qa lecture notes.pdf

7/27/2019 QA Lecture notes.pdf

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QUANTITATIVE ANALYSIS

Lecture Notes,

Past Quiz, Exam, & Home Work

Questions

Prof. Dr. Serhan FTOLUGazimausa 2005


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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, Erhan ERDOAN and Sleyman EFE,

and Gift, have done in organizing and typing my in-class lectures.

Assoc. Prof. Dr. Serhan FTOLU

Department of Business Administration

Eastern Mediterranean University


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Quantitative Analysis MGMT 322

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INDEX

CHAPTER 2: A REVIEW OF PROBABILITY THEORY..................................................- 4 -

A.PROBABILITY......................................................................................................................... -4-B.EVENT..................................................................................................................................... -4-EXAMPLE ..................................................................................................................................... -4-C.EXPERIMENT.......................................................................................................................... -4-D.SAMPLE SPACE ...................................................................................................................... -4-EXAMPLE ..................................................................................................................................... -4-E.MUTUALLY EXCLUSIVE EVENTS .......................................................................................... -5-EXAMPLE ..................................................................................................................................... -5-

ALTERNATIVE APPROACHES IN PROBABILITY THEORY.......................................- 5 -

1.CLASSICAL APPROACH .......................................................................................................... -5-EXAMPLE ..................................................................................................................................... -6-2.RELATIVE FREQUENCY APPROACH...................................................................................... -6-EXAMPLE ..................................................................................................................................... -6-3.SUBJECTIVE APPROACH ........................................................................................................ -6-EXAMPLE ..................................................................................................................................... -6-

PROBABILITY RULES...........................................................................................................- 7 -

A.MARGINAL (UNCONDITIONAL)PROBABILITY: ................................................................... -7-B.THE ADDITION RULE: ........................................................................................................... -7-

1.THE ADDITION RULE FORMUTUALLY EXCLUSIVE EVENTS ................................................... -7-EXAMPLE ..................................................................................................................................... -7-2.THE ADDITION RULE FORNON-MUTUALLY EXCLUSIVE EVENTS .......................................... -7-EXAMPLE ..................................................................................................................................... -8-EXAMPLE ..................................................................................................................................... -8-

TYPES OF EVENTS.................................................................................................................- 9 -

1.STATISTICALLY INDEPENDENT EVENTS ............................................................................... -9-2.STATISTICALLY DEPENDENT EVENTS ................................................................................ -10-TYPES OF PROBABILITIES UNDERINDEPENDENCE ................................................................ -10-1.MARGINAL (UNCONDITIONAL)PROBABILITY ....................................................................... -10-2.JOINT PROBABILITY ............................................................................................................... -10-3.CONDITIONAL PROBABILITY ................................................................................................. -10-TYPES OF PROBABILITIES UNDERDEPENDENCE ................................................................... -11-1.MARGINAL (UNCONDITIONAL)PROBABILITY ....................................................................... -11-2.JOINT PROBABILITY ............................................................................................................... -11-3.CONDITIONAL PROBABILITY ................................................................................................. -11-EXAMPLE ................................................................................................................................... -11-


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BAYES THEOREM ..............................................................................................................- 12 -

EXAMPLE ................................................................................................................................... -12-EXAMPLE ................................................................................................................................... -13-

PROBABILITY DISTRIBUTION.........................................................................................- 15 -

DISCRETE VSCONTINUOUS RANDOM VARIABLES.................................................................... 15DISCRETE PROBABILITY DISTRIBUTION ........................................................................................ 15CONTINUOUS PROBABILITY DISTRIBUTION .................................................................................. 15BINOMIAL PROBABILITY DISTRIBUTION................................................................................ -18-CHARACTERISTICS OF A BERNOULLI TYPE PROCESS ................................................................ -18-BINOMIAL FORMULA: ................................................................................................................ -18-EXAMPLE ................................................................................................................................... -19-NORMAL PROBABILITY DISTRIBUTION .................................................................................. -23-CHARACTERISTICS..................................................................................................................... -24-TWO PARAMETERS:MEAN ()&STANDARD DEVIATION ()................................................... -24-EXAMPLE: .................................................................................................................................. -20-3MATHEMATICAL FACTS ABOUT NORMAL DISTRIBUTION ................................................. -24-EXAMPLE: .................................................................................................................................. -25-

CHAPTER 3: FORECASTING...............................................................................................-29-

THE METHODOLOGY OF FORECASTING................................................................................. -29-TECHNICAL ANALYSISFUNDAMENTALANALYSISBASICSTEPSINFORECASTING

FORECASTING MODELS........................................................................................................... -31-FORNON-SEASONAL PRODUCTS WITH LINEARTREND TYPE OF PATTERN........................ -31-NAVE MODEL ........................................................................................................................... -33-MOVING AVERAGE MODEL..-39-SIMPLE EXPONENTIAL MODEL .................................................................................................. -41-IMPORTANT STEPS ..................................................................................................................... -42-TIME SERIES REGRESSION MODEL ............................................................................................ -42-IMPORTANT POINTS ................................................................................................................... -44-SMOOTHING LINEARTREND MODEL......................................................................................... -46-IMPORTANT STEPS ..................................................................................................................... -46-

CHAPTER 4: ALTERNATIVE DECISION MAKING ENVIRONMENTS....................- 48 -

BASIC STEPS IN DECISION MAKING

ALTERNATIVE DECISION MAKING ENVIRONMENTS

ALTERNATIVE CRITERIA FORDECISION MAKING UNDERUNCERTAINTY .......................... -48-THE MAXI-MAX CRITERION...................................................................................................... -49-THE MAXI-MIN CRITERION ....................................................................................................... -49-THE CRITERION OF REALISM..................................................................................................... -50-THE MINI-MAX REGRET CRITERION ......................................................................................... -50-ALTERNATIVE CRITERIA FORDECISION MAKING UNDERRISK.......................................... -51-


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EXPECTED VALUE CRITERION................................................................................................... -51-EXAMPLE ................................................................................................................................... -51-CRITERION OF RATIONALITY..................................................................................................... -52-THE CRITERION OF MAXIMUM LIKELIHOOD ............................................................................. -52-

SUPPLYING THE NUMBERS: HOW TO OBTAIN MEAN AND STANDARDDEVIATION WHEN DATA IS MISSING OR INCOMPLETE........................................- 57 -

CHAPTER 5: COST VOLUME PROFIT ANALYSIS ................................................. - 61 -

GENERAL DEFINITION AND USE OF COST- VOLUME- PROFIT ANALYSISITS APPLICATIONCOMBINING UNIT MONETARY VALUES AND PROBABILITY DISTRIBUTION.............................. -64-3STEP PROCEDURE TO OBTAIN EXPECTED PROFITS.............................................................. -67-3STEP PROCEDURE TO OBTAIN EXPECTED LOSS................................................................... -68-

Home Works-70-


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Chapter 2: A Review of Probability Theory

A. ProbabilityProbability is the chance or likelihood of something occurring (happening).

We express probabilities as either fractions or decimalsE.g: P(H) = 9/10 (as a fraction)or

P(H) = 0.9 (as a decimal)

Note: The probabilities should neither be < 0, nor >1.

B. Event

An eventis defined as either one or more than one of possible outcomes of doingsomething or an activity.

Example

ACTIVITY: Inflationary process in Turkey in the year 2005:

Events: A: If annual inflation for 2005 is exactly 10%B: If annual inflation for 2005 is less than 8%

C: If annual inflation for 2005 is either 5% or 7%

C. Experiment

Is the activity that generates events.For example, tossing a coin twice.

D. Sample Space

The sample space of an experiment is the list of all possible outcomes of thatexperiment.

Example

Experiment: Tossing a coin.Sample Space: S = {HH, TT, HT, TH}

Events: A: Exactly one outcome, that is, getting 2heads = HHB: 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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E. Mutually Exclusive Events

Mutually exclusive events are those events which can take place only one at atime, and not togetherat the same time.

F. Non Mutually Excusive Events

Are those events which can happen togetherat the same time.

Example 1

Experiment: Tossing a coin twiceEvents: 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 maleC: Selected student is femaleD: Selected student is 22 years old.

Therefore: Events AB are non-mutually exclusive because they can happen togetheratthe same time.Events AD are mutually exclusive because they can not happen together atthe same time.Events AC are non-mutually exclusive because they can happen togetheratthe same time.Events BC are mutually exclusive because they can not happen together atthe same time.

