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CAIIB -Financial Management

Module A -QuantitativeTechniques and Business

Mathematics

Madhav K PrabhuM.Tech, MIM, PMP, CISA, CAIIB, CeISB, MCTS, DCL



Agenda

• Time Value of Money

• Bond Valuation Theory

• Sampling• Regression and Correlation



Time Value of Money



Objectives

• What do we mean by Time value of money

• Present Value, Discounted Value, Annuity



Time Value of Money

• What is Time Value of Money?

– Future Value

– Present Value

• Future Value: Compounding:

Principal P 20,000 20,000 20,000

Interest Rate i 10% 10% 10%

No. of Years n 1 2 3

Future Value FV 22,000 24,200 26,620

Interest Amount 2,000 2,200 2,420

Assuming Compounding Done Annually

How would you

do

Compounding?



Compounding

• Compounding Formula

• What if compounding is done on monthly basis?

n

n i P FV )1(*

t n

nt

i P FV

*

1*

Principal P 20,000 20,000 20,000


No. of Years n 1 2 3

Times Compounding in a Year t 12 12 12

Maturity Value FV 22,094 24,408 26,964

Interest Amount 2,094 4,408 6,964

Assuming Compounding Done Monthly



Compounding Exercise

• Exercise:

– Prepare a table showing compounding as per

following conditions:

– Rate of Interest - 5%, 12% and 15%

– Compounding 2 & 4 times in a year

– Principal Rs.100,000/-



Discounting

• Present Value

– You have an option to receive Rs. 1,000/- either today or after

one year. Which option you will select? Why?

– Decision will depend upon the present value of money; which

can be calculated by a process called Discount ing (oppo si te ofCompound ing)

– Interest Rate and Time of Receipt of money decide Present

Value

– What is the present value of Rs. 1,000/- today and a year later?

To compute Present Value?



Discounting contd…

• Formula to find Present Value of Future Cash Receipt

– Where PV = Present Value, P = Principal, i = Rate of Interest, n = Number

of Years after which money is received

• Assuming Rate of Interest is 10%, value of Rs. 1,000/- to be received

after 1 year will be,

• Whereas the value of money to be received today will be Rs. 1,000/-

nn

i

P PV

1

1101

100009909

%.

What if you were to choose between:

a. Receive Rs. 1,000/- every year for 3 years, OR

b. Receive Rs. 2,500/- today? (assume 10% annual in terest rate)



Discounting of a Series contd…

• How discounting is done for a series of cashflow? e.g.

– Receive Rs. 1,000/- at the end of every year for 3 years OR

– Receive Rs. 2,500/- today

– Assume Rate of Interest @10%

Principal P 20,000 20,000 20,000


Year n 1 2 3

Present Value PV 18,181.82 16,528.93 15,026.30

Assuming Discounting Done Annually

I f cashf low was to occur every 6 months instead of 1 year, what impact

it wil l have on Present Value?



Periodic Discounting

• What if the receipts are over six months’

interval ? Find Present Value of the money

receipts

• Periodic Discounting Formula

– Receive Rs. 1,000/- at the end of every 6 months for 1-1/2 years OR

– Receive Rs. 2,600/- today

– Assume Rate of interest @10%

n

t

i

P PV

1Where, P = Principal, i = Rate of

Interest,

t = Times Payments made in a Year,

n = nth Period (in this case it is half

year)



Periodic Discounting Formula

Principal P 1,000 1,000 1,000


HY n 1 2 3

Times Discounting in a Year t 2 2 2 Discount Factor DF 0.9524 0.9070 0.8638

Present Value PV=P*DF 952.38 907.03 863.84

Sum of Present Value

Assuming Discounting Done Semi-Annually

2,723.25

321

2

%101

1000

2

%101

1000

2

%101

100025.2723

Expressed mathematically, the equation will look like:

Generically expressed,

the formula is:

Here, N = 3

N

nn

n

t

i

x

SUMofPV

11



Charting of Cashflow

• For any financial proposition prepare a chart of cashflow: e.g.

