Lecture Notes Prepared

by

GAZİMAĞUSA- 2005

“Man is one of the best general-purpose computers available and if one designs for man as a moron, one ends up with a system that requires a genius to maintain it. Thus we are not suggesting that we take man out of the system, but we are suggesting that he be properly employed in terms of both his abilities and limitations. Some designers have required that he be a hero as well as a genius.”

E.L. Thomas “Design and Planning”, Hastings

House, NY 1967

2

PREFACE

Production/Operations Management is ubiquitous, that is, ever present. Daily we

come in contact with various goods and services produced by the transformation of inputs

to outputs under the control of production/operations managers. Production/Operations is

a basic function along with marketing and finance functions.

This “Lecture Notes” is complied from various textbooks and intended for use in

undergraduate and graduate courses. I have taught Production/Operations Management

(POM) for more than 35 years to over thousand of students, whose quantitative skills

were quite varied, with many of them exhibiting severe math anxiety. I have omitted in

the notes the theoretical treatments and have concentrated on providing an-easy-to-read,

easy-to-understand treatment of the basic POM knowledge. In other words, my aim was

to give students a solid understanding of the analytical tools necessary to solve

production/operations problems. Also the methods presented and discussed in the notes

are those readily applicable to real-world problems. With this class goal in mind, my

objective was to create a lecture notes that would develop an understanding of POM, i.e.

to give students a basic, but through, working knowledge of Production/Operations

Management.

By necessity, I have written the notes with expectation that the students will have

at least a minimal background in basic mathematics and statistics. Although I tried to

prepare the notes to be read and used without this background, I do not recommend it,

because the student will be better prepared and more receptive if mathematical and

statistical background is present.

These notes may be quite simple for some graduate students. After teaching both

types of students over many years, however, I have not found the level of the text to be an

issue. I have found that graduate students can cover the material more rapidly than the

undergraduates. Extra homework, case studies as examples of real operations problems

can be assigned to graduate students. Supplement to these notes, selected readings from

other textbooks may also be added.

The text includes both quantitative methods and POM principles. It is really the

most difficult task for us to maintain the proper balance and speed of presentation. I

3

recommend the students also to use computer techniques in connection with this course.

There are several sources of programs for general use. These programs give the students

a “feel” for how the computer can be used as an aid to decision-making operations.

I have made every effort to ensure that this “Notes” will help students learn about

the extent, substance and excitement of POM and be a useful ancilla for students

struggling with the difficult texts of POM. They should be aware that many of the

principles and concepts of POM are applicable to other aspects of their professional and

personal life. I wish them “Good Luck”.

I would like to thank a great mentor; Prof.Dr.h.c. Şükrü Fuat ERLAÇİN, who

taught me a great deal and inspired me even more. İ am greatful to the students at EMU

Department of Business Administration, who suffered through typo-ridden drafts of

earlier versions of this lecture notes. I hope this new printing will contribute to the

success of students.

I, therefore, welcome any recommendation for improvements of this lecture notes.

It is my belief that comments or suggestions for additions, deletions, corrections,

rearrangements, etc. from readers will enhance future editions.

January, 2005 M. Hulusi DEMIR, Ph.D,

Dr.Sc.

4

THE PRODUCTION/OPERATIONS FUNCTION IN BUSINESS

The purpose of production/operations is to satisfy people’s wants. Primitive man

no doubt spent a great proportion of his time attempting to satisfy his own fundamental

wants. Want of food, clothing, accommodation, etc., were important then, as they are

now, but the means by which they are satisfied have changed substantially. Initially,

individuals worked entirely for their own purposes to ensure their own survival. Later,

families, tribes and larger groups became the dominant social units, people’s wants

multiplied and the procedure, which attempted to ensure their satisfaction, changed. The

role of the worker began to evolve and groups of such workers were evolved in hunting,

farming, building, etc., to satisfy certain of the community’s needs.

With the continued development of civilization, people’s wants became even

greater and, furthermore, the wants of individuals and of groups, since they were no

longer restricted to the necessities of life, began to differ. This development necessitated

5

more complex methods of production and eventually it was virtually impossible for

individuals themselves to satisfy all or even a majority of their own wants.

The use of monetary system facilitated this development since it enabled people

to concentrate on those activities for which they were best equipped. They were able to

produce goods far in excess of their own demand, to sell these to others, and use the

money obtained to satisfy their other wants. The monetary system also enabled people to

be paid for working, with the money, which they then used for purchases. In other words,

they were engaged in production in order to satisfy the wants of others, and to earn

money by which their own wants and those of their own family might be satisfied-they

were both producers and consumers.

Wants need not to be restricted to the acquisition and use of goods but will also

extend to the use of services. The services currently in demand are varied and extensive,

ranging from professional services such as those of lawyers, architects, barbers, to the use

of libraries, distribution systems, transport, communication, etc. Accordingly, therefore,

the definition of production must also be concerned with the provision of services.

Production in a transport organization, such as State Railways, could be described

as the conversion of certain “inputs” such as rolling stock, railway track, staff, etc., into a

distribution or transport service. One of the production functions of a bank might be

described as the conversion of deposits into loans, and retailers too might be said to be

evolved in production as they convert their bulk orders from wholesalers into the single

commodities wanted by their customers. Bus and taxi services, motels and dentists,

tailors, fire services, hospitals are all production systems. They all, in effect, convert

inputs in order to provide outputs, which are required by a customer. The outputs of a

production system are normally called “products”; these products may be tangible goods,

intangible services or a combination.

This description of the production will be familiar to any reader who has studied

or to read economics and/or business. But at the same time it is unlikely to meet the

approval of some people, who see production as being concerned primarily with creating

or manufacturing goods rather than services. In order to differentiate we prefer to use the

term "manufacture" solely for "fabrication or assembly of a physical object by means of

equipment, men and materials". Since the structure of these short notes to some extent

6

reflects a particular view of the nature of operating systems and operating management,

we shall, for brevity, use the term "production/operations".

A production/operations system is a configuration of resources combined for the

provision of goods or services.

A production/operations system is that part of the organization that exists primarily

to generate and produce the organization products, i.e. goods or services.

Why study POM?

We study POM for four reasons:

a. POM is one of the three major functions of any organisations, and it is

integrally related to all the other business functions. All organisations market

(sell), finance (account), and produce (operate), and it is important to know how

the POM segment functions. Therefore, we study how people organise

themselves for productive enterprise.

b. We study POM because we want to know how goods and services are

produced. The production function is the segment of our society that creates the

products we use.

c. We study POM to understand what production/operations managers do. By

understanding what these managers do, you can develop the skills necessary to

become such a manager. This will help you explore the numerous and lucrative

career opportunities in POM.

d. We study POM because it is such a costly part of an organisation. A large

percentage of the revenue of most firms is spent in the POM function. Indeed,

POM provides a major opportunity for an organisation to improve its profitability

and enhance its service to society.

Resources in Production/Operations Systems

Managers in production and operations functions practice production/operations

management. They do not practice behavioural science, quantitative methods or systems

7

analysis, although they utilise these underlying disciplines. Likewise doctors do not

practice biology, although they know how to use biological methods. While

methodologies are certainly important, they are not essence of POM.

Production/operations managers are principally concerned with the use of

physical resources; therefore we shall take a physical view of production/operations

systems and concentrate on the physical resources used by the system, which for

convenience will be categorized as follows:

1. Materials, i.e. the physical items consumed or converted by the system, e.g. raw

materials, fuel, indirect materials.

2. Capital, i.e. the physical items equipment and facilities, used by the system, e.g.

plant, tools, vehicles, buildings.

3. Human Resources, i.e. the people, workers and managers, who provide or

contribute to the operation of the system, without which neither machine nor

materials are effectively used.

"External Information" may be added as input of a production/operations system in

addition to physical resources.

The production/operations manager's job is to manage the process of converting

inputs into desired outputs. Our definition of production/operations management is,

then, the management of conversion process, which converts land, labour, capital and

management input into desired output of goods and/or services. In conjunction with other

functional areas, it also deals with the management of resources (inputs) and the

distribution of finished goods and services to customers (outputs). In conjunction with

other functional areas, it also deals with the management of resources (inputs) and the

distribution of finished goods and services to customers (outputs).

Production/Operations Management is the systematic direction and control of the

process that transforms inputs into finished goods or services.

As mentioned before, the term "Production/Operations Management" evolved

from factory oriented terms like "manufacturing management", "production

8

management" and "production operations", but its present meaning has been broadened to

embrace service industries and nonprofit activities as well. The underlying theory of

production/operations management is common to both goods and services production.

Forecasting, scheduling, quality control and other managerial activities have much in

common from one type of operation to the next. Thus most of the theory is as applicable

to the management of university hospital or airline operations as it is to the manufacture

of televisions.

Fig.1 represents operations where resources are utilised and transformations

occur. The heart of POM is the management of production systems. A

“Production/Operations System”, as defined above, uses operations resources to

transform inputs into desired output. An “Input” may be a raw material, a customer, or a

finished product from another system. Operations resources consist of what we term the

“Five P’s of POM”: people, plants, parts, processes, and planning and control systems.

“People” are direct and indirect workforce. “Plants” include factories or service branches

where production is carried out. “Parts” include the materials (or, in the case of services

the supplies) that go through the system. “Processes” include the equipment and steps by

which production is accomplished. “Planning and Control Systems” are the procedures

and information management uses to operate the system.

VALUE ADDED

9

CUSTOMEROPERATIONS

2

TRANSFORMATION (CONVERSION) PROCESS

OUTPUTS

GOODS

SERVICES

INPUTS

WORKERS MANAGERS EQUIPMENT FACILITIES MATERIALS ENERGY EXTERNAL

INFORMATION

Information Feedback

on Performance

EXTERNAL ENVIRONMENT

Fig. 1: The Production/Operations Management (POM) System.

Often a product passes through several such operations before being finished. An

operation can be a machine center in a manufacturing plant, a teller station in a bank or a

department in an office. The types of transformation vary widely and include physical or

chemical (a factory), locational (an airline), attitudinal (a theater), educational (a school),

physiological (an emergency room), informational (a computer center), exchange (a

store) and storage (a distribution center).

Customer may come in direct contact with the production system and sometimes

is an active participant in the transformation e.g. the student at the university.

The essence of the operations function is to add value during the transformation

process: “value-added” is the term used to describe the difference between the costs of

inputs and the value or price of outputs. In addition to this if the focus of the firm is the

customer; everything the firm does should improve matters for the customers. This gives

rise to the concept of the value-added. This concept applies not only to essential

10

1

14

3

5

manufacturing steps, but also managerial, administrative and service activities as well.

An improved design process, for example, adds value.

We therefore, may define Production/Operations as follows:

Production/Operations refers to the generation of goods and services, the set

of value-added activities that transform inputs into outputs.

Information feedback can come from external sources, such as reports on

economy, a call from a vendor on past-due shipments, or new customer orders. It can also

come from internal sources such as reports on cost variances, customer service or

inventory levels. Information from both sources is needed to manage the production and

operations system.

Outputs from manufacturing operations are goods produced for consumers or for

industrial firms. The word “manufacture” is derived from Latin factum meaning making

and manus meaning hand, but of course, the term is no longer confined solely to manual

operations, and is now used to cover both manual and machine work, or any combination

of both. Manufacturing implies production of tangible output such as an automobile, a

clock radio, and a refrigerator. Therefore manufactured goods are physical, durable

products. We can see, feel and inspect them. Outputs from service operations vary from

delivered mail for a post office to a recovered patient for a hospital. Therefore services

are usually intangible and perishable. They are often ideas, concepts or information or an

act. You cannot touch a consultancy advice or a haircut. Certainly, a management

consultancy, although it produces reports and documents, would see itself as a service

provider, which uses facilitating goods. Finally, some pure services do not produce

products at all. A psychotherapy clinic, for example, provides therapeutic treatment for

its customers without any facilitating goods. This distinction gets cloudy, when we try to

classify an organization as either a goods producer or a service producer. In reality,

almost all services are a mixture of a service and a tangible product; similarly, the sale of

most goods includes or requires a service. For instance, many products have the service

components of financing and transportation (e.g. automobile sales).

A “Pure Service” does not include a tangible product. Although there are not very

many pure services, one example is counseling. A university lecture is a pure service;

while pad of paper and the pen/pencil with which student takes notes are pure goods. In

11

reality, most products are not pure goods and services but a combination of both.

