Lecture notes

QUANTITATIVE ANALYSIS
Lecture Notes,
Past Quiz, Exam, & Home Work
Questions
Prof. Dr. Serhan ÇİFTÇİOĞLU
Gazimağusa 2005

Foreword
These “Lecture Notes” have been prepared to serve as a study guide for the
students of “Quantitative Analysis” course (MGMT 322), that I offer for business
students. They are designed to outline the critical topics that are covered by the course
but also attempt to give an example about the application of each quantitative and
statistical tool covered by the course. At the end of these “Lecture Notes” I also added
examples of past quizzes, exams, and homework questions, to help interested students
both to improve their analytical capacity regarding the application of the quantitative
tools, that they have learn, and at the same time better prepare for exams.
I would like to express my gratitude for the excellent work that our teaching
assistants and former students of MGMT 322, Erhan ERDOĞAN and Süleyman EFE,
and Gift, have done in organizing and typing my in-class lectures.
Assoc. Prof. Dr. Serhan ÇİFTÇİOĞLU
Department of Business Administration
Eastern Mediterranean University

Quantitative Analysis – MGMT 322
- 1 -
INDEX
CHAPTER 2: A REVIEW OF PROBABILITY THEORY..................................................- 4 -
A. PROBABILITY......................................................................................................................... - 4 -
B. EVENT..................................................................................................................................... - 4 -
EXAMPLE ..................................................................................................................................... - 4 -
C. EXPERIMENT.......................................................................................................................... - 4 -
D. SAMPLE SPACE ...................................................................................................................... - 4 -
EXAMPLE ..................................................................................................................................... - 4 -
E. MUTUALLY EXCLUSIVE EVENTS .......................................................................................... - 5 -
EXAMPLE ..................................................................................................................................... - 5 -
ALTERNATIVE APPROACHES IN PROBABILITY THEORY.......................................- 5 -
1. CLASSICAL APPROACH .......................................................................................................... - 5 -
EXAMPLE ..................................................................................................................................... - 6 -
2. RELATIVE FREQUENCY APPROACH...................................................................................... - 6 -
EXAMPLE ..................................................................................................................................... - 6 -
3. SUBJECTIVE APPROACH ........................................................................................................ - 6 -
EXAMPLE ..................................................................................................................................... - 6 -
PROBABILITY RULES...........................................................................................................- 7 -
A. MARGINAL (UNCONDITIONAL) PROBABILITY: ................................................................... - 7 -
B. THE ADDITION RULE: ........................................................................................................... - 7 -
1. THE ADDITION RULE FOR MUTUALLY EXCLUSIVE EVENTS ................................................... - 7 -
EXAMPLE ..................................................................................................................................... - 7 -
2. THE ADDITION RULE FOR NON-MUTUALLY EXCLUSIVE EVENTS .......................................... - 7 -
EXAMPLE ..................................................................................................................................... - 8 -
EXAMPLE ..................................................................................................................................... - 8 -
TYPES OF EVENTS.................................................................................................................- 9 -
1. STATISTICALLY INDEPENDENT EVENTS ............................................................................... - 9 -
2. STATISTICALLY DEPENDENT EVENTS ................................................................................ - 10 -
TYPES OF PROBABILITIES UNDER INDEPENDENCE ................................................................ - 10 -
1. MARGINAL (UNCONDITIONAL) PROBABILITY....................................................................... - 10 -
2. JOINT PROBABILITY............................................................................................................... - 10 -
3. CONDITIONAL PROBABILITY ................................................................................................. - 10 -
TYPES OF PROBABILITIES UNDER DEPENDENCE ................................................................... - 11 -
1. MARGINAL (UNCONDITIONAL) PROBABILITY....................................................................... - 11 -
2. JOINT PROBABILITY............................................................................................................... - 11 -
3. CONDITIONAL PROBABILITY ................................................................................................. - 11 -
EXAMPLE ................................................................................................................................... - 11 -

- 2 -
BAYES’ THEOREM ..............................................................................................................- 12 -
EXAMPLE ................................................................................................................................... - 12 -
EXAMPLE ................................................................................................................................... - 13 -
PROBABILITY DISTRIBUTION.........................................................................................- 15 -
DISCRETE VS CONTINUOUS RANDOM VARIABLES.................................................................... 15
DISCRETE PROBABILITY DISTRIBUTION........................................................................................ 15
CONTINUOUS PROBABILITY DISTRIBUTION .................................................................................. 15
BINOMIAL PROBABILITY DISTRIBUTION................................................................................ - 18 -
CHARACTERISTICS OF A BERNOULLI TYPE PROCESS ................................................................ - 18 -
BINOMIAL FORMULA:................................................................................................................ - 18 -
EXAMPLE ................................................................................................................................... - 19 -
NORMAL PROBABILITY DISTRIBUTION .................................................................................. - 23 -
CHARACTERISTICS..................................................................................................................... - 24 -
TWO PARAMETERS: MEAN (µ) & STANDARD DEVIATION (σ)................................................... - 24 -
EXAMPLE:.................................................................................................................................. - 20 -
3 MATHEMATICAL FACTS ABOUT NORMAL DISTRIBUTION ................................................. - 24 -
EXAMPLE:.................................................................................................................................. - 25 -
CHAPTER 3: FORECASTING...............................................................................................-29-
THE METHODOLOGY OF FORECASTING................................................................................. - 29 -
TECHNICAL ANALYSIS
FUNDAMENTAL ANALYSIS
BASIC STEPS IN FORECASTING
FORECASTING MODELS........................................................................................................... - 31 -
FOR NON-SEASONAL PRODUCTS WITH LINEAR TREND TYPE OF PATTERN........................ - 31 -
NAÏVE MODEL ........................................................................................................................... - 33 -
MOVING AVERAGE MODEL……………………………………………………………………. .-39-
SIMPLE EXPONENTIAL MODEL.................................................................................................. - 41 -
IMPORTANT STEPS..................................................................................................................... - 42 -
TIME SERIES REGRESSION MODEL............................................................................................ - 42 -
IMPORTANT POINTS ................................................................................................................... - 44 -
SMOOTHING LINEAR TREND MODEL......................................................................................... - 46 -
IMPORTANT STEPS..................................................................................................................... - 46 -
CHAPTER 4: ALTERNATIVE DECISION MAKING ENVIRONMENTS....................- 48 -
BASIC STEPS IN DECISION MAKING
ALTERNATIVE DECISION MAKING ENVIRONMENTS
ALTERNATIVE CRITERIA FOR DECISION MAKING UNDER UNCERTAINTY .......................... - 48 -
THE MAXI-MAX CRITERION...................................................................................................... - 49 -
THE MAXI-MIN CRITERION....................................................................................................... - 49 -
THE CRITERION OF REALISM..................................................................................................... - 50 -
THE MINI-MAX REGRET CRITERION......................................................................................... - 50 -
ALTERNATIVE CRITERIA FOR DECISION MAKING UNDER RISK .......................................... - 51 -

- 3 -
EXPECTED VALUE CRITERION................................................................................................... - 51 -
EXAMPLE ................................................................................................................................... - 51 -
CRITERION OF RATIONALITY..................................................................................................... - 52 -
THE CRITERION OF MAXIMUM LIKELIHOOD............................................................................. - 52 -
SUPPLYING THE NUMBERS: HOW TO OBTAIN MEAN AND STANDARD
DEVIATION WHEN DATA IS MISSING OR INCOMPLETE........................................- 57 -
CHAPTER 5: COST – VOLUME – PROFIT ANALYSIS .................................................- 61 -
GENERAL DEFINITION AND USE OF COST- VOLUME- PROFIT ANALYSIS
ITS APPLICATION
COMBINING UNIT MONETARY VALUES AND PROBABILITY DISTRIBUTION.............................. - 64 -
3 – STEP PROCEDURE TO OBTAIN EXPECTED PROFITS.............................................................. - 67 -
3 – STEP PROCEDURE TO OBTAIN EXPECTED LOSS................................................................... - 68 -
Home Works…………………………………………………………………-70-

- 4 -
Chapter 2: A Review of Probability Theory
A. Probability
Probability is the chance or likelihood of something occurring (happening).
We express probabilities as either fractions or decimals
E.g: P(H) = 9/10 (as a fraction)
or
P(H) = 0.9 (as a decimal)
Note: The probabilities should neither be < 0, nor >1.
B. Event
An event is defined as either one or more than one of possible outcomes of doing
something or an activity.
Example
ACTIVITY: Inflationary process in Turkey in the year 2005:
Events: A: If annual inflation for 2005 is exactly 10%
B: If annual inflation for 2005 is less than 8%
C: If annual inflation for 2005 is either 5% or 7%
C. Experiment
Is the activity that generates events.
For example, tossing a coin twice.
D. Sample Space
The sample space of an experiment is the list of all possible outcomes of that
experiment.
Example
Experiment: Tossing a coin.
Sample Space: S = {HH, TT, HT, TH}
Events: A: Exactly one outcome, that is, getting 2heads = HH
B: More than one outcome, that is, getting at least one tail = (TT, HT, TH)
C: Exactly three outcomes, that is, getting 2tails or getting one head and one tail
= (TT, HT, TH)

- 5 -
E. Mutually Exclusive Events
Mutually exclusive events are those events which can take place only one at a
time, and not together at the same time.
F. Non – Mutually Excusive Events
Are those events which can happen together at the same time.
Example 1
Experiment: Tossing a coin twice
Events: X: Getting two Heads: P(HH)
Y: Getting two Heads or two Tails: P(HH or TT)
Z: Getting Head-Tail or Tail-Head: P(HT or TH)
X and Y are non - mutually exclusive but X and Z are mutually exclusive.
Example 2
Experiment: Randomly selecting a student in this class.
Events: A: Selected student is 21 years old.
B: Selected student is male
C: Selected student is female
D: Selected student is 22 years old.
Therefore: Events AB are non-mutually exclusive because they can happen together at
the same time.
Events AD are mutually exclusive because they can not happen together at
the same time.
Events AC are non-mutually exclusive because they can happen together at
the same time.
Events BC are mutually exclusive because they can not happen together at
the same time.
Alternative Approaches in Probability Theory
1. Classical Approach
# of outcomes favorable to the occurrence of event “A”
P(A) =
Total # of possible outcomes

- 6 -
Example 1
Experiment: Randomly selecting a student in a class of 40 students.
Events: A: Selecting a female student, given total number of females equal to 10.
B: Selecting a male student, given total number of males equal to 30.
The total number of students is 40
P(A) = 10/40
P(B) = 30/4
2. Relative Frequency Approach
The probability of the event found by using the given past data.
Example
Experiment: Selecting a student randomly.
Events: A: Selecting a student who gets “A” from MGMT 322.
Past Data: 20% of all students who have taken MGMT 322 in the past get “A”.
P(A) = 0.20
B: Galatasaray wins the championship in 2005, using the past data obtain how
frequently the similar event occurred in the past.
Past Data: 40% of the last 9 years, Galatasaray was a champion.
P(B) = 0.40
3. Subjective Approach
The probability of the event found by using the personal feeling or experience.
Example
Event: A: Galatasaray wins.
P(A) = 0.90
As a supporter or a football watcher you may have the feeling that Galatasaray
will win 90%.

