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EASTERN MEDITERRANEAN UNIVERSITY
MGMT 322
QUANTITATIVE ANALYSIS
LECTURE NOTES
Prof. Dr. Serhan ÇİFTÇİOĞLU
Quantitative Analysis – MGMT 322
QUANTITATIVE ANALYSIS
Lecture Notes,
Past Quiz, Exam, & Home Work
Questions
Prof. Dr. Serhan ÇİFTÇİOĞLU
Gazimağusa 2007
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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, Meltem İKİNCİ, 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
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 F. NON- 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 .......................................... - 8 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 -
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Quantitative Analysis – MGMT 322
3. CONDITIONAL PROBABILITY ................................................................................................. - 11 BAYES’ THEOREM .............................................................................................................. - 12 EXAMPLE ................................................................................................................................... - 12 EXAMPLE ................................................................................................................................... - 13 EXAMPLE ................................................................................................................................... - 13 PROBABILITY DISTRIBUTION......................................................................................... - 15 DISCRETE VS CONTINUOUS RANDOM VARIABLES.................................................................... 17
DISCRETE PROBABILITY DISTRIBUTION ........................................................................................ 17
CONTINUOUS PROBABILITY DISTRIBUTION .................................................................................. 17
PROBABILITY DISTRIBUTION OF A RANDOM VARIABLE……….................................................... 18
EXAMPLE..................................................................................................................................... 19
EXPECTED VALUE OF A RANDOM VARIABLE................................................................... 20
BINOMIAL PROBABILITY DISTRIBUTION ................................................................................ - 21 CHARACTERISTICS OF A BERNOULLI TYPE PROCESS ................................................................ - 21 BINOMIAL FORMULA: ................................................................................................................ - 21 EXAMPLE ................................................................................................................................... - 22 EXAMPLE ................................................................................................................................... - 22 NORMAL PROBABILITY DISTRIBUTION .................................................................................. - 26 CHARACTERISTICS ..................................................................................................................... - 27 TWO PARAMETERS: MEAN (μ) & STANDARD DEVIATION (σ)................................................... - 27 EXAMPLE: .................................................................................................................................. - 23 3 MATHEMATICAL FACTS ABOUT NORMAL DISTRIBUTION ................................................. - 27 EXAMPLE: .................................................................................................................................. - 28 EXAMPLE: .................................................................................................................................. - 28 EXAMPLE: .................................................................................................................................. - 28 -
CHAPTER 3: FORECASTING ............................................................................................... -32THE METHODOLOGY OF FORECASTING................................................................................. - 32 TECHNICAL ANALYSIS
FUNDAMENTAL ANALYSIS
BASIC STEPS IN FORECASTING
FORECASTING MODELS ........................................................................................................... - 34 FOR NON-SEASONAL PRODUCTS WITH LINEAR TREND TYPE OF PATTERN........................ - 34 NAÏVE MODEL ........................................................................................................................... - 35 MOVING AVERAGE MODEL……………………………………………………………………. . -41SIMPLE EXPONENTIAL MODEL .................................................................................................. - 43 IMPORTANT STEPS ..................................................................................................................... - 43 TIME SERIES REGRESSION MODEL ............................................................................................ - 44 IMPORTANT POINTS ................................................................................................................... - 46 SMOOTHING LINEAR TREND MODEL......................................................................................... - 48 -
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Quantitative Analysis – MGMT 322
IMPORTANT STEPS ..................................................................................................................... - 48 CHAPTER 4: ALTERNATIVE DECISION MAKING ENVIRONMENTS .................... - 50 -
BASIC STEPS IN DECISION MAKING
ALTERNATIVE DECISION MAKING ENVIRONMENTS
ALTERNATIVE CRITERIA FOR DECISION MAKING UNDER UNCERTAINTY .......................... - 50 THE MAXI-MAX CRITERION ...................................................................................................... - 51 THE MAXI-MIN CRITERION ....................................................................................................... - 52 THE CRITERION OF REALISM ..................................................................................................... - 52 THE MINI-MAX REGRET CRITERION ......................................................................................... - 52 ALTERNATIVE CRITERIA FOR DECISION MAKING UNDER RISK .......................................... - 54 EXPECTED VALUE CRITERION ................................................................................................... - 54 EXAMPLE ................................................................................ ERROR! BOOKMARK NOT DEFINED.
CRITERION OF RATIONALITY ..................................................................................................... - 54 THE CRITERION OF MAXIMUM LIKELIHOOD ............................................................................. - 54 -
SUPPLYING THE NUMBERS: HOW TO OBTAIN MEAN AND STANDARD DEVIATION WHEN DATA IS
MISSING OR INCOMPLETE .......................................................................................................... - 59 EXAMPLE: .................................................................................................................................. - 28 EXAMPLE: .................................................................................................................................. - 28 -
CHAPTER 5: COST – VOLUME – PROFIT ANALYSIS ................................................. - 62 GENERAL DEFINITION AND USE OF COST- VOLUME- PROFIT ANALYSIS
ITS APPLICATION
COMBINING UNIT MONETARY VALUES AND PROBABILITY DISTRIBUTION ............................. - 66 3 – STEP PROCEDURE TO OBTAIN EXPECTED PROFITS .............................................................. - 69 3 – STEP PROCEDURE TO OBTAIN EXPECTED LOSS ................................................................... - 70 -
Home Works………………………………………………………………….-72Past Quiz Questions………………………………………………………..-78Past Exam Questions............................................................................-87-
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Quantitative Analysis – MGMT 322
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.
0≤P()≤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 2006:
Events: A: If annual inflation for 2006 is exactly 10%
B: If annual inflation for 2006 is less than 8%
C: If annual inflation for 2006 is either 5% or 7%
C. Experiment
The experiment is the activity that generates events.
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)
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Quantitative Analysis – MGMT 322
E. Mutually Exclusive Events
Mutually exclusive events are those events which if only one of them can take
place 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.
The crucial question we should ask in determining whether events are mutually
exclusive is: Can two or more of these events occur at one time? If we answer yes, the
events are not mutually exclusive.
Alternative Approaches in Probability Theory
1. Classical Approach
P(A) =
# of outcomes favorable to the occurrence of event “A”
Total # of possible outcomes
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Quantitative Analysis – MGMT 322
Example 1
Experiment: Randomly selecting a student in a class of 40 students.