Invested in 10% Bonds 01-Jan-04 (1,000) Outflow

Interest received 30-Jun-04 50 Inflow

Interest received 31-Dec-04 50 Inflow

New Bond Purchased from

Open Market

31-Dec-04 (1,020) Outflow

Interest received 30-Jun-05 100 Inflow

Sold Bond in Open Market 30-Jun-05 2,050 Inflow

Timeline

01.01.0

4

Invested in Bonds

(1,000)

30.06.04

Interest Received +50

31.12.04

Interest Received + 50

New Bond Purchased (1,020)

Net ( 970)

30.06.05


Sold Bond +2,050

Total +2,150



Net Present Value• Net Present Value means the dif ferenc e between the PV of Cash Inf low s &

Cash Outf lows

• How do yo u compute NPV?

– Prepare Cashflow Ch art

– Net of f Inf low & Outf low for each p eriod separately

• I f Inf low > Outf low, pos it ive cash

• I f Inf low < Outf low, negative cash

• Find present values of Inf lows & Outf lows by apply ing Disco unt Factor (or

Present Value Factor)

• NPV = (PV of Inf low s) LESS (PV of Ou tf low s); Result can be +ve OR -ve

• Cont inuing w ith our examp le of Bond Investment:

Timeline01.01.04

Invested in Bonds

(1,000)

30.06.04

Interest Received +50

31.12.04


New Bond Purchased (1,020)

Net ( 970)

30.06.05


Sold Bond +2,050

Total +2,150

Inflow

Outflow



NPV contd…

• If Cashflows are discounted at say 10%, the sum of PV is 25.05, a positivenumber & therefore the IRR has be higher than 10% to make Net Present

Value to zero

What is IRR?

Description Date Amount In / Out PV Outflow PV Inflow

Invested in 10% Bonds 01-Jan-04 (1,000) Outflow (1,000.00)

Interest received 30-Jun-04 50 Inflow 47.62

Interest received 31-Dec-04 50 Inflow 45.35 New Bond Purchased from

Open Market31-Dec-04 (1,020) Outflow (925.17)

Interest received 30-Jun-05 100 Inflow 86.38

Sold Bond in Open Market 30-Jun-05 2,050 Inflow 1,770.87

Sum (1,925.17) 1,950.22

Net Present Value 25.05

How these values are arr ived at?



Internal Rate of Return (IRR)

• Definition: The Rate at which the NPV is Zero. It can also be termed

as “ Effective Rate ”

• If we want to find out IRR of the bond investment cashflow:

Description DateComposit

Flow

Invested in Bonds 01-Jan-04 (1,000)

Interest received 30-Jun-04 50

Interest received + New Bond

Purchased31-Dec-04 (970)

Interest received + Sold Bond 30-Jun-05 2,150

11.38%IRR of entire cashflow



IRR Contd…

• To prove that at IRR of 11.38% the NPV of Investment Cashflow

is zero, see the formula & table:

3210

2

%38.11

1

2150

2

%38.11

1

970

2

%38.11

1

50

2

%38.11

1

10000

Description DateComposit

FlowPV Factor

NPV at

IRR

Invested in Bonds 01-Jan-04 (1,000) 1.00000 (1,000.00)

Interest received 30-Jun-04 50 0.94615 47.31 Interest received +

New Bond Purchased31-Dec-04 (970)

0.89520(868.34)

Interest received +

Sold Bond30-Jun-05 2,150

0.846991,821.04

11.38% Sum of PVs 0.00 IRR of entire cashflow



IRR - Additional Example

• You buy a car costing Rs. 600,000/-

• Banker is willing to finance upto Rs. 500,000/-• The loan is repayable over 3 years, in Equated

Monthly Installments (EMI) of Rs. 15,000/-

• Installments are payable In A rrears

• What is the IRR?

• How do you express this mathematically? What are

the values of each component in the formula?