Alternative Approaches in Probability Theory

1. Classical Approach

# of outcomes favorable to the occurrence of event AP(A) =

Total # of possible outcomes


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Example 1

Experiment: Randomly selecting a student in a class of 40 students.Events: A: Selecting afemale student, given total number of females equal to 10.

B: Selecting a male student, given total number of males equal to 30.The total number of students is 40

P(A) = 10/40P(B) = 30/4

2. Relative Frequency Approach

The probability of the event found by using the givenpast data.

Example

Experiment: Selecting a student randomly.Events: A: Selecting a student who gets A from MGMT 322.

Past Data: 20% of all students who have taken MGMT 322 in the past get A.P(A) = 0.20

B: Galatasaray wins the championship in 2005, using the past data obtain howfrequently the similar event occurred in the past.

Past Data: 40% of the last 9 years, Galatasaray was a champion.P(B) = 0.40

3. Subjective Approach

The probability of the event found by using thepersonal feeling or experience.

Example

Event: A: Galatasaray wins.P(A) = 0.90As a supporter or a football watcher you may have the feeling that Galatasaraywill win 90%.


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Probability Rules

A. Marginal (Uncondit ional) Probability :

Only one event can take place. P(A)

B. The Addit ion Rule:

1. The Addition Rule for Mutually Exclusive Events

If Events A, B, and C are mutually exclusive (meaning they can not happen togetherat the same time.)Then: P(A or B) = P(A) + P(B)

P( A or B or C) = P(A) + P(B) + P(C)

Questions that can be asked and answered by using Addition Rule:Q1: What is the probability of A or B or C taking place?Q2: What is the probability of A or B taking place?Q3: What is the probability of B or C taking place?

Example

Suppose that the number of students in each age group in the class is as follows:

Age Male Female Total

20 0 2 221 1 3 4

22 6 3 923 3 1 424 4 0 4

Total 14 9 23

Events: A: Selecting a random student who is 20 years old.B: Selecting a random student who is 21 years old.C: Selecting a random student who is 22 years old.

D: Selecting a random student who is 23 years old.

P(A or B) = P(A) + P(B) = 2/23 + 4/23 = 6/23P(A or C) = P(A) + P(C) = 2/23 + 9/23 = 11/23P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) = 2/23 + 4/23 + 9/23 + 4/23 = 19/23

2. The Addition Rule for Non-Mutually Exclusive Events

If A and B are non-mutually exclusive events,Then: P(A or B) = P(A) + P(B) P(A and B)


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Example 1

Suppose that the sex and the eye color of students in a class of 40 people is distributed asfollows:

Black Green Total

Male 13 9 22Female 8 10 18Total 21 19 40

Experiment: Selecting a random student.Events:A: Selecting a male student.B: Selecting a female student.

C: Selecting a black eye student.D: Selecting a green eye student.

P(A or C) = P(A) + P(C) P(A and C) = (22/40 + 21/40) 13/40 = 30/40P(B or C) = P(B) + P(C) P(B and C) = (18/40 + 21/40) 8/40 = 31/40

Example 2

Suppose that the Ford Motor Corporation produces 3 different models for Turkishconsumers. Annual volume of production and sales of all 3 model combined is 700 units.However, some of the autos sold turned to be defective. Based on the past data the

production manager estimated the following distribution of defective and non-defectiveautos for each model manufactured.

Models

A B C Total

Defective 10 20 70 100Non-Defective 140 180 280 600

Total 150 200 350 700

a. What is the probability that a randomly selected consumer will buy either modelA or C?

b. What is the probability that the consumer will buy either model B or a Defectiveauto?

c. What is the probability that the consumer will buy either model A or a Non-defective auto?

a. P(A or C) = P(A) + P(C) = 150/700 + 350/700 = 500/700b. P(B or D) = P(B) + P(D) P(B and D) = (200/700 + 100/700) 20/700 = 280/700c. P(A or N) = P(A)+P(N) P(A and N) = (150/700 + 600/700) 140/700= 610/700


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Example 3

There are 500 companies whose stocks are traded on Istanbul stock exchange. Thesecompanies 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 thenumber of companies in each sector that are likely to report positive profits and losses inthe year 2005:

Sectors

Industry Service Mining Total

Positive Profits 150 240 30 420Losses 50 10 20 80Total 200 250 50 500

a. What is the probability of randomly selected company (out of these companies

listed on Istanbul Stock Exchange) to report positive profits in year 2005?b. What is the probability of a randomly selected company to report either losses in

year 2005 or to be from Mining sector?c. What is the probability of a randomly selected company to be either from Industry

or from Service sector?

a. P( Positive profits) = 420/500 = 0.84

b. P( Losses or Mining) = P( Losses) + P( Mining) P( Losses and Mining)= 80/50 + 50/500 20/500 = 110/500 = 0.22

c. P( Industry or Service) = P(Industry) + P(Service)= 200/500 + 250/500 = 450/500 = 0.9

Types of Events

1. Statistically Independent Events

These are those events whereby the probability of occurrence of an event is notaffected by ( or dependent upon) the occurrence of the other event.

Example

Events: A: Izlem goes to Istanbul next TuesdayB: Istanbul stock market decreases by more than 10% next Friday.

Therefore: Event A and B are Independent events because the occurrence of event Adoes not have an impact on the probability of the occurrence of event B.


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2. Statistically Dependent Events

These are those events whereby the probability of the occurrence of an event isaffected by ( or dependent upon) the occurrence of the other event.

Example 1Events: 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 isexpected to increase the probability of the event A

Example 2

Events: A: The quantity of automobiles purchased in Turkey risesB: The quantity of petroleum consumed in Turkey rises

Therefore: Event A and B are Dependent because the probability of the occurrence ofevent B is affected by or dependent upon the occurrence of event A, as the more theautomobiles Turkish people are purchasing, the more the petroleum they will buy to runtheir automobiles.

Types of Probabil ities under Independence

1. Marginal (Unconditional) Probability

P(A): This is interested in the probability of a single event.

2. Joint Probability

P(AB): This is where you are interested in eitherthe probability of event A and event Bhappening together at the same time orin 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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Example

Events: A: It rains in Cyprus todayB: Exchange rate of dollar/yen increases today

Therefore: P(A/B) = P(A)

PB/A) = P(B)

Types of Probabilit ies under Dependence

1. Marginal (Unconditional) Probability

P(A) : Is interested in just the probability of a single event.

2. Joint Probability

P(AB) = P(A/B) * P(B)P(BA) = P(B/A) * P(A)P(AB) = P(BA)

3. Conditional Probability

P(A/B) = P(AB) / P(B)P(B/A) = P(BA) / P(A)

ExampleSuppose that the eye color of students in a class of 40 people as follows;

Black Green Total

Male 20 8 28Female 10 2 12Total 30 10 40

Events:A: Selecting a Male student.B: Selecting a student with Black Eyes.

C: Selecting a Female student.D: Selecting a student with Green Eyes.

Q1. If the selected student is a Male student, what is the probability that he hasgot Green Eyes?

P(D/A) = P(DA) / P(A) = (8/40) / (28/40) = 0.074


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Q2. If the selected student is a Female student, what is the probability that she hasgot Green Eyes.

P(D/C) = P(DC) / P(C) = (2/40) / (12/40) = 0.167

Bayes TheoremIs 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 Type1and the other half is Type2. Given that the probability of getting Ace (1) when you roll aType1 die is 0.3 and that of rolling Ace (1) of Type2 is 0.6. You are also given the initialprobability of Type1 as 0.5 and that of Type2 as 0.5.If you randomly select a die and roll it and you find out that Ace (1) has come up, what isthe probability of this die to be a Type1 die?

NB: These two events (Ace and Type1 or 2) are dependent because the probability ofgetting 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, usethe following five steps below in order to get the Answer:

Step 1: Prepare the following table:

Marginal Probability of ElementaryEvents

Probability of Secondary EventConditional Probability of SecondaryEvent, Given Each of the Elementary

Event

P(Type 1) = 0.5 P(Ace / Type 1) = 0.3

P(Type 2) = 0.5P(Ace)

P(Ace / Type 2) = 0.6

Step 2: Computation of probability of secondary event:P(secondary event) = summation of its joint probabilities of each one of the elementaryevents.