Events: A: Selecting a female student, given that the total number of females equal to 10.
B: Selecting a male student, given that the total number of males is 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, got “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 the champion.
P (B) = 0.40
3. Subjective Approach
The probability of the event is 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%.
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Quantitative Analysis – MGMT 322
Probability Rules
A. Marginal (Unconditional) Probability:
Marginal probability is the probability of the occurrence of a single event. 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
20
21
22
23
24
Total
Male
0
1
6
3
4
14
Female
2
3
3
1
0
9
Total
2
4
9
4
4
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
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Quantitative Analysis – MGMT 322
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)
Example 1
Suppose that the sex and the eye color of students in a class of 40 people are distributed
as follows:
Male
Female
Total
Black
13
8
21
Green
9
10
19
Total
22
18
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 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 out 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.
Defective
Non-Defective
Total
A
10
140
150
Models
B
20
180
200
C
70
280
350
Total
100
600
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 Nondefective auto?
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Quantitative Analysis – MGMT 322
Solution:
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
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 2007:
Sectors
Industry
Service
Mining
Total
240
30
420
Positive Profits 150
50
10
20
80
Losses
200
250
50
500
Total
a. What is the probability of a randomly selected company (out of these companies
listed on Istanbul Stock Exchange) to report positive profits in year 2007?
b. What is the probability of a randomly selected company to report either loss 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.
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Quantitative Analysis – MGMT 322
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”.
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 number 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 purchase, the more the petroleum they will buy to run their
automobiles.
Types of Probabilities under Independence
1. Marginal (Unconditional) Probability
P(A): In this case, we are interested in the probability of the occurrence of only
one 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)
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Quantitative Analysis – MGMT 322
Example
Events: A: The weather is rainy 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) : In this case, we are interested in the probability of the occurrence of only
one 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 given as follows;
Male
Female
Total
Black
20
10
30
Green
8
2
10
Total
28
12
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
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Quantitative Analysis – MGMT 322
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
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.
Assume that you have randomly selected a die and roll it and you found 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
P(Type 1) = 0.5
Conditional Probability of Secondary
Event, Given Each of the Elementary
Event
P(Ace / Type 1) = 0.3
P(Ace)
P(Type 2) = 0.5
P(Ace / Type 2) = 0.6
Step 2: Obtain the MARGINAL probability of the secondary event.
P(secondary event) = summation of the individual joint probabilities of the secondary
event with each one of the elementary events.
P(Ace) = P(Type1, Ace) + P( Type2, Ace)
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Quantitative Analysis – MGMT 322
Step 3: Compute the 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: Apply Baye’s Theorem by using the joint probabilities you obtained in step3 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
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Quantitative Analysis – MGMT 322
Example3
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 suggests 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 suggests 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
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Quantitative Analysis – MGMT 322
Example 4
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 have
been reviewing applicants and noted that the first one had a college degree. What is the
probability that the person comes from a low-income family?
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%
Note: To find the probability of a single event, you need to add up all of its joint
probabilities.
Example 5
A new diagnostic test to detect gout has been introduced into Lincoln County Hospital.
The manufacturer of the test device has determined that the probability of a falsepositive reading (a positive test result for a person who is known not to have gout) is 8%.
The probability of a false- negative reading (a negative reading for a person who has the
disease) is 4%. County health authorities estimate that approximately 15% of the
population in the county has gout. If a test is conducted on a patient from that county and
the test result is positive, what is the probability that gout is present?
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Quantitative Analysis – MGMT 322
Solution:
Events:
H: Person who is healthy.
I: Person who is ill.
P: A positive test
N: A negative test.
P(P/H)= 0.08
P(N/I)=0.04
P(I)=0.15
P(H)=1- P(I)=1- 0.15= 0.85
P(I/P)=?
P(H)=0.85
Step 1:
Marginal Probability of
elementary events
Secondary Event
P(H)=0.85
P(I)=0.15
P
Conditional probability of
secondary event given each
elementary event.
P(P/H)=0.08
P(P/I)=? 0.96
P(N/I)= 0.04
P(P/I)= 1- P(N/I)
P(P/I)= 1- 0.04= 0.96
Step 2: Computation of probability of secondary event.
P (P)= P(HP) + P(IP)
Step 3: Computation of joint probability of secondary event with each of the elementary
events.
P(HP)= P(P/H) x P(H)
P(HP)= 0.08 x 0.85= 0.068
P(IP)= P(P/I) x P(I)
P(IP)= 0.96 x 0.15= 0.144
Step 4: P(P)= P(HP)+ P(IP)
P(P)= 0.068 + 0.144= 0.212
Step 5: Use the Bayes’ Theorem and get the final answer.
P (I/P) = P (IP) / P (P) = 0.144/ 0.212 = 0.679= 67.9%
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Quantitative Analysis – MGMT 322
Probability Distribution
Random Variables
A Random Variable is a variable which takes on values unpredictably as a result
of a random process.
Example: 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 only limited number (less than infinity) 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 an infinite number of possible values over a certain range.
a. Normal Probability Distribution
b. Exponential Probability Distribution
E.g: Possible values of X:
X1 X2 X3 X4 X5 X6
20 21 22 23 24 26
Examples of Continuous Random Variable:
Stock prices, height, exchange rates, demand, profits, interest rates, petroleum prices etc.
E.g.: Height of a randomly selected student in the class:
Xmin=155 cm
Xmax=186 cm
There is infinite number of possibilities between these limits. Therefore, X (height) is
considered to be a continuous random variable.
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Quantitative Analysis – MGMT 322
Probability Distribution of a Random Variable
It 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
Age
Number of Students
20
1
21
5
22
7
23
4
24
3
26
1
Total
21
Step 1: Possible X values:
X1
X2
X3
X3
X5
X6
20
21
22
23
24
26
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Quantitative Analysis – MGMT 322
Step 2: Obtain the probability of each possible X values
x
P(x)
20
1/21
21
5/21
22
7/21
23
4/21
24
3/21
26
1/21
Example
Suppose that you have a portfolio made up of the shares (stocks) of 3 different
companies. Each company’s stock is assumed to follow random walk in terms of its price
behavior; the probability of increase or (decrease) in any day is given by 0.5. Obtain the
probability distribution of the possible number of companies (in your portfolio) whose
stock price may increase in any day.