• What will be the impact on IRR if the EMIs are

payable In Advance ?

• Can we use IRR for computing Interest & Principal

break-up?



IRR - Additional Example contd…

• Plot the cashflow:

– EMI in Arrears

01.01.200

6

+500,000 01.02.200

6

-15,000

01.03.200

6

-15,000

… … … 01.04.200

6

-15,000

01.11.200

8

-15,000

01.12.200

8

-15,000… … …

Begin

End

Value of ‘i ’

to be

determined

36

1n n

12

i1

n15,000500,000

Formula

Expression

Values in Expression

N

nn

n

t

i x P

11

1 2 3 35 36



IRR - Additional Example contd…

• Plot the cashflow:

– EMI in Advance

01.01.200

6

+500,000 01.02.200

6

-15,000

01.03.200

6

-15,000

… … … 01.04.200

6

-15,000

01.12.200

8

-15,000

01.01.200

9

-15,000… … …

Begin

End

Value of ‘i ’

to be

determined

36

2n n

12

i1

n15,000

115,000-500,000

Formula

Expression

Values in Expression

N

n

n

n

t

i

x X P 2

1

1

1 2 3 35 36

-15,000



BOND VALUATION



Objectives

• Distinguish bond’s coupon rate, current

yield, yield to maturity

• Interest rate risk

• Bond ratings and investors demand for

appropriate interest rates



Bond characteristics

• Bond - evidence of debt issued by a body

corporate or Govt. In India, Govt predominantly

– A bond represents a loan made by investors to the

issuer. In return for his/her money, the investorreceives a legaI claim on future cash flows of the

borrower.

– The issuer promises to:

• Make regular coupon payments every period until the bond

matures, and

• Pay the face/par/maturity value of the bond when it matures



How do bonds work?

• If a bond has five years to maturity, an Rs.80 annual coupon, and aRs.1000 face value, its cash flows would look like this:

• Time 0 1 2 3 4 5

• Coupons Rs.80 Rs.80 Rs.80 Rs.80 Rs.80

• Face Value 1000• Market Price Rs.____

• How much is this bond worth? It depends on the level of currentmarket interest rates. If the going rate on bonds like this one is 10%,then this bond has a market value of Rs.924.18. Why?



nr

F I

r

I

r

I PV

bond a for formulaGeneral

)1()1(1

:

2

55432 )10.01(

1000

)10.01(

80

)10.01(

80

)10.01(

80

)10.01(

80

10.01

80)(

bond of price PV

Coupon payments Face valueMaturity

Annuity componentLump sumcomponent



Bond prices and Interest Rates

• Interest rate same as coupon rate

– Bond sells for face value

• Interest rate higher than coupon rate

– Bond sells at a discount

• Interest rate lower than coupon rate

– Bond sells at a premium



Bond terminology

• Yield to Maturity

– Discount rate that makes present value of

bond’s payments equal to its price

• Current Yield

– Annual coupon divided by the current

market price of the bond

Current yield = 80 / 924.18 = 8.66%



Rate of return

• Rate of return

= Coupon income + price change

----------------------------------------

Investment

e.g. you buy 6 % bond at 1010.77 and sell next

year at 1020

Rate of return = 60+9.33/1010.77 = 6.86%



Risks in Bonds

• Interest rate risk

– Short term v/s long term

• Default risk

– Default premium



Bond pricing

• The following statements about bond pricing are always true.

– Bond prices and market interest rates move in oppositedirections.

– When a bond’s coupon rate is (greater than / equal to / lessthan) the market’s required return, the bond’s

market value will be (greater than / equal to / less than) its parvalue.

– Given two bonds identical but for maturity, the price of thelonger-term bond will change more (in percentage terms) than thatof the shorter-term bond, for a given change in market interestrates.

– Given two bonds identical but for coupon, the price of thelower-coupon bond will change more (in percentage terms) thanthat of the higher-coupon bond, for a given change in marketinterest rates.