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P(Ace) = P(Type1, Ace) + P( Type2, Ace)

Step 3: Computation of joint probability of secondary events with each of the elementaryevents:

P(Type 1, Ace) = P(Ace/Type1)* P(Type 1)= 0.3 * 0.5 = 0.15

P(Type 2, Ace) = P(Ace/Type2) * P(Type2)= 0.6 * 0.5 = 0.30

P(Ace) = 0.15 + 0.30 = 0.45

Step 4: Use the Bayes Theorem to get the final answerP(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 thechanges in the petroleum prices. The probability of inflation rate increasing in 2001 isestimated to be 0.60. The probability of petroleum prices rising is 0.40. The probability ofboth inflation rate and the petroleum prices rising is 0.35. On the other hand, theprobability of inflation rate rising and at the same time, the petroleum prices not rising isestimated to be 0.20. If petroleum prices do not rise in 2001, what is the probability ofinflation rate rising?

Events: I: Inflation rate increasing.P: Petroleum prices rising.N: Petroleum prices not rising

Given:

P(I) = 0.6P(P) = 0.4P(IP) = 0.35

P(IN) = 0.2

Solution:

P(N) = 1 P(P) = 1- 0.4 = 0.6P(I / N) = P(IN) / P(N) = 0.2 / 0.6 = 0.333 0.3


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Example2:

Annual profits of construction sector depend on wage rate paid to labor employed by thesector. The past data about the behavior of annual profits suggested that the probability ofprofits rising in a given year is 0.80 and the probability of profits falling is 0.20. The pastdata also suggest that in 60% of all the years during which the profits have increased,

wage rate has declined, and it has increased in the remaining 40%. However, in 80% ofall years during which profits have fallen, wage rate has increased, and the onlyremaining 20% wage rate has decreased. Due to the economic crisis in Turkey, wage ratehas 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 torise in the year?

Events: A: Profits risingB: Profits fallingC: Wage rate decliningD: Wage rate increasing

Given:

P(A) = 0.80P(B) = 0.20P(C/A) = 0.60P(D/A) = 0.40P(D/B) = 0.80P(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) / 0.80P(CA) or P(AC) = 0.60 * 0.80 = 0.48

P(C) = P(CA) + P(CB)= [P(C/A) * P(A)] + [P(C/B) * P(B)]= (0.60 * 0.80) + (0.20 * 0.20)= 0.48 + 0.04 = 0.52

P(A/C) = 0.48 / 0.52 = 0.92

Example 3A child chosen at random in a community school system comes from low-income family15% of the time. Children from low-income families in the community graduate fromcollege only 20% of the time. Children not from low-income families have a 40% chanceof graduating from college. As an employer of people from this community, you arereviewing applicants and note that the first one had a college degree. What is theprobability that the person comes from a low-income family?


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Events: L: Child coming from a low-income familyG: Graduate from collegeH: Child coming from not a low-income family

Given:

P(L) = 0.15

P(H) = 1- P(H) = 1- 0.15 = 0.85P(G/L) = 0.20P(G/H) = 0.40P(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 = 0.37P(L/G) = 0.03 / 0.37 = 0.08 = 8%

Probability Distribution

Random Variables

A Random Variable is a variable which takes on values unpredictably as a resultof a random process.

EG: Events: X: Age of a randomly selected studentY: The value of stock market Index in Istanbul by the end of todays

session.Z: Interest rates on a new Bond issued by the Turkish Government that will

take place next week.

DISCRETE VERSUS CONTINUOS RANDOM VARIABLES

1. Discrete Random Variable

This is allowed to take on only limited number of possible values.

E.g: X = Age of a randomly selected student.a. Binomial Distributionb. Poisson Probability Distribution

2. Continuous Random Variable

This is allowed to take any value over a certain range. They are infinite numbers ofpossibilities.

a. Normal Probability Distribution


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b. Exponential Probability Distribution

E.g: Possible values of X:

X1 X2 X3 X4 X5 X620 21 22 23 24 26

Probability Distribution of a Random Variable

Is the listing of probabilities that correspond to each possible value of that randomvariable.

Two Steps to Obtain the Probability Distribution of X:

Step 1Obtain all the possible values that X can take on.

Step2

Obtain the probability of the occurrence of each possible value of x.

Then, present the results in a form of a Table or a Graph.E.g: Age of a randomly selected student

Step 1: Possible X values:

Age Number of Students

20 121 5

22 7

23 4

24 3

26 1

Total 21


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X120

X221

X322

X323

X524

X626

Step 2: Obtain the probability of each possible X values

Expected Value of a random variable

Its donated as; E(X)It is the weighted value of all the possible values of x.

E(X) for a Discrete Random VariableE(X) = P(Xi)XiP(Xi) = Probability of occurrence of the ith value

E.g: E(X) = (20)(1/21) + (21)(5/21) + (22)(7/21) + (23)(4/21) + (24)(3/21) + (26)(1/21)= 0.95 + 5 + 7.33 + 4.38 + 3.43 + 1.24 22 years

x P(x)

20 1/21

21 5/21

22 7/21

23 4/21

24 3/21

26 1/21


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E(X) for Continuous Random Variable

E(X) = P(X) dx

Example

Suppose that the average growth rates of GDP of countries in the world are given below:

Average Growth Rate Number of Countries

8% 206% 304% 502% 1oo

Total 200

What is the Expected (average) growth rate of a randomly selected country in the world?

Solution:

E(Growth Rate) = (0.08)(20/200) + (0.06)(30/200) + (0.04)(50/200) + (0.02)(100/200)= 0.008 + 0.009 + 0.01 + 0.01= 0.037 = 3.7%

Binomial Probability Distribut ion

It is the probability distribution of a binomial random variable which takes onvalues as a result of Bernoulli Process.It is a type of Discrete Probability Distribution.

Characteristics of a Bernoulli Type Process

i. Its made up fixed number of trials of the same activity.ii. Each trial has two out comes only. (Success or Failure)

iii. The probability of success on each trial remains fixed over all trials.

iv. Each trial is statistically independent.

Binomial Formula:

P(r) = rnrqprnr

N

)!(!

!


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n: Number of trials.r: Number of success outcomes.p: Probability of success outcomes.q = (1 p) : probability of failure.

P(r) = probability of getting exactly r (successes) out of n trials.

Kind of Questions that can be asked and answered using Binomial Formula:

Q1: Obtain the probability distribution of a Binomial random Variable XQ2: What is the probability of getting at leastr* successes out on n trials; P(r r*)Q3: What is the probability of getting at mostr* successes out of n trials; P(r r*)Q4: What is the probability of getting exactly r* successes; P(r*)

General Answers:

Q1:Using the Binomial Formula, find the probability distribution of X

X P(X)X1 P(X1)X2 P(X2)X3 P(X3)Xn P(Xn)

Q2: P(r r*) = P(r*) + P(r* + 1) + P(n)Q3: P(r r*) = P(0) + P(1) + .P(r*)Q4: Use the formula

Example 1

Suppose that we toss a coin 4 times.

Q1. What is the probability of getting exactly 2 Heads?Q2. What is the probability of getting at least 3 Heads?Q3. What is the probability of getting at most 3 Heads?Q4. Using the binomial formula obtain the binomial distribution of the possible

number of tails have may come up.

Solution:Q1. n = 4 p = 1/2 (Head) q = 1/2 (Tail)

4!

P(2) =2!(4 - 2)!

(1/2)2(1/2)4-2 = 0.37


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Q2. P(r3) = P(3) + P(4)

4!

P(3) =3!(4 - 3)!

(1/2)3(1/2)4-3 = 0.25

4!

P(4) =4!(4 - 4)!

(1/2)4(1/2)4-4 = 0.063

P(r3) = 0.313

Q3. P(r3) = P(0) + P(1) + P(2) + P(3)

4!

P(0) =0!(4 - 0)!

(1/2)0(1/2)4-0 = 0.063

4!

P(1) =1!(4 - 1)!

(1/2)1(1/2)4-1 = 0.25

P(r3) = 0.938

Q4.p: probability of getting Tailq: 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. Ifinflation rate is greater than 100%, inflation is high. If it is less than 100%, inflation islow for that year. Sabanci Corp. makes $50 million of positive profits for each year ofinflation and loses $20 million for each year of low inflation. Economists estimated thatthe probability of high inflation is 80% for each of the next 4 years.


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a) What is the probabilityof getting exactly 2 years of high inflation in the next 4years?

n= 4 r= 2 p=0.8 q=0.2

P (2) = 4! / 2! (4-2)! *[(0.8)2 * (0.2)4-2] = 0.1536

b) What is the prob. of getting at least 3 years of HI over the next 4 years?