Solution:
Suppose that we have 3 companies which their names are A, B, C. The probability of
increasing (or decreasing) of these companies’ stock prices is 0.5. Let “increasing” be (+)
sign and “decreasing” be (-) sign.
So;
Sample Space ={(A+,B+,C+)(A+,B+,C-)( A+,B-,C+)( A-,B+,C+)( A+,B-,C-)( A-,B+,C-)
( A-,B-,C+)(A-,B-,C-)}
X: Number of stocks of which its stock price is increasing.
X
0
1
2
3
P(X)
1/8
3/8
3/8
1/8
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Quantitative Analysis – MGMT 322
i.e: When X is zero; there is a decrease in the price of all these three companies ABC.
(A-,B-,C-)
When X is two, we see that (A+,B+,C-)( A+,B-,C+)( A-,B+,C+). In these 3 cases, we
notice that there is a price increase in 2 out of 3 of these companies.
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
E(X) for a 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
8%
6%
4%
2%
Total
Number of Countries
20
30
50
100
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%
Example
Suppose that we have 500 faculty members at EMU. 100 of these are full Professors, 150
of them are Assoc. Professors and the rest 250 of them are Assistant Professors. The
monthly salary of each group is $4000, $3000 and $2000 respectively.
a.)Obtain the probability distribution of the salary of a randomly selected faculty member
at EMU.
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Quantitative Analysis – MGMT 322
b.)What is the expected value of the salary of a randomly selected faculty member at
EMU.
Solution:
X: Salary of a randomly selected faculty member.
a)
X
P(X)
4000
P(4000)= 100/500=0.2
3000
P(3000)=150/500= ~0.30
2000
P(2000)=250/500=0.50
b) E(X)= E(Salary)
E(X)= ∑P(Xi)Xi = (0.2*4000) + (0.3*3000)+(0.5*2000)
E(X)= 800 +900 + 1000= $2,700
Binomial Probability Distribution
It is the probability distribution of a random variable called Binomial Random
Variable. Binomial R.V. takes on values as a result of Bernoulli Process. It is a type of
Discrete Probability Distribution.
Characteristics of a Bernoulli Type Process
i.
ii.
iii.
iv.
It’s made up fixed number of trials of the same activity.
Each trial has two out comes only. (Success or Failure)
The probability of success on each trial remains fixed over all trials.
Each trial is statistically independent of each other.
Binomial Formula:
P(r) =
N!
p r q nr
r!(n  r )!
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.
Questions that can be 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*)
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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*)
Q5: Obtain E(X) or any other variable based on probability distribution of X. i.e. E(Y).
General Answers:
Q1:Using the Binomial Formula, find the probability distribution of X
X
P(X)
X1
X2
X3
Xn
P(X1)
P(X2)
P(X3)
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!
(1/2)2(1/2)4-2 = 0.37
P(2) =
2!(4 - 2)!
Q2. P(r≥3) = P(3) + P(4)
4!
(1/2)3(1/2)4-3 = 0.25
P(3) =
3!(4 - 3)!
P(4) =
(1/2)4(1/2)4-4 = 0.063
4!
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Quantitative Analysis – MGMT 322
4!(4 - 4)!
P(r≥3) = 0.313
Q3. P(r ≤3) = P(0) + P(1) + P(2) + P(3)
4!
(1/2)0(1/2)4-0 = 0.063
P(0) =
0!(4 - 0)!
4!
(1/2)1(1/2)4-1 = 0.25
P(1) =
1!(4 - 1)!
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 Corporation makes $50 million of positive profits for each year
of high 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.
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 probability 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
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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) = 0.4096 + 0.4096 + 0.1536 + 0.0256 =
0.9984
P (0) = 0.4096
P (1) = 0.4096
P (2) = 0.1536
P (3) = 0.0256
d) What is the expected net profit for Sabanci over the next 4 years?
p = 0.8
q = 0.2
X1
X2
X3
X4
X5
X
All 4 years of HI
4*50 = 200
3 years of HI and 1 year of LI [(3X50) – (1*20)] = 130
2 years of HI and 2 years of LI
(2*50) – (2*20) = 60
1 year of HI and 3 years LI
(1*50) – (3* 20) = -10
0 year of HI and 4 years of LI
(0*50) – (4*20) = -80
P (X)
P (200) = P (4)
P (130) = P (3)
P (60) = P (2)
P (-10) = P (1)
P (-80) = P (0)
E(Net Profıt) = (-80 * 0.0016) + (-10 * 0.0256) + (60 * 0.1536) + (130 * 0.4096) + (200 *
0.4096) = 144  $144 Million
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 has 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?
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Quantitative Analysis – MGMT 322
Solution:
n= 3 p = 0.2 (winning) q = 1- 0.2 = 0.8 (losing)
X= Number of horses (on which TT has bet) winning
Y= Earnings (Possible profits)
3!
P(0) =
0.20 0.830 = 0.512
0!3  0!
P(1) =
3!
0.21 0.831 = 0.384
1!3  1!
P(2) =
3!
0.22 0.832
2!3  2!
P(3) =
3!
0.23 0.833 = 0.008
3!3  3!
X
0
1
2
3
Profit
(0*40) – 6 = -6
(1*40) – 6 = 34
(2*40) – 6 = 74
(3*40) – 6 = 114
= 0.096
P(X)
0.512
0.384
0.096
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
Profits (Π )= Revenues- Cost
$2*3= $6 (Cost of a ticket)
i.e.: If X=0 , Y=Π= - 6
X= 1, Y= (1*$40) – 6 = $34
X= 2, Y= (2*$40) – 6 = $74
d.
X
0
1
2
3
Total
Profit
-6
34
74
114
216
P(X)
0.512
0.384
0.096
0.008
1
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Quantitative Analysis – MGMT 322
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
0
(0*10) – (5*5) = -25
1
(1*10) – (4*5) = -10
2
(2*10) – (3*5) = 5
3
(3*10) – (2*5) = 20
4
(4*10) – (1*5) = 35
5
(5*10) – (0*5) = 50
Prob(X)
0.0010
0.0146
0.0878
0.2637
0.3955
0.2373
E(profits) = (-25) * (0.0010) + (-10) * (0.0146) + (5) * (0.0878)+(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.
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Quantitative Analysis – MGMT 322
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: Population Mean (μ) & Standard Deviation (σ)
μ: Located in the center of the population and it shows us the central tendency of
the data. 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
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)
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Quantitative Analysis – MGMT 322
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?
