No, in both cases the full interest is not earned/owed. When the payment period (PP) is less than the compounding period (CP), the interest is prorated based on the time elapsed within the CP. So in the savings account example, the interest earned on the first 2 monthly deposits would be less than the interest earned on the 3rd deposit. And for the credit card, paying early would reduce the interest owed for that period.

1. Chapter 4
Nominal and Effective
Interest Rates
MS291: Engineering Economy

2. Content of the Chapter








Interest Rate: important terminologies
Nominal and Effective Rate of Interest
Effective Annual Interest Rate
Converting Nominal rate into Effective Rate
Calculating Effective Interest rates
Equivalence Relations: PP and CP
Continuous Compounding
Varying Intrest Rates

3. Previous Learning
 Our learning so for is based “one” interest rate that’s
compounded annually

Interest rates on loans, mortgages, bonds & stocks are
commonly based upon interest rates compounded more
frequently than annually
 When amount is compounded more than once annually,
distinction need to be made between nominal and effective
rate of interests

4. Interest Rate:
important terminologies
New time-based definitions to understand and remember
Interest period (t) – period of time over which interest is expressed. For example,
1% per month.
Compounding period (CP) – The time unit over which interest is charged or earned.
For example,10% per year, here CP is a year.
Compounding frequency (m) – Number of times compounding occurs within the interest
period t. For example, at i = 10% per year, compounded
monthly, interest would be compounded 12 times during the
one year interest period.

5. Examples of interest rate
Statements
 Annual interest rate of 8% compounded monthly …
 Here interest period (t)
= 1 year

compounding period (CP)
= 1 month

compounding frequency (m) = 12
 Annual interest rate of 6% compounded weekly …
 Here interest period (t)
= 1 year

compounding period (CP)
= 1 week

compounding frequency (m) = 52

6. Different Interest Statements
Interest rates can be quoted in many ways:
•
•
•
•
Interest equals “6% per 6-months”
Interest is “12%” (annually)
Interest is 1% per month
“Interest is “12.5% per year, compounded monthly”
You must “read” the various ways to state interest and to do
calculations.

7. Nominal Interest Rate
•
A nominal interest rate is denoted by (r)
• It does not include any consideration of the compounding of
interest(frequency)
• It is given as:
r = interest rate per period x number of compounding periods
• Nominal rates are all of the form “r % per time period”

8. Examples: Nominal Interest Rate
• 1.5% per month for 24 months
– Same as: (1.5%)(24)= 36% per 24 months
• 1.5% per month for 12 months
– Same as (1.5%)(12 months) = 18% per year
• 1.5% per month for 6 months
– Same as: (1.5%)(6 months) = 9% per 6 months or semi annual
period
• 1% per week for 1 year
– Same as: (1%)(52 weeks) = 52% per year

9. Summary: Nominal Interest Rate
 A nominal rate do not reference the frequency of compounding
 They all have the format “r% per time period”
 Nominal rates can be misleading…How?
Annual interest of $80 on a $1,000 investment is a nominal rate of 8%
whether the interest is paid in $20 quarterly instalments, in $40 semi-annual
instalments, or in an $80 annual payment?
 We need an alternative way to quote interest rates….
 The true Effective Interest Rate is then applied….

10. Effective Interest Rate (EIR)
 Effective interest rates (i) take accounts of the effect of the
compounding
 EIR are commonly express on an annual basis (however any
time maybe used)
 EIR rates are mostly of the form:
“r % per time period, CP-ly (Compounding Period)”
Nominal rates are all of the form “r % per time period”

11. Examples : Effective Interest
Rates
Quote: “12 percent compounded monthly” is translated as:
12% is the nominal rate
“compounded monthly” conveys the frequency of the compounding
throughout the year
For this quote there are 12 compounding periods within a year.

12. Examples: Effective Interest Rate
 Some times, Compounding period is not mentioned in Interest
statement
 For example, an interest rate of “1.5% per month” ………..
It means that interest is
compounded each month; i.e., Compounding Period is 1 month.
 REMEMBER: If the Compounding Period is not mentioned it
is understood to be the same as the time period mentioned
with the interest rate.

13. Example
Compounding
Period
Statement
1.
10% per year
2.
10 % per year
compounded monthly
3.
4.
3% per quarter
compounded daily
1.5% per month
compounded monthly
What it is ?
1. CP = not stated
but it’s a year
1. Effective rate per year
2.
2. Effective rate per year
CP = Stated,
CP= month
3. CP= stated,
CP= day
3. Effective rate per quarter
4. CP=stated,
CP=month
4.
Effective rate per month
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

14. Some other names for
NIR and EIRs
 Interest on Credit Cards, loans and house mortgages
sometime use term Annual Percentage Rate(APR) for interest
payment….its same as Nominal Interest Rate
For example: An APR of 15% is the same as a nominal 15%
per year or a nominal 1.25% on a monthly basis.
 Returns for investments, certificates of deposits and saving
accounts commonly use Annual Percentage Yield (APY) which
is same as Effective Interest Rate
 Remember: the effective rate is always greater than or equal
to the nominal rate, and similarly APY > APR .. Why ?