P (r 3) = P (3) + P (4) = 0.4096 + 0.4096 = 0.8192

P (3) = 4! / 3! (4-3)! * [(0.8)3 * (0.2)1] = 0.4096

P (3) = 4! / 4! (4-4)! * [(0.8)4 * (0.2)0] = 0.4096

c) What is the probability of getting at most 3 years of LI?

p = 0.2q = 0.8

P (r 3) = P (0) + P (1) + P (2) +P (3) = 0.0016 + 0.0256 + 0.1536 + 0.4096 =0.5904

P (0) = 0.0016P (1) = 0.0256P (2) = 0.1536P (3) = 0.4096

d) What is the expected net profit for Sabanci over the next 4 years?

X P (X)X1 All 4 years of HI

4*50 = 200P (200) = P (4)

X2 3 years of HI and 1 year ofLI [(3X50) (1*20)] = 130

P (130) = P (3)

X3 2 years of HI and 2 years ofLI(2*50) (2*20) = 60

P (60) = P (2)

X4 1 year of HI and 3 years LI

(1*50) (3* 20) = -10

P (-10) = P (1)

X5 0 year of HI and 4 years ofLI(0*50) (4*20) = -80

P (-80) = P (0)


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Example 3

Prof. TT has put $2 on each of the 3 horses running three different races in IstanbulHorse Racing center. He feels that each of the bets he has made has a 0.2 probability ofwinning. A winning ticket on any of the three horses will earn Prof. TT $40. Assuminga Bernoulli process, answer the following questions:

a. What is the probability of at least 2 horses ( on which Prof. TT ha s bet) winning?b. What is theExpected Earnings from all the three horses?c. What is the probability of at most 1 out of the three horses (on which Prof. TT has

bet) winning?d. What is the probability distribution of his possible earnings from all the three

races?

Solution:

n= 3 p = 0.2 q = 1- 0.2 = 0.8x = number of horses

P(0) =( )

( ) ( ) 030 8.02.0!03!0

!3

= 0.512

P(1) =( )!13!1

!3

( ) ( ) 131 8.02.0 = 0.384

P(2) =( )

( ) ( ) 232 8.02.0!23!2

!3

= 0.096

P(3) =( )

( ) ( ) 333 8.02.0!33!3

!3

= 0.008

X Profit P(X)

0 (0*40) 6 = -6 0.5121 (1*40) 6 = 34 0.3842 (2*40) 6 = 74 0.0963 (3*40) 6 = 114 0.008

a. P(r 2) = P(2) + P(3) = 0.096 + 0.008 = 0.104

b. E(X) = -6(0.512) + 34(0.384) + 74(0.096) + 114(0.008)= -3.072 + 13.056 + 7.104 + 0.912 = $18

c. P(r 1) = P(0) + P(1) = 0.512 + 0.384 = 0.896


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d. X Profit P(X)

0 -6 -6/2161 34 34/2162 74 74/216

3 114 114/216Total 216

Example 4

Suppose that the inflationary process in Turkey is a Bernoulli process over the next 5years. The probability of inflation rate to be high (meaning that it exceeds 100%) for eachyear over the next 5 years is estimated to be . Note that the low inflation corresponds tothe 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 yearsHigh Inflation

Years Profits Prob(x)

E(profits) = (-25) * (0.0001) + (-10) * (0.0147) + (5) *(0.0879)+(20)*(0.2637)+(35)*(0.3955)+(50)*(0.2373)

= $31.250 Million

Normal Probability Distribution

This is a type of continuous probability distribution and its range extends from negativeinfinity to positive infinity.

0 (0*10) (5*5) = -25 0.00011 (1*10) (4*5) = -10 0.01472 (2*10) (3*5) = 5 0.08793 (3*10) (2*5) = 20 0.26374 (4*10) (1*5) = 35 0.3955

5 (5*10) (0*5) = 50 0.2373


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Characteristics

1. Bell shaped probability curve.2. It has a single peak3. Mean () lies at the center and the distribution is symmetrical at the vertical line

erected at .

4. Two tails extend indefinitely.

Two Parameters: Mean () & Standard Deviation (): Located in the center of the population and it is the average of the data.: It gives information about relative spread of data around the mean.

Application of Normal Distr ibution

3 Mathematical Facts about Normal Distribution


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1) Approximately 68% of the population lies in the interval ranging 1 below the to 1.

2) 95% of the population lies in an interval ranging 2 below the to 2 above the2.

3) 99% of the population lies in an interval ranging 3 below the to 3 above the

2.X~N (50, 5)

= x/ N = $ 50, = 5Z ~ N (0,1)

Example 1:

Suppose that the stock price of Corp. X (s) is distributed normally with a of 100 TL

and of 10 TL?


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a) What is the probability of s to be more than 120 TL in any day?b) What is the probability of s to be less than 90 TL in any day?

a)

z = 120 100/ 10 = 2

P (s > 120) = P (z > 2) = 1 P (z < 2) = 1 0.97725 = 0.02275

b)

z = 90- 100 / 10 = - 1

P (s < 90) = P (z 1) = 1 0.84134 = 0.15866


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Example 2

Prof. TT believes that the annual profit of Turkish bank is a normal random variablewith a mean of $600,000 and a standard deviation of $100,000. Prof. TT is currentlyanalyzing 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 thatProf.TT will analyze?

b. If Prof. TT randomly selects a bank to analyze, what is the probability that thisbank will have an annual profit volume of more than $400,000?

Solution:

= $600,000 = $100,000

X Y

500,000 600,000 650,000 $

a. P(500,000 $ 650,000) = (z1 z z2)z1 = 500,000 600,000 / 100,000 = -1

z2 = 650,000 600,000 / 100,000 = 0.5

Area X = P(z +1) 0.5 = 0.84134 0.5 = 0.34134Area Y = P(z 0.5) 0.5 = 0.69146 0.5 = 0.19146Area X + Area Y = 0.34134 + 0.19146 = 0.5328

The Approximate number of Banks = 270*0.5328 = 143.856 144

b. P($ > 400,000)


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400,000 600,000 $

Z* = 400,000 600,000 / 100,000 = -200,000 / 100,000 = -2P(z +2) = 0.97725

Example 3

A financial analyst computed the return on stockholders equity for all the companies

listed on the New York Stock Exchange. She found that the mean of this distributionwas 10%, with a standard deviation of 5%. She is interested in examining furtherthose companies whose return on stockholders equity is between 16% and 22%. Ofthe approximately 1,300 companies listed on the exchange, how many are of interestto her?

Solution:

= 10% = 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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CHAPTER 3: FORECASTING

Definition of Forecasting

It is the prediction of the future behavior of a variable. E.g. stock prices, inflation rates,interest rates, exchange rates, profits, gold prices, land prices, etc.The most critical variable that companies are interested to forecast is Demand(sales).Usually the companies are interested to forecast monthly sales (next month), orquarterly sales (next quarter).

Importance of Sales Forecasting

There are many decisions that critically depend on sales forecasts and these include:1) Input ordering decisions: Depends on how much working capital and other inputs

you will need.2) Production decisions: Depends on how much you will produce in a certain time

period, that is, it determines the output.3) The pricing decisions: This depends on whether the market is monopolistic or

oligopolistic.

4)

Advertising decisions: Depends on how much you will spend on advertising.5) Inventory decisions: This is how much you should maintain or change dependingon demand (sales) forecast.

The Methodology of Forecasting

1) Our Methodology (Technical analysis)This methodology uses past sales data only to predict (forecast) future sales.The limitation of this method is that it only depends on past data to predict the futuresales.

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, like Economic theory andRegression Analysis.It also uses past data about all the variables below to forecast future demand.

Formula: QD

x = f (Px, Ps, Pc, I, Expectations, Population)

1) BASIC STEPS IN FORECASTING( Technical Analysis)

Step 1: Determine the type of the product based on its sales pattern.

Step 2: Determine the forecasting models that can be used for this type of product.

Step 3: Apply each model on the actual past data and obtain MSE of the model. Bestmodel is the one which yields smallest MSE (Mean Square Error)

Step 4: Use the BEST MODEL to obtain a forecast for the future.

MSE test is the test of forecasting accuracy

MSE =et2/ n


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TYPES OF PRODUCTS BASED ON THEIR SALES PATTERN

How To Determine The Type Of Product:

First collect past actual sales data (monthly/ quarterly) and plot it against time and seethe pattern.