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?
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Quantitative Analysis – MGMT 322
Solution:
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
Example 2
Prof. TT believes that the annual profit of Turkish banks 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?
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Quantitative Analysis – MGMT 322
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
500,000
Y
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)
400,000
600,000
$
Z* = 400,000 – 600,000 / 100,000 = -200,000 / 100,000 = -2
P(z ≤ +2) = 0.97725
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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
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Quantitative Analysis – MGMT 322
CHAPTER 3: FORECASTING
Definition of Forecasting
Forecasting is predicting the future behavior of a financial or economic 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: How much working capital and other inputs (i.e. labor
or raw materials) you will need.
2) Production decisions: How much you will produce in a certain time period.
3) The pricing decisions: This depends on whether the market is monopolistic or
oligopolistic. At what price you should sell depending on the perceived demand
curve.
4) Advertising decisions: How much you will spend on advertising.
5) Inventory decisions: How much stocks you should have.
The Methodology of Forecasting
1) Our Methodology (Technical analysis)
This methodology uses only past sales data to predict (forecast) future sales.
The drawback of this method is that it only depends on past data to predict the future
sales.
Note: Here, the crucial point is to find a pattern in the past sales data and use it to
forecast the future.
Formula: Ft+1
= f (Xt, X t-1,… X t-n )
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2) The Econometric Methodology (Fundamental Analysis)
This methodology is made up of statistical tools such as Economic theory and
Regression Analysis.
It also uses past data about all the variables given below to forecast future demand.
Formula: QDx = f (Px, Ps, Pc, I, Expectations, Population)
E.g.: QDx = a + b1Px + b2Ps + b3, I : We use regression to estimate the relationships
between these variables and expected demand for our product.
1) BASIC STEPS IN FORECASTING( Technical Analysis)
Step 1: Determine the type of your product based on its sales pattern.
Step 2: List the forecasting models that can be used for this type of product.
Step 3: By using past data, calculate MSE (Mean Squared Error) of each model to
determine the BEST MODEL to use.
Best model is the one which yields smallest MSE (Mean Squared Error).
Step 4: Use the BEST MODEL to obtain a forecast for the next period(s).
MSE test – is the test to measure the forecasting accuracy
MSE =
Σet2/ n
et- It refers to the forecasting accuracy.
n- It refers to the number of error terms.
TYPES OF PRODUCTS BASED ON THEIR SALES PATTERN
How to Determine the Type of Product:
Initially, collect past sales data (monthly/ quarterly) and plot it against time and see
the pattern.
The following two are the possibilities:
NON- SEASONAL
a) Constant level
b) Linear Trend
c) Exponential Smoothing
d) Damped Trend
SEASONAL
a) Constant Level
b) Linear Trend
c) Exponential Trend
d) Damped Trend
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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. There is no change in sales.
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b) Constant-level (Seasonal):
It has fixed amount of sales but with regular ups and downs.
c) Linear Trend (Non-Seasonal):
There is a general increase of sales over time.
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Quantitative Analysis – MGMT 322
d) Linear Trend (Seasonal):
Sales increase at a constant rate. (with seasonal fluctuations).
e) Exponential Trend (Non-Seasonal):
Sales increase over time at an increasing rate.
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Quantitative Analysis – MGMT 322
f) Exponential Trend (Seasonal):
Sales increase over time with regular ups and downs.
g) Damped Trend (Non-Seasonal):
Sales increase over time at a decreasing rate.
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Quantitative Analysis – MGMT 322
h) Damped Trend (Seasonal):
It has regular ups and downs.
Ft+1 = Xt
F April = X March
et =X t - Ft
F2nd Quarter 2002 = X1st Quarter 2002
Steps in Computing the Mean Squared Error (MSE)
Steps 1: Obtain data as much as possible
Steps 2: If you have enough data, meaning that 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
forecasting errors for each time period.
a) If you have enough data, use only errors coming from the Forecasting
Sample (the 2nd half of the data) to compute the MSE.
b) If you have less than enough data, 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.)
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Quantitative Analysis – MGMT 322
Application of Naïve Model
Formula: Ft + 1 = Xt
Ft = Forecast for the next period
Xt = Actual sales in the current period.
Et = Xt – Ft
**** (Remember that et can be positive, negative or zero.)
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 January, 2006 using the Naïve Model and
find the MSE of this model:.
WARM-UP SAMPLE
Month
2005 Jan.
Feb.
March
April
May
June
July
August
Sept.
Oct.
Nov.
Dec.
2006 Jan.
T
1
2
3
4
5
6
7
8
9
10
11
12
13
Xt
28
27
33
25
34
33
35
30
33
35
27
29
Ft
et
28
27
33
25
34
33
35
30
33
35
27
29
-1
+6
-8
+9
-1
+2
-5
+3
+2
-8
+2
FORECASTING SAMPLE
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Quantitative Analysis – MGMT 322
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 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, 2006 (F13) = 29
b) MSE: Since we have enough data, we are going to take only the forecasting errors
from the forecasting sample. Therefore:
12
MSE 
e
t 7
6
2
t

(2) 2  (5) 2  (3) 2  (2) 2  (8) 2  (2) 2
6
= 18.3
Example 2
Suppose that you are the manager of HSBC Bank. You believe that the pattern of
quarterly amount of saving deposits is non-seasonal and exhibits approximately a
constant-level type of behavior. You need a forecast for the saving deposits at your bank
for the 1st quarter of 2007 in order to make better decisions 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 2007 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
2005
2005
2005
2006
2006
2006
2006
2007
T
1
2
3
4
5
6
7
8
Quarter
2nd
3rd
4th
1st
2nd
3rd
4th
1st
Deposits
120
100
110
130
90
110
100
Ft
et
120
100
110
130
90
110
100
-20
+10
+20
-40
+20
-10
- 40 -
Quantitative Analysis – MGMT 322
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.
(20) 2  (10) 2  (20) 2  (40) 2  (20) 2  (10) 2
6
= 500
MSE =
Application of Moving Averages Model
Unweighted M.A Model
2-period (Un.W) M.A
3-period (Un.W) M.A
Weighted M.A Model
2-period (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
- 41 -
Quantitative Analysis – MGMT 322
Example 1
From the data given below, find the forecast for F13 using 3-period (Un.W) MA Model
and obtain the MSE of this model.