- 7 -
Probability Rules
A. Marginal (Unconditional) Probability:
Only one event can take place. P(A)
B. The Addition Rule:
1. The Addition Rule for Mutually Exclusive Events
If Events A, B, and C are mutually exclusive (meaning they can not happen together
at the same time.)
Then: P(A or B) = P(A) + P(B)
P( A or B or C) = P(A) + P(B) + P(C)
Questions that can be asked and answered by using Addition Rule:
Q1: What is the probability of A or B or C taking place?
Q2: What is the probability of A or B taking place?
Q3: What is the probability of B or C taking place?
Example
Suppose that the number of students in each age group in the class is as follows:
Age Male Female Total
20 0 2 2
21 1 3 4
22 6 3 9
23 3 1 4
24 4 0 4
Total 14 9 23
Events: A: Selecting a random student who is 20 years old.
B: Selecting a random student who is 21 years old.
C: Selecting a random student who is 22 years old.
D: Selecting a random student who is 23 years old.
P(A or B) = P(A) + P(B) = 2/23 + 4/23 = 6/23
P(A or C) = P(A) + P(C) = 2/23 + 9/23 = 11/23
P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) = 2/23 + 4/23 + 9/23 + 4/23 = 19/23
2. The Addition Rule for Non-Mutually Exclusive Events
If A and B are non-mutually exclusive events,
Then: P(A or B) = P(A) + P(B) – P(A and B)

- 8 -
Example 1
Suppose that the sex and the eye color of students in a class of 40 people is distributed as
follows:
Black Green Total
Male 13 9 22
Female 8 10 18
Total 21 19 40
Experiment: Selecting a random student.
Events:
A: Selecting a male student.
B: Selecting a female student.
C: Selecting a black eye student.
D: Selecting a green eye student.
P(A or C) = P(A) + P(C) – P(A and C) = (22/40 + 21/40) – 13/40 = 30/40
P(B or C) = P(B) + P(C) – P(B and C) = (18/40 + 21/40) – 8/40 = 31/40
Example 2
Suppose that the Ford Motor Corporation produces 3 different models for Turkish
consumers. Annual volume of production and sales of all 3 model combined is 700 units.
However, some of the autos sold turned to be defective. Based on the past data the
production manager estimated the following distribution of defective and non-defective
autos for each model manufactured.
Models
A B C Total
Defective 10 20 70 100
Non-Defective 140 180 280 600
Total 150 200 350 700
a. What is the probability that a randomly selected consumer will buy either model
A or C?
b. What is the probability that the consumer will buy either model B or a Defective
auto?
c. What is the probability that the consumer will buy either model A or a Non-
defective auto?
a. P(A or C) = P(A) + P(C) = 150/700 + 350/700 = 500/700
b. P(B or D) = P(B) + P(D) – P(B and D) = (200/700 + 100/700) – 20/700 = 280/700
c. P(A or N) = P(A)+P(N) – P(A and N) = (150/700 + 600/700) – 140/700= 610/700

- 9 -
Example 3
There are 500 companies whose stocks are traded on Istanbul stock exchange. These
companies operate in these basic sectors of the economy; Industry, Service, and Mining.
Each one of these companies can report their positive profits and losses for each year.
Based on the past data, financial analysts estimated the following distribution of the
number of companies in each sector that are likely to report positive profits and losses in
the year 2005:
Sectors
Industry Service Mining Total
Positive Profits 150 240 30 420
Losses 50 10 20 80
Total 200 250 50 500
a. What is the probability of randomly selected company (out of these companies
listed on Istanbul Stock Exchange) to report positive profits in year 2005?
b. What is the probability of a randomly selected company to report either losses in
year 2005 or to be from Mining sector?
c. What is the probability of a randomly selected company to be either from Industry
or from Service sector?
a. P( Positive profits) = 420/500 = 0.84
b. P( Losses or Mining) = P( Losses) + P( Mining) – P( Losses and Mining)
= 80/50 + 50/500 – 20/500 = 110/500 = 0.22
c. P( Industry or Service) = P(Industry) + P(Service)
= 200/500 + 250/500 = 450/500 = 0.9
Types of Events
1. Statistically Independent Events
These are those events whereby the probability of occurrence of an event is not
affected by ( or dependent upon) the occurrence of the other event.
Example
Events: A: Izlem goes to Istanbul next Tuesday
B: Istanbul stock market decreases by more than 10% next Friday.
Therefore: Event A and B are Independent events because the occurrence of event “A”
does not have an impact on the probability of the occurrence of event “B”.

- 10 -
2. Statistically Dependent Events
These are those events whereby the probability of the occurrence of an event is
affected by ( or dependent upon) the occurrence of the other event.
Example 1
Events: A: Stock Market Index rises tomorrow
B: Interest rates go down today
Therefore: Event A and B are Dependent because the occurrence of event “B” is
expected to increase the probability of the event “A”
Example 2
Events: A: The quantity of automobiles purchased in Turkey rises
B: The quantity of petroleum consumed in Turkey rises
Therefore: Event A and B are Dependent because the probability of the occurrence of
event “B” is affected by or dependent upon the occurrence of event “A”, as the more the
automobiles Turkish people are purchasing, the more the petroleum they will buy to run
their automobiles.
Types of Probabilities under Independence
1. Marginal (Unconditional) Probability
P(A): This is interested in the probability of a single event.
2. Joint Probability
P(AB): This is where you are interested in either the probability of event A and event B
happening together at the same time or in succession.
P(AB) = P(A) * P(B)
P(BA) = P(B) * P(A)
A and B are independent events.
3. Conditional Probability
P(A/B): Probability of A, given B!
As long as A and B are independent events P(A) is not affected by P(B) so;
P(A/B) = P(A)
P(B/A) = P(B)

- 11 -
Example
Events: A: It rains in Cyprus today
B: Exchange rate of dollar/yen increases today
Therefore: P(A/B) = P(A)
PB/A) = P(B)
Types of Probabilities under Dependence
1. Marginal (Unconditional) Probability
P(A) : Is interested in just the probability of a single event.
2. Joint Probability
P(AB) = P(A/B) * P(B)
P(BA) = P(B/A) * P(A)
P(AB) = P(BA)
3. Conditional Probability
P(A/B) = P(AB) / P(B)
P(B/A) = P(BA) / P(A)
Example
Suppose that the eye color of students in a class of 40 people as follows;
Black Green Total
Male 20 8 28
Female 10 2 12
Total 30 10 40
Events:
A: Selecting a Male student.
B: Selecting a student with Black Eyes.
C: Selecting a Female student.
D: Selecting a student with Green Eyes.
Q1. If the selected student is a Male student, what is the probability that he has
got Green Eyes?
P(D/A) = P(DA) / P(A) = (8/40) / (28/40) = 0.074

- 12 -
Q2. If the selected student is a Female student, what is the probability that she has
got Green Eyes.
P(D/C) = P(DC) / P(C) = (2/40) / (12/40) = 0.167
Bayes’ Theorem
Is used for revising prior (before) estimates of probabilities using limited new
information to obtain posterior ( new or after) probabilities.
Event A gives us P (A); after sometime passes event B occurs which is dependent
on event A.
We can use Bayes’ Theorem to obtain posterior probability of A:
P (A/B) = P (AB)/ P(B)
Example 1
Suppose we have 100 dices in a basket, and we have two types; half of it is Type1
and the other half is Type2. Given that the probability of getting Ace (1) when you roll a
Type1 die is 0.3 and that of rolling Ace (1) of Type2 is 0.6. You are also given the initial
probability of Type1 as 0.5 and that of Type2 as 0.5.
If you randomly select a die and roll it and you find out that Ace (1) has come up, what is
the probability of this die to be a Type1 die?
NB: These two events (Ace and Type1 or 2) are dependent because the probability of
getting Ace is affected by the occurrence of Type1 or 2.
P( Type1/Ace) = P(Type1 Ace) / P(Ace)
Note: When you are not given the values for the formula like in the example above, use
the following five steps below in order to get the Answer:
Step 1: Prepare the following table:
Marginal Probability of Elementary
Events
Probability of Secondary Event
Conditional Probability of Secondary
Event, Given Each of the Elementary
Event
P(Type 1) = 0.5 P(Ace / Type 1) = 0.3
P(Type 2) = 0.5
P(Ace)
P(Ace / Type 2) = 0.6
Step 2: Computation of probability of secondary event:
P(secondary event) = summation of its joint probabilities of each one of the elementary
events.

- 13 -
P(Ace) = P(Type1, Ace) + P( Type2, Ace)
Step 3: Computation of joint probability of secondary events with each of the elementary
events:
P(Type 1, Ace) = P(Ace/Type1)* P(Type 1)
= 0.3 * 0.5 = 0.15
P(Type 2, Ace) = P(Ace/Type2) * P(Type2)
= 0.6 * 0.5 = 0.30
P(Ace) = 0.15 + 0.30 = 0.45
Step 4: Use the Baye’s Theorem to get the final answer
P(Type1/ Ace) = P(Type 1, Ace) / P(Ace)
= 0.15 / 0.45 = 1/3 This is posterior probability
Example 2
Economists believe that the annual inflation rate in Turkey is affected by the
changes in the petroleum prices. The probability of inflation rate increasing in 2001 is
estimated to be 0.60. The probability of petroleum prices rising is 0.40. The probability of
both inflation rate and the petroleum prices rising is 0.35. On the other hand, the
probability of inflation rate rising and at the same time, the petroleum prices not rising is
estimated to be 0.20. If petroleum prices do not rise in 2001, what is the probability of
inflation rate rising?
Events: I: Inflation rate increasing.
P: Petroleum prices rising.
N: Petroleum prices not rising
Given:
P(I) = 0.6
P(P) = 0.4
P(IP) = 0.35
P(IN) = 0.2
Solution:
P(N) = 1 – P(P) = 1- 0.4 = 0.6
P(I / N) = P(IN) / P(N) = 0.2 / 0.6 = 0.333 ≈ 0.3

- 14 -
Example2:
Annual profits of construction sector depend on wage rate paid to labor employed by the
sector. The past data about the behavior of annual profits suggested that the probability of
profits rising in a given year is 0.80 and the probability of profits falling is 0.20. The past
data also suggest that in 60% of all the years during which the profits have increased,
wage rate has declined, and it has increased in the remaining 40%. However, in 80% of
all years during which profits have fallen, wage rate has increased, and the only
remaining 20% wage rate has decreased. Due to the economic crisis in Turkey, wage rate
has been declining since the beginning of 2005. Given this downward trend in wages,
what is our revised estimate for the probability of annual profits of construction sector to
rise in the year?
Events: A: Profits rising
B: Profits falling
C: Wage rate declining
D: Wage rate increasing
Given:
P(A) = 0.80
P(B) = 0.20
P(C/A) = 0.60
P(D/A) = 0.40
P(D/B) = 0.80
P(C/B) = 0.20
Solution: Without using the table
P(A/C) = P(AC) / P(C) or P(CA) / P(A)
0.60 = P(CA) / 0.80
P(CA) or P(AC) = 0.60 * 0.80 = 0.48
P(C) = P(CA) + P(CB)
= [P(C/A) * P(A)] + [P(C/B) * P(B)]
= (0.60 * 0.80) + (0.20 * 0.20)
= 0.48 + 0.04 = 0.52
P(A/C) = 0.48 / 0.52 = 0.92
Example 3
A child chosen at random in a community school system comes from low-income family
15% of the time. Children from low-income families in the community graduate from
college only 20% of the time. Children not from low-income families have a 40% chance
of graduating from college. As an employer of people from this community, you are
reviewing applicants and note that the first one had a college degree. What is the
probability that the person comes from a low-income family?

