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
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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
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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.
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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
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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
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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
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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
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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
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1
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3
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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
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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.
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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
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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
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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.
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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.
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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.
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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
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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,
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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.
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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
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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
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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