The following two are the possibilities:

NON- SEASONAL SEASONALa) Constant level a) Constant Levelb) Linear Trend b) Linear Trendc) Exponential Smoothing c) Exponential Trendd) Damped Trend d) Damped Trend

Forecasting Models

FOR NON-SEASONAL PRODUCTS WITH CONSTANT LEVEL TYPE OFPATTERN

1) Nave Model2) Moving Averages (M.A)3) Simple Exponential Smoothing

FOR NON-SEASONAL PRODUCTS WITH LINEAR TREND TYPE OF PATTERN

1) Nave Model2) Time Series Regression3) Smoothing Linear Trend (A Linear Exponential Smoothing)

a) Constant- level (Non-seasonal):It has a fixed amount of sales.


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


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c) Linear Trend (Non-Seasonal):

General increase of sales over time.

d) Linear Trend (Seasonal):

Sales increase at a constant rate.


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e) Exponential Trend (Non-Seasonal):Sales increase over time at an increasing rate.

f) Exponential Trend (Seasonal):Sales increase over time with regular ups and downs.


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

h) Damped Trend (Seasonal):


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Has regular ups and downs.

Ft+1 = Xt et =X t - Ft

F April = X March F2nd Quarter 2002 = X1st Quarter 2002

Steps in Computing The Mean Squared Error (MSE)

Steps 1: Obtain as many data as possible

Steps 2: If you have enough data, meaning that if you have at least 12 units of data incase of monthly data or at least 8 units of data in case of quarterly data divide the dataset into 2 halves, the 1st half is called the Warm-Up Sample and 2nd half is called theForecasting Sample.

Steps 3: Apply the model from the very beginning and obtain forecasts and errors foreach time period.

a) If you have enough data, apply each model on all the data and obtain

Forecasting Errors, but use only the forecasting errors coming from theForecasting Sample ( the 2nd half of the data) to obtain the MSE.b) If you do nothave enough, use all the available or given data as

forecasting sample and use all the forecasting errors coming from thissample to obtain MSE.

Steps 5: Bestmodel is the one which yields smallest MSE. (Use the BEST MODELto obtain a forecast for the future.)


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Applicat ion of Nave Model

Formula: Ft + 1 = XtFt = Forecast for next periodXt = Actual sales in the current period.

Et = Xt Ft

Et = Forecasting error for period tXt = Actual sales in period tFt = Forecasts in period t

E.g. If FJan, 2005 = 5,500XJan, 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 NorthCyprus, obtain the forecast of sales for the month of January, 2003 using the NaveModel and find the MSE of this model:.

WARM-UP SAMPLE

Month T Xt Ft et2002 Jan. 1 28

Feb. 2 27 28 -1March 3 33 27 +6April 4 25 33 -8May 5 34 25 +9June 6 33 34 -1July 7 35 33 +2

August 8 30 35 -5Sept. 9 33 30 +3Oct. 10 35 33 +2

Nov. 11 27 35 -8Dec. 12 29 27 +2

2003 Jan. 13 29

FORECASTING SAMPLE


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6

)2()8()2()3()5()2(

6

222222

12

7

2

+++++==

=t

te

MSE

Solution:

Nave Models Formula = Ft + 1 = XtNB: There is no forecast and no error for the first month (F1 and e1), but we obtain theforecasts and the errors for the next and the following months (periods) according to the

Nave Model. Thus the forecast for the next period (month) is the forecast for theprevious period.

F2 = 28, e2 = X2 F2 = 27 28 = -1F3 = 27, e3 = X3 F3 = 33 27 = +6F4 = 33, e4 = X4 F4 = 25 33 = -8Follow the same procedure up to F12.

a) Forecasts for January, 2003 (F13) = 29

b) MSE: Since we have enough data, we are going to take only the forecasting errors

from the forecasting sample. Therefore:

= 18.3

Example 2

Suppose that you are the manager of HSBC Bank. You believe that the pattern ofquarterly amount of savings deposits is non-seasonal and exhibits approximately aconstant-level type of behavior. You need a forecast for the savings deposits at your bankfor the 1st quarter of 2005 in order to make better decision about the portfolio investmentand interest rate policies of your bank.Using the Nave Model, obtain a forecast for the amount of savings deposits for the 1 stquarter of 2005 and obtain the MSE of the model, given the following past quarterly dataabout the amount of savings deposits ( in million of dollars) of your bank:

Year T Quarter Deposits Ft et

2003 1 2nd 1202003 2 3rd 100 120 -202003 3 4th 110 100 +102004 4 1st 130 110 +202004 5 2nd 90 130 -402004 6 3rd 110 90 +202004 7 4th 100 110 -102005 8 1st 100


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Solution:

F2 = 120, e2 = 100 120 = -20F3 = 100, e3 = 110 100 = +10

Follow the same procedure up to F7.

a) Forecasts for 1st quarter, 2005 (F8) = 100b) MSE: Since we do nothave enough data, use all the forecasting errors to obtain theMSE.

MSE =6

)10()20()40()20()10()20( 222222 +++++

= 500

Applicat ion of Moving Averages Model

Unweighted M.A Model Weighted M.A Model

2-period (Un.W) M.A 2-period (W) M.A3-period (Un.W) M.A 3-period (W) M.A

2- Period (Unw) M.A Model:Ft+1 = Xt+ Xt-1/ 2

3- Period (Unw) M.A Model:Ft+1 = Xt + Xt-1 + Xt-2 / 3

2- Period (W) M.A Model:Ft+1 = wt Xt + wt-1 Xt-1

3- Period (W) Ft+1 = wt Xt + wt-1 Xt-1 + wt-2 Xt-2

wt + wt-1 + wt-2 = 1 wt >wt-1 > wt-2


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6

)7.2()7.5()3.2()3.0()4()3.4(

6

222222

12

7

2

+++++==

=t

te

MSE

Example 1

From the data given below, find the forecast for F13 using 3-period (Unw) MA Modeland obtain the MSE of this model.

WARM-UPSAMPLE

t Xt Ft et1 282 273 334 25 29.3 -4.35 34 28.3 +5.7

6 33 30.7 +2.37 35 30.7 +4.38 30 34 -49 33 32.7 +0.3

10 35 32.7 +2.311 27 32.7 -5.712 29 31.7 -2.713 30.3

Solution:

3-Period (Unw) MA Models Formula:Ft + 1 = Xt + Xt-1 + Xt 2 / 3

F4 = 33 + 27 + 28 / 3 = 29.3, e4 = 25 29.3 = -4.3F5 = 25 + 33 + 27 / 3 = 28.3, e5 = 34 28.3 = +5.7

F13 = 29 +27 +35 / 3 = 30.3

= 13.3

FORECASTING SAMPLE


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Example 2

Given the Dow-Jones Index of stock market prices as below, use a 4-period (W) MAmodel to find the MSE of the model using the following weights:0.5, 0.3, 0.1 and 0.1respectively.

Month t Dow-Jones

Index (Xt)

Ft et

May 1 820June 2 840July 3 857August 4 823September 5 784 835 -51October 6 793 809 -16November 7 817 800 17December 8 820 807 13

January 9 826 813 13February 10 824 820 4March 11 830 824 6April 12 842 827 15May 13 840 835 5June 14 839 838 1

MSE: 6.91

7

)1()5()15()6()4()13()13(

7

2222222

14

8

2

=++++++

=

=t

et

Application of Simple Exponential Model

Formula: Ft+1 = Ft +et 0 <


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Important Steps

Step 1: To obtain F1, we take the mean (average) of the warm-up sample.

If we do not have enough data, we take the MEAN of all data as F1.Step 2: To find the Best Value of, we apply the model with each possible value ofon the warm-up sample (or on all of the data in case we do not have enough data) andobtain MSE for each case. Best value of is the one which yields smallest MSE.Step 3: Then apply the model with the best value of on the forecasting sample toget MSE of the model and obtain Forecast for the future.

If we did not have enough data, smallest data MSE obtained in step 2 becomes theMSE of the model.