WARM-UPSAMPLE
t
1
2
3
4
5
6
7
8
9
10
11
12
13
Xt
28
27
33
25
34
33
35
30
33
35
27
29
Ft
et
29.3
28.3
30.7
30.7
34
32.7
32.7
32.7
31.7
30.3
-4.3
+5.7
+2.3
+4.3
-4
+0.3
+2.3
-5.7
-2.7
FORECASTING SAMPLE
Solution:
3-Period (Unw) MA Model’s Formula:
Ft + 1 = Xt + Xt-1 + Xt – 2 / 3
F4 = 33 + 27 + 28 / 3 = 29.3,
F5 = 25 + 33 + 27 / 3 = 28.3,
e4 = 25 – 29.3 = -4.3
e5 = 34 – 28.3 = +5.7
F13 = 29 +27 +35 / 3 = 30.3
12
MSE 
e
t 7
6
2
t
(4.3) 2  (4) 2  (0.3) 2  (2.3) 2  (5.7) 2  (2.7) 2

6
= 13.3
- 42 -
Quantitative Analysis – MGMT 322
Example 2
Dow-Jones Stock Market Price Index is given below. Use 4-period (W) MA model to
find MSE of the model using the following weights: 0.5, 0.3, 0.1 and 0.1 respectively.
Month
t
May
June
July
August
September
October
November
December
January
February
March
April
May
June
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Dow-Jones
Index (Xt)
820
840
857
823
784
793
817
820
826
824
830
842
840
839
Ft
et
835
809
800
807
813
820
824
827
835
838
-51
-16
17
13
13
4
6
15
5
1
14
 et
MSE:
t 8
7
2

(13) 2  (13) 2  (4) 2  (6) 2  (15) 2  (5) 2  (1) 2
 91.6
7
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) The best value of α
Note = Possible values of α = 0.1, 0.2, 0.3,……., 0.9
Important Steps
Step 1: To obtain F1, we take the MEAN (average) of the warm-up sample.
If we do not have enough data, F1 is taken as the MEAN of ALL of the data.
Step 2: To find the Best Value of α, we need to experiment with each possible value
of α on the warm-up sample (or on all of the data if we do not have enough data) and
obtain MSE for each case. Best α value is the one which yields smallest MSE.
- 43 -
Quantitative Analysis – MGMT 322
Step 3: If we have enough data, we apply the model with the best value of α on the
forecasting sample to find MSE of the model and obtain Forecast for the future.
If we do 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.
WARM-UPSAMPLE
t
1
2
3
4
5
6
7
8
9
10
11
12
13
Xt
28
27
33
25
34
33
35
30
33
35
27
29
Ft
30
29.8
29.5
29.9
29.4
29.9
30.2
30.7
30.6
30.8
31.2
30.8
30.6
Forecasting
As
Sample
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
F3 = F2 + α et2
= 29.8 + 0.1(-2.8) = 29.5
F13 = F12 + α et12
= 30.8 + 0.1(-1.8) = 30.6
- 44 -
et
-2
-2.8
3.5
-4.9
4.6
3.1
4.8
-0.7
2.4
4.2
-4.2
-1.8
Quantitative Analysis – MGMT 322
12
MSE 
e
t 7
2
t
6

(4.9) 2  (0.7) 2  (2.4) 2  (4.2) 2  (4.2) 2  (1.8) 2
 11.3
6
Example 2
You are the production manager of Arcelik Corporation in Turkey which is specialized in
production of TV sets among other kinds of durable goods. You need a forecast for the
next quarter in order to plan production process and determine working capital needs.
Apply Simple Exponential Smoothing Model to obtain the forecast for the 2nd Qtr of
2007. 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
2006
2006
2006
2006
2007
2007
Qtr
1
2
3
4
1
2
Data
5000
4800
5200
5100
4900
t
1
2
3
4
5
6
Ft
F1 = 5000
F2 = 5000
F3 = 4900
F4 = 5050
F5 = 5075
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
5
 et
MSE =
t 1
5
2

(0) 2  (200) 2  (300) 2 (50) 2 (175) 2
 32625
5
- 45 -
Et
0
-200
300
50
-175
Quantitative Analysis – MGMT 322
Application of Time Series Regression Model
Salest = a + b (time)…………….Ft = a +bt
“Sales” is a dependent variable while “Time” is an independent variable.
Least-Square Formulas
b
 tx  nt x
 t  n(t )
2
2
a  x  bt
b= slope of the linear trend line
a= intercept of the linear trend line
X= values of dependent variable (sales)
t = values of the independent variable (time)
Least- square formulas utilize past data about dependent and independent variable.
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 only the warm-up sample data for X and t to get b
and a. If we do not have enough data, we use all of the data.
2) First we obtain b and then a.
3) After we get b and a, we apply Ft = a + bt, on FORECASTING SAMPLE to get
MSE of the model and obtain forecasts for the future periods.
- 46 -
Quantitative Analysis – MGMT 322
Example
Given the following sales data, find forecasts for January and February, 2007 (F13 and
F14) using Time-Series Regression Model and obtain the MSE of this model.
WARM-UP SAMPLE
Months
2006 Jan.
Feb.
March
April
May
June
July
August
Sept.
Oct.
Nov.
Dec.
2007 Jan.
T
1
2
3
4
5
6
7
8
9
10
11
12
13
Xt
60
55
64
51
69
66
83
90
76
95
72
88
Ft
et
66.8
68.5
70.2
71.9
73.6
75.3
77
16.2
21.5
5.8
23.1
-1.6
12.7
FORECASTING SAMPLE
Formula for Time Series Regression Model:
Ft = a + bt
b
 tx  nt x
 t  nt
2
2
Σtx = (1*60) + (2*55) +( 3*64) + (4*51) +( 5*69) + (6*66) = 1307
Σt = 1 2 3 4 5 +6 = 91
2
2+ 2+ 2+ 2+ 2
2
Σ
t- = t/ 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
- 47 -
Quantitative Analysis – MGMT 322
12
MSE 
e
2
t
t7
6
(16.2) 2  (215
. ) 2  (58
. ) 2  (231
. ) 2  (  16
. ) 2  (12.7) 2
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: Obtain the value of F1.
