CAIIB -Financial Management
Module A -Quantitative
Techniques and Business
Mathematics
Madhav K Prabhu
M.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:
Assuming Compounding Done Annually
Principal
P
Interest Rate
i
No. of Years
n
Future Value
FV
Interest Amount
20,000
10%
20,000
10%
20,000
10%
1
2
3 How would you
22,000
24,200
2,000
2,200
26,620 do
2,420 Compounding?
Compounding
• Compounding Formula
FVn  P * (1  i ) n
• What if compounding is done on monthly basis?
 i
FVn  P * 1  
 t
n*t
Assuming Compounding Done Monthly
Principal
P
20,000
i
No. of Years
n
1
2
3
Times Compounding in a Year
t
12
12
12
22,094
24,408
26,964
2,094
4,408
6,964
Interest Amount
FV
10%
20,000
Interest Rate
Maturity Value
10%
20,000
10%
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 Discounting (opposite of
Compounding)
– 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
PVn 
P
1  i n
– 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,
1000
909.09 
1  10%1
• Whereas the value of money to be received today will be Rs. 1,000/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 interest 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%
Assuming Discounting Done Annually
Principal
P
20,000
20,000
Interest Rate
i
10%
10%
Year
n
1
2
Present Value
PV
18,181.82
16,528.93
20,000
10%
3
15,026.30
If cashflow was to occur every 6 months instead of 1 year, what impact
it will have on Present Value?
Periodic Discounting
• What if the receipts are over six months’
interval ? Find Present Value of the money
receipts
– 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%
• Periodic
Discounting
Formula
P
PV 
 i
1  
 t
n
Where, 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
Expressed mathematically, the equation will look like:
2723 .25 
1000
1
 10 % 
1 

2 


1000
 10 % 
1 

2 

2

1000
 10 % 
1 

2 

3
Generically expressed,
the formula is:
SUMofPV
N
xn
Assuming Discounting Done Semi-Annually

n
Principal
P
1,000
1,000
1,000
i
n 1 

1



Interest Rate
i
10%
10%
10%
t


HY
n
1
2
3
Times Discounting in a Year
t
2
2
2 Here, N = 3
Discount Factor
DF
0.9524 0.9070 0.8638
Present Value
PV=P*DF 952.38 907.03 863.84
2,723.25
Sum of Present Value

Charting of Cashflow
• For any financial proposition prepare a chart of cashflow: e.g.
Invested in 10% Bonds
Interest received
Interest received
New Bond Purchased from
Open Market
Interest received
Sold Bond in Open Market
01-Jan-04
30-Jun-04
31-Dec-04
(1,000) Outflow
50 Inflow
50 Inflow
31-Dec-04
(1,020) Outflow
30-Jun-05
30-Jun-05
100
2,050
Inflow
Inflow
Interest Received +50
01.01.0
4
+ 100
+2,050
+2,150
31.12.04
Timeline
30.06.04
Invested in Bonds
(1,000)
Interest Received
Sold Bond
Total
30.06.05
Interest Received
New Bond Purchased
Net
+ 50
(1,020)
( 970)
Net Present Value
•
•
Net Present Value means the difference between the PV of Cash Inflows &
Cash Outflows
How do you compute NPV?
–
–
Prepare Cashflow Chart
Net off Inflow & Outflow for each period separately
•
•
•
•
•
If Inflow > Outflow, positive cash
If Inflow < Outflow, negative cash
Find present values of Inflows & Outflows by applying Discount Factor (or
Present Value Factor)
NPV = (PV of Inflows) LESS (PV of Outflows); Result can be +ve OR -ve
Continuing with our example of Bond Investment:
Interest Received
Sold Bond
Total
Inflow
Interest Received +50
01.01.0
4
31.12.04
Timeline
30.06.04
Invested in Bonds
(1,000)
Outflow
30.06.05
Interest Received
New Bond Purchased
Net
+ 100
+2,050
+2,150
+ 50
(1,020)
( 970)
NPV
•
contd…
If Cashflows are discounted at say 10%, the sum of PV is 25.05, a positive
number & therefore the IRR has be higher than 10% to make Net Present
Value to zero
Description
Invested in 10% Bonds
Interest received
Interest received
New Bond Purchased from
Open Market
Interest received
Sold Bond in Open Market
How these values are arrived at?
Date
Amount In / Out
01-Jan-04
(1,000) Outflow
30-Jun-04
50
Inflow
31-Dec-04
50
Inflow
31-Dec-04
30-Jun-05
30-Jun-05
(1,020) Outflow
100
2,050
PV Outflow PV Inflow
(1,000.00)
47.62
45.35
(925.17)
Inflow
Inflow
Sum
(1,925.17)
Net Present Value
86.38
1,770.87
1,950.22
25.05
What is IRR?
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
Composit
Flow
01-Jan-04 (1,000)
30-Jun-04
50
Date
Invested in Bonds
Interest received
Interest received + New Bond
31-Dec-04
Purchased
Interest received + Sold Bond
30-Jun-05
IRR of entire cashflow
(970)
2,150
11.38%
IRR
Contd…
• To prove that at IRR of 11.38% the NPV of Investment Cashflow
is zero, see the formula & table:
0
1000
 11 .38 % 
1 