- 15 -
Events: L: Child coming from a low-income family
G: Graduate from college
H: Child coming from not a low-income family
Given:
P(L) = 0.15
P(H) = 1- P(H) = 1- 0.15 = 0.85
P(G/L) = 0.20
P(G/H) = 0.40
P(L/G) = ?
Solution:
P(L/G) = P(LG) / P(G)
P(G/L) * P(L) / P(G)
P(G) = [P(G/L) * P(L)] + [P(G/H) * P(H)]
= (0.20 * 0.15) + (0.40 * 0.85)
= 0.03 + 0.34 = 0.37
P(L/G) = 0.03 / 0.37 = 0.08 = 8%
Probability Distribution
Random Variables
A Random Variable is a variable which takes on values unpredictably as a result
of a random process.
EG: Events: X: Age of a randomly selected student
Y: The value of stock market Index in Istanbul by the end of today’s
session.
Z: Interest rates on a new Bond issued by the Turkish Government that will
take place next week.
DISCRETE VERSUS CONTINUOS RANDOM VARIABLES
1. Discrete Random Variable
This is allowed to take on only limited number of possible values.
E.g: X = Age of a randomly selected student.
a. Binomial Distribution
b. Poisson Probability Distribution
2. Continuous Random Variable
This is allowed to take any value over a certain range. They are infinite numbers of
possibilities.
a. Normal Probability Distribution

- 16 -
b. Exponential Probability Distribution
E.g: Possible values of X:
X1 X2 X3 X4 X5 X6
20 21 22 23 24 26
Probability Distribution of a Random Variable
Is the listing of probabilities that correspond to each possible value of that random
variable.
Two Steps to Obtain the Probability Distribution of X:
Step 1
Obtain all the possible values that X can take on.
Step2
Obtain the probability of the occurrence of each possible value of x.
Then, present the results in a form of a Table or a Graph.
E.g: Age of a randomly selected student
Step 1: Possible X values:
Age Number of Students
20 1
21 5
22 7
23 4
24 3
26 1
Total 21

- 17 -
X1
20
X2
21
X3
22
X3
23
X5
24
X6
26
Step 2: Obtain the probability of each possible X values
Expected Value of a random variable
It’s donated as; E(X)
It is the weighted value of all the possible values of x.
E(X) for a Discrete Random Variable
E(X) = ∑P(Xi)Xi
P(Xi) = Probability of occurrence of the ith value
E.g: E(X) = (20)(1/21) + (21)(5/21) + (22)(7/21) + (23)(4/21) + (24)(3/21) + (26)(1/21)
= 0.95 + 5 + 7.33 + 4.38 + 3.43 + 1.24
≈ 22 years
x P(x)
20 1/21
21 5/21
22 7/21
23 4/21
24 3/21
26 1/21

- 18 -
E(X) for Continuous Random Variable
E(X) = ∫P(X) dx
Example
Suppose that the average growth rates of GDP of countries in the world are given below:
Average Growth Rate Number of Countries
8% 20
6% 30
4% 50
2% 1oo
Total 200
What is the Expected (average) growth rate of a randomly selected country in the world?
Solution:
E(Growth Rate) = (0.08)(20/200) + (0.06)(30/200) + (0.04)(50/200) + (0.02)(100/200)
= 0.008 + 0.009 + 0.01 + 0.01
= 0.037 = 3.7%
Binomial Probability Distribution
It is the probability distribution of a binomial random variable which takes on
values as a result of Bernoulli Process.
It is a type of Discrete Probability Distribution.
Characteristics of a Bernoulli Type Process
i. It’s made up fixed number of trials of the same activity.
ii. Each trial has two out comes only. (Success or Failure)
iii. The probability of success on each trial remains fixed over all trials.
iv. Each trial is statistically independent.
Binomial Formula:
P(r) = rnr
qp
rnr
N −
− )!(!
!

- 19 -
n: Number of trials.
r: Number of success outcomes.
p: Probability of success outcomes.
q = (1 – p) : probability of failure.
P(r) = probability of getting exactly r (successes) out of n trials.
Kind of Questions that can be asked and answered using Binomial Formula:
Q1: Obtain the probability distribution of a Binomial random Variable X
Q2: What is the probability of getting at least r* successes out on n trials; P(r ≥ r*)
Q3: What is the probability of getting at most r* successes out of n trials; P(r≤ r*)
Q4: What is the probability of getting exactly r* successes; P(r*)
General Answers:
Q1:Using the Binomial Formula, find the probability distribution of X
X P(X)
X1 P(X1)
X2 P(X2)
X3 P(X3)
Xn P(Xn)
Q2: P(r ≥ r*) = P(r*) + P(r* + 1) + ……P(n)
Q3: P(r≤ r*) = P(0) + P(1) + …….P(r*)
Q4: Use the formula
Example 1
Suppose that we toss a coin 4 times.
Q1. What is the probability of getting exactly 2 Heads?
Q2. What is the probability of getting at least 3 Heads?
Q3. What is the probability of getting at most 3 Heads?
Q4. Using the binomial formula obtain the binomial distribution of the possible
number of tails have may come up.
Solution:
Q1. n = 4 p = 1/2 (Head) q = 1/2 (Tail)
4!
P(2) =
2!(4 - 2)!
(1/2)2
(1/2)4-2
= 0.37

- 20 -
Q2. P(r≥3) = P(3) + P(4)
4!
P(3) =
3!(4 - 3)!
(1/2)3
(1/2)4-3
= 0.25
4!
P(4) =
4!(4 - 4)!
(1/2)4
(1/2)4-4
= 0.063
P(r≥3) = 0.313
Q3. P(r ≤3) = P(0) + P(1) + P(2) + P(3)
4!
P(0) =
0!(4 - 0)!
(1/2)0
(1/2)4-0
= 0.063
4!
P(1) =
1!(4 - 1)!
(1/2)1
(1/2)4-1
= 0.25
P(r ≤3) = 0.938
Q4. p: probability of getting Tail
q: probability of getting Head.
P(0) = 1/16 P(1) = 4/16 P(2) = 6/16 P(3) = 4/16 P(4) = 1/16
Example 2:
Suppose there are 2 outcomes for inflation in Turkey for each year: High or Low. If
inflation rate is greater than 100%, inflation is high. If it is less than 100%, inflation is
low for that year. Sabanci Corp. makes $50 million of positive profits for each year of
inflation and loses $20 million for each year of low inflation. Economists estimated that
the probability of high inflation is 80% for each of the next 4 years.

- 21 -
a) What is the probability of getting exactly 2 years of high inflation in the next 4
years?
n= 4 r= 2 p=0.8 q=0.2
P (2) = 4! / 2! (4-2)! *[(0.8)2
* (0.2)4-2
] = 0.1536
b) What is the prob. of getting at least 3 years of HI over the next 4 years?
P (r ≥ 3) = P (3) + P (4) = 0.4096 + 0.4096 = 0.8192
P (3) = 4! / 3! (4-3)! * [(0.8)3
* (0.2)1
] = 0.4096
P (3) = 4! / 4! (4-4)! * [(0.8)4
* (0.2)0
] = 0.4096
c) What is the probability of getting at most 3 years of LI?
p = 0.2
q = 0.8
P (r ≤ 3) = P (0) + P (1) + P (2) +P (3) = 0.0016 + 0.0256 + 0.1536 + 0.4096 =
0.5904
P (0) = 0.0016
P (1) = 0.0256
P (2) = 0.1536
P (3) = 0.4096
d) What is the expected net profit for Sabanci over the next 4 years?
X P (X)
X1 All 4 years of HI
4*50 = 200
P (200) = P (4)
X2 3 years of HI and 1 year of
LI [(3X50) – (1*20)] = 130
P (130) = P (3)
X3 2 years of HI and 2 years of
LI
(2*50) – (2*20) = 60
P (60) = P (2)
X4 1 year of HI and 3 years LI
(1*50) – (3* 20) = -10
P (-10) = P (1)
X5 0 year of HI and 4 years of
LI
(0*50) – (4*20) = -80
P (-80) = P (0)

- 22 -
Example 3
Prof. TT has put $2 on each of the 3 horses running three different races in Istanbul
Horse Racing center. He feels that each of the bets he has made has a 0.2 probability of
winning. A winning ticket on any of the three horses will earn Prof. TT $40. Assuming
a Bernoulli process, answer the following questions:
a. What is the probability of at least 2 horses ( on which Prof. TT ha s bet) winning?
b. What is the Expected Earnings from all the three horses?
c. What is the probability of at most 1 out of the three horses (on which Prof. TT has
bet) winning?
d. What is the probability distribution of his possible earnings from all the three
races?
Solution:
n= 3 p = 0.2 q = 1- 0.2 = 0.8
x = number of horses
P(0) =
( )
( ) ( ) 030
8.02.0
!03!0
!3 −
−
= 0.512
P(1) =
( )!13!1
!3
−
( ) ( ) 131
8.02.0
−
= 0.384
P(2) =
( )
( ) ( ) 232
8.02.0
!23!2
!3 −
−
= 0.096
P(3) =
( )
( ) ( ) 333
8.02.0
!33!3
!3 −
−
= 0.008
X Profit P(X)
0 (0*40) – 6 = -6 0.512
1 (1*40) – 6 = 34 0.384
2 (2*40) – 6 = 74 0.096
3 (3*40) – 6 = 114 0.008
a. P(r ≥ 2) = P(2) + P(3) = 0.096 + 0.008 = 0.104
b. E(X) = -6(0.512) + 34(0.384) + 74(0.096) + 114(0.008)
= -3.072 + 13.056 + 7.104 + 0.912 = $18
c. P(r ≤ 1) = P(0) + P(1) = 0.512 + 0.384 = 0.896

- 23 -
d. X Profit P(X)
0 -6 -6/216
1 34 34/216
2 74 74/216
3 114 114/216
Total 216
Example 4
Suppose that the inflationary process in Turkey is a Bernoulli process over the next 5
years. The probability of inflation rate to be high (meaning that it exceeds 100%) for each
year over the next 5 years is estimated to be ¾. Note that the low inflation corresponds to
the case when the annual inflation rate is less than or equal to 100%.
If the company XYZ earns approximately $10 million in a typical year of high inflation,
and loses $5 million in a typical year of low inflation, what is the expected profit of XYZ
for the next 5 years?
Solution:
N = 5 years
High Inflation
Years Profits Prob(x)
E(profits) = (-25) * (0.0001) + (-10) * (0.0147) + (5) *
(0.0879)+(20)*(0.2637)+(35)*(0.3955)+(50)*(0.2373)
= $31.250 Million
Normal Probability Distribution
This is a type of continuous probability distribution and its range extends from negative
infinity to positive infinity.
0 (0*10) – (5*5) = -25 0.0001
1 (1*10) – (4*5) = -10 0.0147
2 (2*10) – (3*5) = 5 0.0879
3 (3*10) – (2*5) = 20 0.2637
4 (4*10) – (1*5) = 35 0.3955
5 (5*10) – (0*5) = 50 0.2373

- 24 -
Characteristics
1. Bell shaped probability curve.
2. It has a single peak
3. Mean (µ) lies at the center and the distribution is symmetrical at the vertical line
erected at µ.
4. Two tails extend indefinitely.
Two Parameters: Mean (µ) & Standard Deviation (σ)
µ: Located in the center of the population and it is the average of the data.
σ: It gives information about relative spread of data around the mean.
Application of Normal Distribution
3 Mathematical Facts about Normal Distribution

- 25 -
1) Approximately 68% of the population lies in the interval ranging 1σ below the µ
to 1σ µ.
2) 95% of the population lies in an interval ranging 2σ below the µ to 2σ above the
2σ µ.
3) 99% of the population lies in an interval ranging 3σ below the µ to 3σ above the
2σ µ.
X~N (50, 5)
µ = Σx/ N = $ 50, σ = 5
Z ~ N (0,1)
Example 1:
Suppose that the stock price of Corp. X (s) is distributed normally with a µ of 100 TL
and σ of 10 TL?