If we have enough data, we apply the least-square formulas on the warm-up sample to
obtain the values for a and b which are used as estimates for S0 and T0.
Therefore;
F1 = S0 + T0 = a+b
If we do not have enough data, we obtain a and b by using all of the data.
Step 2: To find the best α1, α2 values, 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.
Step 3: Then we apply the model with the 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 do not
have enough data, the smallest MSE obtained in Step 2 becomes the MSE of the model.
- 48 -
Quantitative Analysis – MGMT 322
Example
Given the following sales data, find the value of F13 and obtain MSE of this model.
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
Months
2001 Jan.
Feb.
March
April
May
June
July
August
Sept.
Oct.
Nov.
Dec.
2002 Jan.
12
T
1
2
3
4
5
6
7
8
9
10
11
12
13
Xt
60
55
64
51
69
66
83
90
76
95
72
88
Ft
56.6
58.6
59.9
62
62.5
64.9
66.5
70.1
74.2
76.5
80.7
84
86.8
(16.5) 2  (19.9) 2  (1.8) 2  (18.5) 2  (8.7) 2  (4) 2
6
= 184.2
MSE =  et 2 
t 7
- 49 -
et
3.4
-3.6
4.1
-11
6.5
-1.1
16.5
19.9
1.8
18.5
8.7
4
Quantitative Analysis – MGMT 322
CHAPTER 4: DECISION MAKING UNDER UNCERTAINITY
AND RISK
BASIC STEPS IN DECISION MAKING
Step 1: Specify and define your problem; and then list the feasible decision alternatives
regarding the solution of that problem.
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 will affect the payoffs (benefits) that can be
obtained from the decision alternatives.
Step 3: Compute the PAY-OFF MATRIX (Table).
Pay-off table shows the amount of benefits that can be obtained from each possible
combination of state of nature and decision alternative. i.e.: Pay-offs: profits, revenues
etc.
Step 4: Apply a criterion (of your choice) that is designed for making decisions in the
type of environment you have to make a decision.
Alternative Decision Making Environments
 Certainty: Manager (decision-maker) knows which state of nature which will
take place. (for sure)
 Uncertainty: Manager knows only the possible states of nature.
 Risk: Manager knows also the probability of occurrence 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 a rising demand
beyond its capacity, what is the best (optimal) way of accommodating this increasing
demand or what is the optimal way of increasing our supply to the market?
Solution:
Step 1: List the feasible decision alternatives:
A
Expand the plant
B
Build a new plant
- 50 -
C
Subcontract out extra
production to the other
companies
Quantitative Analysis – MGMT 322
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
States of Nature
High Demand
Moderate Demand
Low Demand
Failure
Expand
$ 500,000
$ 250,000
- $ 250,000
- $ 450,000
Decision Alternatives
Build
$ 700,000
$ 300,000
- $ 400,000
- $ 800,000
Subcontract
$ 300,000
$ 150,000
- $ 10,000
- $ 100,000
N.B: This company measures pay-offs as simple profits over the next five years.
Step 4: Apply an appropriate criterion to make your decision.
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 with each decision alternative.
Max-pay off
Expand
$ 500,000
Build
$ 700,000
Subcontract
$ 300,000
Step2: The optimal alternative is the one which yields the MAXIMUM payoff in the list
obtained in step 1.
The optimum choice is to build a new plant. Because; $ 700,000>$ 500,000>$ 300,000
- 51 -
Quantitative Analysis – MGMT 322
2) The Maxi-Min Criterion
This criterion is preferred by pessimistic managers.
Step1: Identify the minimum payoff that can be obtained with each decision alternative.
Min-pay off
Expand
- $ 450,000
Build
- $ 800,000
Subcontract
- $ 100,000
Step2: The optimal alternative is the one 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 representing your degree of
optimism about the future states of nature. Relatively higher values of α represent higher
degrees of optimism.
Step2: Based on the following, compute MEASURE OF REALISM value for each
decision alternative.
Measure of Realism = (α*max payoff) + ((1-α)*min payoff)
Step3: Optimal alternative is the one 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 of each pay-off value = Max Pay off value in the same row – That pay-off
Step2: Identify the MAX REGRET value for each decision alternative.
Step3: The optimal choice is the one which yields MINIMUM in the list obtained in
step2.
- 52 -
Quantitative Analysis – MGMT 322
Example
Step 1: Rewrite 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
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
States of Nature
High Demand
Moderate Demand
Low Demand
Failure
Expand
$ 200,000
$ 050,000
$ 240,000
$ 350,000
Decision Alternatives
Build
0
0
$ 390,000
$ 700,000
Subcontract
$ 400,000
$ 150,000
0
0
Step 2: Identify the maximum regret value for each decision alternative:
Max-Regret
Table
Expand
Build
Subcontract
$ 350,000
$ 700,000
$ 400,000
Step 3: Choosing to Expand is the optimal alternative. Because; $ 700,000>$ 400,000>$
350,000
- 53 -
Quantitative Analysis – MGMT 322
Application of Alternative Criteria for Decision Making under
Risk
1) Expected Value Criterion
Step1: List the feasible decision alternatives. When state of nature (demand) 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 based on the past sales data.
Step3: Using the cost and price data, obtain the CONDITIONAL PROFIT TABLE. (payoff table)
Step4: Using the probabilities, calculate expected value of profits for each decision
alternative.
Step5: Optimal alternative is the one which yields MAXIMUM expected profits.
2) Criterion of Rationality
Sometimes we may not trust the past data that are used to obtain probabilities for states of
nature. Under such 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 estimate the prob. of
each state of nature.
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 a significantly
higher probability than others.
In this case:
Step 1: Assume that the state of nature with the MAXIMUM likelihood (highest prob.)
will take place.
Step 2: Given this assumption, OPTIMAL ALTERNATIVE is the one which yields
maximum conditional profits.
- 54 -
Quantitative Analysis – MGMT 322
Example 1
Mid Town Food store stocks mangoes during early summer season. These are flown in
from Meritt Island, Florida, each Monday and must be sold within the week following. In
the past, the store experienced the following sales of mangoes:
Quantities Buyers Bought
20
25
40
60
# of weeks this occurred
10
30
50
10
Total = 100
Food Store buys mangoes for $2 and sells them for $4.