SAMPLING



Objectives

• Distinguish sample and population

• Sampling distributions

• Sampling procedures• Estimation – data analysis and

interpretation

• Testing of hypotheses – one sample data• Testing of hypotheses – two sample data



Pouplation and Sample

Population Sample

Definition Collection of items being

considered

Part or portion of

population chosen forstudy

Characteristics and

Symbols

Parameters

Population size = N

Population mean = m Population standard

deviation = s

Statistics

Sample size = n

Sample mean = xSample standard deviation

= S



Types of sampling

• Non random or judgement

• Random or probability



Methods of sampling

• Sampling is the fundamental method of inferring

information about an entire population without going to

the trouble or expense of measuring every member of

the population. Developing the proper sampling

technique can greatly affect the accuracy of your results.



Random sampling

• Members of the population are chosen in

such a way that all have an equal chance

to be measured.

• Other names for random sampling include

representative and proportionate

sampling because all groups should be

proportionately represented .



Types of Random sampling

• Simple random sampling• Systematic Sampling: Every k th member of the

population is sampled.

• Stratified Sampling: The population is divided into twoor more strata and each subpopulation is sampled(usually randomly).

• Cluster Sampling: A population is divided into clustersand a few of these (often randomly selected) clusters areexhaustively sampled.

• Stratified v/s cluster – Stratified when each group has small variation withn itself but if

there is wide variation between groups

– Cluster when there is considerable variation within each groupbut groups are similar to each other



Sampling from Normal Populations

• Sampling Distribution of the mean

• the probability distribution of

sample means, with all

samples having the same sample size n.

• Standard error of mean for infinite populations

s

x

= sn

1/2

• Standard Normal probability distribution



• Density Curve (or probability density function)the graph of a continuous probability distribution

– The total area under the curve must equal 1.

– Every point on the curve must have a vertical height that is 0 orgreater.

Definitions



Because the total area under the

density curve is equal to 1,

there is a correspondence betweenarea and probability.



Definition

Standard Normal Deviation a normal probability distribution that has a

mean of 0 and a standard deviation of 1



Definition

Standard Normal Deviationa normal probability distribution that has a

mean of 0 and a standard deviation of 1

0 1 2 3-1-2-3 0 z = 1.58

Area = 0.3413Area

0.4429

Score (z )