Consider the local McDonald’s, a hospital operating room or a custom homebuilder.

None are strictly service or manufacturing operations; in each case a strong service

element and a strong production element go hand in hand.

As a summary, services are “those economic activities that typically produce an

intangible product (such as education, entertainment, lodging, government and health

services).

Table 1 shows the continuum of characteristics of goods and services producers.

More Like A Goods Producer More Like A Services Producer

Physical, durable products Intangible, perishable products

Product can be resold Reselling a service is unusual

Output can be inventoried Many Outputs cannot be inventoried

Low customer contact High customer contact

Long response time to demand Short response time to demand

Regional, national or international markets Local Markets

Large facilities with economies of scale Small facilities (often difficult to automate)

Capital Intensive Labor Intensive

Quality easily measured

Site of the facility is important for cost

Selling is distinct from production

Product is transportable

Quality not easily measured

Site of the facility is important for

customer contact.

Selling is often a part of the service

Provider, not product, is often transportable

Table 1: Continuum of Characteristics of Goods and Services Producers.

Here are some examples to describe the inputs, the transformation and outputs of

the productive system:

University Library

Inputs……. Librarians, staff, library facilities and equipment, energy, capital.

Transformation Process… Organising information, arranging materials for access,

interacting with library users.

12

Outputs… … Students and faculty provided with research and study materials.

Hotel

Inputs… … Facilities, staff, materials for housekeeping and food preparation,

communications equipment, energy, capital.

Transformation Process… Taking reservations, check-in and checkout procedures,

Providing other services.

Outputs… … Customers satisfied with lodging and related services.

Small Manufacturing Firm

Inputs… … Raw materials, workers, supervisors, management, warehouse,

manufacturing facilities and equipment, energy, capital.

Transformation Process… Ordering raw materials, transforming and assembling

raw materials into final product, packaging product, taking

supply orders.

Outputs… … Customers satisfied with the way the product appears and operates.

Therefore, POM is about the way organizations produce goods and services. As

mentioned above, everything you wear, eat, sit on, use, read etc. comes to you courtesy of

P/O managers who organized production. Every book you borrow from the library, every

treatment you receive at the hospital, and every lecture you attend at university - all have

been produced.

The Evolution and Growth of Production/Operations Management

The production/operations management, POM, has an interesting and rich past

and challenging future. Although the formal study of POM is relatively young, i.e. began

only in the twentieth century, the activity itself is ancient. The Egyptian pyramids, The

Great Wall of China, The City of Ephesus represent significant production achievements.

The Egyptian pyramids, for example, vividly illustrate that tens of thousands of people

worked for many years on large-scale construction projects. These certainly required

substantial organisational and planning abilities, as well as skills in directing and

controlling the actual construction. Surely management skills must have been necessary

13

for ancient Egyptians to communicate with thousands of people every day without

modern technology of today. Decisions about the design of products, location of

facilities, scheduling of personnel and acquisition of materials clearly had to be made.

Some of the modern principles of POM were known and utilised in ancient societies. The

ancient Egyptians and Chinese developed such specialisation, with workers confining

themselves to particular crafts or trades.

Production of goods remained at a handicraft level until the industrial revolution

took hold. 1764 is often quoted as the year in which the Industrial Revolution began…

This date is an important date, because during this year James Watt invented the steam

engine and advanced the use of mechanical power to increase productivity.

The introduction of interchangeable parts by Eli Whitney (1798) allowed the

manufacture of firearms, clocks, watches, sewing machines and other goods to shift from

customised one-at-a-time production to volume production of standardised parts. This

meant the factory needs a system of measurements and inspection, a standard method of

production and supervisors to check the quality of worker’s production.

Soon afterward, steam engine was used to create the first train (by Richard

Trevithick in1802) and the first steam-boat (by Robert Fulton in 1807). This began a long

stream of applications whereby human and animal powers were replaced by engine

power.

The Industrial Revolution was the transformation of a society from peasant and

local occupations into a society with worldwide connections by means of great use of

machinery and large-scale commercial operations. It is beginning of factory system.

The traditional manufacturing system of independent skilled workers individually

pursuing their specialties was replaced by a factory system that mass-produced items by

bringing together large numbers of semi-skilled workers. The factory system profited

from savings created by large-scale production. e.g. Raw materials could often be

purchased more cheaply in large lots. Another savings came from the specialisation of

labour. Each worker concentrated on one specific task or job. Production efficiency

improved substantially and the factory system revolutionised business.

Adam Smith’s “The Wealth of Nations” (1776) publicized the advantages of the

division of labour, in which the production process was broken down into series of small

14

tasks, each performed by a different worker. His idea to increase productivity a system of

specialization or division of labor, which included:

(1) skill development on the part of workers,

(2) avoidance of lost time due to changing jobs,

(3) the use of specialized machines,

was a radical concept in 1776. It was quickly accepted and became one of the principal

factors, which enabled the Industrial Revolution to occur, and the present system of

production to evolve.

Growth of the factory system was rapid; there was no well-established craft

system to supplant, and unskilled labor was available. Specialization of jobs and division

of labor began to take place Charles Babbage, a prominent mathematician and engineer,

promoted an economic analysis of work and pay on the basis of skill requirements.

The large scale of production introduced by the Industrial Revolution created

more complexity and with it the need for better management. In the early 1900s a new

management philosophy called Scientific Management emerged. The basis for the

scientific management was the belief that was laws governing production systems just as

laws for natural systems. If those laws could be identified, they could be used to find the

best way to perform any job and the best way to make a product. The scientific laws of

natural systems were discovered through observation and experimentation. Proponents of

scientific management believed the same approach would work for discovering the laws

of production systems.

Frederick W. Taylor was dissatisfied worker in Midvale Steel Company,

Philadelphia, in the late 1800s. Advancing through the ranks to foreman master

mechanic, and chief engineer, he came to know, and deplore, the boondoggling, loafing,

and general inefficiencies that existed in his company. Taylor refused to accept such

practices. Fortunately he was advanced to a position where he could experiment with

some ideas for improvement. Believing that a scientific approach to management could

improve labor efficiency, he proposed the actions outlined in Table 2. Taylor’s

philosophy was to replace “subjective management” by “objective management” based

on science and became widely known through his consulting work and his book

“Principles of Scientific Management”, published in 1915.

15

His “shop system”, which included attention to training and instruction,

specifications, standards (by stopwatch studies), and incentive pay systems, brought him

the title “Father of Scientific Management”.

Frank and Lillian Gilbreth developed motion economy studies. They introduced

“Therbligs”. Therbligs are the basic physical elements of motion include such activities

as select, grasp, position, assemble, reach, hold, rest and inspect. Henry Gantt instituted a

charting system for scheduling production. Henry Ford, one of the Taylor’s biggest

advocates, inaugurated assembly-line mass production for automobiles. He and Charles

Sorenson combined what they knew about standardized parts with the quasi-assembly

lines of meatpacking and mail-order industries and added the revolutionary concept of the

assembly line where men stood still and material moved. Henry Ford’s focus was largely

on manufacturing efficiency;

By adopting fixed work-stations,

Increasing task specialisation,

Moving work to the worker.

So he applied scientific management to the production of the Model T in 1913 and

reduced the time required assembling a car from high of 728 hours to 1.5 hours. A model

chassis moved slowly down a conveyor belt with six workers walking along beside it,

picking up parts from carefully spaced piles on the floor and fitting them to the chassis.

The short assembly time per car allowed Model be produced T to in high volumes, or “En

masse”, yielding the name “Mass Production”.

Mass Production is high-volume production of a standardised product for

a mass market.

Ford increased productivity and lowered prices. In doing so, he also made the

automobile affordable for the average person.

Taylor and his associates concentrated on the problems of foreman,

superintendents, and lower middle managers in factories; because it was here that most

was mass production and efficiency in the factories to respond to the great western

markets.

16

The key premise of Scientific Management era was that any operation could be

improved by breaking it down into components, measuring the work content, and seeking

ways to improve work methods. Taylor’s philosophy was to replace “subjective”

management by “objective” management based on science. It centred on three ideas:

1. Scientific laws govern how much a worker can produce per year;

2. It is management’s function to discover and apply these laws to productive

operations systems; and

3. It is the worker’s function to carry out decisions without question.

In the factory a middle-level production department gained much of the control

over manufacturing issues formerly handled by the president and foreman. Therefore the

basis of scientific management is a focus on economic efficiency at the production core

of the organization. Of central importance is the belief that rationality in the part of

management will obtain economic efficiency. Economic Efficiency refers to the ratio of

outputs to input. Organizational Efficiency typically is a ratio of product or service

outputs to land, capital or labor inputs.

Efficiency (%) = (Output/Input) * 100%

Example:

The standard in a cafeteria is the preparation of 200 cheeseburgers per hour. If

labor input produces 150 cheeseburgers per hour, how efficient is the operation?

Solution:

Labor Efficiency (%) = (Labor Output/Labor Input) * 100% = (150/200) *100%

= 75%

Compared with the standard, this operation is 75% efficient in the preparation of

cheeseburgers.

17

1. Collect data on each element of work and develop standardized procedures for

workers,(i.e. establish proper work methods and tools),

2. Scientifically select, train, and develop workers instead of letting them train

themselves, (i.e. provide the proper training),

3. Strive for a spirit of cooperation between management and the workers so that

high production at good pay is fostered,(i.e. establish legitimate incentives for

work to be done, and to develop a hearty cooperation between management and

the workers),

4. Divide the work between management and labor so that each group does the work

for which it is best suited, (i.e. to match employees to the right job).

Table 2: Taylor’s Philosophy of Scientific Management

The process school of management by Henry Fayol, which is with scientific

management referred as part of classical management, was developed in the early

1900s. Management was viewed as a continuous process involving the functions of

planning, organizing and controlling by a manager who influences others through the

functions he or she performs.

The behavioural school of management began in 1920s with a human relations

movement that emerged quite unexpectedly from some research studies intended to

examine the effects of changes in the physical environment on production output – a

typical scientific management study. Elton Mayo directed attention to behavioural

factors. The famous study at Western Electric’s Hawthorne plant showed productivity

depends not only on the physical environment but also on social norms and personal

feelings. These studies indicated that worker motivation – along with the physical and

technical work environment – is a crucial element in improving productivity. This led to

a moderation of the scientific management school, which emphasized “humanizing the

work place”, as well as improving productivity.

The quantitative school of management is concerned with decision making,

mathematical modeling and system theory. Decision models can be used to represent a

18

productive system in mathematical terms. A decision model is expressed in terms of

performance measures, constraints and decision variables. The purpose of such a model is

to find optimal or satisfactory values of decision variables which improve systems

performance within the applicable constraints. These models can then help guide

management decision making. In 1915 F.W. Harris developed an Economic Order

Quantity formula for inventory management. In 1931 Shewhart developed quantitative

decision models for use in statistical quality control work.

During World War II operations researchers used mathematical equations to

simulate and analyze the effects of various warfare decision strategies. After World War

II mathematical and statistical models applied increasingly to the solution of management

problems. Linear programming (George Dantzig, 1947), inventory models. PERT/CPM,

simulation are few examples of these models. Many of these techniques would not be

feasible without the computer, which became a practical reality in the 1950s. The field of

POM as it is now known had emerged by the 1950s.

As computers became available in the 1950s, the power of OR was multiplied.

The speed and capacity of computers made them ideal for applying OR methods, such as

linear programming and simulation, to complex business problems. By the late 1960s

MRP and CRP were introduced by Joseph Orlicky, Oliver White and others. The 1970s

and 1980s witnessed continued development of MRP II systems and JIT plus TQM and

Kanban systems.

Today we are having an electronic revolution. Manufacturers are installing chips

(microprocessors) and computers in virtually all types of production and operations.

Robots are now doing much of the monotonous, dirty, and possibly dangerous work that

can be done by machines. In factories, they perform assembly, painting, welding and

other repetitive tasks. The movement now is toward more fully automated factories and

service systems. In service systems traditional ways of doing things (e.g. delivering mail)

are being replaced by more efficient methods (electronic mail). Our society is in

transaction.

