Chapter 4
Nominal and Effective Interest Rates
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EGR 312 - 6
Interest Rate Statements
he terms ‘nominal’ and ‘effective’ enter into consideration
hen the interest period is less than one year.
New time-based definitions to understand and
remember
est period (t) – period of time over which interest is expressed. For examp
1% per month.
nding period (CP) – Shortest time unit over which interest is charged or e
For example,10% per year compounded monthly.
Compounding frequency (m) – Number of times compounding
occurs within the interest
period t. For example, at i = 10%
per year, compounded
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2
monthly, interest would be
compounded 12 times during the
Understanding Interest Rate Terminology
A nominal interest rate (r) is obtained by multiplying
an interest rate that is
expressed over a short time period by the number of
compounding periods in a longer time period: That is:
r = interest rate per period x number of
compounding
Example: Ifperiods
i = 1% per month, nominal rate per year is
r = (1)(12)
12%
per year
This rate is often referred
to as =the
Annual
Percentage Rate (APR).
Effective interest rates (i) take compounding into account
(effective rates can be obtained from nominal rates via a formula
to be discussed later).
IMPORTANT: Nominal interest rates are essentially simple interest
rates. Therefore, they can never be used in interest formulas.
Effective rates must always be used hereafter in all interest formulas.
This rate is often referred to as the Annual Percentage Yield (APY).
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EGR 312 - 6
More About Interest Rate
Terminology
There are 3 general ways to express interest
rates as shown below
Sample Interest Rate Statements
Comment
When no compounding period
i
=
2%
per month
(1)
i = 12% per year
is given, rate is effective
i (2)
= 10% per year, comp’d semiannually
When compounding period is give
i = 3% per quarter, comp’d monthlyand it is not the same as interest
period, it is nominal
(3)
4
i = effective 9.4%/year, comp’d semiannuallyWhen
compounding period is give
i = effective 4% per quarter, comp’d monthlyand rate is specified as effective,
rate is effective over stated period
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Examples
Identify the following interest rate
statements as either nominal or
effective:
(a) 1.3% per month;
(b) 1 % per week, compounded
weekly;
(c) nominal 15% per year,
compounded monthly;
(d) effective 1.5% per month,
compounded daily; and
(e) 15% per year, compounded
semiannually.
5
(a) effective
(b) effective
(c) nominal
(d) effective
(e) nominal
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Nominal -vs- Effective Interest Rates
Corresponding effective annual interest rates:
iCP)m –1
In general, ie = (1+
__________________
Let compounding frequency, m, be the number of time
the compounding occurs within the time period, t.
1)
12% interest per year, compounded monthly.
m = 12
Effective rate per CP, iCP = r/m = 1% (per month).
(1+ iCP)12 –1 = 12.68%
Effective annual rate, ie = ________________
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EGR 312 - 6
Nominal -vs- Effective Interest Rates
Corresponding effective annual interest rates:
2)
12% interest per year, compounded quarterly.
Compounding frequency: m = 4
3% (per quarter).
Effective rate per CP, iCP = r/m = _____________
Effective annual rate , ie
3)
4
(1+
.03)
–1 = 12.55%
= _____________________
3% interest per quarter, compounded monthly.
3
m = _____
r/m = 1% (per month).
Effective rate per CP, iCP = __________________
(1+ .01)12 –1 = 12.68%
Effective annual rate, ie = ___________________________
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EGR 312 - 6
Nominal -vs- Effective Interest Rates
Example: You are purchasing a new home and have
been quoted a 15 year 6.25% APR loan. If you
take out a $100,000 mortgage using the above
rates, what is your monthly payment?
m = 12
Compound period  __________
6.25% / 12 = .521% (per month)
iCP = __________________________
15(years) x 12(months/year) = 180 months
n = ____________________________
A = $100,000(A/P, .521%, 180) = $857.42.
A = ___________________________
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EGR 312 - 6
Determining Compounding Frequency, m

Given a stated APR and APY can you determine the compounding
frequency?