- 26 -
a) What is the probability of s to be more than 120 TL in any day?
b) What is the probability of s to be less than 90 TL in any day?
a)
z = 120 – 100/ 10 = 2
P (s > 120) = P (z > 2) = 1 – P (z < 2) = 1 – 0.97725 = 0.02275
b)
z = 90- 100 / 10 = - 1
P (s < 90) = P (z <-1) = 1- P (z > 1) = 1 – 0.84134 = 0.15866

- 27 -
Example 2
Prof. TT believes that the annual profit of Turkish bank is a normal random variable
with a mean of $600,000 and a standard deviation of $100,000. Prof. TT is currently
analyzing those banks whose annual profit volume lies between $500,000 and
$650,000.
a. If total number of bank is 270, what is the approximate number of bank that
Prof.TT will analyze?
b. If Prof. TT randomly selects a bank to analyze, what is the probability that this
bank will have an annual profit volume of more than $400,000?
Solution:
µ = $600,000
σ = $100,000
X Y
500,000 600,000 650,000 $
a. P(500,000 ≤ $ ≤ 650,000) = (z1 ≤ z ≤ z2)
z1 = 500,000 – 600,000 / 100,000 = -1
z2 = 650,000 – 600,000 / 100,000 = 0.5
Area X = P(z ≤ +1) – 0.5 = 0.84134 – 0.5 = 0.34134
Area Y = P(z ≤ 0.5) – 0.5 = 0.69146 – 0.5 = 0.19146
Area X + Area Y = 0.34134 + 0.19146 = 0.5328
The Approximate number of Banks = 270*0.5328 = 143.856 ≈ 144
b. P($ > 400,000)

- 28 -
400,000 600,000 $
Z* = 400,000 – 600,000 / 100,000 = -200,000 / 100,000 = -2
P(z ≤ +2) = 0.97725
Example 3
A financial analyst computed the return on stockholder’s equity for all the companies
listed on the New York Stock Exchange. She found that the mean of this distribution
was 10%, with a standard deviation of 5%. She is interested in examining further
those companies whose return on stockholders’ equity is between 16% and 22%. Of
the approximately 1,300 companies listed on the exchange, how many are of interest
to her?
Solution:
µ = 10%
σ = 5%
n = 1,300 (Total Population)
10% 16% 22% r
Z1 = 16% - 10% / 5% = 1.2
Z2 = 22% - 10% / 5% = 2.4
P(1.2 ≤ Z ≤ 2.4)
For: Z2 = P(Z ≤ 2.4) = 0.99180
Z1 = P( Z ≤ 1.2) = 0.88493
Number of Companies of interest to her = (0.99180 – 0.88493) * 1,300
= 0.10687 * 1300 = 139

- 29 -
CHAPTER 3: FORECASTING
Definition of Forecasting
It is the prediction of the future behavior of a variable. E.g. stock prices, inflation rates,
interest rates, exchange rates, profits, gold prices, land prices, etc.
The most critical variable that companies are interested to forecast is Demand
(sales).Usually the companies are interested to forecast monthly sales (next month), or
quarterly sales (next quarter).
Importance of Sales Forecasting
There are many decisions that critically depend on sales forecasts and these include:
1) Input ordering decisions: Depends on how much working capital and other inputs
you will need.
2) Production decisions: Depends on how much you will produce in a certain time
period, that is, it determines the output.
3) The pricing decisions: This depends on whether the market is monopolistic or
oligopolistic.
4) Advertising decisions: Depends on how much you will spend on advertising.
5) Inventory decisions: This is how much you should maintain or change depending
on demand (sales) forecast.
The Methodology of Forecasting
1) Our Methodology (Technical analysis)
This methodology uses past sales data only to predict (forecast) future sales.
The limitation of this method is that it only depends on past data to predict the future
sales.
Formula: Ft+1 = f (Xt, X t-1,… X t-n )

- 30 -
2) The Econometric Methodology (Fundamental Analysis)
This methodology is made up of statistical tools, like Economic theory and
Regression Analysis.
It also uses past data about all the variables below to forecast future demand.
Formula: QD
x = f (Px, Ps, Pc, I, Expectations, Population)
1) BASIC STEPS IN FORECASTING( Technical Analysis)
Step 1: Determine the type of the product based on its sales pattern.
Step 2: Determine the forecasting models that can be used for this type of product.
Step 3: Apply each model on the actual past data and obtain MSE of the model. Best
model is the one which yields smallest MSE (Mean Square Error)
Step 4: Use the BEST MODEL to obtain a forecast for the future.
MSE test – is the test of forecasting accuracy
MSE = Σet2
/ n

- 31 -
TYPES OF PRODUCTS BASED ON THEIR SALES PATTERN
How To Determine The Type Of Product:
First collect past actual sales data (monthly/ quarterly) and plot it against time and see
the pattern.
The following two are the possibilities:
NON- SEASONAL SEASONAL
a) Constant level a) Constant Level
b) Linear Trend b) Linear Trend
c) Exponential Smoothing c) Exponential Trend
d) Damped Trend d) Damped Trend
Forecasting Models
FOR NON-SEASONAL PRODUCTS WITH CONSTANT LEVEL TYPE OF
PATTERN
1) Naïve Model
2) Moving Averages (M.A)
3) Simple Exponential Smoothing
FOR NON-SEASONAL PRODUCTS WITH LINEAR TREND TYPE OF PATTERN
1) Naïve Model
2) Time Series Regression
3) Smoothing Linear Trend (A Linear Exponential Smoothing)
a) Constant- level (Non-seasonal):
It has a fixed amount of sales.

- 32 -
b) Constant-level (Seasonal):
It has fixed amount of sales but with regular ups and downs.

- 33 -
c) Linear Trend (Non-Seasonal):
General increase of sales over time.
d) Linear Trend (Seasonal):
Sales increase at a constant rate.

- 34 -
e) Exponential Trend (Non-Seasonal):
Sales increase over time at an increasing rate.
f) Exponential Trend (Seasonal):
Sales increase over time with regular ups and downs.

- 35 -
g) Damped Trend (Non-Seasonal):
Sales increase over time at a decreasing rate.
h) Damped Trend (Seasonal):

- 36 -
Has regular ups and downs.
ş
Ft+1 = Xt et =X t - Ft
F April = X March F2
nd
Quarter 2002 = X1st Quarter 2002
Steps in Computing The Mean Squared Error (MSE)
Steps 1: Obtain as many data as possible
Steps 2: If you have enough data, meaning that if you have at least 12 units of data in
case of monthly data or at least 8 units of data in case of quarterly data divide the data
set into 2 halves, the 1st
half is called the Warm-Up Sample and 2nd
half is called the
Forecasting Sample.
Steps 3: Apply the model from the very beginning and obtain forecasts and errors for
each time period.
a) If you have enough data, apply each model on all the data and obtain
Forecasting Errors, but use only the forecasting errors coming from the
Forecasting Sample ( the 2nd
half of the data) to obtain the MSE.
b) If you do not have enough, use all the available or given data as
forecasting sample and use all the forecasting errors coming from this
sample to obtain MSE.
Steps 5: Best model is the one which yields smallest MSE. (Use the BEST MODEL
to obtain a forecast for the future.)

- 37 -
Application of Naïve Model
Formula: Ft + 1 = Xt
Ft = Forecast for next period
Xt = Actual sales in the current period.
Et = Xt – Ft
Et = Forecasting error for period t
Xt = Actual sales in period t
Ft = Forecasts in period t
E.g. If FJan, 2005 = 5,500
XJan, 2005 = 4,000
EJan, 2005 = 4,000 – 5,000 = -1,500
Example 1
Suppose that we are given the following past sales data of Televisions in North
Cyprus, obtain the forecast of sales for the month of January, 2003 using the Naïve
Model and find the MSE of this model:.
WARM-UP SAMPLE
Month T Xt Ft et
2002 Jan. 1 28
Feb. 2 27 28 -1
March 3 33 27 +6
April 4 25 33 -8
May 5 34 25 +9
June 6 33 34 -1
July 7 35 33 +2
August 8 30 35 -5
Sept. 9 33 30 +3
Oct. 10 35 33 +2
Nov. 11 27 35 -8
Dec. 12 29 27 +2
2003 Jan. 13 29
FORECASTING SAMPLE

- 38 -
6
)2()8()2()3()5()2(
6
222222
12
7
2
+−+++−+
==
∑=t
te
MSE
Solution:
Naïve Model’s Formula = Ft + 1 = Xt
NB: There is no forecast and no error for the first month (F1 and e1), but we obtain the
forecasts and the errors for the next and the following months (periods) according to the
Naïve Model. Thus the forecast for the next period (month) is the forecast for the
previous period.
F2 = 28, e2 = X2 – F2 = 27 – 28 = -1
F3 = 27, e3 = X3 – F3 = 33 – 27 = +6
F4 = 33, e4 = X4 – F4 = 25 – 33 = -8
Follow the same procedure up to F12.
a) Forecasts for January, 2003 (F13) = 29
b) MSE: Since we have enough data, we are going to take only the forecasting errors
from the forecasting sample. Therefore:
= 18.3
Example 2
Suppose that you are the manager of HSBC Bank. You believe that the pattern of
quarterly amount of savings deposits is non-seasonal and exhibits approximately a
constant-level type of behavior. You need a forecast for the savings deposits at your bank
for the 1st
quarter of 2005 in order to make better decision about the portfolio investment
and interest rate policies of your bank.
Using the Naïve Model, obtain a forecast for the amount of savings deposits for the 1st
quarter of 2005 and obtain the MSE of the model, given the following past quarterly data
about the amount of savings deposits ( in million of dollars) of your bank:
Year T Quarter Deposits Ft et
2003 1 2nd
120
2003 2 3rd
100 120 -20
2003 3 4th
110 100 +10
2004 4 1st
130 110 +20
2004 5 2nd
90 130 -40
2004 6 3rd
110 90 +20
2004 7 4th
100 110 -10
2005 8 1st
100

- 39 -
Solution:
F2 = 120, e2 = 100 – 120 = -20
F3 = 100, e3 = 110 – 100 = +10
Follow the same procedure up to F7.
a) Forecasts for 1st
quarter, 2005 (F8) = 100
b) MSE: Since we do not have enough data, use all the forecasting errors to obtain the
MSE.
MSE =
6
)10()20()40()20()10()20( 222222
−++−+++−
= 500
Application of Moving Averages Model
Unweighted M.A Model Weighted M.A Model
2-period (Un.W) M.A 2-period (W) M.A
3-period (Un.W) M.A 3-period (W) M.A
2- Period (Unw) M.A Model:
Ft+1 = Xt+ Xt-1/ 2
3- Period (Unw) M.A Model:
Ft+1 = Xt + Xt-1 + Xt-2 / 3
2- Period (W) M.A Model:
Ft+1 = wt Xt + wt-1 Xt-1
3- Period (W) Ft+1 = wt Xt + wt-1 Xt-1 + wt-2 Xt-2
wt + wt-1 + wt-2 = 1 wt >wt-1 > wt-2

- 40 -
6
)7.2()7.5()3.2()3.0()4()3.4(
6
222222
12
7
2
−+−+++−+
==
∑=t
te
MSE
Example 1
From the data given below, find the forecast for F13 using 3-period (Unw) MA Model
and obtain the MSE of this model.
WARM-UPSAMPLE
t Xt Ft et
1 28
2 27
3 33
4 25 29.3 -4.3
5 34 28.3 +5.7
6 33 30.7 +2.3
7 35 30.7 +4.3
8 30 34 -4
9 33 32.7 +0.3
10 35 32.7 +2.3
11 27 32.7 -5.7
12 29 31.7 -2.7
13 30.3
Solution:
3-Period (Unw) MA Model’s Formula:
Ft + 1 = Xt + Xt-1 + Xt – 2 / 3
F4 = 33 + 27 + 28 / 3 = 29.3, e4 = 25 – 29.3 = -4.3
F5 = 25 + 33 + 27 / 3 = 28.3, e5 = 34 – 28.3 = +5.7
F13 = 29 +27 +35 / 3 = 30.3
= 13.3
FORECASTING SAMPLE

- 41 -
Example 2
Given the Dow-Jones Index of stock market prices as below, use a 4-period (W) MA
model to find the MSE of the model using the following weights:0.5, 0.3, 0.1 and 0.1
respectively.
Month t Dow-Jones
Index (Xt)
Ft et
May 1 820
June 2 840
July 3 857
August 4 823
September 5 784 835 -51
October 6 793 809 -16
November 7 817 800 17
December 8 820 807 13
January 9 826 813 13
February 10 824 820 4
March 11 830 824 6
April 12 842 827 15
May 13 840 835 5
June 14 839 838 1
MSE: 6.91
7
)1()5()15()6()4()13()13(
7
2222222
14
8
2
=
++++++
=
∑=t
et
Application of Simple Exponential Model
Formula: Ft+1 = Ft + αet 0 < α <1 α = smoothing parameter
To apply this model, we need;
1) A value for F1
2) 2) The best value of α
Note = Possible values of α = 0.1,0.2, 0.3,……., 0.9

- 42 -
Important Steps
Step 1: To obtain F1, we take the mean (average) of the warm-up sample.
If we do not have enough data, we take the MEAN of all data as F1.
Step 2: To find the Best Value of α, we apply the model with each possible value of α
on the warm-up sample (or on all of the data in case we do not have enough data) and
obtain MSE for each case. Best value of α is the one which yields smallest MSE.
Step 3: Then apply the model with the best value of α on the forecasting sample to
get MSE of the model and obtain Forecast for the future.
If we did not have enough data, smallest data MSE obtained in step 2 becomes the
MSE of the model.
Example 1
Given the following sales data, find the forecast for F13 using Simple exponential
Smoothing Model and obtain the MSE of the model.
t Xt Ft et
1 28 30 -2
2 27 29.8 -2.8
3 33 29.5 3.5
4 25 29.9 -4.9
5 34 29.4 4.6
6 33 29.9 3.1
7 35 30.2 4.8
8 30 30.7 -0.7
9 33 30.6 2.4
10 35 30.8 4.2
11 27 31.2 -4.2
12 29 30.8 -1.8
13 30.6
As
Assume α = 0.1
F1 = 28 + 27 + 33 + 25 + 34 + 33 / 6 = 30
Ft + 1 = Ft + α et
F2 = F1 + α et1
= 30 + 0.1(-2) = 29.8
WARM-UPSAMPLE
Forecasting
Sample

