This document discusses nominal and effective interest rates, including continuous compounding. It defines nominal and effective interest rates, and how to calculate effective rates for different compounding periods. It provides examples of calculating effective rates for annual, semi-annual, quarterly, and continuous compounding. It also discusses how to handle calculations when the payment period is equal to, longer than, or shorter than the compounding period. This includes using the appropriate effective rate and number of periods in calculations.

1. 1
Nominal and Effective Interest
Rates and Continuous
Compounding
Chapter 4Chapter 4
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2. 2
Items Covered in this ChapterItems Covered in this Chapter
Nominal and Effective Interest Rates
Continuous Compounding
Equivalence calculations for payment
periods equal to or longer than the
compounding period.
Equivalence calculations for payment
periods shorter than the compounding
period.
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3. 3
Nominal Versus Effective Interest Rates
Nominal Interest Rate: Interest rate
quoted based on an annual period
Effective Interest Rate:
Actual interest earned or paid in a year
or some other time period

4. 4
Why Do We Need an Effective
Interest Rate per Payment Period?
Payment period
Interest period
Payment period
Interest period
Payment period
Interest period

5. 5
Nominal and Effective Interest RateNominal and Effective Interest Rate
 Compounding at other intervals than yearly; e.g.,
daily, monthly, quarterly, etc. The two terms are used
when the compounding period is less than 1 year
 Nominal also called Annual Percentage Rate (APR)
means not actual or genuine, it must be adjusted or
converted into effective rate in order to reflect time
value considerations.
 Nominal interest rate, r, is an interest rate that does
not include any consideration of compounding.
 Nominal interest rate is equal to the interest rate per
period multiplied by the number of periods:
Nominal rate (r) per period= i per period * number
of periods.
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6. 6
Nominal and Effective Interest Rate (cont.)Nominal and Effective Interest Rate (cont.)
 A nominal rate can be found for any time period longer than
the originally stated period.
 A nominal rate of 1.5%per month is expressed as a nominal
4.5% per quarter, or 9% per semiannual period, or 18% per
year, or 36% per two years, etc.
 Effective interest rate is the actual rate that applies for a
stated period of time. The compounding of interest during
the time period of the corresponding nominal rate is
accounted for by the effective interest rate.
 An effective rate has the compounding frequency attached
to the nominal rate statement.
 Only effective interest rates can be used in the time value
equations or formulas.
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7. 7
Nominal and Effective Interest Rate (cont.)Nominal and Effective Interest Rate (cont.)
 Time period-the period over which the interest is expressed.
This is the t in the statement of r% per time period t, for
example, 1 % per month. The time unit of 1 year is by far the
most common. It is assumed when not stated otherwise.
 Compounding period (CP)- the shortest time unit over which
interest is charged or earned. This is defined by the
compounding term in the interest rate statement, for example,
8% per year compounded monthly. If not stated, it is assumed to
be 1 year.
 Compounding frequency- the number of times that m
compounding occurs within the time period t. If the
compounding period CP and the time period t are the same, the
compounding frequency is 1, for example, 1% per month
compounded monthly.

8. 8
Nominal and Effective Interest Rate-ExampleNominal and Effective Interest Rate-Example
Consider the rate 8% per year, compounded
monthly. It has a time period t of 1 year, a
compounding period CP of 1 month, and a
compounding frequency m of 12 times per
year.
A rate of 6% per year, compounded weekly,
has t = 1 year, CP = 1 week, and m = 52,
based on the standard of 52 weeks per year.

9. 9
Nominal and Effective Interest Rate-ExampleNominal and Effective Interest Rate-Example
 The different bank loan rates for three separate electric generation
equipment projects are listed below. Determine tbe effective rate on the
basis of the compounding period for each quote.
 (a) 9% per year, compounded quarterly.
 (b) 9% per year, compounded monthly.
 (c) 4.5% per 6-montbs, compounded weekly.