2


Description
0

50
1
 11 .38 % 
1 

2


Date
Invested in Bonds
01-Jan-04
Interest received
30-Jun-04
Interest received +
31-Dec-04
New Bond Purchased
Interest received +
30-Jun-05
Sold Bond
IRR of entire cashflow

970
 11 .38 % 
1 

2


Composit
Flow
(1,000)
50
(970)
2,150
11.38%
2

PV Factor
2150
 11 .38 % 
1 

2


NPV at
IRR
1.00000 (1,000.00)
0.94615
47.31
0.89520
0.84699
Sum of PVs
(868.34)
1,821.04
0.00
3
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 Arrears
• 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
Begin
1
2
3
+500,000
01.02.200
6
01.03.200
6
01.04.200
6
01.01.200
6
-15,000
-15,000
-15,000
n 1
………
36
01.11.200
8
01.12.200
8
-15,000
-15,000
End
Formula
Expression
N
P
………
35
Values in Expression
xn
 i
1  
 t
n
36 15,000n
500,000  
n
n 1
 i 
1  
 12 
Value of ‘i’
to be
determined
IRR - Additional Example
contd…
• Plot the cashflow:
– EMI in Advance
Begin
1
2
3
+500,000
01.02.200
6
01.03.200
6
01.04.200
6
-15,000
-15,000
-15,000
-15,000
………
………
35
36
01.12.200
8
01.01.200
9
-15,000
-15,000
End
01.01.200
6
Formula
Expression
N
P  X1  
n2
Values in Expression
xn
 i
1  
 t
n
36
500,000- 15,000  
1 n2
15,000n
 i 
1  
 12 
n
Value of ‘i’
to be
determined
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 investor
receives 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 a
Rs.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 current
market 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?
Coupon payments
PV ( price ) of bond 
Maturity Face value
80
80
80
80
80
1000





1  0.10 (1  0.10) 2 (1  0.10)3 (1  0.10) 4 (1  0.10)5 (1  0.10)5
Annuity component
General formula for a bond :
I
I
I F
PV 