- 43 -
F3 = F2 + α et2
= 29.8 + 0.1(-2.8) = 29.5
F13 = F12 + α et12
= 30.8 + 0.1(-1.8) = 30.6
Example 2
You are the production manager of Arcelik Corporation in Turkey which specialized in
production of TV sets among other kinds of durables. You need a forecast for next
quarter in order the plan production process and determine working capital needs. Apply
Simple Exponential Smoothing Model to obtain the forecast for the 2nd
Qtr of 2005. What
is the MSE of the model? Assume that the best value of α = 0.5
The past actual sales data about TV sales are given below:
Year Qtr Data t Ft Et
2004 1 5000 1 F1 = 5000 0
2004 2 4800 2 F2 = 5000 -200
2004 3 5200 3 F3 = 4900 300
2004 4 5100 4 F4 = 5050 50
2005 1 4900 5 F5 = 5075 -175
2005 2 6 F6 = 4987.5
Solution:
F1 = 5000 + 4800 + 5200 + 5100 + 4900 / 5 = 5000
F2 = 5000 + 0.5(0) = 5000
F3 = 5000 + 0.5(-200) = 4900
F4 = 4900 + 0.5(300) = 5050
F5 = 5050 + 0.5(50) = 5075
F6 = 5075 + 0.5(-175) = 4987.5
MSE = 32625
5
)175()50()300()200()0(
5
22222
5
1
2
=
−+−+
=
∑=t
et
3.11
6
)8.1()2.4()2.4()4.2()7.0()9.4(
6
222222
12
7
2
=
−+−+++−+
==
∑=t
te
MSE

- 44 -
Application of Time Series Regression Model
Salest = a + b (time)…………….Ft = a +b
Sales is dependent variable while Time is an independent variable.
Least-Square Formulas
∑
∑
−
−
= 22
)(tnt
xtntx
b tbxa −=
b= slope of the linear trend line
a= intercept of the linear trend line
X= values of the dependent variable (sales)
t = values of the independent variable (time)
Note: You can start the values of t from any number as long as they are consecutive.
X = mean of the X values
t = mean of the t values
Important Points
1) If we have enough data, we use the warm-up sample data for X and t in least-
squares formulas to get a and b. If we do not have enough data, we use all the data
to get a and b.
2) First we obtain b and then a.
3) After we get Ft = a + bt, we apply this equation on the Forecasting Sample to get
MSE of the model. Obtain forecasts for the future periods.
Example
Given the following sales data, find forecasts for January and February, 2002 (F13 and
F14) using Time-Series Regression Model and obtain the MSE of this model.

- 45 -
WARM-UP SAMPLE
Months T Xt Ft et
2001 Jan. 1 60
Feb. 2 55
March 3 64
April 4 51
May 5 69
June 6 66
July 7 83 66.8 16.2
August 8 90 68.5 21.5
Sept. 9 76 70.2 5.8
Oct. 10 95 71.9 23.1
Nov. 11 72 73.6 -1.6
Dec. 12 88 75.3 12.7
2002 Jan. 13 77
Formula for Time Series Regression Model:
Ft = a + bt
∑
∑
−
−
= 22
ntt
xtntx
b
Σtx = (1*60) + (2*55) +( 3*64) + (4*51) +( 5*69) + (6*66) = 1307
Σt2
= 12+
22+
32+
42+
52
+62
= 91
t-
= Σt/ n = 1+2+3+4+5+6 / 6 = 3.5
X-
= 60+55+64+51+69+66 / 6 = 60.8
b = 1.7 a = X-
- bt-
= 60.8 – (1.7)(3.5) = 54.9
Therefore:
F1 = 54.9 + 1.7(1) = 56.6
F7 = 54.9 + 1.7 (7) = 66.8
F13 = 54.9 + 1.7(13) = 77
F14 = 54.9 + 1.7(14) = 78.7
FORECASTING SAMPLE

- 46 -
MSE
e t
t
= =
+ + + + − +=
∑ 2
7
12
2 2 2 2 2 2
6
162 215 58 231 16 12 7
6
( . ) ( . ) ( . ) ( . ) ( . ) ( . )
= 242.6
Application of Smoothing Linear Trend Model (Linear
Exponential Smoothing)
1) Ft+1 = St + Tt 2) St=Ft + α1et 3) Tt = Tt-1 + α2 et
Note = “0<α1<1”, “0< α2<0.1”
α1, α2 = smoothing parameters
α1 = 0.1, 0.2, 0.3,………, 0.9
α2 = 0.01, 0.02, 0.03, 0.09
St = smoothed level at the end of period
Tt = smoothed Trend at the end of period t+1
Important Steps
Step 1: To get F1, we use least-squares formulas on warm-up sample. If we have enough
data and get a and b which have used as estimates for S0 and T0 to get
F1 = S0 + T0 = a+b
If we do not have enough data, we obtain a and b using all of the data.
Step 3: Then we apply model with best (α1, α2) values on the forecasting sample to get
MSE of the model and obtain forecast for the next period. Note that if we did not have
enough data smallest MSE obtained in Step 2 becomes the MSE of the model.
Example
Given the following sales data, find F13 and obtain MSE of this model.

- 47 -
Assume α1 = 0.1 and α2 = 0.01
F1 = S0 + T0 = a +b = 54.9 + 1.7 = 56.6
F2 = S1 + T1 = 56.9 + 1.7 = 58.6
S1 = F1 + α1 (e1) = 56.6 + 0.1 (3.4) = 56.94=56.9
T1 = T0 + α2 (e1) = 1.7 + 0.01(3.4) = 1.734=1.7
F13 = S12 + T12
S12 = F12 + α1 (e12) = 84 + 0.1(4) = 84.4
T12 = T11 + α2 (e12) = 2.4 + 0.01(4) = 2.4
F13 = 84.4 + 2.4 = 86.8
MSE = =∑=
12
7
2
t
et
6
)4()7.8()5.18()8.1()9.19()5.16( 222222
+++++
= 184.2
Months T Xt Ft et
2001 Jan. 1 60 56.6 3.4
Feb. 2 55 58.6 -3.6
March 3 64 59.9 4.1
April 4 51 62 -11
May 5 69 62.5 6.5
June 6 66 64.9 -1.1
July 7 83 66.5 16.5
August 8 90 70.1 19.9
Sept. 9 76 74.2 1.8
Oct. 10 95 76.5 18.5
Nov. 11 72 80.7 8.7
Dec. 12 88 84 4
2002 Jan. 13 86.8

- 48 -
CHAPTER 4: DECISION MAKING UNDER UNCERTAINITY
AND RISK
BASIC STEPS IN DECISION MAKING
Step 1: List the feasible Decision Alternatives
Step 2: List the states of nature.
States of nature are the most important (critical factors) future events that are not under
the control of decision-maker but which arte likely to affect the payoffs (benefits) that
can be obtained from the decision alternatives.
Step 3: Calculate the payoffs of the matrix.
Pay-off table shows the amount of pay-offs that can be obtained for each possible
combination of state of nature and decision alternatives.
Alternative Decision Making Environments
Certainty: A manager (decision-maker) knows for sure the state of nature which
will take place.
Uncertainty: This is where the decision-maker knows only the possible states of
nature.
Risk: This is where the decision-maker knows also the probability of each state of
nature.
Alternative Criteria for Decision Making under Uncertainty
Example
Problem: We have a Tape and CD manufacturing company. It is facing increased
demand beyond its capacity, what is the best (optimal) way of accommodating this
increasing demand or what is the optimal way of increasing supply to the market?
Solution:
Step 1: List the feasible decision alternatives:
A B C
Expand the Plant Build a new plant Subcontract out extra
production to the other
companies

- 49 -
Step 2: List the states of nature to this problem:
1. High Demand e.g 2million units of sales volume each year.
2. Moderate Demand e.g 1million units of sales volume each year.
3. Low Demand e.g ½ million units of sales volume each year.
4. Failure e.g almost zero sales.
Step 3: Prepare the Pay-off Table or pay-off Matrix:
Payoff Matrix
Decision Alternatives
States of Nature Expand Build Subcontract
High Demand $ 500,000 $ 700,000 $ 300,000
Moderate Demand $ 250,000 $ 300,000 $ 150,000
Low Demand - $ 250,000 - $ 400,000 - $ 10,000
Failure - $ 450,000 - $ 800,000 - $ 100,000
N.B: This company measures pay-offs as simple profits over the next five years.
Pay-off Table and Decision Making Criteria
1) The Maxi-Max Criterion
This criterion is preferred by optimistic managers.
Step1: Identify the maximum payoff that can be obtained for each decision alternative.
Max-pay off Expand Build Subcontract
$ 500,000 $ 700,000 $ 300,000
Step2: The optimum choice is the alternative which yields the maximum payoff in this
list obtained in step 1.
The optimum choice is to build a new plant. Because; $ 700,000>$ 500,000>$ 300,000
2) The Maxi-Min Criterion
This criterion is preferred by pessimistic managers.
Step1: Identify the minimum payoff that can be obtained for each decision alternative.
Min-pay off Expand Build Subcontract
- $ 450,000 - $ 800,000 - $ 100,000

- 50 -
Step2: The optimum choice is the alternative which yields the maximum payoff in this
list obtained in step 1.
The optimum choice is to Subcontract. Because; -$ 100,000>-$ 450,000>-$ 800,000
3) The Criterion of Realism
Step1: Choose a value of α such that 0<α<1, α is an index of your optimism about future.
Relatively large value of α represents higher degree of optimism about the most
optimistic state of nature.
Step2: Calculate the measure of realism value for each decision alternative.
Measure of Realism = (α*max payoff) + ((1-α)*min payoff)
Step3: The optimal choice is the decision alternative which yields maximum measure of
realism value.
Example:
Step1: Take α = 0.7
Step2: MoR Expand = (0.7*500,000) + (0.3*(-450,000)) = 215,000
MoR Build = (0.7*700,000) + (0.3*(-800,000)) = 250,000
MoR Subcontract = (0.7*300,000) + (0.3*(-100,000)) = 180,000
Step3: The optimal choice is to Build a New Plant. Because; 250,000>215,000>180,000
4) The Mini-Max Regret Criterion
Step1: Obtain the regret table:
Regret Value = Max Pay off of the row – The payoff
Step2: Identify the max regret value for each decision alternative.
Step3: The optimal choice is the one which yields minimum in the list, which has been
obtained in step2.
Example:
Step 1: Re-write the Decisions Alternative Table to come up with the Regret Table:
In the 1st
Row:
Regret Value for $500,000 = 700,000 – 500,000 = $200,000
Regret Value for $700,000 = 700,000 – 700,000 = 0
Regret Value for $300,000 = 700,000 – 300,000 = $400,000
In the 2nd
Row:
Regret Value for $250,000 = 300,000 – 250,000 = $50,000

- 51 -
Regret Value for $300,000 = 300,000 – 300,000 = 0
Regret Value for $150,000 = 300,000 – 150,000 = $150,000
N.B: Follow the same procedure to obtain the regret values for the 3rd
and 4th
rows.
Regret Table
Decision Alternatives
States of Nature Expand Build Subcontract
High Demand $ 200,000 0 $ 400,000
Moderate Demand $ 050,000 0 $ 150,000
Low Demand $ 240,000 $ 390,000 0
Failure $ 350,000 $ 700,000 0
Step 2: Write the maximum decision value for each alternative:
Max-Regret
Table
Expand Build Subcontract
$ 350,000 $ 700,000 $ 400,000
The optimum choice is to Expand. Because; $ 700,000>$ 400,000>$ 350,000
Application of Alternative Criteria for Decision Making under
Risk
1) Expected Value Criterion
Step1: List the feasible decision alternatives. When state of nature is given as a discrete
random variable; feasible decision alternatives are exactly the same as the states of
nature.
Step2: Obtain the probability of each state of nature using the past data.
Step3: Using the cost and price data, obtain the conditional profit table.
Step4: Using the probabilities obtain the expected value of profits for each decision
alternative.
Step5: Optimal choice is the decision alternative which yields maximum expected profit.