Table A-2

Standard Normal Distributionµ = 0

s

= 1

0 x

z

Table for Standard Normal (z) Distribution



.0239

.0636

.1026

.1406.1772

.2123

.2454

.2764

.3051

.3315

.3554

.3770

.3962.4131

.4279

.4406

.4515

.4608

.4686

.4750

.4803

.4846.4881

.4909

.4931

.4948

.4961

.4971

.4979

.4985

.4989

0.00.10.2

0.30.40.50.60.70.80.91.01.1

1.21.31.41.51.61.71.81.92.0

2.12.22.32.42.52.62.72.82.9

3.0

.0000

.0398

.0793

.1179.1554

.1915

.2257

.2580

.2881

.3159

.3413

.3643

.3849.4032

.4192

.4332

.4452

.4554

.4641

.4713

.4772

.4821.4861

.4893

.4918

.4938

.4953

.4965

.4974

.4981

.4987

.0040

.0438

.0832

.1217.1591

.1950

.2291

.2611

.2910

.3186

.3438

.3665

.3869.4049

.4207

.4345

.4463

.4564

.4649

.4719

.4778

.4826.4864

.4896

.4920

.4940

.4955

.4966

.4975

.4982

.4987

.0080

.0478

.0871

.1255.1628

.1985

.2324

.2642

.2939

.3212

.3461

.3686

.3888.4066

.4222

.4357

.4474

.4573

.4656

.4726

.4783

.4830.4868

.4898

.4922

.4941

.4956

.4967

.4976

.4982

.4987

.0120

.0517

.0910

.1293.1664

.2019

.2357

.2673

.2967

.3238

.3485

.3708

.3907.4082

.4236

.4370

.4484

.4582

.4664

.4732

.4788

.4834.4871

.4901

.4925

.4943

.4957

.4968

.4977

.4983

.4988

.0160

.0557

.0948

.1331.1700

.2054

.2389

.2704

.2995

.3264

.3508

.3729

.3925.4099

.4251

.4382

.4495

.4591

.4671

.4738

.4793

.4838.4875

.4904

.4927

.4945

.4959

.4969

.4977

.4984

.4988

.0199

.0596

.0987

.1368.1736

.2088

.2422

.2734

.3023

.3289

.3531

.3749

.3944.4115

.4265

.4394

.4505

.4599

.4678

.4744

.4798

.4842.4878

.4906

.4929

.4946

.4960

.4970

.4978

.4984

.4989

.0279

.0675

.1064

.1443.1808

.2157

.2486

.2794

.3078

.3340

.3577

.3790

.3980.4147

.4292

.4418

.4525

.4616

.4693

.4756

.4808

.4850.4884

.4911

.4932

.4949

.4962

.4972

.4979

.4985

.4989

.0319

.0714

.1103

.1480.1844

.2190

.2517

.2823

.3106

.3365

.3599

.3810

.3997.4162

.4306

.4429

.4535

.4625

.4699

.4761

.4812

.4854.4887

.4913

.4934

.4951

.4963

.4973

.4980

.4986

.4990

.0359

.0753

.1141

.1517.1879

.2224

.2549

.2852

.3133

.3389

.3621

.3830

.4015.4177

.4319

.4441

.4545

.4633

.4706

.4767

.4817

.4857.4890

.4916

.4936

.4952

.4964

.4974

.4981

.4986

.4990

*

*

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09z

Table for Standard Normal (z ) Distribution



Example: If a data reader has an average (mean)

reading of 0 units and a standard deviation of 1 unit and if

one data reader is randomly selected, find the probabilitythat it gives a reading between 0 and 1.58 units.

That is 44.29% of the readings between 0 and1.58 degrees.

0 1.58

Area = 0.4429

P ( 0 < x < 1.58 ) = 0.4429



Central Limit Theorem

1. The random variable x has a distribution (which

may or may not be normal) with mean µ and

standard deviations

.

2. Samples all of the same size n are randomly

selected from the population of x values.



Central Limit Theorem

1. The distribution of sample x will, as the

sample size increases, approach a normal

distribution.2. The mean of the sample means will be the

population mean µ.

3. The standard deviation of the sample meanswill approach s

n



Practical Rules Commonly Used:

1. For samples of size n larger than 30, the distribution of

the sample means can be approximated reasonably well

by a normal distribution. The approximation gets better

as the sample size n becomes larger.

2. If the original population is itself normally distributed,

then the sample means will be normally distributed for

any sample size n (not just the values of n larger than 30).



REGRESSION -

CORRELATION



Objectives

• Relationship between two or more

variables

• Scatter diagrams

• Regression analysis

• Method of least squares

Regression



RegressionDefinition• Regression Equation

Regression



RegressionDefinition• Regression Equation

Given a collection of paired data, the regressionequation

• Regression Line(line of best fit or least-squares line)

the graph of the regression equation

y = b0 + b1x^

algebraically describes the relationship between thetwo variables

Th R i E ti



The Regression Equation

x is the independent variable(predictor variable)

y is the dependent variable

(response variable)

^

y = b 0 +b 1x^

y = mx +b

b 0 = y - intercept

b 1 = slope



Notation for Regression

Equation

y-intercept of regression equation

0 b0

Slope of regression equation1 b1

Equation of the regression line y =0 +

1 x y = b0 + b1

PopulationParameter

SampleStatistic

x^



Assumptions

1. We are investigating only linear relationships.

2. For each x value, y is a random variable

having a normal (bell-shaped) distribution.