More recent development include:

1. Extension of management concepts, principles and methods to the service sectors

(e.g. banks, hospitals, restaurants etc.),

2. Growing recognition of the importance of the strategic side of operations,

19

3. Rapidly changing technologies,

4. Intensive pressure of foreign competition,

5. Emergence of new management philosophy such as Total Quality Managament,

Benchmarking, Contingency Approach etc.,

6. Knowledge is becoming the most critical input into the transformation process for

firms in both manufacturing and service sector. In the future “Intelligence

Manufacturing Systems”(IMSs) may be used routinely to collect, store, and

disseminate knowledge.

Production/Operations Management has changed and continue to change. This

makes production/operations one of the most exciting areas of the firm.

OR,computerized

System & High Technology

Scientific Management

Industrial Revolution

Handicraft Era

Fig 2: Key Individuals and Events in the Development of POM Systems.

20

Issues and Concerns in Production/Operations Management

As demonstrated in Table 3 in many situation production/operations managers in both the manufacturing and service sectors has to make similar types of decisions. Whatever the firm, there is a wide range of issues for which the operations function assume leadership.

_______________________ ______________________ ______________

Inputs Value-Added Activities Outputs

Materials, labour, information, Performed with tools, machines, Goods and

Technology, environmental and techniques, and human skills. Services.

Government constraints.

___________________________ ____________________________ _________________

_______________________________________________________________________________

F e e d b a c k

___________________________________________________________________________________________________

* What types of skills do the * How will the firm use its * Who are the customers,

employees need? How will resources to produce its where are they ,and

the firm use these skills? products? what are their needs?

* What type of equipment does *How many facilities does *What mix of products

the firm need? the firm need, where should will be produced?

they be, and what should

they produce?

*What type of information does *How will the firm arrange the *How customized will

the firm need? How will the firm people and equipment in its each product be?

store, access and update this facilities?

information?

* What type of materials does *How can the firm improve its

the firm need? Who will provide operations?

them? How much should be kept

on hand?

*How will the firm match capacity

21

with anticipated demand?

*How will the firm schedule work

and prioritize customer orders?

Table 3: Issues and concerns in production/operations management.

Types of Production/Operations Processes

Effective production/operations process is essential to the company’s continuing

success. Not only are there numerous types of production, there are also many ways of

classifying or grouping them for descriptive purposes. Classifying production/operations

processes by their characteristics can provide valuable insights into how they should be

managed.

In general, the processes by which goods and services are produced can be

categorised in two traditional ways. Firstly, we can identify continuous, repetitive,

intermittent and job shop production process.

Job shop (jumbled flow). A wide variety of customized products are

made by a highly skilled workforce using general-purpose equipment.

These processes are referred to as jumbled-flow processes because there

are many possible routings through the process.

Examples: Mayo Clinic, home renovating firm, stereo repair shop,

gourmet restaurant.

Intermittent (batch) flow. A mixture of general-purpose and special-

purpose equipment is used to produce small to large batches of products.

Examples: clothing and book manufacturers, winery, caterer.

Repetitive flow (mass production). The product or products are

processed in lots, each item of production passing through the same

sequence of operations, i.e. several standardized products follow a

predetermined flow through sequentially dependent work centers. Workers

typically are assigned to a narrow range of tasks and work with highly

specialised equipment.

Examples: automobile and computer assembly lines, insurance home

office.

22

Continuous flow (flow shop). Commodity like products flow

continuously through a linear process. This type of process will

theoretically run for 24 hrs/day, 7 days/week and 52 weeks/year and,

whilst this is often the objective, it is rarely achieved.

Examples: chemical, oil, and sugar refineries, power and light utilities.

These four categories represent points on continuum of process organisations.

Processes that fall within a particular category share many characteristics that

fundamentally influence how a process should be managed.

The second and similar classification divides production processes into

Process Production. This type of process involves the continuous

production of a commodity in bulk, often by chemical rather than

mechanical means.

Mass Production. Is conceptually similar to process production, except

that discrete items such as motorcars and domestic appliances are usually

involved. A single or a very small range of similar items is produced in

very large numbers.

Batch Production. Occurs where the number of discrete items to be

manufactured in a period is insufficient to enable mass production to be

used. Similar items are, where possible, manufactured together in batches.

Jobbing Production. Although strictly consisting of the manufacture of

different products in unit quantities (in practice corresponds to the

intermittent process mentioned above).

Increasing quantity of Increasing variety of Production products made

23

Jobbing Type Production

Batch Type Production

Mass Type Production

Process Production

Strict Jobbing ContinuousProduction Process production

Fig.3: The types of production.Frequently a firm has more than one type of operating process in its production

system. e.g. a firm may use a repetitive-flow process to produce high-volume parts but

use an intermittent-flow process for lower-volume parts.

A link often exists between a firm’s product line and its operating processes. Job

shop organisations are commonly utilised when a product or family of products is first

introduced. As sales volumes increase and the product's design stabilises, the process

tends to move along the continuum toward a continuous-flow shop. Thus, as products

evolve, the nature of the operating processes used to produce them evolves as well.

Productions/Operations Management Problems

POM is a functional field of business with clear line management responsibilities.

Problems of management in the production/operations function basically concerns two

types of decision:

i. those relating to the design or establishment of the production/operations

system.

ii. Those relating to the operation, performance and running of the

production/operations system.

Problems in the design of production/operations system are as follows:

i. Design/specification of goods/service,

ii. Location of facilities,

iii. Layout of facilities/resources and materials handling,

iv. Determination of capacity/capability,

v. Design of works or jobs,

vi. Involvement in determination of renumeration system and work standards.

Problems in the operation of system are:

24

1. Planning and scheduling of activities,

2. Inventory (Stock) control,

3. Quality control,

4. Maintenance and replacement,

5. Involvement in performance measurement.

Every business organization will embrace these problems areas to a greater or

lesser extent. The relative emphasis will differ between companies and industries, and

also over a period of time. Problems in the first section are of long term nature and will

assume considerable importance at only infrequent intervals. Problems in the second

section will be of a resurring nature, i.e. they are of short term nature.

This simple classification of problems enables us to offer a definition of POM as follows:

Production/Operations Management is concerned with the design, operation

and improvement of the systems for manufacture, transport, supply or service.

The Productivity Challenge

The creation of goods and services requires changing resources into goods and

services. The more efficiently we make this change the more productive we are.

Productivity; is the ratio of outputs (goods and service) divided by one or more

inputs ( such as labour, capital or management).

The production/operations manager’s job is to enhance (improve) this ratio of

outputs to inputs.

Productivity is a measure of operational performance. Thus improving

productivity means improving efficiency. This improvement can be achieved in two

ways:

1. a reduction in inputs while output remains constant ,

2. an increase in output while inputs remain constant.

Both represent an improvement in productivity. Production is the total goods and

services produced. High Production may imply only that more people are working and

25

that employment levels are high (low unemployment), but it does not imply high

productivity.

To judge the success of an economic system in meeting its goals, economists use

one or more of the following measures:

Gross National Product, GNP,

Gross Domestic Product, GDP,

Balance of Trade,

National Debt,

Productivity.

Productivity in this sense, is the measure of economic growth that compares how

much a system produces with regard to the resources needed to produce it.

Measurement of productivity is an excellent way to evaluate a country’s ability

to provide an improving standard of living for its people. Only through increases in

productivity can the standard of living improve. Moreover, only through increases in

productivity can labour,capital and management receive additional payments. If returns to

labour, capital, or management are increased without increased productivity, prices rise.

On the other hand, downward pressure is placed on prices when productivity increases,

more is being produced with the same resources.

e.g. If units produced =1000 units

labour hours used =250 hrs.

Productivity = 1000/250 = 4units/labour-hour.

Many measures of productivity are possible and all are rough approximations, e.g.

value of output can be measured by what customer pays or simply by the number of units

produced or customers served. Value of inputs can be judged by their cost or simply by

the number of hours worked. Managers usually pick several reasonable measures and

monitor trends to spot areas needing improvement. A manager at an insurance firm might

measure office productivity as the number of insurance policies issued/processed per

26

employee each week. A manager at a carpet company might measure the productivity of

installers as the number of square meters of carpet installed per hour. Both reflect

“labour productivity”, which is an index of “output/person” or “output/hrs worked”.

Productivity measures can be based on a single input (Single-Factor Productivity

or Partial Productivity) or on more than one input (Multi-Factor Productivity) or on all

inputs. The choice depends on the purpose of the measurement.

Single-factor Productivity: Indicates the ratio of one resource (input) to the

goods and services produced (outputs).

Productivity = {Output of a specific Product}/ {Input of a specific Resource}

Example 1.

Three employees process 600 insurance policies in a week. They work 8 hrs. per

day, 5-days per week. Find labour productivity.

Solution:

Labour Productivity = [Policies Issued]/[Employee Hours]

Plabour = 600 policies/[(3 employees)(40 hrs/employee)]

Plabour = 5 Policies/hr.

Multi-factor Productivity: Indicates the ratio of many or all resources (inputs) to

the goods and services produced (outputs).

Example 2.

A team of workers make 400 units of a product, which is valued by its standard

cost of 10 MU each (before markups for other expenses and profit). The accounting

department reports that for this job the actual costs are:

400 MU for labour,

1 000 MU for materials and

300 MU for overhead.

27

Calculate multi-factor productivity.

Solution:

Multi-Factor Productivity = [Quantity at standard cost]/[Labour cost + Materials

cost +

Overhead Cost]

Pmf = [400 Units x 10 MU]/[400MU+1000MU+300MU] = 4 000MU / 1 700 MU

Pmf = 2.35

These measures must be compared with both performance levels in prior periods

and with future goals. If they are not living up to expectations, the process should be

investigated for improvement opportunities. When tracking performance at the

department or at an individual process level, productivity measures may often be

insufficient. The smart manager monitors multiple measures of performance, setting

goals for future and seeking better ways to design and operate processes.

Example 3.

Azim Title company has a staff of 4 each working 8 hours/day (for a payroll cost

of 640MU/day) and overhead expenses of 400MU/day, Azim processand closes on 8

titles each day. The company recently purchased a computerised title-search system that

will allow the processing of 14 titles/day, although the staff, their work hours, and pay

will be the same , the overhead expenses are now 800MU/day.

Labour-productivity with the old system = = 0.25titles/labour-hr

Labour-productivity with the new system = = 0.4375titles/lab.hr.

Multi-factor productivity with the old system = = 0.0077 titles/MU

Multi-factor productivity with the new system = = 0.0097titles/MU

28

Labour productivity has increased from 0.25 to 0.4375.

The change is = 1.75 or 75% increase in labour productivity.

Multi-factor productivity has increased from 0.0077 to 0.0097. This change is

0.0097/0.0077 = 1.259 or a 25.9% increase in multi-factor productivity.

Example 4.

a. Productivity can be measured in a variety of ways, such as labour, capital,

energy, material usage, and so on. At Modern Lumper, Inc. Ali Caliskan, president and

producer of apple crates sold to growers, has been able, with his current equipment, to

produce 240 crates per 100 logs, the current purchases 100 logs per day and each log

requires 3 labour-hrs to process.

He believes that he can hire a professional buyer who can buy a better-quality log

at the same cost. If this is the case, he can increase his production to 260 crates/100logs,

this labour-hours will increase by 8 hrs per day.

What will be the impact on productivity(measured in crates per labour-hour) if

the buyer is hired?

b. Ali Caliskan has decided to look at his productivity from a multifactor (total

factor productivity) perspective. To do so, he has determined his labour, capital, energy

and material usage and decided to use money units (MU for dollars or TL) as the

common denominator

His total labour-hours are now 300 hrs/day and will increase to 308 hrs/day. His

capital and energy costs will remain constant at 350MU and 150MU per day,

respectively. Material costs for the 100 logs per day are 1000MU and will remain the

same.

Because he pays an average of 10 MU/hr (with fringes), Caliskan wants to

determine his productivity increase?

Solution:

a. aa) Current Labour Poductivity =

29

ab) Labour Productivity with buyer = = 0.844

crates/lab-hr.

Using current productivity (0.8 from (a)) as a base, the increase will be 5.5%

(0.844/0.8=1.055 or a 5.5% increase)

b. Current System System with Professional Buyer

Labour 300hrs@10MU=3.000 308hrs@10MU = 3.080 MU

Material 100logs/day 1.000 1.000

Capital 350 350

Energy 150 150

Total Cost 4.500 MU 4.580MU

Productivity of current system =

Productivityof proposed system

Using current productivity (0.0533) as a base, the increase will be 0.047. That is,

or 6.4% increase.