Example 1

Given the following sales data, find the forecast for F13 using Simple exponentialSmoothing Model and obtain the MSE of the model.

t Xt Ft et1 28 30 -22 27 29.8 -2.83 33 29.5 3.54 25 29.9 -4.95 34 29.4 4.66 33 29.9 3.17 35 30.2 4.88 30 30.7 -0.79 33 30.6 2.4

10 35 30.8 4.211 27 31.2 -4.212 29 30.8 -1.813 30.6

As

Assume = 0.1

F1 = 28 + 27 + 33 + 25 + 34 + 33 / 6 = 30

Ft + 1 = Ft + etF2 = F1 + et1

= 30 + 0.1(-2) = 29.8

WARM-UPSAMPLE

ForecastingSample


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F3 = F2 + et2= 29.8 + 0.1(-2.8) = 29.5

F13 = F12 + et12= 30.8 + 0.1(-1.8) = 30.6

Example 2

You are the production manager of Arcelik Corporation in Turkey which specialized inproduction of TV sets among other kinds of durables. You need a forecast for nextquarter in order the plan production process and determine working capital needs. Apply

Simple Exponential Smoothing Model to obtain the forecast for the 2nd Qtr of 2005. Whatis the MSE of the model? Assume that the best value of = 0.5The past actual sales data about TV sales are given below:

Year Qtr Data t Ft Et

2004 1 5000 1 F1 = 5000 02004 2 4800 2 F2 = 5000 -2002004 3 5200 3 F3 = 4900 3002004 4 5100 4 F4 = 5050 502005 1 4900 5 F5 = 5075 -1752005 2 6 F6 = 4987.5

Solution:

F1 = 5000 + 4800 + 5200 + 5100 + 4900 / 5 = 5000F2 = 5000 + 0.5(0) = 5000F3 = 5000 + 0.5(-200) = 4900F4 = 4900 + 0.5(300) = 5050F5 = 5050 + 0.5(50) = 5075F6 = 5075 + 0.5(-175) = 4987.5

MSE= 326255

)175()50()300()200()0(

5

22222

5

1

2

=++

=

=t

et

3.116

)8.1()2.4()2.4()4.2()7.0()9.4(

6

222222

12

7

2

=+++++

==

=t

te

MSE


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Applicat ion of Time Series Regression Model

Salest = a + b (time).Ft = a +bSales is dependent variable while Time is an independent variable.

Least-Square Formulas

=

22 )(tnt

xtntxb tbxa =

b= slope of the linear trend linea= intercept of the linear trend lineX= values of the dependent variable (sales)t = values of the independent variable (time)

Note: You can start the values of t from any number as long as they are consecutive.

X = mean of the X values

t= mean of the t values

Important Points

1) If we have enough data, we use the warm-up sample data for X and t in least-squares formulas to get a and b. If we do not have enough data, we use all the datato get a and b.

2) First we obtain b and then a.3) After we get Ft = a + bt, we apply this equation on the Forecasting Sample to get

MSE of the model. Obtain forecasts for the future periods.

Example

Given the following sales data, find forecasts for January and February, 2002 (F13 andF14) using Time-Series Regression Model and obtain the MSE of this model.


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WARM-UP SAMPLE

Months T Xt Ft et

2001 Jan. 1 60Feb. 2 55March 3 64April 4 51May 5 69June 6 66July 7 83 66.8 16.2

August 8 90 68.5 21.5Sept. 9 76 70.2 5.8Oct. 10 95 71.9 23.1

Nov. 11 72 73.6 -1.6

Dec. 12 88 75.3 12.72002 Jan. 13 77

Formula for Time Series Regression Model:Ft = a + bt

=

22 ntt

xtntxb

tx = (1*60) + (2*55) +( 3*64) + (4*51) +( 5*69) + (6*66) = 1307t2 = 12+ 22+32+42+52 +62= 91

t- =t/ n = 1+2+3+4+5+6 / 6 = 3.5X- = 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) = 56.6F7 = 54.9 + 1.7 (7) = 66.8F13 = 54.9 + 1.7(13) = 77F14 = 54.9 + 1.7(14) = 78.7

FORECASTING SAMPLE


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MSE

e tt

= =+ + + + +

=

27

12

2 2 2 2 2 2

6

16 2 215 58 231 16 12 7

6

( . ) ( . ) ( . ) ( . ) ( . ) ( . )

= 242.6

Application of Smoothing Linear Trend Model (LinearExponential Smoothing)

1) Ft+1 = St + Tt 2) St=Ft + 1et 3) Tt = Tt-1 + 2 et

Note = 0


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Assume 1 =0.1 and 2 = 0.01

F1 = S0 + T0 = a +b = 54.9 + 1.7 = 56.6

F2 = S1 + T1 = 56.9 + 1.7 = 58.6

S1 = F1 + 1 (e1) = 56.6 + 0.1 (3.4) = 56.94=56.9T1 = T0 + 2 (e1) = 1.7 + 0.01(3.4) = 1.734=1.7

F13 = S12 + T12S12 = F12 + 1 (e12) = 84 + 0.1(4) = 84.4

T12 = T11 + 2 (e12) = 2.4 + 0.01(4) = 2.4

F13 = 84.4 + 2.4 = 86.8

MSE = ==

12

7

2

t

et 6

)4()7.8()5.18()8.1()9.19()5.16( 222222 +++++

= 184.2

Months T Xt Ft et2001 Jan. 1 60 56.6 3.4Feb. 2 55 58.6 -3.6March 3 64 59.9 4.1April 4 51 62 -11May 5 69 62.5 6.5June 6 66 64.9 -1.1July 7 83 66.5 16.5August 8 90 70.1 19.9Sept. 9 76 74.2 1.8Oct. 10 95 76.5 18.5Nov. 11 72 80.7 8.7Dec. 12 88 84 42002 Jan. 13 86.8


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CHAPTER 4: DECISION MAKING UNDER UNCERTAINITY

AND RISK

BASIC STEPS IN DECISION MAKINGStep 1: List the feasible Decision Alternatives

Step 2: List the states of nature.States of nature are the most important (critical factors) future events that are not underthe control of decision-maker but which arte likely to affect the payoffs (benefits) thatcan be obtained from the decision alternatives.

Step 3: Calculate the payoffs of the matrix.Pay-off table shows the amount of pay-offs that can be obtained for each possiblecombination of state of nature and decision alternatives.

Alternative Decision Making Environments

Certainty: A manager (decision-maker) knows for sure the state of nature whichwill take place.

Uncertainty: This is where the decision-maker knows only the possible states ofnature.

Risk: This is where the decision-maker knows also the probability of each state of

nature.

Al ternative Cri teria for Decision Making under Uncertainty

Example

Problem: We have a Tape and CD manufacturing company. It is facing increaseddemand beyond its capacity, what is the best (optimal) way of accommodating thisincreasing demand or what is the optimal way of increasing supply to the market?

Solution:

Step 1: List the feasible decision alternatives:

A B C

Expand the Plant Build a new plant Subcontract out extraproduction to the other

companies


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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 million units of sales volume each year.

4. Failure e.g almost zero sales.

Step 3: Prepare the Pay-off Table or pay-off Matrix:

Payoff Matrix

Decision Alternatives

States of Nature Expand Build SubcontractHigh Demand $ 500,000 $ 700,000 $ 300,000

Moderate Demand $ 250,000 $ 300,000 $ 150,000

Low Demand - $ 250,000 - $ 400,000 - $ 10,000Failure - $ 450,000 - $ 800,000 - $ 100,000

N.B: This company measures pay-offs as simple profits over the next five years.

Pay-off Table and Decision Making Criteria

1) The Maxi-Max Criterion

This criterion is preferred by optimistic managers.

Step1: Identify the maximum payoff that can be obtained for each decision alternative.

Max-pay off Expand Build Subcontract$ 500,000 $ 700,000 $ 300,000

Step2: The optimum choice is the alternative which yields the maximum payoff in thislist obtained in step 1.The optimum choice is to build a new plant. Because; $ 700,000>$ 500,000>$ 300,000

2) The Maxi-Min CriterionThis criterion is preferred by pessimistic managers.

Step1: Identify the minimum payoff that can be obtained for each decision alternative.

Min-pay off Expand Build Subcontract- $ 450,000 - $ 800,000 - $ 100,000


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Step2: The optimum choice is the alternative which yields the maximum payoff in thislist 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 such that 0180,000

4) The Mini-Max Regret CriterionStep1: Obtain the regret table:Regret Value = Max Pay off of the row The payoff

Step2: Identify the max regret value for each decision alternative.

Step3: The optimal choice is the one which yields minimum in the list, which has beenobtained in step2.

Example:

Step 1: Re-write the Decisions Alternative Table to come up with the Regret Table:

In the 1st Row:Regret Value for $500,000 = 700,000 500,000 = $200,000Regret Value for $700,000 = 700,000 700,000 = 0Regret 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


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Regret Value for $300,000 = 300,000 300,000 = 0Regret Value for $150,000 = 300,000 150,000 = $150,000

N.B: Follow the same procedure to obtain the regret values for the 3 rd and 4th rows.