2
1  r (1  r )
(1  r ) n
Lump sum
component
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 opposite
directions.
– When a bond’s coupon rate is (greater than / equal to /
less
than) the market’s required return, the bond’s
market value will be (greater than / equal to / less than) its par
value.
– Given two bonds identical but for maturity, the price of
the
longer-term bond will change more (in percentage terms) than that
of the shorter-term bond, for a given change in market interest
rates.
– Given two bonds identical but for coupon, the price of
the
lower-coupon bond will change more (in percentage terms) than
that of the higher-coupon bond, for a given change in market
interest 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 for
study
Characteristics and
Symbols
Parameters
Population size = N
Population mean = m
Population standard
deviation = s
Statistics
Sample size = n
Sample mean = x
Sample 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 kth member of the
population is sampled.
• Stratified Sampling: The population is divided into two
or more strata and each subpopulation is sampled
(usually randomly).
• Cluster Sampling: A population is divided into clusters
and a few of these (often randomly selected) clusters are
exhaustively 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 group
but 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
 sx = s/n1/2
• Standard Normal probability distribution
Definitions
• 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 or
greater.
Because the total area under the
density curve is equal to 1,
there is a correspondence between
area 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 Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Area = 0.3413
Area
0.4429
-3
-2
-1
0
1
Score (z )
2
3
0
z = 1.58
Table A-2
Standard Normal Distribution
s=1
µ=0
0
x
z
Table for Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
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
.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
.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
*
*
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 probability
that it gives a reading between 0 and 1.58 units.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
That is 44.29% of the readings between 0 and
1.58 degrees.
Central Limit Theorem
1. The random variable x has a distribution (which
may or may not be normal) with mean µ and
standard deviation s.
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 means
n
will approach s/
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
Definition
• Regression Equation
Regression
Definition
• Regression Equation
Given a collection of paired data, the regression
equation
y^ = b0 + b1x
algebraically describes the relationship between the
two variables
• Regression Line
(line of best fit or least-squares line)
the graph of the regression equation
The Regression Equation
x is the independent variable
(predictor variable)
^y is the dependent variable
(response variable)
y^ = b0 +b1x
b0 = y - intercept
y = mx +b
b1 = slope
Notation for Regression
Equation
Population
Parameter
Sample
Statistic
y-intercept of regression equation
0
b0
Slope of regression equation
1
b1
Equation of the regression line
y = 0 + 1 x
^y = b + bx
0
1
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 same
variance. Also, for a given value of x, the
distribution of y-values has a mean that lies
on the regression line. (Results are not
seriously 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 is
plotted as a single point.
Positive Linear Correlation
y
y
y
(a) Positive
x
x
x
(b) Strong
positive
(c) Perfect
positive
Negative Linear Correlation
y
y
y
(d) Negative
x
x
x
(e) Strong
negative
(f) Perfect
negative
No Linear Correlation
y
y
x
(g) No Correlation
x
(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 or
technical factors.
• Cyclical fluctuation
– The term business cycle or economic cycle refers to the
fluctuations of economic activity (business fluctuations) around
its long-term growth trend. The cycle involves shifts over time
between periods of relatively rapid growth of output (recovery
and 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
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.
b.
c.
d.
31.66
33.66
36.66
39.66
1. If A invests Rs. 24 at 7 % interest rate for
5 years, total value at end of five years is
a.
b.
c.
d.
31.66
33.66
36.66
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 linear
relationship 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 linear
relationship 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 using
statistical sampling?
a. Statistical sampling will be looked upon by the
courts as providing superior audit evidence.
b. Statistical sampling requires the auditor to
make fewer judgmental decisions.
• c. Statistical sampling aids the auditor in
evaluating results.
d. Statistical sampling is more convenient to use
than nonstatistical sampling.
• Which of the following is an example of using
statistical sampling?
a. Statistical sampling will be looked upon by the
courts as providing superior audit evidence.
b. Statistical sampling requires the auditor to
make fewer judgmental decisions.*
c. Statistical sampling aids the auditor in
evaluating results.
d. Statistical sampling is more convenient to use
than nonstatistical sampling.
• Which of the following best illustrates the concept of
sampling risk?
a. An auditor may select audit procedures that are not
appropriate 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 representative
of the population as a whole.
d. An auditor may fail to recognize deviations in the
documents examined.
• Which of the following best illustrates the concept of
sampling risk?
a. An auditor may select audit procedures that are not
appropriate 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 representative
of the population as a whole.*
d. An auditor may fail to recognize deviations in the
documents examined.
• The advantage of using statistical sampling
techniques 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 sampling
techniques 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 methods
a. 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 methods
a. 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 a
long period of time is called
a. periodicity.
b. cycle.
c. regression.
d. trend.
• Gradual shifting of a time series over a
long period of time is called
a. periodicity.
b. cycle.
c. regression.
d. trend. *
• Seasonal components
a. cannot be predicted.
b. are regular repeated patterns.
c. are long runs of observations above or
below the trend line.
d. reflect a shift in the series over time.
• Seasonal components
a. cannot be predicted.
b. are regular repeated patterns. *
c. are long runs of observations above or
below the trend line.
d. reflect a shift in the series over time.
• Short-term, unanticipated, and
nonrecurring 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, and
nonrecurring 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 to
smooth
a. the irregular component.
• b. wide seasonal variations.
c. significant trend effects.
d. long range forecasts.
• The focus of smoothing methods is to
smooth
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. 11.25
b. 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. 11.25
b. 28.50
c. 39.75*
d. 44.25
• The forecasting method that is appropriate
when 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
• The forecasting method that is appropriate
when 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