All of these y distributions have the samevariance. Also, for a given value of x, the

distribution of y-values has a mean that lies

on the regression line. (Results are notseriously affected if departures from normal

distributions and equal variances are not too

extreme.)



Definition

• Correlation

exists between two variables

when one of them is related to

the other in some way



Assumptions

1. The sample of paired data (x ,y ) is a

random sample.

2. The pairs of (x ,y ) data have a

bivariate normal distribution.



Definition

• Scatterplot (or scatter diagram)

is a graph in which the paired (x ,y )

sample data are plotted with a

horizontal x axis and a vertical y

axis. Each individual (x ,y ) pair isplotted as a single point.

P iti Li C l ti



Positive Linear Correlation

x x

y y y

x

(a) Positive (b) Strongpositive

(c) Perfectpositive

Negati e Linear Correlation



Negative Linear Correlation

x x

y y y

x

(d) Negative (e) Strongnegative (f) Perfectnegative

No Linear Correlation



No Linear Correlation

x x

y y

(g) No Correlation (h) Nonlinear Correlation



TIME SERIES



Objectives

• Understanding four components of time

series

• Compute seasonal indices

• Regression based techniques



Time series

• Group of data or statistical information

accumulated at regular intervals



Variations in Time series

• Secular trend – A persistent trend in a single direction. A market movement over

the long term which does not reflect cyclical seasonal ortechnical factors.

• Cyclical fluctuation

– The term business cycle or economic cycle refers to thefluctuations of economic activity (business fluctuations) aroundits long-term growth trend. The cycle involves shifts over timebetween periods of relatively rapid growth of output (recoveryand prosperity), and periods of relative stagnation or decline(contraction or recession).

• Seasonal variation – Pattern of change within a year

• Irregular variation – Unpredictable, changing in a random manner

http://glossary.reuters.com/?title=Cyclical

http://en.wiktionary.org/wiki/cycle

http://en.wikipedia.org/wiki/Recession

http://en.wikipedia.org/wiki/Recession

http://en.wiktionary.org/wiki/cycle

http://glossary.reuters.com/?title=Cyclical



Trend analysis

• To describe historical patterns

• Past trends will help us project future



LINEAR PROGRAMMING



Objectives

• Understanding Linear programming basics

• Graphic and Simplex methods



Linear Programming

• Problem formulation if

– All equations are linear

– Constraints are known and deterministic

– Variables should have non negative values

– Decision values are also divisible



Types of LP problems

• Maximisation

• Minimisation

• Transportation

• Decision making



Multiple Choice Questions



1. If A invests Rs. 24 at 7 % interest rate for

5 years, total value at end of five years is

a. 31.66

b. 33.66

c. 36.66

d. 39.66



1. If A invests Rs. 24 at 7 % interest rate for

5 years, total value at end of five years is

a. 31.66

b. 33.66

c. 36.66

d. 39.66



• What is the effective annual rate of 12%

compounded semiannually?

A) 11.24%

B) 12.00%

C) 12.36%

D) 12.54%



• What is the effective annual rate of 12%compounded semiannually?

A) 11.24%

B) 12.00%

C) 12.36% *

D) 12.54%



• What is the effective annual rate of 12%compounded continuously?

A) 11.27%

B) 12.00%

C) 12.68%

D) 12.75%



• What is the effective annual rate of 12%compounded continuously?

A) 11.27%

B) 12.00%

C) 12.68%

D) 12.75% *



• A study is done to see if there is a linearrelationship between the life expectancy of

an individual and the year of birth. The

year of birth is the ______________.• A. Unable to determine

• B. dependent variable

• C. independent variable



• A study is done to see if there is a linearrelationship between the life expectancy of

an individual and the year of birth. The

year of birth is the ______________.• A. Unable to determine

• B. dependent variable

• C. independent variable *



• Which of the following is an example of usingstatistical sampling?a. Statistical sampling will be looked upon by thecourts as providing superior audit evidence.b. Statistical sampling requires the auditor tomake fewer judgmental decisions.