Regret Table

Decision Alternatives

States of Nature Expand Build SubcontractHigh Demand $ 200,000 0 $ 400,000

Moderate Demand $ 050,000 0 $ 150,000Low Demand $ 240,000 $ 390,000 0

Failure $ 350,000 $ 700,000 0

Step 2: Write the maximum decision value for each alternative:

Max-RegretTable

Expand Build Subcontract

$ 350,000 $ 700,000 $ 400,000

The optimum choice is to Expand. Because; $ 700,000>$ 400,000>$ 350,000

Applicat ion of A lternative Cr iter ia for Decision Making underRisk

1) Expected Value Criterion

Step1: List the feasible decision alternatives. When state of nature is given as a discreterandom variable; feasible decision alternatives are exactly the same as the states ofnature.

Step2: Obtain the probability of each state of nature using the past data.

Step3: Using the cost and price data, obtain the conditional profit table.

Step4: Using the probabilities obtain the expected value of profits for each decisionalternative.

Step5: Optimal choice is the decision alternative which yields maximum expected profit.


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2) Criterion of Rationality

Sometimes we may not trust the past data that are used to obtain probabilities for states ofnature, under those circumstances we can use this criterion which assumes that all statesof nature are equally likely.We use this criterion when we have limited or unreliable data to calculate probabilities.

The rest of the steps are exactly the same as those of the Expected Value Criterion.

3) The Criterion of Maximum Likelihood

We can use this criterion particularly when one of the states of nature has significantlyhigher probability then others.In this case:Step 1: We assure that this state of nature with the highest probability will actually occur.

Step 2: Given this assumption we chose the decision alternative which yields maximumconditional profits.

Example 1

The Mid Town Food store stocks mangoes during early summer season. These are flownin from Meritt Island, Florida, each Monday and just be sold within the week following.In the past, the store has been experiencing the following sales of mangoes:

Quantities Buyers Bought # of weeks this occurred

20 1025 3040 5060 10

Total = 100

Food Store buys mangoes for $2 and sells them for $4.Given this, apply each of the following criteria to determine the optimal quantity ofmangoes that must be stocked per week:

a. Expected Value Criterionb. Criterion of Maximum Likelihoodc. Criterion of Rationality


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a) Using the Expected Value Criterion:

Step 1: Obtain the probabilities of states of nature

Probabilities of

weekly demand

Quantities Buyers

BoughtStates of Nature 10/100 = 0.1 2030/100 = 0.3 2550/100 = 0.5 4010/100 = 0.1 60

Step 2: Obtain the Feasible Stocking Actions (Decision Alternative)N.B: Obtain exactly the number of quantity demanded from the supplier.

Step 3: Obtain the Conditional Profit tableObtain the weekly profit values for each possible combination of Stocks and Demand:Profit = TR TCTR = P * QsoldTC = C * QstockedP = 4 and C = 2

a) Weekly profits when Stocks = 20units/week and Demand = 20Profit = TR TC = (20*4) (20*2) = 40

b) Weekly profits when Stocks = 25units/week and Demand = 20Profit = TR TC = (20*4) (25*2) = 30c) Weekly profits when Stocks = 40units/week and Demand = 20

Profit = TR TC = (20*4) (40*2) = 0d) 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 andDemands.

Feasible Stocking ActionProb. Demand 20 25 40 60

0.1 20 $ 40 $ 30 0 -$ 40States of nature 0.3 25 $ 40 $ 50 $ 20 -$ 20

0.5 40 $ 40 $ 50 $ 80 $ 400.1 60 $ 40 $ 50 $ 80 $ 120

20 25 40 60


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Loss = 4 2 = $2 / unit

Step 4: Calculate the Expected Values:

E()20 = (0.1*40) + (0.3*40) + (0.5*40) + (0.1*40) = $40E()25 = (0.1*30) + (0.3*50) + (0.5*50) + (0.1*50) = $48E()40 = (0.1*0) + (0.3*20) + (0.5*80) + (0.1*80) = $54E()60 = (0.1*-40) + (0.3*-20) + (0.5*40) + (0.1*120) = $22

Therefore: The optimum choice is to stock 40 units because it yields max. expectedprofit of $54.

b) Using the Criterion of Rationality

Taking = 0.25 as our probability

Feasible Stocking Actions

Prob. Demand 20 25 40 60

0.25 20 $ 40 $ 30 0 -$ 40States of Nature 0.25 25 $ 40 $ 50 $ 20 -$ 20

0.25 40 $ 40 $ 50 $ 80 $ 400.25 60 $ 40 $ 50 $ 80 $ 120

Expected value of profits:

E()20 = (0.25*40) + (0.25*40) + (0.25*40) + (0.25*40) = $40E()25 = (0.25*30) + (0.25*50) + (0.25*50) + (0.25*50) = $45E()40 = (0.25*0) + (0.25*20) + (0.25*80) + (0.25*80) = $45E()60 = (0.25*-40) + (0.25*-20) + (0.25*40) + (0.25*120) = $25

Therefore: The optimal choice is to either stock 25 or 40 units because they yield themaximum expected profit of $45.

c) Using the Maximum Likelihood CriterionWe take the highest probability of 0.5 from the conditional probability table

Feasible Stocking Actions

Prob. 20 25 40 60

0.1 20 $ 40 $ 30 0 -$ 400.3 25 $ 40 $ 50 $ 20 -$ 200.5 40 $ 40 $ 50 $ 80 $ 400.1 60 $ 40 $ 50 $ 80 $ 120


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The optimum choice is to stock 40 units because it yields the maximum expected profitof $80.

Example 2

Suppose that you are the sales manager of Migros supermarket chain in Istanbul. You aretrying to make a decision about the optimal number of bottles of Efes beer that should bestocked for each month. The cost of each bottle to Migros is $6o and the beer is sold for arevenue of $95 per beer. For each bottle that is unsold by the end of the week andreturned back to Efes, Migros receives $20.The sales data for the last 50 weeks is givenas follows:

Quantities Buyers Bought Number of Months this occured20 20

21 1522 1023 5

a. Using the Expected Value Criterion, obtain the optimal number of Efes bottles thesales manager of Migros should stock per month.

b. Using the maximum Likelihood Criterion, obtain the optimal number of Efesbottles the sales manager of Migros should stock per month.

a) Using The Expected Value Criterion

Bottles of Beer Demanded Number of weeks Probability20 20 20/50 = 0.4021 15 15/50 = 0.3022 10 10/50 = 0.2023 5 5/50 = 0.10

Total 50 1.00

Conditional Profit Table:

Profit = TR TC

TR = P * QsoldTC = C * QstockedP = $95 and C = $60Profit Received from the Unsold Bottles = $20

a. Weekly Profit when Stock = 20 bottles/week and Demand = 20Profit = TR TC = (20*95) (20*60) = 700

b. Weekly Profit when Stock = 21 bottles/week and Demand = 20


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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 = 20Profit = (TR TC) + (# of unsold bottles*Profit Received)

= [(20*95) (22*60)] + (2*20) = 620

d. Weekly Profits when Stock = 23 bottles/week and Demand = 20Profit = (TR TC) + (# of unsold bottles*Profit Received)= [(20*95) (23*60)] + (3*20) = 580

Feasible Stock Action

Probabilities Demand 20 21 22 23

0.40 20 700 660 620 580States of Nature 0.30 21 700 735 695 655

0.20 22 700 735 770 7300.10 23 700 735 770 805

E()20 = (0.40*700) + (0.30*700) + (0.20*700) + (0.10*700) = $700E()21 = (0.40*660) + (0.30*735) + (0.20*735) + (0.10*735) = $705E()22 = (0.40*620) + (0.30* 695) + (0.20*770) + (0.10*770) = $687.50E()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. expectedprofit of $705.

b) Using The Maximum Likelihood CriterionWe take the highest probability of 0.40 from the conditional probability table

Feasible Stock Action

Probabilities Demand 20 21 22 23

0.40 20 700 660 620 580States of Nature 0.30 21 700 735 695 655

0.20 22 700 735 770 7300.10 23 700 735 770 805

Therefore: The optimum choice is to stock 20 bottles because it yields max. expectedprofit of $700.


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Supplying the Numbers: How to Obtain Mean andStandard Deviation when Data are Missing or IncompleteThis is a technique used to obtain estimates for mean and standard deviation (ofparticularly demand for a product) when we dont have past data to use in formula for

mean and standard deviation.

X~N (x ,x)

This technique involves asking questions to some experts in the company and thenproceeding their answers in a scientific manner to get estimates for mean and standarddeviation.

For Example:

Q1: In your opinion, what is expected annual volume of sales? Or on average how muchof this product can be sold per year?