- 52 -
2) Criterion of Rationality
Sometimes we may not trust the past data that are used to obtain probabilities for states of
nature, under those circumstances we can use this criterion which assumes that all states
of nature are equally likely.
We use this criterion when we have limited or unreliable data to calculate probabilities.
The rest of the steps are exactly the same as those of the Expected Value Criterion.
3) The Criterion of Maximum Likelihood
We can use this criterion particularly when one of the states of nature has significantly
higher probability then others.
In this case:
Step 1: We assure that this state of nature with the highest probability will actually occur.
Step 2: Given this assumption we chose the decision alternative which yields maximum
conditional profits.
Example 1
The Mid Town Food store stocks mangoes during early summer season. These are flown
in from Meritt Island, Florida, each Monday and just be sold within the week following.
In the past, the store has been experiencing the following sales of mangoes:
Quantities Buyers Bought # of weeks this occurred
20 10
25 30
40 50
60 10
Total = 100
Food Store buys mangoes for $2 and sells them for $4.
Given this, apply each of the following criteria to determine the optimal quantity of
mangoes that must be stocked per week:
a. Expected Value Criterion
b. Criterion of Maximum Likelihood
c. Criterion of Rationality

- 53 -
a) Using the Expected Value Criterion:
Step 1: Obtain the probabilities of states of nature
Probabilities of
weekly demand
Quantities Buyers
Bought
States of Nature 10/100 = 0.1 20
30/100 = 0.3 25
50/100 = 0.5 40
10/100 = 0.1 60
Step 2: Obtain the Feasible Stocking Actions (Decision Alternative)
N.B: Obtain exactly the number of quantity demanded from the supplier.
Step 3: Obtain the Conditional Profit table
Obtain the weekly profit values for each possible combination of Stocks and Demand:
Profit = TR – TC
TR = P * Qsold
TC = C * Qstocked
P = 4 and C = 2
a) Weekly profits when Stocks = 20units/week and Demand = 20
Profit = TR – TC = (20*4) – (20*2) = 40
b) Weekly profits when Stocks = 25units/week and Demand = 20
Profit = TR – TC = (20*4) – (25*2) = 30
c) Weekly profits when Stocks = 40units/week and Demand = 20
Profit = TR – TC = (20*4) – (40*2) = 0
d) Weekly profits when Stocks = 60units/week and Demand = 20
Profit = TR – TC = (20*4) – (60*2) = -40
N.B: Follow the same procedure to obtain the conditional profits for other Stockings and
Demands.
Feasible Stocking Action
Prob. Demand 20 25 40 60
0.1 20 $ 40 $ 30 0 -$ 40
States of nature 0.3 25 $ 40 $ 50 $ 20 -$ 20
0.5 40 $ 40 $ 50 $ 80 $ 40
0.1 60 $ 40 $ 50 $ 80 $ 120
20 25 40 60

- 54 -
Loss = 4 – 2 = $2 / unit
Step 4: Calculate the Expected Values:
E(п)20 = (0.1*40) + (0.3*40) + (0.5*40) + (0.1*40) = $40
E(п)25 = (0.1*30) + (0.3*50) + (0.5*50) + (0.1*50) = $48
E(п)40 = (0.1*0) + (0.3*20) + (0.5*80) + (0.1*80) = $54
E(п)60 = (0.1*-40) + (0.3*-20) + (0.5*40) + (0.1*120) = $22
Therefore: The optimum choice is to stock 40 units because it yields max. expected
profit of $54.
b) Using the Criterion of Rationality
Taking ¼ = 0.25 as our probability
Feasible Stocking Actions
Prob. Demand 20 25 40 60
0.25 20 $ 40 $ 30 0 -$ 40
States of Nature 0.25 25 $ 40 $ 50 $ 20 -$ 20
0.25 40 $ 40 $ 50 $ 80 $ 40
0.25 60 $ 40 $ 50 $ 80 $ 120
Expected value of profits:
E(п)20 = (0.25*40) + (0.25*40) + (0.25*40) + (0.25*40) = $40
E(п)25 = (0.25*30) + (0.25*50) + (0.25*50) + (0.25*50) = $45
E(п)40 = (0.25*0) + (0.25*20) + (0.25*80) + (0.25*80) = $45
E(п)60 = (0.25*-40) + (0.25*-20) + (0.25*40) + (0.25*120) = $25
Therefore: The optimal choice is to either stock 25 or 40 units because they yield the
maximum expected profit of $45.
c) Using the Maximum Likelihood Criterion
We take the highest probability of 0.5 from the conditional probability table
Feasible Stocking Actions
Prob. 20 25 40 60
0.1 20 $ 40 $ 30 0 -$ 40
0.3 25 $ 40 $ 50 $ 20 -$ 20
0.5 40 $ 40 $ 50 $ 80 $ 40
0.1 60 $ 40 $ 50 $ 80 $ 120

- 55 -
The optimum choice is to stock 40 units because it yields the maximum expected profit
of $80.
Example 2
Suppose that you are the sales manager of Migros supermarket chain in Istanbul. You are
trying to make a decision about the optimal number of bottles of Efes beer that should be
stocked for each month. The cost of each bottle to Migros is $6o and the beer is sold for a
revenue of $95 per beer. For each bottle that is unsold by the end of the week and
returned back to Efes, Migros receives $20.The sales data for the last 50 weeks is given
as follows:
Quantities Buyers Bought Number of Months this occured
20 20
21 15
22 10
23 5
a. Using the Expected Value Criterion, obtain the optimal number of Efes bottles the
sales manager of Migros should stock per month.
b. Using the maximum Likelihood Criterion, obtain the optimal number of Efes
bottles the sales manager of Migros should stock per month.
a) Using The Expected Value Criterion
Bottles of Beer Demanded Number of weeks Probability
20 20 20/50 = 0.40
21 15 15/50 = 0.30
22 10 10/50 = 0.20
23 5 5/50 = 0.10
Total 50 1.00
Conditional Profit Table:
Profit = TR – TC
TR = P * Qsold
TC = C * Qstocked
P = $95 and C = $60
Profit Received from the Unsold Bottles = $20
a. Weekly Profit when Stock = 20 bottles/week and Demand = 20
Profit = TR – TC = (20*95) – (20*60) = 700
b. Weekly Profit when Stock = 21 bottles/week and Demand = 20

- 56 -
Profit = (TR – TC) + (# of unsold bottles*Profit Received)
= [(20*95) – (21*60)] + (1*20) = 660
c. Weekly Profit when Stock = 22 bottles/week and Demand = 20
= [(20*95) – (22*60)] + (2*20) = 620
d. Weekly Profits when Stock = 23 bottles/week and Demand = 20
= [(20*95) – (23*60)] + (3*20) = 580
Feasible Stock Action
Probabilities Demand 20 21 22 23
0.40 20 700 660 620 580
States of Nature 0.30 21 700 735 695 655
0.20 22 700 735 770 730
0.10 23 700 735 770 805
E(п)20 = (0.40*700) + (0.30*700) + (0.20*700) + (0.10*700) = $700
E(п)21 = (0.40*660) + (0.30*735) + (0.20*735) + (0.10*735) = $705
E(п)22 = (0.40*620) + (0.30* 695) + (0.20*770) + (0.10*770) = $687.50
E(п)23 = (0.40*580) + (0.30*655) + (0.20*730) + (0.10*805) = $655
Therefore: The optimum choice is to stock 21 bottles because it yields max. expected
profit of $705.
b) Using The Maximum Likelihood Criterion
We take the highest probability of 0.40 from the conditional probability table
Feasible Stock Action
Probabilities Demand 20 21 22 23
0.40 20 700 660 620 580
States of Nature 0.30 21 700 735 695 655
0.20 22 700 735 770 730
0.10 23 700 735 770 805
Therefore: The optimum choice is to stock 20 bottles because it yields max. expected
profit of $700.

- 57 -
Supplying the Numbers: How to Obtain Mean and
Standard Deviation when Data are Missing or Incomplete
This is a technique used to obtain estimates for mean and standard deviation (of
particularly demand for a product) when we don’t have past data to use in formula for
mean and standard deviation.
X~N (µx ,σx)
This technique involves asking questions to some experts in the company and then
proceeding their answers in a scientific manner to get estimates for mean and standard
deviation.
For Example:
Q1: In your opinion, what is expected annual volume of sales? Or on average how much
of this product can be sold per year?
Answer: 1500 units/year, your estimate for µx Annual Demand
Before you ask the 2nd
question, form an interval around µx (1,500) by taking any
percentage of estimates for µx such as 20%:
20% * 1500 = 300
Then add and subtract 300 from 1500 to get an interval around 1500 given below:
Q2: In your opinion, what are the odds (chances) for the annual sales volume to lie
between 1200 and 1800?

- 58 -
Answer: Odds are 3 to 1 (3:1)
P(1200< X <1800) = (3+3)/(1+3+3+1) = 6/8 = 0.75
Step 1: Obtain the probability of X being less than or equal to the right hand value of the
interval.
P(X < 1800) = 7/8 = 0.875=
1331
331
+++
++
Step 2: Obtain the Z value that corresponds to the probability value obtained in step 1 (or
closest probability value in the normal table).
Probability as close as possible to 0.875 = 0.87493 (from the table)→ Z* = 1.15
Step 3:
X*- µx
Z* =
σx

- 59 -
1800-1500
1.15=
σx
σx= 261
Therefore estimates for µ and σ = X~N (1500, 261)
Example
Suppose that the production manager of Phillips Electronics company in Netherlands
believes that the expected annual volume of sales of the new product models that they
plan to introduce to the market in 2003 is 200,000.Furthermore he believes that the odds
(chances) are 4 to 3 that the annual sales volume will lie between 100,000 and 300,000.
What is your best estimate for the standard deviation of annual sales volume of this
product model of Phillips?
Solution:
µ = 200,000
Odds = 4:3
Range = 100,000:300,000
σ = ?
3 4 4 3
100,000 200,000 300,000 X

- 60 -
P(X ≤ 300,000) = 3+4+4 / 3+4+4+3 = 11/14 = 0.7857
Z* = 0.79
Z* = X- µ / σ
0.79 = 300,000 – 200,000 / σ
0.79 = 100,000 / σ
σ = 126582
Therefore: σ = 126582

- 61 -
Chapter 5: Cost – Volume – Profit Analysis
We use Cost-Volume Analysis also called Break-Even Analysis to determine the
profitability and riskness of an investment project. This technique involves asking
probability questions and assessing the risk faced by the company in investing in this
project, based on the answers to those questions.
It also involves computing the expected value of net profit from this project which also
helps the managers to make more informed investment decisions.
Application: We build our example we used earlier in estimating the mean and the
standard deviation of annual demand for a product that a company is considering to
produce.
Note that annual demand (X) was estimated to have a mean(µ) of 1,500 and a standard
deviation(σ) of 261.
X~N (1500, 261)
Data About The Project
If: The cost of capital investment = $100,000 (initial capital)
TFC = $50,000
P/unit = $10 (Price that the company expects to sell).
AVC/unit = $5 (Average variable cost is assumed to be approximately constant for the
range of output that the company expects to purchase).
Questions That Managers Can Ask Specific To This Project
Q1: What is the probability of at least breaking even on this project for each year?
Q2: What is the probability of earning more than $900 profit for each year?
Q3: What is the probability of losing more than $200 for each year?
Q4: What is the probability of earning minimum 2% annual rate of return for each dollar
invested?
Q5: What is the expected value of Net Profit on this project?