• c. Statistical sampling aids the auditor inevaluating results.d. Statistical sampling is more convenient to use

than nonstatistical sampling.



• Which of the following is an example of usingstatistical sampling?a. Statistical sampling will be looked upon by thecourts as providing superior audit evidence.b. Statistical sampling requires the auditor tomake fewer judgmental decisions.*c. Statistical sampling aids the auditor inevaluating results.d. Statistical sampling is more convenient to use

than nonstatistical sampling.



• Which of the following best illustrates the concept ofsampling risk?a. An auditor may select audit procedures that are notappropriate to achieve the specific objective.b. The documents related to the chosen sample may not

be available for inspection.c. A randomly chosen sample may not be representativeof the population as a whole.d. An auditor may fail to recognize deviations in thedocuments examined.



• Which of the following best illustrates the concept ofsampling risk?a. An auditor may select audit procedures that are notappropriate to achieve the specific objective.b. The documents related to the chosen sample may not

be available for inspection.c. A randomly chosen sample may not be representativeof the population as a whole.*d. An auditor may fail to recognize deviations in thedocuments examined.



• The advantage of using statistical samplingtechniques is that such techniques

a. Mathematically measure risk.

• b. Eliminate the need for judgmental decisions.c. Are easier to use than other sampling

techniques.

d. Have been established in the courts to be

superior to nonstatistical sampling.



• The advantage of using statistical samplingtechniques is that such techniques

a. Mathematically measure risk. *

b. Eliminate the need for judgmental decisions.

c. Are easier to use than other sampling

techniques.

d. Have been established in the courts to be

superior to nonstatistical sampling.



• Time series methodsa. discover a pattern in historical data and

project it into the future.

b. include cause-effect relationships.c. are useful when historical information is

not available.

d. All of the alternatives are true.



• Time series methodsa. discover a pattern in historical data and

project it into the future.

b. include cause-effect relationships.c. are useful when historical information is

not available.

d. All of the alternatives are true.



• Gradual shifting of a time series over along period of time is called

a. periodicity.

b. cycle.c. regression.

d. trend.



• Gradual shifting of a time series over along period of time is called

a. periodicity.

b. cycle.c. regression.

d. trend. *



• Seasonal componentsa. cannot be predicted.

b. are regular repeated patterns.

c. are long runs of observations above orbelow the trend line.

d. reflect a shift in the series over time.



• Seasonal componentsa. cannot be predicted.

b. are regular repeated patterns. *

c. are long runs of observations above orbelow the trend line.

d. reflect a shift in the series over time.



• Short-term, unanticipated, andnonrecurring factors in a time series

provide the random variability known as

a. uncertainty.b. the forecast error.

c. the residuals.

d. the irregular component.



• Short-term, unanticipated, andnonrecurring factors in a time series

provide the random variability known as

a. uncertainty.b. the forecast error.

c. the residuals.

d. the irregular component.*



• The focus of smoothing methods is tosmooth

a. the irregular component.

• b. wide seasonal variations.c. significant trend effects.

d. long range forecasts.



• The focus of smoothing methods is tosmooth

a. the irregular component. *

b. wide seasonal variations.c. significant trend effects.

d. long range forecasts.



• . Linear trend is calculated as Tt = 28.5+ .75t. The trend projection for period 15

is

a. 11.25b. 28.50

c. 39.75

d. 44.25



• . Linear trend is calculated as Tt = 28.5+ .75t. The trend projection for period 15

is

a. 11.25b. 28.50

c. 39.75*

d. 44.25



• The forecasting method that is appropriatewhen the time series has no significanttrend, cyclical, or seasonal effect isa. moving averages

• b. mean squared errorc. mean average deviationd. qualitative forecasting methods



• The forecasting method that is appropriatewhen the time series has no significant

trend, cyclical, or seasonal effect is

a. moving averages *b. mean squared error

c. mean average deviation

d. qualitative forecasting methods



Thank You

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