Example: A Certificate of Deposit has a stated APR of 8% with an
Annual Yield of 8.3%. What is the compounding period?
Compound
Period
m
Effective Annual Interest
1 day
1 week
1 month
6 months
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Effective interest rates for any time period

Let PP represent the payment period
(period of time between cash flows.)
And m is the number of compounding
periods per payment period.


10
Effective i = (1+r/m)m –1
Where,
r = nominal interest rate per payment
period, PP.
m = number of compounding periods per
payment period.
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Other Examples

If cash flows are received on a semi-annual
basis, what is the effective semi-annual
interest rate under the following conditions:
a)
9% per year, compounded quarterly:
(1+4.5%/2)2 – 1 = 4.55%
Effective isa = ____________________
b)
3% per quarter, compounded
quarterly:
2
(1+6%/2) – 1 = 6.09%
Effective isa = ____________________
c)
11
6
8.8% per year, (1+4.4%/6)
compounded
– 1 =monthly.
4.48%
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Equivalence Relations: PP and CP
definition: Payment Period (PP) – Length of time between cash
In the diagram below, the compounding period (CP) is semiannual and the
payment period (PP) is monthly
$1000
or the diagram below, the CP is quarterly and the payment period (PP) is semiannual
F=?
i = 10% per year, compounded quarterly
0
1
2
3
Years
4
0
6
5
1
7
2
8
3
4
5
Semi-annual periods
A = $8000
Semi-annual PP
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EGR 312 - 6
Equivalence Relations

Example: Consider the following cash flow. Find
the present worth if the cash flows earn a) 10% per
year compounded quarterly, or b) 9% per year
compounded monthly.
1
2
3
4
5
6
7
(Time in Years)
$300
$500
$700
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EGR 312 - 6
Equivalence Relations
Example: a) 10% per year compounded quarterly
1
2
3
(Time
4
5in Years)
6
7
$300
$500
$700
4
i
=
(1+10%/4)
–1 = 10.38%
a
ia = _______________________
P = $500(P/F,10.38%,3) + $700 (P/F,10.38%,6)
+ $300 (P/F,10.38%,7) = 909.11
____________________________
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EGR 312 - 6
Equivalence Relations
Example: b) 9% per year compounded monthly.
(Time in Years)
1
2
3
4
5
6
7
$300
$500
$700
ia = (1+9%/12)12 –1 = 9.38%
ia = ____________________
P = $500(P/F,9.38%,3) + $700 (P/F,9.38%,6)
+ $300 (P/F,9.38%,7) = 951
_____________________
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EGR 312 - 6
Single Amounts with PP
> CP
ms involving single amounts, the payment period (PP) is usually
n the compounding period (CP). For these problems, there are a
f i and n combinations that can be used, with only two restrictio
(1) The i must be an effective
interest
rate,units
and on n must be the same as those of i
(2) The time
(i.e., if i is a rate per quarter, then n is the number of
quarters between P and F)
There are two equally correct ways to determine i and n
1: Determine effective interest rate over the compounding period
set n equal to the number of compounding periods between P a
2: Determine the effective interest rate for any time period t, and
set n equal to the total number of those same time periods.
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EGR 312 - 6
Series with PP ≥ CP
s cash flows, first step is to determine relationship between PP a
Determine if PP ≥ CP, or if PP < CP
n PP ≥ CP, the only procedure (2 steps) that can be used is as fol
(1)First, find effective i per PP
Example: if PP is in quarters, must find
i/quarter
(2) effective
Second,
determine n, the number of A
values involved
Example: quarterly payments for 6 years
Note: Procedure when PP < CP is discussed later
yields
n = 4×6 = 24
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EGR 312 - 6
Example: Series with
PP
≥
CP
How much money will be accumulated in 10 years from a deposit
of $500 every 6 months if the interest rate is 1% per month?