Given this information, 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
Solution:
a) Using the Expected Value Criterion:
Step 1: Obtain the probabilities of states of nature
Probabilities of
weekly demand
States of Nature 10/100 = 0.1
30/100 = 0.3
50/100 = 0.5
10/100 = 0.1
Quantities Buyers
Bought
20
25
40
60
Step 2: Obtain the Feasible Stocking Actions (Decision Alternative)
N.B: Obtain exactly the number of quantity demanded from the supplier.
20
25
40
60
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
- 55 -
Quantitative Analysis – MGMT 322
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.
States of nature
Prob.
0.1
0.3
0.5
0.1
Feasible Stocking Action
20
25
$ 40
$ 30
$ 40
$ 50
$ 40
$ 50
$ 40
$ 50
Demand
20
25
40
60
40
0
$ 20
$ 80
$ 80
60
-$ 40
-$ 20
$ 40
$ 120
40
0
$ 20
$ 80
$ 80
60
-$ 40
-$ 20
$ 40
$ 120
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: Since $54 is the maximum profit value; optimal choice is to stock 40
units/week.
b) Using the Criterion of Rationality
Taking ¼ = 0.25 as our probability
States of Nature
Prob.
0.25
0.25
0.25
0.25
Demand
20
25
40
60
Feasible Stocking Actions
20
25
$ 40
$ 30
$ 40
$ 50
$ 40
$ 50
$ 40
$ 50
- 56 -
Quantitative Analysis – MGMT 322
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: Since $45 is the maximum expected profit value, the optimal choice is to
either stock 25 or 40 units.
c) Using the Maximum Likelihood Criterion
We take the highest probability of 0.5 from the conditional probability table
Feasible Stocking Actions
Prob.
0.1
0.3
0.5
0.1
20
25
40
60
20
$ 40
$ 40
$ 40
$ 40
25
$ 30
$ 50
$ 50
$ 50
40
0
$ 20
$ 80
$ 80
60
-$ 40
-$ 20
$ 40
$ 120
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
going 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 $60 and then, it is sold at a
price of $95 per unit. Migros receives $20 for each bottle that is unsold by the end of the
week and returned back to Efes. The sales data for the last 50 weeks is given as follows:
Quantities Buyers Bought
20
21
22
23
Number of Months this occured
20
15
10
5
a. Using the Expected Value Criterion, obtain the optimum number of Efes bottles
the sales manager of Migros should stock per month.
b. Using the maximum Likelihood Criterion, obtain the optimum number of Efes
bottles the sales manager of Migros should stock per month.
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Quantitative Analysis – MGMT 322
a) Using The Expected Value Criterion
Bottles of Beer Demanded
20
21
22
23
Total
Number of weeks
20
15
10
5
50
Probability
20/50 = 0.40
15/50 = 0.30
10/50 = 0.20
5/50 = 0.10
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
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
Profit = (TR – TC) + (# of unsold bottles*Profit Received)
= [(20*95) – (22*60)] + (2*20) = 620
d. Weekly Profits when Stock = 23 bottles/week and Demand = 20
Profit = (TR – TC) + (# of unsold bottles*Profit Received)
= [(20*95) – (23*60)] + (3*20) = 580
States of Nature
Probabilities
0.40
0.30
0.20
0.10
Demand
20
21
22
23
Feasible Stock Action
20
21
700
660
700
735
700
735
700
735
22
620
695
770
770
23
580
655
730
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.
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Quantitative Analysis – MGMT 322
b) Using The Maximum Likelihood Criterion
We take the highest probability of 0.40 from the conditional probability table
States of Nature
Probabilities
0.40
0.30
0.20
0.10
Demand
20
21
22
23
Feasible Stock Action
20
21
700
660
700
735
700
735
700
735
22
620
695
770
770
23
580
655
730
805
Therefore: The optimum choice is to stock 20 bottles because it yields max. expected
profit of $700.
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 (particularly
of 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 method involves asking questions to some experts in the company and utilizing 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 as given below:
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Quantitative Analysis – MGMT 322
Q2: In your opinion, what are the odds (chances) for the annual sales volume to lie
between 1200 and 1800?
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.
1 3  3
P(X < 1800) = 7/8 = 0.875=
1 3  3 1
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Quantitative Analysis – MGMT 322
Step 2: Obtain the Z value that corresponds to the probability value obtained in step 1 (or
closest probability value in the Normal ProbabilityTable).
Probability as close as possible to 0.875 = 0.87493 (from the table) → Z* = 1.15
Step 3:
Z* =
1.15=
X*-
μx
σx
1800-1500
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 2007 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=?
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Quantitative Analysis – MGMT 322
3
100,000
4
4
200,000
3
300,000
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
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X
Quantitative Analysis – MGMT 322
Chapter 5: Cost – Volume – Profit Analysis
Cost-Volume Analysis, also called Break-Even Analysis, is the feasibility analysis that is
used to determine the profitability and riskiness of an investment project. This technique
involves asking probability questions to other experienced people in the company and
assessing the risk faced by the company in investing in this project, based on the answers
obtained.
It also involves computing the expected value of net profit from this project which
enables the managers to make prudent investment decisions.
Application: We will use our previous example for estimating the mean and the
standard deviation of annual demand for the product that the company is planning 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; 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 constant for the range of
output that the company expects to purchase).
Questions That Managers Can Ask Specific About This Project
Q1: What is the probability of at least breaking even on this project for each year?
P(Π≥0)
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?
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Quantitative Analysis – MGMT 322
At XBE
Profit = TR = TC
TFC
XBE =
P V
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)
Z* =
1000 – 1500
261
Z* = -1.91
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Quantitative Analysis – MGMT 322
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)
Z* =
1180 – 1500
261
Z* = -1.23
P(X > 1180) = 0.89065
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Quantitative Analysis – MGMT 322
Answer 3: P(Losing more than $200 for each year)
P(Profit < -200) = P(X < X*)
X* = 1000 + (-200 / 5) = 960
960 – 1500
261
Z* =
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)
1400  1500
Z* =
261
Z* = -0.38
1,400
Z*=-0.38
1,500
X
0
Z
P(Profit ≥ $2000) = 0.64803
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Quantitative Analysis – MGMT 322
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 about 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
7,000
8,000
P(X < 9,000) = 2+2+3/3+2+2+3 = 0.70
Z = 0.52
3
9,000
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X
Quantitative Analysis – MGMT 322
Z=X-µ/σ
0.52 = 9,000 – 8,000 / σ
σ = 1923.
a) P(Losing Money)
6,000
Z*=?