Example 5:

Ilhan Bal makes wooden boxes in which to ship bikes. Ilhan and his three

employees invest 40 hours per day making the 120 boxes.

a. What is their productivity?

b. Ilhan and his employees have discussed redesigning the process to improve

efficiency. If they can increase the rate to 125 per day. What would be their

new productivity?

c. What would be their increase in productivity?

Solution:

30

a. Plabour = output/input = 120 boxes/40 hours = 3.0 boxes/hour

b. Plabour = output/input = 125 boxes/40 hours = 3.125 boxes/hour

c. Change in productivity = 0.125 boxes/hour

Percentage change = 0.125 boxes/hour/3.0 boxes/hour = 4.166%

Example 6.

Magusa Metal Works produces cast bronze valves on a 12 person assembly line.

On a recent day, 240 valves produced during an 8 hour shift. Calculate the labour

productivity.

Solution:

Total labour hours = 12 persons @ 8 hours = 96 hours

Labour productivity = 240 valves/96 hours = 2.5 valves/hour

Example 7.

Gaye produces “Final Exam Care Packages’ for resale by the sorority. She is

currently working a total of 6 hours a day to produce 120 care packages.

a. What is Gaye’s productivity?

b. Gaye thinks that by redesigning the package she can increase her total

productivity to 150 care packages per day. What would be her new

productivity?

c. What will be the increase in productivity if Gaye makes the change?

Solution:

a. P = units produced/input = 120 pkgs/6hrs = 20 pkgs/hr

b. P = units produced/input = 150 pkgs/6hrs = 25 pkgs/hr

c. Increase in productivity = {25 pkgs/hr – 20 pkgs/hr}/20 pkgs/hr = 25 %

Example 8.

Sergio Farmerson makes billiard balls in his famous Boston plant. With recent

increases in his costs, he has a new-found interest in efficiency. Sergio is interested in

determining the productivity of his organisation. He would like to know if his

organisation is maintaining the manufacturing average of 3% increase in productivity. He

31

has the following data representing a month from last year and an equivalent month this

year.

Last Year This Year

Units produced 1.000 1.000

Labour (hours) 300 275

Resin (kgs) 50 45

Capital invested (MU) 10.000 11.000

Energy (kw) 3.000 2.850

a. Show the productivity change for each category and then determine the

improvement for labour hours, the typical standard for comparison.

b. Sergio determines his cost to be as follows:

Labour 10 MU/hour

Resin 5 MU/kg

Capital 1% per month of investment

Energy 0.50 MU/kw

Show the productivity change, for one month last year versus one month this

year, on a multifactor basis with money units (MU) as the common denominator.

Solution:

a.

Resource Last Year This Year Change Percent

Change

Labour 1000/300 = 3.33 1000/275 = 3.64 0.31 0.31/3.33 =

9.3%

Resin 1000/50 = 20 1000/45 = 22.22 2.22 2.22/20 =

11.1%

Capital 1000/10000 = 0.1 1000/11000 = 0.09 -0.01 -0.01/0.1= -

10.0%

Energy 1000/3000 = 0.33 1000/2850 = 0.35 0.02 0.02/0.33 =

6.1%

32

b.

Last Year This Year

Production 1.000 units 1.000 units

Labour hrs@10 MU 3.000 MU 2.750 MU

Resin@5 MU 250 MU 225 MU

Capital cost/month 100 MU 110 MU

Energy@ 0.50 MU 1.500 MU 1.425 MU

TOTAL……………….. 4.850 MU 4.510 MU

Percent change in productivity = {1000/4850 – 1000/4510}/ 1000/4850 = -0.05 fewer

resources =

5% improvement

Example 9:

The manager of a carpet store is trying to determine optimal installation crew size.

He has tried various crew sizes with the results shown below. Compute the average

labour productivity for each crew size. Which crew size do you recommend?

Crew Size Meters Installed

2 706

4 1308

3 1017

3 1002

4 1288

2 692Solution:

Crew Size Meters Installed Labour Productivity

2 706 706/2 = 353 meters/ workers

4 1308 1308/4 = 327meters/ workers

3 1017 1017/3 = 339 meters/ workers

3 1002 1002/3 = 334 meters/ workers

33

4 1288 1288/4 = 322 meters/ workers

2 692 692/2 = 346 meters/ workers

Crew Size Average Labour Productivity

2 (353 + 346)/2 = 349.5 meters/ workers

3 (339 +334 )/2 = 336.5 meters/ workers

4 (327 + 322)/2 = 328.5 meters/ workers

Recommend optimal crew size = 2 workers.

Example 10:

The weekly output of a production process is shown below, together with data for

labour and material inputs. The standard inventory value of the output is 125 MU/unit.

Overhead is charged weekly at the rate of 1500 MU plus 0.5 times direct labour cost.

Assume a 40-hr/ week and an hourly wage of 16 MU. Material cost is 10 MU per

running meter. Compute the average multi-factor productivity for this process.

Week Output # workers Material (meters)

1 412 6 2840

2 364 5 2550

3 392 5 2720

4 408 6 2790

Solution:

Week 1 = 412 (125) MU = 1.444

[6*40*16]MU+[2840*10]MU+ [0.5*6*40*16]MU + 1500 MU

Week 2 = 365 (125) = 1.431

34

5*40*15 MU+ 2550*10 MU+ 0.5*5*40*16 MU + 1500 MU

Week 3 = 392(125) = 1.463

5*40*16 MU + 2720*10 MU+ 0.5*5*40*16 MU+1500 MU

Week 4 = 408 (125) = 1.457

6*40*16 MU + 2790*10 MU+ 0.5*6*40*16 MU+ 1500 MU

Average = [1.444 + 1.431 + 1.463 + 1.451] / 4 = 1.447

Example 11:

A company has introduced a process improvement that reduces processing time

for each unit, so that output increased by 25% with less material, but one additional

worker required.

Under the old process, five workers could produce 60 units/ hours. Labour costs

are 12 MU/ hours, and Material costs (input) was previously 16 MU/unit. For the new

process, material is now 10 MU / unit. Overhead is charged at 1.6 times direct labour

cost. Finished units sell for 31 MU each. What increase in productivity is associated with

the process improvement?

Solution:

Before= 60 units/hr * 31 MU/units = 1860 =

5*12 MU/hour + 60 units/hr *16 MU/units + 1.6[5*12 MU/hr] 1116

35

=1.667

After = 60 units/hr * 31 MU/units*1.25 = 2.325

6*12 MU/hr + 75 units/hr*10 MU/unit + 1.6 [6*12 MU /hr] 937.2

= 2.481

Productivity increase = [2.481 - 1.667] /1.667 * 100 = 48.83%

Example 12:

Suzan has a part-time “cottage industry” producing seasonal plywood yard

ornaments for resale at local craft fairs and bazaars. She currently works a total of 4 hours

per day to produce 10 ornaments.

a) What is her productivity?

b) She thinks that by redesigning the ornaments and switching from use of wooden

glue to a hot-glue gun she can increase her total production to 20 ornaments per

day. What is her new productivity?

c) What is her percentage increase in productivity?

Solution:

a) Productivity = 10 ornaments/day = 2.5 ornaments/hrs 4 hrs/day

b) Productivity = 20 ornaments /day = 5 ornaments/hrs 4 hrs/day

c) Change in productivity = 5 – 2.5 = 2.5 ornaments /hrs

Percent change = 2.5/2.5 * 100 = 100%.

Example 13:

36

Suzan’s Ceramics spent 3000 MU on a new kiln last year, in the belief that it

would cut energy usage 25% over the old kiln. This kiln is an oven that turns “green

ware” into finished pottery. Suzan is concerned that the new kiln requires extra labour

hours for its operation. Suzan wants to check the energy savings of the new oven and also

to look other measures of their productivity to see if the change really was beneficial.

Suzan has the following data to work with:

Were the modifications beneficial?

Solution:

Resource Last year This year Change Percent Change

Labour 4000/350 = 11.43 4000/375 = 10.67 = - 0.76 = - 6.7

Capital 4000/15000 = 0.27 4000/18000 = 0.22 = - 0.04 = - 16.7

Energy 4000/3000 = 1.33 4000/2600 = 1.54 = - 0.21 = 15.4

The energy modifications did not generate the expected savings; labour and capital

productivity decreased.

“Study the past if you would

devine the future”

Last Year This year

Production (finished units) 4000 4000

Greenware(kgs) 5000 5000

Labor (hours) 350 375

Capital (MU) 15000 18000

Energy (kWh) 3000 2600

37

Confucius

(

490 B.C.)

38

FORECASTING DEMAND

Introduction

Forecasting is the art and science of predicting future events. Forecasting is an

integral part of all managerial planning. Every manager considers some kind of forecast

in every decision he/she makes. Some of these forecasts are quite simple; e.g. take the

case of the office manager, who, on Thursday, forecasts the workload he anticipates for

Friday in order to give some of his employees time off.

Other forecasts are much more complex; consider the vice-president of finance

for a computer producing company trying to forecast, a year in advance, and the

company’s seasonal needs for working capital.

We can distinguish among these different kinds of forecasting needs by

considering how far into the future they focus. Detailed forecasts for individual items are

used to plan the short-run use of the system. These are up to one-year; usually less than 3-

months forecasts, such as job scheduling and worker assignments. Some forecasts have

exceptionally long time horizons and deal with issues much more difficult to quantify,

e.g. consider the company which produces heating and cooling systems for houses and

consider the long-term outlook for energy, energy related technologies, and government

constraints on energy production etc. New product planning and plant location and layout

issues need much longer time horizon forecasts.

Sophisticated mathematical tools and methods aid the manager today. When

managers plan, they determine in the present what courses of action their organizations

will take in the future. The first step in planning is therefore forecasting the future

demand for products/services and the resources necessary to produce these outputs. No

one forecasts with the accuracy that the users of the forecast would like. Still decisions

must be made every day, and they get made with the best information that is available,

not with perfect forecasts.

Forecasting Objectives and Uses

39

Forecasts are estimates of the occurrence, timing or magnitude of future events.

They give production/operation managers a rational basis planning and scheduling

activities, even though actual demand is quite uncertain. Forecasting deals with what we

think will happen in the future. Planning deals with what should happen in the future.

Thus, through planning, we consciously attempt to alter future events, while we use

forecasting only to predict them. Good planning utilizes a forecast as an input.

Forecasting is one input to all types of planning and scheduling activities. Fig. 1 shows

that forecasting includes a complex set of inputs before a decision is reached. The general

economic climate of the firm, assessment of past and present events, analysis of future

conditions, recommendation of others, and legal constraints constitute the bases for a

rational approach to developing forecasts. Decision-makers’ emotions, intuition and

personal motives and values continuously modify this logic as they exercise judgement in

reaching forecasting decision.

Decision Maker

Quantitative/Qualitative Analysis

General Economic Trends

Recommendations of others

Past History

Analysis of Future Conditions

Assessment of Present Conditions

Legal Constraints

Emotions and Intuition

Personal Motives and Values

Social and Cultural Values

Other Factors

Fig. 1: The Forecasting Decision.

40

Table 1 lists some advantages of good forecasts.

Improved employee relations,

Improved materials management,

Better use of capital and facilities,

Improved customer service.

Table 1: Advantages of good forecasts.

Management scientists have developed many forecasting techniques to help managers

to handle the increasing complexity in management decision-making. Each of these

techniques has a special use. There is no universal forecasting method for all situations.

Whether you use one forecasting technique or another, the forecasting processes are

same. Here are six steps in the forecasting process:

i. Determine the objective of the forecast. (What is its use?)

ii. Select the period over which the forecast will be made. (What are your

information needs over what time period?)

iii. Select the forecasting approach you will use. (Which forecasting technique

will most likely produce forecasts of greater use to you?)

iv. Collect the information to be used in the forecasting process. (Which data

will most likely produce forecasts of greatest use to you?)

v. Make the forecast.

vi. Validate and implement results.

41

Types of Forecasts

There are two basic kinds of forecasts: Qualitative forecasts (judgmental forecasts) and

quantitative forecasts.

________________________________________________________________________

______

I. Qualitative (Judgmental) Forecasts II. Quantitative Forecasts

i. Delphi Technique A. Extensions of Past History

ii. Panel of Experts i. Moving Averages

ii. Exponential Smoothing

iii. Trend Analysis

iv. Moving Total

B. Causal Forecasting

i. Regression Analysis

ii. Correlation analysis

________________________________________________________________________

______

42

Table 2: Types of Forecasts.