10. 10
Effective interest Rates for Any PeriodEffective interest Rates for Any Period
Effective i =[1+ r/m]m
– 1
◦ i: effective interest rate per year (or certain period)
◦ m: number of compounding periods per payment period
◦ r: nominal interest rate per payment periods
it is possible to take a nominal rate (r% per
year or any other time period) and convert it
to an effective rate i for any time basis, the
most common of which will be the PP time
period.
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11. 11
Nominal and effective Interest rate-Nominal and effective Interest rate-
 Consider 18 % per compounded at several periods.
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12. 12
ExamplesExamples
1) Nominal rate of 18% compounded yearly
with time interval of one year (m=1)
i=[1+0.18/1]1
– 1=18% per year
2) Nominal rate of 18% compounded semi-
annual with a time interval of one year
i=[1+0.18/2]2
– 1= 18.81% per year
3) Nominal rate of 18% compounded quarterly
with a time interval of 1 year i=[1+0.18/4]4
-1= 19.252% per 1 year
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13. 13
Effective interest Rate Problem 1Effective interest Rate Problem 1
 A company wants to buy new machine. The company received
three bids with interest rates. The company will make
payments on semi-annual basis only. The engineer is confused
about the effective interest rates –what they are annually and
over the payment period of 6 months.
 Bid #1: 9% per year, compounded quarterly
 Bid #2: 3% per quarter, compounded quarterly
 Bid #3: 8.8% per year, compounded monthly
◦ (a) Determine the effective rate for each bid on the basis of
semiannual payments, and construct cash flow diagrams
similar to Figure 4-3 for each bid rate.
◦ (b) What are the effective annual rates? These are to be a
part of the final bid selection.
◦ (c) Which bid has the lowest effective annual rate?

14. 14

15. 15

16. 16
Effective Interest Rate Problem 2Effective Interest Rate Problem 2
 The interest rate on a credit card is 1% per month. Calculate
the effective annual interest rate and use the interest factor
tables to find the corresponding P/F factor for n=8years?
1) 1% is an effective interest rate (Not nominal!!!!)
 Nominal rate = 0.01per month*12months/year
= 0.12
 i=[1+0.12/12]12
-1= 0.1268 = 12.68%
2) P/F = 1/ [1+0.1268]8
= 0.3848
3) by interpolation:
◦ 12% 0.4039
◦ 12.68% P/F
◦ 14% 0.3506
(P/F, 12.68%, 8) = 0.4039-0.0181= 0.3858
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17. 17
Effective Interest Rate for ContinuousEffective Interest Rate for Continuous
CompoundingCompounding
useful for modeling simplifications
If an interest rate r is compounded m times per
year, after m periods, the result is
i= lim m—∞ (1 +r/m)m
-1
Since lim m-> ∞ (1 +r/m)m
= er
, where e ≈ 2.7818
Further,
ia=effective continuous interest rate= er
-1
Example: if the nominal annual r = 15% per year,
the effective continuous rate per year is
i% = e0.15
-1=16.183%
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18. 18
Calculations of Effective continuousCalculations of Effective continuous
compounding of IRcompounding of IR
For a IR of 18% per year compounded
continuously, calculate the effective monthly
and annual interest rates?
Solution:
◦ r= 0.18/12=0.015 per month, the effective monthly
rate = i per month= er
– 1= e0.015
-1= 1.511%
◦ The effective annual rate for a nominal rate
r= 18% per year
i per year = e0.18
– 1= 19.72%
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19. 19
Calculations of Effective continuousCalculations of Effective continuous
compounding of IRcompounding of IR
If an investor requires an effective return of at least
15% on his money, what is the minimum annual
nominal rate that is acceptable if continuous
compounding takes place?
Solution
◦ r =?=er
-1= 0.15
er
= 1.15
lner
= ln 1.15
r = 0.1376 = 13.976%
A rate of 13.976 per year compounded continuously will
generate an effective 15% per year return.
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20. 20
Calculations for payment periodsCalculations for payment periods equal to orequal to or
longerlonger than the compounding periodsthan the compounding periods
For uniform series and gradients:
For uniform series and gradient factors, there are
three cases:
◦ Case 1 PP=CP
◦ Case 2 PP>CP
◦ Case 3 PP<CP
For cases 1 and 2 follow the following steps:
◦ Step 1: count the number of payments and use that number
as n, i.e., payments made quarterly for 5 years…then n is 20
quarters
◦ Step 2: find the effective interest rate over the same time
period as n in step 1. i.e., n is expressed in quarters…then
the effective rate per quarter should be found and used.
◦ Step 3: use these values of n and i in the tables
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21. 21
Calculations for payment periodsCalculations for payment periods equal to orequal to or
longerlonger than the compounding periods (Sec 4.6)than the compounding periods (Sec 4.6)
For single payment factors:
if the compounding period (CP) and payment period
(PP) do not agree (coincide) then interest tables
cannot be used until appropriate corrections are
made.
For Single payment factors:
◦ An effective rate must be used for i
◦ The units on n must be the same as those on i
◦ If the IR is per X, then n should be in terms of X
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22. 22
PP≥ CP example
a quality manager will pay $500 every 6 months for
the software maintenance contract. What is the
equivalent amount after the last payment, if these
funds are taken from a pool that has been returning
20% per year, compounded quarterly?