Solution:
First, find relationship between PP and CP
PP = six months, CP = one month; Therefore, PP > CP
Since PP > CP, find effective i per PP of six months
Step 1.
i /6 months = (1 + 0.06/6)6 – 1 = 6.15%
Next, determine n (number of 6-month periods)
Step 2:
n = 10(2) = 20 six month periods
Finally, set up equation and
solve for F
F = 500(F/A,6.15%,20) = $18,692
(by factor or spreadsheet)
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EGR 312 - 6
Equivalence Relations (PP ≥ CP)
Find P for the following in standard factor
expressions :
Cash Flow
$500 semi–annually
for 5 years
$75 monthly for
3 years
$180 quarterly for
15 years
$25 per month
increase for 4 years
$5000 per quarter
for 6 years
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Interest Rate Standard Notation
16% per year,
compounded
semi-annually
24% per year,
compound monthly
5% per quarter
P = $500(P/A,8%,10)
P = $75(P/A,2%,36)
P = $180(P/A,5%,60)
1% per month
P = $25(P/G,1%,48)
+ $25(P/A,1%,48)
1% per month
P = $5000(P/A,3%,24)
EGR 312 - 6
Equivalence Relations (PP ≥ CP)
 Over the past 10 years, Gentrack has placed
varying sums of money into a special capital
accumulation fund. The company sells compost
produced by garbage-to-compost plants in the
United States and Vietnam. Figure below is the
cash flow diagram in $1000 units. Find the
amount in the account now (after 10 years) at an
interest rate of 12% per year, compounded
semiannually.
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EGR 312 - 6
Equivalence Relations (PP ≥ CP)
 Method 1:
 Method 2:
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EGR 312 - 6
Series with PP < CP
olicies: (1) interperiod cash flows earn no interest (most co
(2) interperiod cash flows earn compound interes
For policy (1), positive cash flows are moved to
beginning
of the interest period in which they occur
and negative
are
moved tointerest
the end
The condition
of PPcash
< CP flows
with no
interperiod
is of
the only
the interest
period
n which
the actual
cash flow diagram is changed
For policy (2), cash flows are not moved and
equivalent P, F, and A values are
determined using the effective
interest rate per payment period
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Example: Series with
PP
<
CP
A person deposits $100 per month into a savings
account for 2 years. If $75 is withdrawn in months
5, 7 and 8 (in addition to the deposits), construct
the cash flow diagram to determine how much will
be in the account after 2 years at i = 6% per year,
compounded quarterly. Assume there is no
interperiod
interest.
Solution:
Since PP
< CP with no inter-period interest, the cash
flow diagram must be changed using quarters as the
F=?
time75periods
75 75
fro
m
this
0
1 2
8 9 10
3
4
5
6
7
23
24
10
0
F=?
75
150
to 0 1 2 3 4 5 6 7 8 9 10
this
23
1
2
300
3
300
300
21
24 7
300
Months
8
Quarters
300
EGR 312 - 6
Equivalence Relations (PP < CP)
No compounding within compound period:
 All deposits are regarded as occurring at the
end of the compounding period.
 All withdrawals are regarded as occurring at the
beginning of the period.
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EGR 312 - 6
Equivalence Relations (PP < CP)
No compounding within compound period:
Example: A company has the following monthly cash
flows. If the company expects an ROR of 12% per year,
compounded quarterly, what is the present value of the
cash flows?