Judgmental Forecasts

We tend to use these kinds of forecasts when “good” data are not readily

available. i.e. used when situation is vague and little data exist (new products, new

technology). With this kind of forecast, we are trying to alter subjective opinion into a

quantitative forecast that we can use. Outside experts can be consulted. We can convene a

panel of experts to make a combined forecast. In each case we rely on human judgment

to interpret past data and make projections about the future, i.e. involves intuition,

experience (e.g. forecasting sales on internet).

Delphi Technique

As forecasts of the social and economic environment become more and more

necessary for managerial decision-making, expert opinion becomes more widely used to

keep us informed about what is likely to happen. One technique, which uses expert’s

opinion, is the Delphi Technique. The Delphi method originated in the Rand

Corporation in 1948 where it was used to assess the potential impact of an Atomic Bomb

attack on the U.S. Since that time, it has been applied to a variety. This method works by

circulating a series of questionnaires among individuals who possess the knowledge and

ability to contribute meaningfully. Each new questionnaire is developed using the

information extracted from the previous one, thus enlarging the scope of information on

which participants can base judgements. The goal is to achieve a consensus forecast. Of

course, experts are wrong from time to time, just like the rest of us. When the first great

hydroelectric plant at Niagara Falls was being designed, proposal was received for direct

and alternating current equipment. The project promoters consulted one of the world’s

best-known power experts; Lord Kelvin. He advised them strongly not to use alternating

current. However the decision-makers disregard the expert and choose the alternating

current, which worked out well since then.

43

For the most part, these are long term, single time forecast, which usually have

very little hard information to go by or data are costly to obtain, so the problem does not

lend itself to analytic techniques. Rather, judgments of experts or others who possess

sufficient knowledge to make predictions are used.

Panel of Experts

This technique differs from the Delphi Technique in that there is no secrecy and

full communication among panel members is encouraged. It is a group estimates by

working together. It is a quick technique. However, as a disadvantage, the group process

tends to influence the outcome (social pressure, majority view etc.).

Quantitative Forecasts

Quantitative forecasts are used when situation is “stable” and historical data exist.

Extensions of Past History

When we take history as our beginning point for forecasting, it does not mean that

we think March will be just like January and February. It simply means that over the

short run we believe that future patterns tend to be extensions of past ones and that we

can make some useful forecasts by studying past behaviour.

Moving Averages

Averages that are updated as new information is received are generally called

moving averages. A “Moving Average” forecast uses a number of the most recent actual

data values in generating a forecast. This tends to dampen or smooth out, the random

increase and decreases of a forecast that uses only one period. The speed of the response

is controlled by adjusting the number of periods we include in the moving average and

44

the weighting we assign to each period. The simple moving average forecast can be

computed using the following equation:

where

wherei = “Age” of the data (i =1,2,…,n)n = Number of periods in the moving average

Ai = Actual value with age i e.g. MA3 = A three-period moving average.

The simple moving average is useful for forecasting demand that is stable and does not display any pronounced demand behaviour, such as a trend or seasonal pattern.

The simplest moving average weighs each period equally, e.g. if we want to

forecast sales for April with a simple 3-month moving average, we would average the

sales for January, February and March. May’s forecast would drop January’s and add

April’s figures. Table 3 illustrates a simple example. In this simple example 3 month

moving average is better than 4 month moving average.

One may think that better forecast may respond faster than these two and use

different weights for each month, because he may believe that newer information is more

reflective of the trend of sales. In the example, he could try to weigh the latest month as

heavily as the preceding 2 months, and the next to last month twice as heavily as the one

3 months ago, the forecast will be as follows:

F= (3M1+2M2+M3)/6

Where

M1 = latest month’s information.

M2 = information from 2 months ago.

M3 = information from 3 months ago.

Month Actual Sales 3-month moving Error 4-month moving Error

45

MAn =

average forecast average forecast

January 10

February 12

March 13

April 16 (10+12+13)/3=11.67 4.33

May 19 (12+13+16)/3=13.67 5.33 (10+12+13+16)/4=12.75 6.25

June 23 (13+16+19)/3=16.00 7.00 (12+13+16+19)/4=15.00 8.00

July 26 (16+19+23)/3=19.33 6.67 (13+16+19+23)/4=17.75 8.25

August 30 (19+23+26)/3=22.67 7.33 (16+19+23+26)/4=21.00 9.00

September 28 (23+26+30)/3=26.33 1.67 (19+23+26+30)/4=24.50 3.50

October 18 (26+30+28)/3=28.00 10.00 (23+26+30+28)/4=26.75 8.75

November 16 (30+28+18)/3=25.33 9.33 (26+30+28+18)/4=25.50 9.50

December 14 (28+18+16)/3=20.67 6.67 (30+28+18+16)/4=23.00 9.00

Total 58.33 62.25

Averages, Ф 6.48 7.78

Table 3: Example for 3 and 4 Month Moving Average Forecasts. (Sales: 000 Monetary

Units)

When we compare the results of the weighted moving average in Table4 with the

simple moving average in Table 3, we see that weighting the latest information/ data

more heavily generated a much more accurate forecast.

The exact weighting to use and the best number of periods to include in the

forecast are both matters of trial-and error (i.e. experimentation).

Month Actual Sales (000MU) 3-Month Weighted Moving Average

Forecast

Error

January 10

February 12

March 13

April 16 [(3*13)+(2*12)+(10)] / 6 = 12.17 3.83

46

May 19 [(3*16)+(2*13)+(12)]/6=14.33 4.67

June 23 [(3*19)+(2*16)+(13)]/6=17.00 6.00

July 26 [(3*23)+(2*19)+(16)]/6=20.50 5.50

August 30 [(3*26)+(2*23)+(19)]/6=23.83 6.17

September 28 [(3*30)+(2*26)+(23)]/6=27.50 0.50

October 18 [(3*28)+(2*30)+(26)]/6=28.33 10.33

November 16 [(3*18)+(2*28)+(30)]/6=23.33 7.33

December 14 [(3*16)+(2*18)+(28)]/6=18.67 4.67

Total 49.00

Average, Ф 5.44

Table 4: 3-Month Weighted Moving Average Forecast

Exponential Smoothing

A technique called “Exponential Smoothing” eliminates some of the

computational disadvantages of forecasting with a weighted moving average. Exponential

Smoothing is a type of moving average used for forecasting. It uses a single weighting

factor called alpha, symbolized “α”. It weighs past data in an exponential manner. “α”

Reflects the extend to which the forecast reflects past average versus most recent

demand.

A low αgives more weight to the past average, a less weight to the recent

demand. ά= 1 would result in a forecast equal to latest demand. It is possible to

experiment with different alphas to improve forecasting accuracy.

Smoothed Forecast

for this month’s sales = α (Sales Last Month) + [(1- α) (Previous Forecast of month’s sales)]

Or

Smoothed Forecast = Forecast for previous + α [Last period’s actual – Forecast for previous]

For this month time period sales time period

47

Example:

Last Month’s Actual Sales = 15.000MU

Last Month’s Forecasted Sales = 16.000MU ά = 0.4

Forecast for this month’s sales = 0.4(15.000) + [(1-0.4) (16.000)]= 15.600MU

Month Actu

al

Sales

Sales Last

Month

α.(Sales

Last

Month)

(1- α) Previous Forecast

of Last Month’s

Sales

(1- α)(Previous

Forecast of

Last Month’s

Sales

Smoothed

Forecast for

This Month

Error

(1) (2) (3) (4) (4)*(3) (5) (6) (5)*(6) (4)*(3)+(5)*(6)

January 10

February 12 10 0.4 4.0 0.6 11.0 A beginning

“Guess”

6.6 10.6 1.4

March 13 12 0.4 4.8 0.6 10.6 6.4 11.2 1.8

April 16 13 0.4 5.2 0.6 11.2 6.7 11.9 4.1

May 19 16 0.4 6.4 0.6 11.9 7.1 13.5 5.5

June 23 19 0.4 7.6 0.6 13.5 8.1 15.7 7.3

July 26 23 0.4 9.2 0.6 15.7 9.4 18.6 7.4

August 30 26 0.4 10.4 0.6 18.6 11.2 21.6 8.4

September 28 30 0.4 12.0 0.6 21.6 13.0 25.0 3.0

October 18 28 0.4 11.2 0.6 25.0 15.0 26.2 8.2

November 16 18 0.4 7.2 0.6 26.2 15.7 22.9 6.9

December 14 16 0.4 6.4 0.6 22.9 13.7 20.1 6.1

Total 60.1

Average, Ф 5.46

Table 5: Exponentially Smoothed Forecast of Sales. (α = 0.4, Sales 000MU)

In the Table 5, notice that even though actual sales turned down in September, the

forecast did not turn down until November. A larger α (say 0.7) might do a better job of

forecasting. In Table 6, we have shown the computations with an alpha of 0.7. Not only

are the forecast sales nearer to actual sales, but also the forecast turned down this time in

October, only a month after actual sales declined.

Whether 0.7 is the best weighting factor to use is a matter of experimentation.

Larger alphas do not always make for better forecasts. Some analysts recommend

beginning with an alpha of 0.2 or 0.3.

48

Some analysts picking an alpha value that approximate “a length of moving

average that makes sense”. An approximate equivalent to an arithmetic moving average,

in terms of the degree of smoothing, can be estimated by

α= 2/(n+1) Thus, a 7-year moving average would correspond,

roughly, to an alpha value of 0.25.

As in the case of moving averages, the selection of a level of ά would be based on

knowledge of management’s needs and the nature of the particular forecasting situation.

Month Actual

Sales

Sales

Last

Month

Α*(Sales

Last

Month)

(1- α) Previous

Forecast

of Last

Month’s

Sales

(1-α)(Previous

Forecast of Last

Month’s Sales

Smoothed Forecast

for This Month

Error

(1) (2) (3) (4) (4)*(3) (5) (6) (5)*(6) (4)*(3)+(5)*(6)

January 10

February 12 10 0.7 7.0 0.3 11 (Guess) 3.3 10.3 1.7

March 13 12 0.7 8.4 0.3 10.3 3.1 11.5 1.5

April 16 13 0.7 9.1 0.3 11.5 3.5 12.6 3.4

May 19 16 0.7 11.2 0.3 12.6 3.8 15.0 4.0

June 23 19 0.7 13.3 0.3 15.0 4.5 17.8 5.2

July 26 23 0.7 16.1 0.3 17.8 5.3 21.4 4.6

August 30 26 0.7 18.2 0.3 21.4 6.4 24.6 5.4

September 28 30 0.7 21.0 0.3 24.6 7.4 28.4 0.4

October 18 28 0.7 19.6 0.3 28.4 8.5 28.1 10.1

November 16 18 0.7 12.6 0.3 28.1 8.4 21.0 5.0

December 14 16 0.7 11.2 0.3 21.0 6.3 17.5 3.5

Total 44.8

Average, Ф 4.07

Table 6: Exponentially Smoothed Forecast of Sales. (α = 0.7, Sales 000MU)

Trend Analysis

49

Trend Analysis is a mathematical method, which fits a trend line to a data set of

past observations and then projects this line into the future for purposes of estimating.

Sometimes it is called as “Time Series”, because time-order sequence of observations

taken at regular intervals over a period of time. Analysis of the series data requires the

analyst to identify the underlying behaviour of the series. Merely plotting the data and

visually examining the plot can often accomplish this. Random and irregular variations

might appear due to unusual circumstance such as severe conditions, strikes, or a major

change in a product or service. They do not reflect typical behaviour, and inclusion in the

series can distort the overall picture, whenever possible, these should be identified and

removed from the data. The trend components of a time series reflect the effects of any

long-term factors on the series. Analysis of trend involves searching for an equation that

will suitably describe the trend. The trend component may be linear, or it may not. The

discussion here will focus exclusively on linear trend because trends are fairly common

and because they are the easiest to work with.

Let’s illustrate this process with an example.

Example 1:

Ali Kazik bought some land from the south of Izmir some years ago and has

developed it himself, building houses/villas on it each year. During the last 6 years, his

construction record has been as follows:

YEARS No. of Houses

1998 28

1999 34

2000 36

2001 42

2002 50

2003 52

If this trend continues Ali Kazik would like to estimate the number of

houses/villas he will probably build in 2006.