- 62 -
At XBE Profit = TR = TC
XBE =
VP
TFC
−
Where p – v = unit contribution to profits
For: X > XBE, Profit >0
X = XBE, Profit = 0
X < XBE, Profit < 0
Profit = TR – TC
TR = PQ
TC = TVC + TFC
TVC = VQ
Profit = PQ – (VQ+TFC)
QBE = TFC / P – V
Answer 1: P(Profit ≥ 0) =P(Profit ≥ XBE)
XBE = TFC / P – V = 5000 / 10 – 5 = 1000
P(X > XBE) = P(X > 1000)
1000 – 1500
Z* =
261
Z* = -1.91

- 63 -
P(Q > 1000) = 0.97193 The probability of reaching at least break even is 97.193 %
Answer 2: P(Earnings at least more than $900 per year)
= P(Profit > 900) = P(X > X*)
X* = 1000 + 900 / 5 = 1180
P(X > X*) = P(X > 1180)
1180 – 1500
Z* =
261
Z* = -1.23
P(X > 1180) = 0.89065

- 64 -
Answer 3: P(Losing more than $200 for each year)
P(Profit < -200) = P(X < X*)
X* = 1000 + (-200 / 5) = 960
960 – 1500
Z* =
261
Z* = -2.07
P(Profit <- 200) = 0.01923
Answer 4: P(Earning minimum 2% annual rate of return for each dollar invested)
P( Profit ≥ Profit*)
Profit* = 2% * $100,000 = $2000
X* = XBE + Profit* / P – V = 1,000 + 2000/10 – 5 =1,400 units
P(Profit ≥ $2000) = P(X ≥ X*) = P(X ≥ 1,400)
Z* =
261
15001400 −
Z* = -0.38
1,400 1,500 X
Z*=-0.38 0 Z

- 65 -
P(Profit ≥ $2000) = 0.64803
Answer 5: Expected Net Profits
E(Net Profit) = E(Annual Profit) – E(Annual Loss)
The area < 1000 is Potential Loss [ P(Profit < 0)], and the area > 1000 is Potential Gain
[ P(Profit ≥ 0), while 1000 is a Break-Even Point.
Example
TT Corporation is discussing to produce a new product. It has been estimated that the
average annual volume of sales will be 8,000. The manager of the sales department
believes that the odds (chances) are 2 to 3 that the annual volume of sales will lie
between 7,000 and 9,000. If they choose to produce the product, they expect to incur an
annual fixed cost of an amount equal to 300,000. The sales price is estimated to be $100
and the variable cost per unit of output is estimated to be $50.
a. What is the probability of losing money on this product for each year?
b. What is the Expected Annual Profit for this project?
c. What is the probability of at least breaking even for each year?
Solution:
µ = 8,000
Odds = 2:3
TFC = $30,000
P = $100
VC/unit = $50
3 2 2 3
7,000 8,000 9,000 X
P(X < 9,000) = 2+2+3/3+2+2+3 = 0.70

- 66 -
Z = 0.52
Z = X - µ / σ
0.52 = 9,000 – 8,000 / σ
σ = 1923.
a) P(Losing Money)
6,000 8,000 X
Z*=? 0 Z
XBE = TFC / P-V = 300,000 / 100-50 = 6000
Z* = 6000 – 8000 / 1923.08 = -1.04
P(X < XBE) = P(Z < -1.04) = 1 – 0.85083 (from the probability table)
= 0.14917
b) E(Annual Profit)
Since 6,000 < 8,000 (XBE < µ)
E(Annual Profit) = k * σ * (UNLI + Z)
= 50*1923.08*(o.7716 +1.04) = $174192.59
c) P(At least Breaking-Even)
P(X ≥ 6,000)

- 67 -
6,000 8,000 X
Z* =-1.04 0 Z
P(X ≤ 6,000) = P(Z > 1.04) = 0.85083
Combining Unit Monetary Values and Probability Distribution
Procedure:
a. 3-Step procedure to obtain in Expected Profit
b. 3-Step procedure to obtain in Expected Loss
c. Obtain Expected Net profit [E(Profit) – E(Loss)]
Given:
X~N (1500, 261)
TFC = 5000
Standard Deviation = 261
Mean = 1500
P = 10
V = 5
XBE = TFC / P – V = 1000
a. 3 – Step Procedure to Obtain Expected Profits
Step1: Determine the Z value for XBE in absolute terms |Z|
Step2: Determine unit normal loss integral value for this z value using UNLI table.
Step3: a) If XBE < Mean, E(Profit) = k*σ*(UNLI +z)
b) If XBE > Mean, E(Profit) = k*σ*UNLI
Where k is the profits per unit of output above XBE.

- 68 -
Note: Sometimes K and C can be different from each other, but if you are given P and V
values, then always P-V = K = C
Example
Step 1: |Z| =
261
15001000 −
= |-1.91| = 1.91
Step 2: For Z = 1.91, UNLI = 0.01077, using the UNLI table.
Step 3: Since 1000 < 1500 (XBE < µ)
E(Profit) = k*σ*(UNLI +z)
= $5 * 261 * (0.01077 + 1.91) = $2506.6
b. 3 – Step Procedure to Obtain Expected Loss
Step1: Determine the Z value for QBE in absolute terms |Z|
Step2: Determine unit normal loss integral value for this z value using UNLI table.
Step3: a) If XBE < Mean, E(Profit) = c*σ*(UNLI +z)
b) If XBE > Mean, E(Profit) = c*σ*UNLI
Where c is the loss per unit of output above XBE.
|(P - V) = k = c|
Example
Step 1: |Z| = |-1.91| = 1.91
Step 2: UNLI = 0.01077 for Z = 1.91
Step 3: Since 1000 < 1500 (XBE < µ)
E(Loss) = c*σ*UNLI
= 5 * 261 * 0.01077 = $14.05
c. Obtain Expected Net Profits
E(Net Profit) = E(Profit) – E(Loss)
= $2506.6 - $14.05 = $2492.55

- 69 -
Example
Girne Septic Tank Company operates a truck called the Honey Cart which is used to
pump out the capacitated septic tanks. The company has estimated that usage of the truck
is normally distributed with a mean of 150 hours per month and a standard deviation of
40 hours per month. In order to break-even the company must operate for 90 hours per
month.
a. Find the expected loss for Honey Cart if the company loses $30 for each hour of
operation below the break even point.
b. Find the expected Profit for Honey Cart if the company earns $28 for each hour of
operation above the break-even point.
c. Find the Expected Net Profit for the operation of the Honey Cart for this company
Solution:
µ = 150
σ = 40
C = $30
K = $28
XBE = 90
|Z| = |XBE - µ / σ| = |90 – 150 / 40| = |-1.5| = 1.5
For Z = 1.5, UNLI = 0.02931
a) Expected Loss = C * σ * UNLI
= 30 * 40 * 0.02931 = $35.17 per month
b) Expected Profit = K * σ * (UNLI + Z)
= 28 * 40 * (0.02931 + 1.5) = $1,712.83 per month
c) Expected Net profit = E(Profit) – E(Loss)
= $1,712.83 - $35.17 = $1,677.66

- 70 -
Home works
Home Work 1
Questions from the textbook:
Chapter 2: Ex 2-7, 2-8, 2-9, 2-12, 2-40, 2-44*
Home work 2
1. Suppose we toss a coin 5times, and it is an unbiased coin
a. What is the probability of at least 4heads that may come up?
b. What is the Expected Value of number of tails that may come up?
2. A Financial Analyst computed the return on stockholder’s equity for all the companies
listed on the New York Stock Exchange. She found that the mean of this distribution was
10%, with a standard deviation of 5%. She is interested in examining further those
companies whose return on stockholder’s equity is between 8% and 13%. Of the
approximately 13,000 companies listed on the exchange, how many are of interest to her?
3. Questions from the textbook:
Chapter 2: Ex 2-32, 2-36, 2-37, 2-38
Home Work 3
1. Following is the sales data gathered for the past 12 months of a company:
Years Months Sales
1999 August 28
1999 September 27
1999 October 33
1999 November 25
1999 December 34
2000 January 33
2000 February 35
2000 March 30
2000 April 33
2000 May 35
2000 June 27
2000 July 29
Apply the following to the above data set to obtain MSE and develop a forecast for
August 2000 for each case:
a. 3-period (W) MA model, taking Wt = 0.5, Wt-1 = 0.3, Wt-2 = 0.2
b. 4-period (UW) MA model

- 71 -
c. Use Simple Exponential Smoothing model, taking α = 0.5 and obtain MSE of the
model and develop a forecast for August 2000.
d. Which specification above said has the Highest Forecasting Accuracy?
Chapter 3: Ex 3-1 (Just MSE Criteria), 3-4, 3-5,
Ex 3-21(a) - Please take note that the warm-up sample should be 4
periods not 6 periods.
Ex 3-25
Home Work 4
1. Use Smoothing Linear Trend model to develop a forecast for August 2000 for
Question 1 in Homework-3, taking α1 = 0.1 and α2 = 0.01. Also obtain MSE of this
model.
Chapter 3: Ex 3-21(b)
Ex 3-22, 3-23 (The warm-up Sample in both Ex 3-22 and 3-23
should be 4periods)
Ex 3-26
Home work 5
Chapter 4: Ex 4-1, 4-2, 4-3, 4-4, 4-5, 4-15, 4-33, 4-3
2. Beth Perry, who sells strawberries in the market environment where “tomorrow’s
demand for strawberries” is a discrete random variable. Beth purchased strawberries for
$3 a case and sells them for $8 a case. The rather high mark up reflects the perishability
of the item and the great risk of stoking it; the product has no value after the first day it is
offered for sale. Beth faces the problem of how many to order today for tomorrow’s
business.
A 90-day observation of past demand gives the information shown in the table below:
Daily Demand Number of days Demanded
10 18
11 36
12 27
13 19
What quantity should Beth Perry buy for tomorrow’s sale to maximize the Expected
Profit?
N.B: Read the textbook; page 165-168.

- 72 -
Home Work 6
Questions from the textbook:
Chapter 5: Ex 5-2, 5-3, 5-5, 5-8, 5-10, 5-12, 5-14

- 73 -
The Graph areas to be shaded:
• Page 27, Example 2 Graph1:
Shade X and Y area.
• Page 28, Example 2 Graph2:
Shade from 400,000 going right, up to the end.
• Page 28, Example 3 Graph:
Shade the area between 16% and 22%
• Page 64, Graph:
Shade the Graph from 1400 going right, up to the end
• Page 66, Graph:
Shade the Graph from 6000 going left up to the end
• Page 67, Graph:
Shade From 6000 going right, up to the end

- 74 -
QUIZ QUESTIONS

- 75 -
1. Suppose we are playing a game, we draw 3 cards with replacement from an
ordinary deck. We need to pay $2 for each drawing and will earn $20 dollar if the
heart will be drawn. What is our expected value of net profit for 4 drawings?
2. Weekly sales of automobiles at S.C’s Auto Gallery is normally distributed with a
mean of 200 and a standard deviation of 50 automobiles.
a. What is the probability that sales will be less than-170 in a given week?
b. What is the probability that sales will be in excess of 260?
c. What is the probability that sales will be between 175 and 230?
3. The probability that CIMSA’s stock price will rise in a given day is estimated to
be 0.60. When CIMSA’s stock price rises, there is a 0.70 probability that the
exchange rate of TL/$ will fall in that day. When CIMSA’s stock price falls, there
is a 0.4 probability that the exchange rate of TL/$ will fall in that day. Tomorrow
what is the probability that CIMSA’s price will rise and also the exchange rate of
TL/$ will fall?
4. Suppose we have three machines for making a particular part. The first machine
produces 4% defectives, the second machine produces 6% defectives, and the
third machine produces 1% defectives. Suppose also that the first machine
supplies 30%, the second machine 20%, and the third machine 50%, of the parts.
If a part is selected at random, what is the probability that it is defective? Given
that the part is defective, what is the probability that it came from the first
machine?

