and latest event times, float times

and project completion times.

-Estimate optimistic, pessimistic and most likely time.

-Calculation of floats and interpretation of their values


1. A project consists of the following activities and time estimates:

Predecessor

event

Successor

event

Most

optimistic

time

Most

likely

time

Most

pessimistic

time

1 2 4 3 12

1 3 8 11 12

1 4 8 14 24

2 5 5 9 10

3 4 2 6 8

3 5 2 4 8

4 5 6 10 14

5 6 1 3 6

You are required to:

i. Construct the operational network.

ii. Determine the critical path.

iii. Calculate the mean time of the critical path and standard

deviation of the critical path

iv. Determine the project duration time.

2. The data given below relates o a project as estimated by the

operational manager of a multinational company based in Lagos,

Nigeria.

114

Activities Preceding activities Duration (months)

A - 10

B - 7

C A 6

D B 8

E A 9

F G,D 12

G E 8

H G,F 9

i. Draw an activity-on-arrow diagram to represent the project.

ii. Identify the non-critical activities in the network.

iii. Identify the critical path.

iv. Calculate the critical path duration.


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in Decision Analysisand Production Management.Pumark Nigeria

Limited, Lagos


Linear Programming.

Akinbade, F.(1996). Basic Operational Research Techniques.Panaf

Publishing Inc., Bariga,Lgaos, Nigeria



115


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116

CHAPTER SIX

INVENTORY CONTROL AND MANAGEMENT



• Know the concept of inventory control

• Identify different types of inventory

• Understand inventory control terminology

• Understand basic Economic Order Quantity (EOQ) model

and its assumptions

• Describe and solve problems on some other inventory

models.

6.1 Introduction

Inventory is generally described as useable resources that bring in

earnings when it is dynamics. In an organisation various types of

items are usually stored as inventory. Inventories are kept in order to

ensure that shortages of needed goods are averted. It is necessary to

hold adequate stock of materials in a firm or any organization in

order to minimize production hold – ups and win customer

satisfaction. As materials are requested for, there is need for

constant reviews of inventory in order to reduce the capital tied down

without affecting the production and customer goodwill. Hence,

inventory management has received much attention in these modern

days. In short form, the inventory control and management deals

with when and how much to order an inventory item and emphasis is

on the minimization of the total cost associated with inventories.

6.2 Inventory Control Terminology

117

Brief definitions of common inventory control terms are

given below:

i. Lead or procurement time: The period of time,

expressed in days, weeks, months, etc between

ordering and replenishment.

ii. Demand: The amount required by sales, production,

etc. usually expressed as a rate of demand per week,

month, etc.

iii. Economic Ordering Quantity (EOQ): This is a

calculated ordering quantity which minimizes the

balance of costs between inventory holding costs

and re–order costs.

iv. Physical stock: The number of items physically in

stock at a given time.

v. Free stock: Physical stock plus outstanding

replenishment orders minus unfulfilled

requirements.

vi. Buffer stock or Minimum stock or safety stock:

A stock allowance to cover errors in forecasting the

lead time or the demand during the lead time.

vii. Maximum stock: A stock level selected as the

maximum desirable which is used as an indicators to

show when stocks have risen too high.

viii. Re-order level: The level of stock at which a

further replenishment order should be placed.

ix. Re-order quantity: The quantity of the

replenishment order.

6.3 Economic Order Quantity (E.O.Q) Analysis

118

The EOQ has been previously defined as the ordering

quantity minimizes the balance of cost between inventory

holding costs and re–order costs.

To calculate a basic EOQ certain assumptions are necessary

i. That there is a known, constant stock holding cost

ii. That there is a known, constant ordering cost

iii. That rates of demand are known

iv. That there is a known, constant price per unit;

v. That replenishment is made instantaneously

vi. No stockouts allowed

The EOQ formula is given below,

Where Co = ordering cost per order

D = Demand per annum

Cc = Carrying cost per item per annum

The method gives an exact answer and the calculation is

based on estimates of costs, demands etc which are of

course, subject to error.