50

To get a quick idea of trend, fig.2 is called “scatter diagram”. When we view all

these data points together, we can see the relationship that exists between the two

variables “Time” and “No. Of Homes Built”. As a result we draw a straight line to show

the relationship. It is more usual to fit a trend line more precisely by using an equation

that relates the two variables in this situation.

To a statistician, a line fitted to scatter diagram will have a good fit, if it

minimizes the sum of squares of the errors between the estimated points on the line and

the actual observed points. The farther away a point is from the line, the more serious is

the error. We are searching for the trend line that gives us the minimum squares of the

errors; this method of fitting a trend line is called “The Least Squares Method”.

The equation for a fitted straight line is:

Y= a + bX

Where,

Y = Forecast value in the trend equation (the dependent variable),

X = Projected value of the independent variable value associated with Y,

b = The rate of change (slope) of the trend line,

x = The independent variable (time in the case),

y = Values of the dependent variable (homes in this case),

= The mean value of the independent variable,

= The mean value of the dependent variable,

a = the point at which the trend line intercepts the Y-axis,

n = the number of periods of data..

51

Fig. 2: Straight-line fitted through scatter diagram.

Table 7: Simple linear trend analysis formulas.

52

I. a = (Σx2 Σy-ΣxΣxy)/ nΣx2- (Σx)2

Ь = (nΣxy-ΣxΣy)/ nΣx2- (Σx)2

II. a = - b

Ь = (Σxy-n )/ Σx2- n( )2

III Σy = na + bΣx

Σxy = aΣx + bΣ x2

IV.

a = - b

Data Point Year Homes Built

N x y x.y x2

1 1998 0 28 0 0

2 1999 1 34 34 1

3 2000 2 36 72 4

4 2001 3 42 126 9

5 2002 4 50 200 16

6 2003 5 52 260 25

Σx = 15 Σy = 242 Σxy = 692 Σx2 = 55

= 2.5 = 40.33

Ь= (Σxy-n )/ Σx2- n( )2 = [692-6(2.5)(40.33)]/[55-6(2.5) 2] = 4.974

a= - b = 40.33-4.974(2.5)= 27.895

Y= 27.895 + 4.974X

or

Σy= na + bΣx 242 = 6a + 15b (i)

Σxy= aΣx + bΣ x2 692 = 15a +55b (ii)

(i) * 5 1210 = 30a+75b (i)

(ii)*2 1384 =30a+110b (ii)

174 = 35 Ь

Ь = 4,974

Substitute b value in equation (i)

242= 6a+15b

242=6a+15(4,974)

242=6a+74,57

a = 27,895

Y= 27,895+4,974X

Now we can forecast the number of homes to be built in the year 2006.

Y 2006 = 27.9+4.97(8) = 67.66 = ~ approx. 68 homes .

Example 2:

53

Year, x Sales (000 units),y x.y x2

1997 0 380 0 0

1998 1 354 354 1

1999 2 430 860 4

2000 3 468 1404 9

2001 4 522 2088 16

2002 5 484 2420 25

2003 6 556 3336 36

2004 7 606 4242 49

Σx = 28 Σy = 3800 Σxy = 14704 Σx2 = 140

Σy= na + bΣx 3800= 8a + 28b (1)

Σxy= aΣx + bΣ x2 14704= 28a+ 140b (2)

(1) x 7 26600=56a+196b (1)

(2) x 2 29400=56a+280b (2)

(2) - (1) 2808=84b

b = 33.43

Substitute b value in equation (1)

3800=8a+28(33.43)

a=358

Sales forecast for the year 2006 will be:

Y=358+33.43X

Y2006 = 358+33.43(9) = 685.87(000) units

Suppose we have performed the forecasting analysis in Example 1 as part of our

assignment. We found 658.870 units for 2005 forecasted sales. “How sure are we of this

forecasted figure?” and “How high and low do we estimate the annual sales could be in

year 2006?”

When time series analysis generates forecasts for future periods, we must

recognize that these are only estimates and that the actual annual sales to be subsequently

realized may differ substantially from the forecasts. The presence of forecasting error’s

54

or chance variations is a fact of life for forecasters. It is a process permeated with

uncertainty. How do forecasters deal with this uncertainty?

The formula below gives the upper and lower limits or ranges of forecasts:

Syx =

The following equation may look more complex, but it is actually an easier-to-use

version of the above equation. Either formula provides the same answer and can be used

in setting up prediction (confidence) intervals around the point estimate.

Syx = the standard error/deviation of the forecast.

“Syx” is a measure of how historical data points have been dispersed about the

trend line. If “Syx” is large, the historical data points have been spread widely about the

trend line. The upper and lower limits are far away. If “Syx” is small, past data points

have been tightly grouped about the trend line and the upper and lower limits are close

together. The interval between upper and lower limits is called “Confidence Interval”.

As stated before, the distribution of forecast values for a future time period has a

standard deviation (Syx), which is a relative measure of how the distribution is dispersed.

The distributions of all future time periods are assumed to be normal distributions if “n”

(number of observations) is large (usually ≥30) or if student-t distributions “n” is small

(usually<30).

Since we rarely have 30 or more observations in our data, and t and normal

distributions tend to converge when n is large, it is assumed that we are dealing with t

distributions.

So;

Upper Limit = Y+ t.Syx Lower Limit = Y – t.Syx

Where t is the number of standard deviations out from the mean of the distribution

to provide a given probability of exceeding these upper and lower limits through chance.

For example, say that we wish to set the limits so that there is only a %10 probability of

exceeding the limits by chance. We look at the student-t value table. Since the degrees of

55

freedom = n-2 and the level of significance is 0.10, the t-value equals 1.943. If “n” is

large (usually ≥30), the distribution is assumed to be normal.

We usually set limits as follows;

This gives us that for any given value of X, the value of Y can be expected to lie

within the interval of designated probability.

Example 2 (continued):

Year x Sales (000

units),y

xy x2 y2

1997 0 380 0 0 144400

1998 1 354 354 1 125316

1999 2 430 860 4 184900

2000 3 468 1404 9 219024

2001 4 522 2088 16 272484

2002 5 484 2420 25 234256

2003 6 556 3336 36 309136

2004 7 606 4242 49 367236

Σx = 28 Σy = 3800 Σxy = 14704 Σx2 = 140 Σy2 =1856752

Y=358+33.43X

Y2006 = 358+33.43(9)= 658.87(000) units.

Now let us compute the value of Syx;

Now that we have the value of Syx, let us compute the upper and lower limits of the

forecast for the year 2006.

Upper limit Y2006 + t Syx = 658.87+28.28(1.943)= 658.87+54.948

56

= 713.818(000) units.

Lower limit Y2006 - t Syx = 658.87-28.28(1.943)= 658.87-54.948

= 603.922(000) units.

There is a 90% probability that our annual sales for the year 2006 will be between

713.818(000) units and 603.922(000) units. There is only a 10% probability that our sales

will fall outside these limits. Our best estimate is 658.870 units.

Example 3:

Azim Motors produces motors for farms. Azim Motors production plant has

operated at near capacity for over a year now. The chief executive of Azim Motors thinks

that the growth in sales will continue, and he wants to develop a long run forecast to be

used to plan facility requirements for the next three years. Sales records for the past 10

years have been accumulated.

Year Annual Sales (000units)

1995 1000

1996 1300

1997 1800

1998 2000

1999 2000

2000 2000

2001 2200

2002 2600

2003 2900

2004 3200

Solution:

Let us now solve for the a and b values;

Year x Annual x.y x2

57

Sales(000units),

y

1995 0 1000 0 0

1996 1 1300 1300 1

1997 2 1800 3600 4

1998 3 2000 6000 9

1999 4 2000 8000 16

2000 5 2000 10000 25

2001 6 2200 13200 36

2002 7 2600 18200 49

2003 8 2900 23200 64

2004 9 3200 28800 81

Σx = 45 Σy = 21000 Σxy = 112300 Σx2 =285

a= (Σx2 Σy-ΣxΣxy)/[nΣx2-(Σx)2]

=[(285(21000)-(45)(112300)]/[(10(285)-(45)2 ]= 1129.09

Ь= (nΣxy-ΣxΣy)/ [nΣx2- (Σx)2] = [(10(112300)-45(21000)]/ [(10(285)-(45)2 ]= 215.758

or alternatively, the other set of formulas for a and b may be used:

Ь= (Σxy - n )/ [Σx2- n( )2] = [112300-10(4.5)(2100)]/[285-10(4.5) 2] = 215.758

a = - b = 2100-215.758(4.5)= 1129.09

or, alternatively, the equations may be used for a and b:

Σy= na + bΣx 2100= 10a + 45b (1)

Σxy= aΣx + bΣ x2 112300= 45a+ 285b (2)

(1) x 4.5 94500=45a+202.5b (1)

(2) x 1 112300=45a+285b (2)

(2)-(1) 17800=82.5b

b =215.758

Substitute b in (1),

21000 = 10a+45(215.758)

a = 1129.09

Now we know the values of a and b, the trend equation;

Y= a + b.X

58

Y= 1129.09 + 215.758X

If we wish to forecast sales in thousands of units for the next three years, we

would substitute 10,11,12, the next three values for X, into the trend equation for X;

Y2005 = 1129.09+215.758(10)= 3286.7(000) units

Y2006 = 1129.09+215.758(11)= 3502.4(000) units

Y2007 = 1129.09+215.758(12)= 3718.2(000) units

In order to find Syx, we have to calculate y2 values.

y y2

1000 1000000

1300 1690000

1800 3240000

2000 4000000

2000 4000000

2000 4000000

2200 4840000

2600 6760000

2900 8410000

3200 10240000

Σy2 = 48180000

The value of Syx is found as follows:

Since we have calculated the value of Syx, let us now compute the upper and lower

limits of the forecast for the year 2005.

Upper limit Y2005 + t Syx = 3286.7+173.02(1.86)= 3608.5(000) units.

Lower limit Y2005 - t Syx = 3286.7-173.02(1.86)= 2964.9(000) units.

There is 90% probability that our annual sales for the year 2005 will be between

3608.5 and 2964.9 thousand units. There is only 10% probability that our sales will fall

outside these limits. Our best estimate is 3286.7 thousand units.

59

Moving Total Method

Moving total method is a forecasting method, which seasonally adjusts the given data

by moving cumulative and then extends them to future by means of trend line. The

procedure is:

(1) Take the sum of first year quarterly sales basis,

(2) Use this basis as a starting point of moving cumulative,

(3) Find the trend line which represents the moving cumulative,

(4) Using trend line find forecasted future moving cumulative for quarters,

(5) Seasonalize the forecasted cumulative to forecasted quarter sales.

Example 4:

Quarterly Sales records of Izmir Beer Factory for the past 3 years are given

below:

Years Quarters Sales (million bottles)

2002 I 182

II 292

III 222

IV 174

2003 I 192

II 308

III 242

IV 182

2004 I 216

II 332

III 262

IV 208

60

Solution:

Now let us calculate the moving total/cumulatives for the second and third years.

But first calculate the sum of first year’s quarters.

The sum is 870 million bottles.

Year 2003 Quarter I Moving Total = 870-182+192=880

Quarter II Moving Total = 880-292+308=896

Quarter III Moving Total = 896-222+242=916

…

…

…

Years Period x Sales(million

bottles)

Moving

Total, y

x2 xy

2002 I 182

II 292

III 222

IV 174

2003 I 0 192 880 0 0

II 1 308 896 1 896

III 2 242 916 4 1832

IV 3 182 924 9 2772

2004 I 4 216 948 16 3792

II 5 332 972 25 4860

III 6 262 992 36 5952

IV 7 208 1018 49 7126

Σx = 28 Σy = 7546 Σ x2 = 140 Σ xy = 27230

Σy= na + bΣx 7546= 8a + 28b (1)

Σxy= aΣx + bΣ x2 27230= 28a+ 140b (2)

By solving the equations we find :a=875 and b=19.5

And then the trend line;

61

Y=a+bX

Y=875+19.5X

Now we wish to forecast moving totals for the next year’s quarters, we would

substitute 8,9,10,11 the next four values for X, into the trend equation for X;

Y2005/I= 875+19.5(8)= 1031(million) bottles

Y2005/II= 875+19.5(9)= 1050.5(million) bottles

Y2005/III= 875+19.5(10)= 1070(million) bottles

Y2005/IV= 875+19.5(11)= 1089.5(million)bottles

Seasonalize the forecasted cumulatives to forecasted quarter sales.