Answer: 1500 units/year, your estimate forx Annual Demand

Before you ask the 2nd question, form an interval around x (1,500)bytaking anypercentage of estimates for x such as 20%:

20% * 1500 = 300Then add and subtract 300 from 1500 to get an interval around 1500 given below:

Q2: In your opinion, what are the odds (chances) for the annual sales volume to liebetween 1200 and 1800?


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Answer: Odds are 3 to 1 (3:1)

P(1200< X


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1800-15001.15= xx= 261

Therefore estimates for and = X~N (1500, 261)

Example

Suppose that the production manager of Phillips Electronics company in Netherlandsbelieves that the expected annual volume of sales of the new product models that theyplan to introduce to the market in 2003 is 200,000.Furthermore he believes that the odds

(chances) are 4 to 3 that the annual sales volume will lie between 100,000 and 300,000.What is your best estimate for the standard deviation of annual sales volume of thisproduct model of Phillips?

Solution:

= 200,000Odds = 4:3Range = 100,000:300,000 = ?

3 4 4 3

100,000 200,000 300,000 X


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P(X 300,000) = 3+4+4 / 3+4+4+3 = 11/14 = 0.7857Z* = 0.79Z* = X- /

0.79 = 300,000 200,000 / 0.79 = 100,000 / = 126582

Therefore: = 126582


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Chapter 5: Cost Volume Profit AnalysisWe use Cost-Volume Analysis also called Break-Even Analysis to determine theprofitability and riskness of an investment project. This technique involves asking

probability questions and assessing the risk faced by the company in investing in thisproject, based on the answers to those questions.It also involves computing the expected value of net profit from this project which alsohelps the managers to make more informed investment decisions.

Application: We build our example we used earlier in estimating the mean and thestandard deviation of annual demand for a product that a company is considering toproduce.Note that annual demand (X) was estimated to have a mean() of 1,500 and a standarddeviation() of 261.

X~N (1500, 261)

Data About The Project

If: The cost of capital investment = $100,000 (initial capital)TFC = $50,000P/unit = $10 (Price that the company expects to sell).AVC/unit = $5 (Average variable cost is assumed to be approximately constant for therange of output that the company expects to purchase).

Questions That Managers Can Ask Specific To This ProjectQ1: What is the probability of at least breaking even on this project for each year?Q2: What is the probability of earning more than $900 profit for each year?Q3: What is the probability of losing more than $200 for each year?Q4: What is the probability of earning minimum 2% annual rate of return for each dollar

invested?Q5: What is the expected value of Net Profit on this project?


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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 = 1180P(X > X*) = P(X > 1180)

1180 1500Z* =

261Z* = -1.23

P(X > 1180) = 0.89065


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Answer 3: P(Losing more than $200 for each year)

P(Profit < -200) = P(X < X*)X* = 1000 + (-200 / 5) = 960

960 1500Z* =

261Z* = -2.07

P(Profit


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P(Profit $2000) = 0.64803

Answer 5: Expected Net Profits

E(Net Profit) = E(Annual Profit) E(Annual Loss)

The area < 1000 is Potential Loss [ P(Profit < 0)], and the area > 1000 is Potential Gain[ P(Profit 0), while 1000 is a Break-Even Point.

Example

TT Corporation is discussing to produce a new product. It has been estimated that theaverage annual volume of sales will be 8,000. The manager of the sales departmentbelieves that the odds (chances) are 2 to 3 that the annual volume of sales will liebetween 7,000 and 9,000. If they choose to produce the product, they expect to incur anannual fixed cost of an amount equal to 300,000. The sales price is estimated to be $100and 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:

= 8,000Odds = 2:3TFC = $30,000P = $100VC/unit = $50

3 2 2 3

7,000 8,000 9,000 X

P(X < 9,000) = 2+2+3/3+2+2+3 = 0.70


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Z = 0.52

Z = X - / 0.52 = 9,000 8,000 / = 1923.

a) P(Losing Money)

6,000 8,000 X

Z*=? 0 Z

XBE = TFC / P-V = 300,000 / 100-50 = 6000Z* = 6000 8000 / 1923.08 = -1.04

P(X < XBE) = P(Z < -1.04) = 1 0.85083 (from the probability table)= 0.14917

b) E(Annual Profit)

Since 6,000 < 8,000 (XBE < )E(Annual Profit) = k * * (UNLI + Z)

= 50*1923.08*(o.7716 +1.04) = $174192.59

c) P(At least Breaking-Even)P(X 6,000)


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6,000 8,000 X

Z* =-1.04 0 Z

P(X 6,000) = P(Z > 1.04) = 0.85083

Combining Unit Monetary Values and Probability Distribution

Procedure:a. 3-Step procedure to obtain in Expected Profitb. 3-Step procedure to obtain in Expected Lossc. Obtain Expected Net profit [E(Profit) E(Loss)]

Given:

X~N (1500, 261)

TFC = 5000Standard Deviation = 261Mean = 1500P = 10V = 5XBE = 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**(UNLI +z)b) If XBE > Mean, E(Profit) = k**UNLI

Where k is the profits per unit of output above XBE.


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Note: Sometimes K and C can be different from each other, but if you are given P and Vvalues, then always P-V = K = C

Example

Step 1: |Z| =261

15001000 = |-1.91| = 1.91

Step 2: For Z = 1.91, UNLI = 0.01077, using the UNLI table.

Step 3: Since 1000 < 1500 (XBE < )

E(Profit) = k**(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**(UNLI +z)b) If XBE > Mean, E(Profit) = c**UNLI

Where c is the loss per unit of output above XBE.

|(P - V) = k = c|

Example

Step 1: |Z| = |-1.91| = 1.91

Step 2: UNLI = 0.01077 for Z = 1.91

Step 3: Since 1000 < 1500 (XBE < )

E(Loss) = c**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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Example

Girne Septic Tank Company operates a truck called the Honey Cart which is used topump out the capacitated septic tanks. The company has estimated that usage of the truckis normally distributed with a mean of 150 hours per month and a standard deviation of40 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:

= 150 = 40C = $30

K = $28XBE = 90

|Z| = |XBE - / | = |90 150 / 40| = |-1.5| = 1.5

For Z = 1.5, UNLI = 0.02931

a) Expected Loss = C * * UNLI= 30 * 40 * 0.02931 = $35.17 per month

b) Expected Profit = K * * (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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Home works

Home Work 1

Questions from the textbook:Chapter 2: Ex 2-7, 2-8, 2-9, 2-12, 2-40, 2-44*

Home work 2

1. Suppose we toss a coin 5times, and it is an unbiased coina. 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 stockholders equity for all the companieslisted on the New York Stock Exchange. She found that the mean of this distribution was10%, with a standard deviation of 5%. She is interested in examining further thosecompanies whose return on stockholders equity is between 8% and 13%. Of theapproximately 13,000 companies listed on the exchange, how many are of interest to her?

3. Questions from the textbook:Chapter 2: Ex 2-32, 2-36, 2-37, 2-38

Home Work 3

1. Following is the sales data gathered for the past 12 months of a company:

Years Months Sales

1999 August 281999 September 271999 October 331999 November 251999 December 342000 January 332000 February 352000 March 302000 April 332000 May 352000 June 27

2000 July 29

Apply the following to the above data set to obtain MSE and develop a forecast forAugust 2000 for each case:

a. 3-period (W) MA model, taking Wt = 0.5, Wt-1 = 0.3, Wt-2 = 0.2b. 4-period (UW) MA model


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c. Use Simple Exponential Smoothing model, taking = 0.5 and obtain MSE of themodel and develop a forecast for August 2000.

d. Which specification above said has theHighest 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 4periods not 6 periods.

Ex 3-25

Home Work 4

1. Use Smoothing Linear Trend model to develop a forecast for August 2000 forQuestion 1 in Homework-3, taking 1 = 0.1 and 2 = 0.01. Also obtain MSE of thismodel.

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 3-23should 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 tomorrowsdemand 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 perishabilityof the item and the great risk of stoking it; the product has no value after the first day it isoffered for sale. Beth faces the problem of how many to order today for tomorrowsbusiness.

A 90-day observation of past demand gives the information shown in the table below:

Daily Demand Number of days Demanded10 18

11 3612 2713 19

What quantity should Beth Perry buy for tomorrows sale to maximize the ExpectedProfit?

N.B: Read the textbook; page 165-168.


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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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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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QUIZ QUESTIONS


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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.Cs 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 CIMSAs stock price will rise in a given day is estimated to

be 0.60. When CIMSAs stock price rises, there is a 0.70 probability that theexchange rate of TL/$ will fall in that day. When CIMSAs 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 CIMSAs 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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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?

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