Example 6.1

A company uses 50,000 certain material per annum which

areN10 each to purchase. The ordering and handling costs

are N150 per order and carrying costs are 15% per annum,

calculate the EOQ.

Solution:

119

here, CO = N150

D = 50,000 materials

Cc =N10 X 15% = N1.5 per material

= 3,162 materials

6.4 EOQ Where Stockouts Are Permitted

The overall objective of stock control is to minimize the

balance of the three main areas of cost i.e. holding costs,

ordering costs and stockout costs.

Stockout costs are difficult to quantify but may be significant

where stockout costs are known than they can be

incorporated into the EOQ formula which thus becomes

Where Cs = stockout costs per item per annum

Example 6.2

Assume the same data as in example 6.1 above except that

stockouts are now permitted. When a stockout occurs and an

order is received for materials the firm has agreed to retain

the order and when replenishment is received, with delivery

cost of N0.75 per material. Other costs associated with

stockouts are estimated as N0.25 per unit. Find the EOQ.

120

Solution:

Co = N150

D = 50,000

Cc = N1.5

Cs = N0.75 + N0.25 = N1

Hence,

= 3162 x 1.58

= 4,996

6.4.1 EOQ with Shortage

Al the assumptions made in section 6.3are upheld. It is also

assumed that shortages are allowed in addition to other

conditions as earlier stated. It is for this reason that more

notations are added to the ones previously used. Therefore,

let t = the total cycle which is equal to t1 +t2

t1 = the time when no shortage exists

t2 = the time when shortage exists

S = the maximum shortage

Sc = shortage cost / penalty cost

Q = the order quantity

M = the remaining quantity when shortage is taken from

order quantity i.e. M = Q – S

Inventory level

M

121

t1 to O

O

Time

Maximum

Figure 5.1: The pattern of inventory with shortages

In the figure above, the changes in the inventory level with time can

be seen. However, it should be mentioned that in this inventory,

except from the purchase cost C, which will be fixed, all other types

of costs will be affected by the decision concerning Q and M.

Therefore, to obtain the optimal value of order quantity (Q*) and

optimal stock level (M) along with optimal shortage level S; the

following formulae are given below:

S* = Q* = M*

Other results include:

Example 6.3

A ford and beverage company based in Lagos usually supplies its

product at a constant rate of 2000 units per month. If the company

can supply any amount of its product at any required time with each

122

ordering cost of N500, holding cost of N20 per unit per month and

the penalty cost of delaying supply of N100 per unit per month, find:

i. The EOQ

ii. Optimal stock level

iii. Optimal shortage level

iv. Total cycle time

Solution:

D = 2000 cost per month

K = N500 per order

h = N20 per unit per month

Sc = N100 per unit per month

(i)

(ii)

= 288.7 units

(iii) S* = Q* - M*

= 346.4 – 288.7

= 57.7 units

123

= 0.17 of month

6.5 Summary

The chapter discussed extensively inventory as stock stored for

future use. Also, discussed in the chapter are methods for minimizing

cost of keeping inventory and maintaining prompt supply of goods

are key. Economic Order Quantity (EOQ) which minimizes the cost

of managing the inventory system to determine the optimal order

quantity is equally discussed in the chapter.


1. A particular item has a demand of 9000 units per year. The

cost of one procurement is N100 and the holding cost per

unit is ₦2.40 per year. The replacement is instantaneous and

no shortages are allowed.

Determine:

(i) The number of orders per year

(ii) The time between orders

(iii) The total cost per year if the cost of one unit is ₦10

2. Given the following data for an item of uniform demand,

instantaneous delivery time and back order facility:

Annual demand : 800 units

Cost of an item : ₦40

Ordering cost : ₦800

Inventory carrying cost : 40%

Back order cost : ₦10

Calculate

124

(iii) Minimum cost order quantity

(iv) Maximum inventory level

(v) Maximum number of back orders

(vi) Time between orders

(vii) Total annual cost


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