Year Quarter Sales (million bottles)

2005 I 1031-1018+216=229

II 1050.5-1031+332=351.5

III 1070-1050.5+262=281.5

IV 1089.5-1070+208=227.5

Example 5:

Year Quarter Sales(million MU)

2003 I 40

II 60

III 100

IV 40

2004 I 60

II 80

III 140

IV 40

Find 2005-quarter sales.

Solution:

62

Years Quarters

x

Sales

(mill. MU)

Moving

Total, y

x2 xy

2003 I 40

II 60

III 100

IV 40

2004 I 0 60 260 0 0

II 1 80 280 1 280

III 2 140 320 4 640

IV 3 40 320 9 960

Σx = 6 Σy = 1180 Σ x2 = 14 Σ xy = 1880

Σy= na + bΣx 1180= 4a + 6b (1)

Σxy= aΣx + bΣ x2 1880= 6a+ 14b (2)

(1) x 3 3540=12a+18b

(2) x 2 3760=12a+28b

(2)-(1) 220=10b

b =22

Substitute b value in equation (1)

1180=4a+6(22)

a=262

The trend line is therefore;

Y=a+bX

Y=262+22X

Now we wish to forecast moving totals for the next year’s quarters, we would

substitute 4,5,6,7 the next four values for X, into the trend equation for X;

Y2005/I = 262+22(4)= 350

Y2005/II = 262+22(5)= 372

Y2005/III = 262+22(6)= 394

Y2005/IV = 262+22(7)= 416

Now let us find the quarter sales for 2005,

63

Year Quarter Sales

2005 I 350-320+60=90

II 372-350+80=102

III 394-372+140=162

IV 416-394+40=62

Causal Forecasts

Regression Analysis

Trend analysis describes the action of some variables over time, as if the variable

were a function of time. Although this often is a useful relationship, it is sometimes more

meaningful to relate the variable we are trying to forecast to other variables that are more

suggestive of a casual relationship. Regression and correlation analysis is means of

describing the association between two or more such variables. They merely quantify the

statistical dependence or extent to which the two or more variables are related.

Regression is used for forecasting by attempting to establish a mathematical relationship

between two or more variables. “Regression” means “Dependence” and involves

estimating the value of a dependent variable, Y, from an independent variable, X. In

simple regression only one independent variable is used, whereas in multiple regression

two or more independent variables are involved.

The simple linear model takes the form

Y= a + bX.

A multiple linear regression model may be one of the forms;

Y=a+bX1+cX2+dX3

or

Y= a+bx+cx2 + d x3

etc.

We shall limit considerations to simple linear regressions, which are often

satisfactory for forecasting purposes. The forecasting procedure using regression is

similar to that of trend analysis. We use the same formulas to that of trend analysis. The

variables are not necessarily related on a time basis.

Correlation Analysis

64

Correlation is a means of expressing the degree of relationship between two or

more variables. It tells how well a linear (or other) equation describes the relationship.

Therefore “correlation” in a linear equation is a measure of the strength of the

relationship between the independent and dependent variables. The correlation coefficient

“r” is a number between –1 and +1 and is designated as positive if y increases with

increases in x and negative if y decreases with increases in x. If “r = 0”, this means there

is no relationship between the two variables.

Y y

X x

Perfect perfect

Negative none positive

(high) (moderate) (low) (low) (moderate) (high)

-1.0 -0.8 -0.6 -0.4 -0.2 0 +0.2 +0.4 +0.6 +0.8 1.0

fig.3: Interpretation of correlation coefficient.

The correlation coefficient is related to the percentage of the variation in y that is

explained by the regression line.

Coefficient of correlation = r =

or

The ratio of explained to total variation is called “coefficient of determination”, r2. It is

effectively the percentage of variation in the dependent variable that is explained by the

regression line.

65

Coefficient of Determination = r2

Example 6:

The general manager of building materials production plant feels the demand for

plasterboard shipments may be related to the number of construction permits issued in the

town during the previous quarter. The manager has collected the data shown in the table

below.

Construction permits

(x)

Plasterboard

shipments (y)

15 6

9 4

40 16

20 6

25 13

25 9

15 10

35 16

a) Derive a regression forecasting equation

b) Determine a point estimate for plasterboard shipments when the number of

construction permits 30.

c) Compute the standard deviation of regression.

d) Compute the upper and lower limits for construction when it is 30.

e) Compute the correlation coefficient, determination coefficient and interpret both.

f) Test the hypothesis that “r = 0” at 10% level of significance.

g) Using correlation coefficient find regression forecasting equation.

Solution:

66

x y x.y x2 y2

15 6 90 225 36

9 4 36 81 16

40 16 640 1600 256

20 6 120 400 36

25 13 325 625 169

25 9 225 625 81

15 10 150 225 100

35 16 560 1225 256

184 80 2146 5006 950

n= 8 pair of observations

=184/8=23 =80/8=10

Σy= na + bΣx 80= 8a + 184b (1)

Σxy= aΣx + bΣ x2 2146= 184a+ 5006b (2)

Multiply (1)*-23 -1840=-184a-4232b (3)

306=774b (4)

Adding (2) and (3) b=306/774= 0.395

Substituting in (1) 80=8a+284(0.395)

8a=80-72.7

a=0.91

Alternatively,

b= (Σxy-n )/ [nΣx2- n( )2] = (2146-8(23)10)/[(5006-8(23)(23)] = 0.395

a= - b = 10-0.395(23)= 0.91

Thus;

Y = a + bX

Y = 0.91 + 0.395X

Regression equation is;

Y = 0.91 + 0.395X

67

Where X = permits

Y = shipments

b) Letting X = 30

Y=0.91+0.395(30)=12.76=~13 shipments.

c) The standard deviation of regression:

d) Prediction intervals:

There is 90% probability that shipment for 30 permits will fall between 17 and 9

shipments. There is only 10% probability that shipment will fall outside these limits.

e) Compute the correlation coefficient, determination coefficient and interpret.

x- (x- )2 (x- ) (y- )

-8 -4 64 16 32

-14 -6 196 36 84

17 6 289 36 102

-3 -4 9 16 12

2 3 4 9 6

2 -1 4 1 -2

-8 0 64 0 0

12 6 144 36 72

0 0 774 150 306

r = = (306)/ =306/340.73=0.8981

r=0.90

There is a high relationship between x and y (strong positive relationship)

r2 = (0.90) 2 =0.81

81% of variation in y can be explained by x and remaining 19% can be explained

by other factors.

68

f) Because the regression equation is developed from sample data that include

random variability, there is always risk that no real relationships exists between the

variables. The significance of any value of r can, however, be tested under a hypothesis

that there is no correlation. The computed statistical-t value is compared with a

theoretical-t value. If the computed value exceeds the tabled value, the hypothesis is

rejected, and the correlation is deemed significant at the specified level. Usually 10% or

5% level of significance (), i.e. 90% or 95% confidence interval are used for testing.

Statistical – t:

n-2= degrees of freedom.

The value from student-t table is the probability of r=0.

Level of significance ()=0.10; Degrees of freedom n-2= 8-2=6. From student-t

tables;

tk=1.943 (Since t-distribution table gives /2 values, we look for 0.05 in the

table).

If tc>tk r>0

If tc < tk r=0

Since tc= 5.06 and tk=1.943.

tc>tk, i.e. 5.06> 1.943

The hypothesis r = 0 is rejected. The computed r is meaningful.

g) Using correlation coefficient formula find regression forecasting equation.

We can alternatively find a and b values for regression equation.

b= (Σ(x- )(y- ))/ (Σ(x- )2)

b= 306/774= 0.395

and

a= - b = 10-0.395(23)=10-9.093= 0.91

69

Therefore the regression equation is as follows:

Y= 0.91 + 0.395X

SOLVED PROBLEMS

1. A firm uses a moving average to forecast next month’s demand. Past actual demand

(in units) is shown below.

a) Compute a simple 5-month moving average to forecast demand for month 52.

70

b) Compute a weighted 3-month moving average where the weights are the highest for

the latest months and descend in order of 32.1.

Month Actual Demand

43 105

44 106

45 110

46 110

47 114

48 121

49 130

50 128

51 137

52

Solution:

a) MA = Σx/ (no. of periods) = (114+121+130+128+137)/5= 126 units

b) MAwt= Σ(wt)X/ Σwt

Where,

wt .(value)= total

3(137)=411

2(128)=256

1(130)=130

797

MAwt= 797/6= 133 units.

2. Izmir Sağlık Hospital has used a 9-month moving average forecasting method to

predict drug and surgical dressing inventory requirements. The actual demand for one

item is as shown in the accompanying table.

a) Using the previous moving-average data, convert to an exponential smoothing forecast

for month 33.

Month Demand

24 78

71

25 65

26 90

27 71

28 80

29 101

30 84

31 60

32 73

b) Compute a 3-month moving average (MA)

c) If Syx= 16.6, find 90% control limits. (Hint: control limits: +(-)t Sf , degree of

freedom n-2. Also round the figures to the nearest integer.)

Solution:

MA = Σx/ no. of periods = (78+65+…+73)/9=78 Units

Assume forecast of the last month = 78

Estimate = 2/(n+1)= 2/(9+1)= 0.2

Smoothed forecast for this = (sales last month)+(1-)(previous forecast of last

month month’s sales sales)

= 0.2(73)+(1-0.2)(78)= 14.6+62.4 = 77 units.

b)

Month Actual Demand 3 Month MA Forecast Demand

MA3

24 78

25 65

26 90

27 71 77.7= ~ 78

28 80 75.3= ~ 75

72

29 101 80.3= ~ 80

30 84 84.0= ~ 84

32 60 88.3= ~ 88

32 73 82

= (78+75+80+84+88+82)/6= 81.2

Control Limits = +(-)tsf = 81.2+(-)2.132(16.6)

= 116.6Units to 45.8 Units

3. Demir Trucking Company is to haul local freight in the southern cities of Aegean

District. During the past five years the annual demand for Demir’s services has grown

steadily, requiring substantial outlays for trucking equipment. The chief executive of

the company has just managed to scrimp enough capital funds over the past five years

to keep his head above water. CEO realizes the capital funds planning, since he

expects the past pattern of growth to continue.

Analyze the past 5 years of data by trend analysis to forecast the next two years’

requirements for capital funds.

Year Capital Funds (millions

of MU)

2000 100

2001 110

2002 130

2003 140

2004 160

Solution:

Year x Capital Funds

(millions of

MU), y

x2 xy y2

2000 0 100 0 0 10000

2001 1 110 1 110 12100

2002 2 130 4 260 16900

73

2003 3 140 9 420 19600

2004 4 160 16 640 25600

Totals 10 640 30 1430 84200

Σy= na + bΣx 640= 5a + 10b (1)

Σxy= aΣx + bΣ x2 1430= 10a+ 306b (2)

Multiply (1)by 2 1280=10a-20b (3)

(2)-(3) 150=10b

b=15

Substituting b in (1) 640=5a+10(15)

640=5a+150

490=5a

a=98

Y=98+15X

Y2005 = 98+15(5)= 98+75= 173 million MU

Y2006 = 98+15(6)= 98+90= 188 million MU

Standard deviation of the trend line;

Syx =

= 3.16 (millions) MU

Assume n>30 (i.e. sample is large), the 95.5% confidence limits are found as

follows;

Y2006 +/- 2syx 188 +/- 2(3.16) 194.32 181.68

There is a 95.5 % probability that capital funds requirement for the year 2006 will

be between 194.32 million MU and 181.68 million MU. There is only 10% probability

that the requirement will fall outside these limits.

4. Azim Kitchen’s Inc. has collected the following data to learn if the number of

building permits might be a useful predictor of their kitchen cabinet demand.

74

Building

Permits, x

(00)

Cabinet

Sales, y

($000)

xy x2 y2

(x-x) (y-y) (x-x) 2 (y-y) 2 (x-x).(y-y)

2 3 6 4 9 -1 -2 1 4 2

5 5 25 25 25 2 0 4 0 0

1 5 5 1 25 -2 0 4 0 0

2 6 12 4 36 -1 1 1 1 -1

5 7 35 25 49 2 2 4 4 4

4 6 24 16 36 1 1 1 1 1

3 5 15 9 25 0 0 0 0 0

4 5 20 16 25 1 0 1 0 0

1 3 3 1 9 -2 -2 4 4 4

27 45 145 101 239 0 0 20 14 10

=3 =5

a) Use the normal equations to derive a regression forecasting equation.