23. 23
PP≥ CP example –cont.
 PP= 6 months, CP is quarterly = 3 months, so PP > CP.
 based on PP (every 6 months), r=20% per year is converted to
semi-annual, r = 0.20/2=0.10,
 m based on r = 6/3=2
 Use Equation (4.8) with r = 0.10 per 6-month period and 2 CP
periods per semiannual period.
 Effective i semi-annual =[1+ r/m]m
– 1= [1+0.10/2]2
-1=10.25%
 Total number of semi-annual payments = 7 yrs*2 = 14
 F=A(F/A,10.25%,14)= 500(28,4891)=14,244.50

24. 24
PP=CP Example
 Suppose you plan to purchase a car and carry a loan of $12,500
at 9% per year, compounded monthly. Payments will be made
monthly for 4 years. Determine the monthly payment. Compare
the computer and hand solutions.
 Soln:
 CP =monthly, PP= monthly, so PP=CP.
 Effective i per month=9%/12= 0.75,
 n= 4 yr x 12 = 48
 Manual:
A = $12,500(A/ P,0.75%,48) = 12,500(0.02489) = $31
1.13

25. 25
PP=CP Example – cont.
Spreadsheet:
Enter PMT(9%/ 12,48, - 12500) into any cell
to display $3 11.06.

26. 26
Calculations for payment periodsCalculations for payment periods ShorterShorter thanthan
the compounding periodsthe compounding periods
Payments are made on shorter periods than
Compounding Interest.
Three possible scenarios:
◦ There is no interest paid on the money deposited or
withdrawn between compounding periods
◦ The money deposited or withdrawn between compounding
periods earns simple interest.
◦ All interperiod transactions earn compound interest
Scenario number 1 is only considered.
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27. 2708/07/14 27
0 1 2 1211109876543
Year
Month
$150
$200
$75 $100
$90
$120
$50
$45
Compounding period is quarterly at 3% interest rate
PP < CP examplePP < CP example

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•P= -150- 200(P/F, 3%, 1)- 175(P/F, 3%,2)+ 210(P/F, 3%,3) -
5(P/F,3%,4(
0 1 2 1211109876543
Year
Month
$150
$200
$175
$210
$50
1 2 3 40
$45
PP < CP examplePP < CP example

29. 29
Non-standard Annuities and GradientsNon-standard Annuities and Gradients
 Treat each cash flow individually
 Convert the non-standard annuity or gradient to standard form by
changing the compounding period
 Convert the non-standard annuity to standard by finding an equal
standard annuity for the compounding period
 How much is accumulated over 20 years in a fund that pays 4%
interest, compounded yearly, if $1,000 is deposited at the end of
every fourth year?
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0 4 8 12 16 20
$1000
F= ?

30. 30
Non-standard Annuities and Gradients-Non-standard Annuities and Gradients-
ExamplesExamples
 Method 1: consider each cash flows separately
F = 1000 (F/P,4%,16) + 1000 (F/P,4%,12) + 1000
(F/P,4%,8) + 1000 (F/P,4%,4) + 1000 = $7013
 Method 2: convert the compounding period from
annual to every four years
ie = (1+0.04)4
-1 = 16.99%
F = 1000 (F/A, 16.99%, 5) = $7013
 Method 3: convert the annuity to an equivalent
yearly annuity
A = 1000(A/F,4%,4) = $235.49
F = 235.49 (F/A,4%,20) = $7012
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