8,000
X
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 Normal 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
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Quantitative Analysis – MGMT 322
c) P(At least Breaking-Even)
P(X ≥ 6,000)
6,000
Z* =-1.04
8,000
X
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 amount of profits per unit of output above XBE.
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Quantitative Analysis – MGMT 322
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| =
1000  1500
= |-1.91| = 1.91
261
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 amount of 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
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Quantitative Analysis – MGMT 322
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 the usage of the
truck is normally distributed with a mean of 150 hours per month and has 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
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Quantitative Analysis – MGMT 322
Home works
Home Work 1
Questions from the MAIN textbook:
Chapter 2: Ex 2-7, 2-8, 2-9, 2-10, 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
4) Suppose that the probability of Turkish economy growing at a higher rate for each year
over the next 4 years is estimated to be 0.70. Assume that the economic growth process is
a Bernoulli type of process. Answer the following:
a) What is the probability distribution of the possible number of years of high
growth over the next 4 years?
b) What is the probability of Turkish economy growing at a higher rate in at most 2
of the next 4 years?
c) What is the expected number of years of low growth over the next 4 years?
5) Suppose that there are 2 possible outcomes for the annual growth rate for Turkish
economy; high or low. The past data suggested that in each year of high growth, Istanbul
Stock Market Index grows approximately by 10 YTL, whereas in each year of low
growth, it grows on average by 5YTL. Economists believe that the probability of high
growth for each year over the next 3 years is 2/3. Assuming that the economic growth
process of Turkey is like a Bernoulli process, answer the following questions:
a) Probability distribution of possible number of years of high growth over the next 3
years.
b) What is the probability of Turkish economy growing at a high rate in exactly 2
years over the next 3 years?
c) What is the probability of Turkish economy growing at a high rate at least 2 years?
d) What is the expected net increase in the value of Istanbul Stock Market over the
next 3 years?
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Quantitative Analysis – MGMT 322
e) What is the probability of Turkish economy growing at a high rate in at most 2
years?
Home Work 3
1. Following is the sales data gathered for the past 12 months of a company:
Years
1999
1999
1999
1999
1999
2000
2000
2000
2000
2000
2000
2000
Months
August
September
October
November
December
January
February
March
April
May
June
July
Sales
28
27
33
25
34
33
35
30
33
35
27
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
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?
2. Questions from the textbook:
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
3. Apply each of the following models to the data set given below and obtain a forecast
for the 2nd Qtr. of 2007. Compute MSE of each model.
a) Naïve model
b) 2-period (UnW.) M.A
c) 3-period (W) M.A; take wt=0.5, wt-1 =0.3, w t-2 =0.2
d) 4-period (UnW.) M.A
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Quantitative Analysis – MGMT 322
Year
2006
2006
2006
2006
2007
2007
Quarter
1st
2nd
3rd
4th
1st
2nd
Data
500
520
490
510
480
?
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.
2. Questions from the textbook:
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
1. Questions from the textbook:
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 the past demand gives the information shown in the table below:
Daily Demand
10
11
12
13
Number of days Demanded
18
36
27
19
What quantity should Beth Perry buy for tomorrow’s sale to maximize the Expected
Profit?
N.B: Read the textbook; page 165-168.
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Quantitative Analysis – MGMT 322
Home Work 6
Questions from the textbook:
Chapter 5: Ex 5-2, 5-3, 5-5, 5-8, 5-10, 5-12, 5-14
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Quantitative Analysis – MGMT 322
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
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Quantitative Analysis – MGMT 322
QUIZ QUESTIONS
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Quantitative Analysis – MGMT 322
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?
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Quantitative Analysis – MGMT 322
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 lowincome 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
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Quantitative Analysis – MGMT 322
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 interest 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 turned 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 2006:
D
N
A
10
140
B
20
180
C
70
280
a. What is the probability that a randomly selected consumer (who has 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?
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Quantitative Analysis – MGMT 322
c. What is the probability that the automobile that he buys will be either model A or nondefective?
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 nondefective, 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 pay-off table of Mr. X
looks like this:
Type of investment
Inflation
High
Stock
$300,000
Moderate
Low
Treasury Bills
$100,000
$ Saving deposits
$ 150,000
$100
$150,000
$ 150,000
$50,000
$200,000
$ 150,000
a). What should we suggest Mr. X to invest by using the criterion of Maxi-Max?
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:
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Quantitative Analysis – MGMT 322
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.
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Quantitative Analysis – MGMT 322
15.
Linda, the 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 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.
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Quantitative Analysis – MGMT 322
17.
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 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 α= 0.6.
c. Compare above-said two models, which one is more accurate?
18.
Mr. Y 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 pay-off table of Mr. Y looks like this:
Type of investment Inflation Stock Treasury Bills $ saving deposits.
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Quantitative Analysis – MGMT 322
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 of
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)
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Quantitative Analysis – MGMT 322
EXAM QUESTIONS
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Quantitative Analysis – MGMT 322
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 2006. (10 pts)
Sales in the past quarter have been as follows:
Years
Qtr.
Sales
2004
2nd Qtr.
50
3rd Qtr.
52
4th Qtr.
47
1st Qtr.
55
2nd Qtr.
53
3rd Qtr.
50
4th Qtr.
54
1st Qtr.
52
2005
2006
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Quantitative Analysis – MGMT 322
Los Bichos does not believe there is a seasonal pattern.
Using 2-period Moving Average model obtain a forecast for the 2nd Qtr. of 2006 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
States of Nature
Gold
Real Estate
Stocks
High Inf.
$100000
$150000
$15000
Medium Inf.
$50000
$20000
$30000
Low Inf.
$10000
$-5000
$80000
A. Using the MINI-MAX REGRET criterion obtain the optimal investment alternative.
(15 pts)
B. Using the MAXI-MIN 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
MIGROS is 10 $, whereas the price that MIGROS charges from the customers is
15$ per bottle. Furthermore, MIGROS receives $5 only for each bottle that is unsold by
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Quantitative Analysis – MGMT 322
the end of the week and returned back to the producer of EFES. 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 2008 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 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
estimated that the cost of capital investment to undertake this project is $ 9 M. The
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Quantitative Analysis – MGMT 322
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 1500 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?
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Quantitative Analysis – MGMT 322
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