- 76 -
5. It has been suggested that we implement a procedure for testing for child abuse.
Suppose that when a child has been abused a doctor can correctly identify the
abuse with probability 0.99, and that when a child has not been abused the doctor
can correctly identify the non-abuse with probability 0.9 Suppose also that the
probability that a child has been abused is 0.05. If the doctor says that a child has
been abused, what is the probability that the child ha actually been abused?
6. The probability that CIMSA’s stock price will rise in a given day is estimated to
be 0.70. When CIMSA’s stock price rises, there is a 0.75 probability that the
exchange rate of TL/$ will fall in that day. When CIMSA’s stock price falls, there
is a 0.5 probability that the exchange rate of TL/$ will fall in that day. Tomorrow
what is the probability that CIMSA’s price will rise and also the exchange rate of
TL/$ will fall?
7. Weekly sales of automabiles at S.C’s Auto Gallery are normally distributed with a
mean of 300 and a standard deviation of 40 automobiles.
a. What is the probability that sales will be less than 270 in a given week?
b. What is the probability that sales will be in excess of 370?
c. What is the probability that sales will be between 284 and 328?
8. Our professor has put $3 on each of the three horses running in three different
races at ISTANBUL HORSE RACING CENTER He feels that each of the three
bets he has made, has a 0.3 probability of winning. A winning ticket on any of the
three will earn our professor $15. Assuming a Bernoulli process, what is the
expected value of his net profit?
9. A child chosen at random in a community school system comes from a low-
income family 15 percent of the time. Children from low-income families in the
community graduate from college only 20 percent of the time. Children not from
low-income families have a 40 percent chance of graduating from college. As an
employer of people from this community, you are reviewing applicants and note

- 77 -
that the first one had a college degree. What is the probability that person comes
from a low-income family?
10. Suppose that the past annual data about Istanbul Stock Market Index suggests that
the probability of index rising in a given year is 0.60, the probability of no change
is 0.05, and the probability of index falling in a year is 0.35. Furthermore analysts
have found out that for all the years during which index have risen, 80% of the
time interest rates have fallen and 20% of the time interest rates increased. For all
the years index have not changed, 50% of the time interest rated have fallen and
50 % of the time interest rates have increased. And for all the years’ index have
fallen, 40% of the time interests rates have fallen and 60% of the time interest
rates have increased. Latest developments in Turkey show that interest rates
started rising and will continue to rise in year 2001. Given this positive upward
trend in interest rates in 2001, what are your revised estimates for the probability
of index rising, not changing, and falling in year 2001?
11. Suppose that Ford Corporation produce 3 different models for Turkish consumers,
in its manufacturing plant located in Turkey. Annual volume of sales and
production of all the 3 models combined is 700 units: However, some of the
automobiles manufactured and sold mm out to be defective and lead to consumer
complaints. Based on the past data, production manager estimated the following
distribution of defective and non-defective automobiles for each model
manufactured in year 2001:
A B C
D 10 20 70
N 140 180 280
a. What is the probability that a randomly selected consumer (who decided to buy a Ford)
will buy either model A or model C?
b. What is the probability that the automobile that he buys will be either model B or
defective?

- 78 -
c. What is the probability that the automobile that he buys will be either model A or non-
defective?
d. If the automobile purchased by a randomly selected consumer is found to be defective,
what is the probability that it is a Model A automobile?
e. If the automobile purchased by a randomly selected consumer is found to be non-
defective, what is the probability that it is Model C automobile?
12. Mr X has $150,000 available for one of three investment alternatives: stock,
treasury bills, $ denominated saving deposits. The investment environment can
assume any one of three states depending on the rate of inflation (which will
determine the rate of depreciation of TL against $). The payoff table of Mr. Y
looks like this:
Type of investment
Inflation Stock Treasury Bills $ Saving deposits
High $300,000 $100,000 $ 150,000
Moderate $100 $150,000 $ 150,000
Low $50,000 $200,000 $ 150,000
a). What should we suggest Mr. X to invest by using the criterion of Maximax?
b). What should we suggest Mr. X to invest by using the criterion of realism? Assume
ά = 0.6.
13. The Province of Quebec is planning to issue hunting licenses for moose through a
province-wide lottery. They have found that by harvesting some animals during
the hunting season, they can increase the number of animals which survive
through the winter months because of the limited food supply. ‘Their estimates of
moose population (in thousands) conditional upon the number of licenses issued
and the severity of the winter are as follows:

- 79 -
Number of licenses Issued
Severity of winter 5,000 6,000 7,000 8,000 9,000
Mild 38 36 34 33 25
Moderate 35 33 33 29 22
Severe 28 30 32 27 20
Harsh 22 26 30 25 16
During the years when records have been kept, 20% of the winters have been mild
while 30%, 40%, and 10% have been moderate, severe, and harsh, respectively. How
many licenses should be issued in order to maximize the expected moose population?
How large will the expected springtime moose population be with this decision?
14. Monthly demand for product A at company B has been as follows:
Month Demand
Jan. 59
Feb. 53
Mar. 57
Apr. 48
May 43
Jun 37
Ju1 38
a. Use Naive Model to develop forecast for August and compute MSE of
this model.
b. Use Smoothing Linear Trend Model to develop forecast for August and
obtain MSE of the model. Choose ά1= 0.2, ά2= 0.02
c. Compare above 2 models and decide which one is more accurate.

- 80 -
15. Linda, the managing editor of a magazine, needs to develop a forecasting
system for monthly sales in order to schedule press runs. Sales (in
thousands of copies) for the first 8 months of 1999 (the first year of
publication) were:
Month Demand
Jan. 50
Feb. 45
Mar. 60
Apr 52
May 69
Jun. 60
Jul. 47
Aug. 53
Linda does not believe there is a seasonal pattern.
a. Use 3-period Weighted Moving Average Model to obtain a forecast for
September of 1999. Wt = 0.5, Wt-l = 0.3, Wt-2 = 0.2
b. What is the MSE of this model?
16. Monthly demand for product A at company B has been as follows:
Month Demand
Jan. 59
Feb. 53
Mar. 57
Apr. 48
May 43
Jun. 37
Jul. 38
Use linear regression model to develop a forecast for August and compute
MSE.

- 81 -
17. Linda, the managing editor of a magazine, needs to develop a forecasting C
f10 system for monthly sales in order to schedule press runs. Sales (in
thousands of copies) for the first 8 months of 1999 (the first year of
publication) were
Month Demand
Jan. 50
Feb. 45
Mar. 60
Apr. 52
May 69
Jun. 60
Jul. 47
Aug. 53
Linda does not believe there is a seasonal pattern.
a. Use 2-period Un-weighted Moving Average Model to obtain a forecast
for September of 1999. What is the MSE of this model?
b. Use Simple Exponential Smoothing Model to forecast the sales of
September and compute the MSE of this model. Take a = 0.6.
c. Compare above-said two models, which one is more accurate?
18. Mr X has $100,000 available for one of three investment alternatives: stock,
treasury bills, $ denominated saving deposits. The investment environment
can assume any one of three states depending on the rate of inflation (which
will determine the rate of depreciation of TL against $). The payoff table of
Mr. Y looks like this:
Type of investment Inflation Stock Treasury Bills $ saving deposits.

- 82 -
Type of Investment
Inflation Stock T. Bills $ Saving Deposits
High $250,000 $50,000 $100,000
Moderate $0 $100,000 $100,000
Low -$100,000 $150,000 $100,000
We have learned that Mr. Y chose treasury bills to invest. According to the
criterion of realism, find a range of alpha values that characterize the level of
optimism implied by this decision maker.
19. The ACME Company is contemplating a new product that would sell for
$11 a unit, the per-unit variable cost for this product is $8, and the fixed
cost per year allocated to this product is $ 240,000. The sales manager for
ACME estimates that annual sales for this product would have a mean of
200,000 with a standard deviation of 85,000 unit& Using this information,
answer the following questions:
a). What is the probability that ACME would loose money on this product
next year?
b). If the cost of capital investment necessary to undertake this project is
estimated to be $3 million, what is the probability of earning at least 5
percent rate of return annually?
c). What are the expected annual profit and expected annual loss from the
production of this product? (Assume that unit loss equals to unit profit)

- 83 -
EXAM QUESTIONS

- 84 -
1. Monthly Demand for TV sets at TT’s Electronics has been as follows:
Months Demand
Jan. 138
Feb. 137
March 143
Apr. 148
May 157
June 153
July 159
Using Linear Regression obtain forecast for August and September and compute MSE of
the model. (15 pts)
2. Los Bichos, the managing editor of ASTROLOGY magazine, needs to develop a
forecast for the 2nd Quarter of 2002. (10 pts)
Sales in the past quarter have been as follows:
Years Qtr. Sales
2000 2nd
Qtr. 50
3rd
Qtr. 52
4th
Qtr. 47
2001 1st
Qtr. 55
2nd
Qtr. 53
3rd
Qtr. 50
4th
Qtr. 54
2002 1st
Qtr. 52

- 85 -
Los Bichos does not believe there is a seasonal pattern.
Using 2-period Moving Average model obtain a forecast for the 2’” Qtr. of 2002 and
compute MSE of the model. (15 pts)
3. Prof. Demir is planning to invest his 1 M $ in anyone of the three investment
alternatives including GOLD, REAL ESTATE and STOCK. He believes that the
ANNUAL RETURNS (pay-offs) from each one of these investment alternatives will
critically depend on the inflation rate. Furthermore he believes that, there are 3 possible
outcomes for inflation. His corresponding pay-off table is given below:
Investment Alternatives
Gold Real Estate Stocks
High Inf. $100000 $150000 $15000
Medium Inf. $50000 $20000 $30000
States of Nature
Low Inf. $10000 $-5000 $80000
A. Using the MINIMAX REGRET criterion obtain the optimal investment alternative.
(15 pts)
B. Using the MAXIMIN criterion obtain the optimal investment alternative. (5 pts)
4. Suppose that you are the sales manager of MIGROS supermarket chain in
Istanbul. You are trying to make a decision about the optimal number of bottles of
EFES beer that should be stocked for each month. The cost of each bottle to

- 86 -
MIGROS is 10 S. whereas the price that MIGROS charges from the customers is
15$ per bottle. For each bottle that is unsold by the end of the week and returned back to
the producer of EFES, MIGROS receives $5 only. The sales data for the last 200 months
is given below:
QUANTITES BUYERS
BOUGHT
NUMBER OF MONTHS
THIS OCCURRED
5000 30
3500. 70
3000 60
2500 40
A. Using the EXPECTED VALUE criterion obtain the optimal number of EFES bottles
that the sales manager of MIGROS should stock per month. (15 pts)
B. Using the MAXIMUM LIKELIHOOD criterion obtain the optimal number of EFES
bottles that the sales manager of MIGROS should stock per month. (5 pts)
5. Suppose that the production manager of HONDA automobiles in Japan believes that
the expected annual volume of sales of the new model that they plan to introduce to the
market in 2003 is 200000. Furthermore he believes that the odds (chances) are 4 to 3 that
the annual sales volume will lie between 100000 and 300000. What is your best estimate
for the standard deviation of annual sales volume of this model of HONDA? (10 pts)
6. The project manager of SABANCI Corporation has been analyzing the feasibility of an
investment project, which involves the production of computer chips in Turkey. They

- 87 -
estimated that the cost of capital investment to undertake this çroject is $ 9 M. The selling
price of each computer chip is believed to be $ 500. The variable cost per unit is
estimated to be approximately $ 200. Annual fixed costs are likely to be around $
300000. The expected annual volume of sales is 1 500 and standard deviation of sales is
estimated to be 250.
A. What is the probability of at least breaking-even for each year? (5 pts)
B. What is the probability of earning less than $75000 profits for each year? (5 pts)
C. What is probability of losing at most $ 90000 for each year? (5 pts)
D. What is the probability of earning more than 2 % annual rate of return for each dollar
invested in this project? (5 pts)
7. Prof. TT has put $2 on each of the 3 horses running m three different races in Istanbul
Horse Racing Center He feels that each of the bets he has made has a 02 probability of
winning. A winning ticket on any of the three horses will earn Prof TT $40 Assuming a
Bernoulli process, answer each of the following questions:
a. What is the probability of at least 2 of the horses(on Prof. TT has bet) winning?
b. What is the EXPECTED EARNINGS from all three races?
c. What is the probability of at most 1 out of three horses (on which Prof TT has bet)
winning?
d. What is the probability distribution of his possible earnings from all three races?

Lecture notes

More Related Content

Viewers also liked (13)

Similar to Lecture notes (20)

Recently uploaded (20)

Lecture notes