i = 12%/yr cpd. qtly
$800
P
$600
1
2
3
$400
4
5
$500
$500
6
7
8
9
10
11
$300
$500
$700
25
$500
withdrawals
12
$500
deposits
$700
EGR 312 - 6
Equivalence Relations (PP < CP)
No compounding within compound period:
i = 12%/yr cpd. qtly
$800
P
$600
1
2
3
$400
4
5
$500
$500
withdrawals
6
7
8
9
10
11
$300
$500
$500
$700
=>
$500
deposits
$700
P
1
26
12
2
3
4
5
6
7
8
9
10
11
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EGR 312 - 6
Equivalence Relations (PP < CP)
No compounding within compound period:
$1400
i = 12%/yr cpd. qtly
P
$500
$400
1
2
3
4
5
6
$500
7
8
9
10
11
12
$500
$700
$1000
$1000
$1400 - $100(P/F,3%,1) - $200(P/F,3%,2)
P=
- $500(P/F,3%,3) - $1000(P/F,3%,4)
____________________________________
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(Hint: we can now look at quarters …)
EGR 312 - 6
Equivalence Relations (PP < CP)
 Last year AllStar Venture Capital agreed to invest funds in
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Clean Air Now (CAN), a start-up company in Las Vegas that
is an outgrowth of research conducted in mechanical
engineering at the University of Nevada–Las Vegas. The
product is a new filtration system used in the process of
carbon capture and sequestration (CCS) for coal-fired power
plants. The venture fund manager generated the cash flow
diagram in Figure Below in $1000 units from AllStar’s
perspective. Included are payments (outflows) to CAN made
over the first year and receipts (inflows) from CAN to AllStar.
The receipts were unexpected this first year; however, the
product has great promise, and advance orders have come
from eastern U.S. plants anxious to become zero-emission
coal-fueled plants. The interest rate is 12% per year,
compounded quarterly, and AllStar uses the no-interperiodinterest policy. How much is AllStar in the “red” at the end of
the year?
EGR 312 - 6
Equivalence Relations (PP < CP)
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EGR 312 - 6
Equivalence Relations (PP < CP)
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EGR 312 - 6
Equivalence Relations (PP < CP)
With interperiod compounding:
 If interest is compounded within the period,
treat interest on cash flows the same as
the treatment of nominal interest rates.
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EGR 312 - 6
Equivalence Relations (PP < CP)
With interperiod compounding:
Example: A company has the following monthly
cash flows. If the company expects an ROR of 12%
per year, compounded quarterly, what is the present
value of the cash flows?
$800
P
$600
1
2
3
$400
4
5
$500
$500
6
7
8
9
10
11
$300
$500
$700
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$500
12
$500
$700
EGR 312 - 6
Equivalence Relations (PP < CP)
With interperiod compounding:
For our example: Interest is compounded monthly at
the rate of 1%.
P = $800(P/F,1%,1) + $600(P/F,1%,2) - $500(P/F,1%,3) + etc….
$800
P
$600
1
2
3
$400
4
5
$500
$500
6
7
8
9
10
11
$300
$500
$700
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$500
12
$500
$700
EGR 312 - 6
Continuous Compounding
When the interest period is infinitely small, interest
is
compounded continuously. Therefore, PP > CP and
m to
increases.
Take limit as m → ∞
find the effective interest
rate equation
r
i
=
e
–1
e: If a person deposits $500 into an account every 3 month
terest rate of 6% per year, compounded continuously,
uch will be in the account at the end of 5 years?
Solution: Payment Period: PP = 3 months
Nominal rate per three months: r = 6%/4 = 1.50%
Effective rate per 3 months: i =
e0.015 – 1 = 1.51%
F = 500(F/A,1.51%,20) = $11,573
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EGR 312 - 6
Varying Rates
en interest rates vary over time, use the interest rate
ociated with their respective time periods to find P
Example: Find the present worth of $2500 deposits in
1 shown
through
if the
r theyears
cash flow
below,8find
the future worth (in year 7) at i = 10% per year.
interest rate is 7% per year for the first five years
Solution:
P =year
2,500(P/A,7%,5)
and
10% per
thereafter. +
2,500(P/A,10%,3)(P/F,7%,5)
= $14,683
valent annual worth value can be obtained by replacing each ca
ount with ‘A’ and setting the equation equal to the calculated P v
14,683 = A(P/A,7%,5) + A(P/A,10%,3)(P/F,7%,5)
35
A = $2500 per year
EGR 312 - 6