Σy = na + bΣx 45 = 9a + 27b (1)

Σxy = aΣx + bΣ x2 145b = 27a + 101b (2)

Multiple (1) by (-3) -135 = -27a - 81b (1)

145=27a+101b (2)

add (1)&(2) 10=20b b = ½ = 0.5

Substitute b = 0.5 in (1) 45 = 9a + 27(0.5)

9a = 45-13.5 = 3.5

Y = 3.5 + 0.5 X

b) Compute the standard deviation of regression..

75

Syx = = = $1.13 (000)

c) Assume our regression equation has been derived from a sufficiently large sample that the confidence limits estimate from equal to Y+ 2Syx may be used. Establish a 95.5% confidence limits estimate for the specific amount of cabinet sales ($000). When permits number is 4.4 (00).

Confidence intervals [3.5+0.5(4.4)] + 2(1.13)= 5.7 + 2.26= 3.44 to 7.96

There is 95.5 % probability that our cabinet sales for 440 permits will between $3440-$7960. There is only 4.5 % probability that sales will fall outside these limits.

d) Confidence limits of 90%

Y +/- t.syx = 5.76 +/- 1.895(1.13) = 5.76 +/- 2.141 = 7.901 3.619

There is 90 % probability that the cabinet sales for 440 permits will fall between $3619 - $ 7901. There is only 10 % chance that sales may fall outside these limits.

e) Correlation Analysis

r = ∑[(x-x)(y-y)]/ = 10 / = 0.5977 = ~ 0.60

There is a moderate relation between the building permits and cabinet sales

h) Coefficient of determination

r2 = (0.60)2 = 0.36

Since r2 = 0.36, we can say that 36% of the variation in kitchen cabinet demand is explained by the number of building permits, and remaining 64% is explained by other factors. 64% is an important portion that has to be discussed in detail.

g) The significance of the value of r = 0.60 can, however, be tested under a hypothesis that are no correlation between the number of permits and demand (sales) for kitchen cabinet, that is Ho r=0.

The computed statistical t- value of r is compared with a theoretical t- value of r for a given size (n=9) and significance level of 5%. If the statistical computed t- value of r, (tc), is greater than the theoretical tabled value of the hypothesis, tk, Ho r=0 is rejected, the correlation is deemed significant at the specific level.

tk = │r│ =│0.60│ = 1.984

76

tk=2.365

tc (1.984)< tk (1.895)

The hypothesis r = 0 is accepted. The computed r is not meaningful.

5. Azim Ticaret’s long-range sales were closely tied to national sector sales. The

following are seven years of historical data for Azim Ticaret.

Year A.T. Sales (000MU) Sector Sales (million MU)

1998 9.5 120

1999 11.0 135

2000 12.0 130

2001 12.5 150

2002 14.0 170

2003 16.0 190

2004 18.0 220

a) Develop a simple linear regression analysis between firm’s sales and sector sales.

Forecast firm’s sales for the next 3 years if the sector sales (estimated) are 250, 270,

300 (million) MU.

b) What percentage of variation in firm’s sales is explained by sector sales?

Solution:

a)

X y x2 Xy y2

120 9.5 14400 1140 90.25

135 11.0 18225 1485 121.00

130 12.0 16900 1560 144.00

77

150 12.5 22500 1875 156.25

170 14.0 28900 2380 196.00

190 16.0 36100 3040 256.00

220 18.0 48400 3960 324.00

Total: 1115 Total: 93.0 Total: 185425 Total: 15440 Total: 1287.50

a = (Σx2Σy-ΣxΣxy)/[nΣx2-(Σx)2] = [185425(93)-(1115)(15440)]/[7(185425)-(1115)2] =

0.528 Ь = (nΣx Σy-ΣxΣy)/ [nΣx2- (Σx)2 ] = [7(15440)-1115(93)]/ (54750) = 0.0801

Y= 0.528+ 0.0801X

Y2005 = 0.528+ 0.0801(250)= 20.55(000) MU

Y2006 = 0.528+ 0.0801(270)= 22.16(000) MU

Y2007 = 0.528+ 0.0801(300)= 24.56(000) MU

b)

r = (nΣxy-Σx Σy))/

= (7(15440)-1115(93))/

= 4385/4461.12 = 0.9829

r2 = 0.966 or 96.6%

96.6% of the variation in firm’s sales is explained by sector sales.

c) Coefficient of determination: r2 = (0.9829) 0.966=~97%

97% of the variation in AT sales can be explained by the other sector sales, and 3% by

the other factors.

d) Test the correlation at 10% level of significance.

tc = │r│ = │0.98│ = 0.98(12.91) = 12.65

degree of freedom = n - 2 = 7 – 2 = 5; level of significance = 10%.

Using student-t table;

tk = 2.015.

The hypothesis H0r = 0 is rejected, because

tc 12.65) > tk (2.015)

Therefore the correlation coefficient, that is found is meaningful.

e) Using correlation formula find regresssion forecast equation

78

b = [n∑xy-∑x∑y]/[n∑x2-(∑x)2] = [7(15440)-1115(93)]/[7(185425)-(1115)2] =

0.0801

a = y – bx = 13.29 – (0.0801)(159.29) = 0.528

Y = 0.528 + 0.0801X

There is no difference between the regression forecasting equations found in (a) and (e).

f) Compute the forecast of AT sales for the year 2005 depending on the sector sales

forecast for that year.

Year x Sector Sales,y xy x 2

1998 0 120 0 0

1999 1 135 135 1

2000 2 130 260 4

2001 3 150 450 9

2002 4 170 680 16

2003 5 190 950 25

2004 6 220 1320 36

21 1115 3795 91

∑y = n.a + ∑x 1115 = 7a + 21b (1)

∑xy = a∑x + b∑x2 3795 = 21a + 91 b (2)

Solving these simultaneous equations for a and b values, we get

a = 111.07 b = 16.07

and the trend equation will be as follows;

Y = 111.07 + 16.07X

Therefore

Y2005 = 111.07 + 16.07 ( 7) = 223.56 million MU

Substituting this value in the regression forecast equation

Y2005 = 0.528 + 0.0801 (223.56) = 18.435 (000) MU

g) Compute the standard deviation of regression;

Syx = =

79

= ~ 0.57 (000) MU

h) Confidence limits of 90% for X = 223.56(mill.MU) and Y = 18.435 (000) MU

Y +/- t.syx 18.435 +/- 2.015 (0.57) 19.584 17.286

Assuming sectors sales of year 2005 be 223.56 (mill.)MU, AT sales will fall

between 19 584 MU and 17 286 MU.

6. Ali Caliskan is a buyer in the purchasing department at Azim Industries Ltd. His

speciality is nonferrous metals. Ali is attempting to develop a system for forecasting

monthly copper prices. He has accumulated 16 months of historical price data:

Month Copper Price (MU/kg)

1 0.85

2 0.82

3 0.90

4 0.79

5 0.83

6 0.85

7 0.89

8 0.81

9 0.95

10 0.90

11 0.90

12 0.85

13 0.83

14 0.81

15 0.87

16 0.85

Ali Caliskan wishes to compare two forecasting systems to forecast copper prices:

Moving averages. (AP= 3) and exponential smoothing (= 0.5).

a) Compute the two sets of monthly forecasts over the past 10 months. (7 thru 16).

80

b) Which forecast system has the least forecasting error?

c) Select the best system and forecast the copper prices for month 17.

Solution:

a)

Month Price Moving

Forecast

Average

Forecast

error

(AP=3)

Exponential

Forecast

Smoothing

Error

(= 0.5)

4 0.79

5 0.83

6 0.85 0.832

7 0.89 0.823 0.067 0.841 0.079

8 0.81 0.857 0.047 0.866 0.056

9 0.95 0.850 0.100 0.838 0.112

10 0.90 0.883 0.017 0.894 0.006

11 0.90 0.887 0.013 0.897 0.003

12 0.85 0.917 0.067 0.899 0.049

13 0.83 0.883 0.053 0.875 0.045

14 0.81 0.860 0.050 0.853 0.043

15 0.87 0.830 0.040 0.832 0.038

16 0.85 0.837 0.013 0.851 0.001

Total errors 0.467 Total errors 0.402

b) Exponential Smoothing has the least forecasting error.

c) The smoothing forecast (= 0.5) seems to have a quicker impulse response from

period to period and yet does not reach the extreme values as the moving average,

therefore it appears to have a better noise dampening ability,

F17= F16+(A16-F16) = 0.851+ 0.5(0.85-0.851)= 0.851 MU.

7. The historical demand for Mother Evin’s Pies is, in thousands of dozens: Month Demand

January 14

81

February 12

March 12

April 11

May 17

June 16

a. Use a weighted moving average with weights of 0.6, 0.3 and 0.1 to find the July

forecast.

b. Use a simple 3-month moving average to find the July forecast.

c. Use single exponential smoothing with α = 0.1 and a June forecast of 14 to find

the July forecast.

Solution:

a. Jjuly = 0.6(AJune) + 0.3 (AMay) + 0.1 (AApril) = 0.6(16) + 0.3(17) + 0.1(11) = 15.8

b. JJuly = (16+17 + 11) : 3 = 14.7

c. FJuly = FJune + α (AJune – Fjune) = 14 + 0.1(16-14) = 14.2

8. The demand for electrical power at Magusa over the period 1998 – 2004 is shown

below, in megawatts:

Year Electrical Power Demand

1998 74

1999 79

2000 80

2001 90

2002 105

2003 142

2004 122

Forecast the electrical power demand in 2005 and 2006.

Solution:

82

Year Time, x Demand, y x2 xy

1998 0 74 0 0

1999 1 79 1 79

2000 2 80 4 160

2001 3 90 9 270

2002 4 105 16 420

2003 5 142 25 710

2004 6 122 36 732

Σx= 21 Σy = 692 Σx2= 91 Σxy = 2371

Σy = na + bΣy 692 = 7a + 21b (1)

Σxy = aΣx + bΣx2 2371 = 21a + 91b (2)

(1)*-3 : -2076 = -21a - 63b

2371 = 21a + 91b 295 = 28b b = 10.5

Substituting b=10.5 in Eq.1

692 = 7a + 21 (10.5)

7a = 171.5

a = 67.36

So Equation is : Y = 67.36 + 10.5X

Y 2005 = 67.36 + 10.5 (7) = 136.86 ~ 137 megawatts

Y 2006 = 67.36 + 10.5 (8) = 151.36 ~ 151 megawatts

9. Levent Construction Co. renovates old homes in TRNC. Over time, the company has

found that its MU volume of renovation work is dependent on the TRNC payroll. The

83

following table lists Levent`s revenues and the amount of money earned by wage earners

in TRNC during the past six years.

Levent`s Sales (000000 MU)

Local Payroll (000000000 MU)

2 1

3 3

2.5 4

2 2

2 1

3.5 7

a. Estimate sales for Levent if the TRNC payroll will be 600 million MU next year

b. Find the standard error in sales

c. Calculate correlation coefficient, coefficient of determination and interpret them

Solution:

Sales, y Payroll, x x2 xy y2

2 1 1 2 4

3 3 9 9 9

2.5 4 16 10 6.25

2 2 2 4 4

2 1 1 2 4

3.5 7 49 24.5 12.25

Σy = 15 Σx = 18 Σx2 = 80 Σxy = 51.5 Σy2= 39.5

ŷ = 2.5 x = 3

b = Σxy – nxŷ = 51.5 -6(3)(2.5) = 0.25

Σx2 – nx2 80-6(32)

84

a = ŷ – bx = 2.5 – 0.25(3) = 1.75

Y= 1.75 + 0.25X

a. Sales in hundred thousands = Y= 1.75 + 0.25 (6) = 3.25 = 325 000 MU

b. Syx = Σy 2 – aΣy – bΣxy = 39.5- (1.75)(150) – 0.25(51.5) = 0.306

n – 2 6- 2

The standard error of the estimate is then 30 600 MU in sales.

c. Left to the student. ( r = 0.901, r2 = 0.81 )

85