15. Converting Nominal rate into
Effective Rate per CP
• So for, we always used t and CP values of 1 year so compounding
frequencies was always m=1, which make them all effective rate of
interest (For EIR , Interest period and Compounding period should be
same)
• But that’s not always the case, we may have situation in which t has different
value than CP in that case we need to find effective rate of interest per
compounding period
• Effective rate per CP can be determined from nominal rate by using following
relation
Effective rate per CP =
r % per time period t
=
r
m compounding periods per t
m
Where: CP is compounding period, t is the basic time unit of the interest, m is the frequency of
compounding and r is nominal interest rate

16. Example: Calculating Effective
Interest rates
 Three different bank loan rates for electric generation
equipment are listed below. Determine the effective
rate on the basis of the compounding period for
each rate
(a) 9% per year, compounded quarterly
(b) 9% per year, compounded monthly
(c) 4.5% per 6 months, compounded weekly

17. Example: Calculating Effective
Interest rates per CP
a. 9% per year, compounded quarterly.
b. 9% per year, compounded monthly.
c. 4.5% per 6 months, compounded weekly.
The principle amount change in
each period for EIR here, …in case
of nominal it will remain same in each
case

18. More About Interest Rate
Terminology
 Sometimes it is not obvious whether a stated rate is a nominal
or an effective rate.
 Basically there are three ways to express interest rates

19. Effective Annual Interest Rates
When we talk about “Annual” we consider year as the interest
period t , and the compounding period CP can be any time unit less
than 1 year
Nominal rates are converted into Effective Annual Interest Rates (EAIR)
via the equation:
ia = (1 + i)m – 1
where
ia = effective annual interest rate
i = effective rate for one compounding period (r/m)
m = number times interest is compounded per year

20. Example
For a nominal interest rate of 12% per year, determine the nominal and effective
rates per year for
(a) quarterly, and
ia = (1 + i)m – 1
(b) monthly compounding
Solution:
where ia = effective annual interest rate
i = effective rate for one compounding period (r/m)
m = number times interest is compounded per year
(a) Nominal r per year = 12% per year
Nominal r per quarter = 12/4 = 3.0% per quarter
Effective i per year = (1 + 0.03)4 – 1 = 12.55% per year
(b) Nominal r per month = 12/12 = 1.0% per month
Effective I per year = (1 + 0.01)12 – 1 = 12.68% per year

21. r = 18% per year,
compounded CP-ly

22. Effective Interest Rates for any
Time Period
 The following relation of Effective Annual Interest
Rates ia = (1+i)m – 1 can be generalize for
determining the effective interest rate for any
time period (shorter or longer than 1 year).
i = (1 + r / m)m – 1
where
i = effective interest rate for any time period
r = nominal rate for same time period as i
m = no. times interest is comp’d in period specified for i

23. Example: Effective Interest Rates
For an interest rate of 1.2% per month, determine the nominal
and effective rates
(a) per quarter, and
i = (1 + r / m)m – 1
(b) per year
Solution:
(a) Nominal rate (r) per quarter = (1.2)(3) = 3.6% per quarter
Effective rate (i) per quarter = (1 + 0.036/3)3 – 1 = 3.64% per quarter
(b) Nominal rate (r) per year = (1.2)(12) = 14.4% per year
Effective rate (i) per year = (1 + 0.144 / 12)12 – 1 = 15.39% per year

24. Economic Equivalence:
From Chapter 1
 Different sums of money at different times may be equal in
economic value at a given rate
Rate of return = 10% per year
$110
Year
0
1
$100 now
1
 $100 now is economically equivalent to $110 one year from
now, if the $100 is invested at a rate of 10% per year
 Economic Equivalence: Combination of interest rate (rate of
return) and time value of money to determine different amounts
of money at different points in time that are economically
equivalent ….. Compounding/Discounting (F/P, P/F, F/A, P/G etc.)

25. Equivalence Relations: Payment
Period(PP) & Compounding
Period(CP)
 The payment period (PP) is the length of time
between cash flows (inflows or outflows)
r = nominal 8% per year, compounded semi-annually
CP
6 months
0
1
│PP │
1 month
2
3
CP
6 months
4
5
6
7
8
9
10
11
12
Months

26. Equivalence Relations: Payment
Period(PP) and Compounding Period
•
It is common that the lengths of the payment period and
the compounding period (CP) do not coincide
•
To do correct calculation …Interest rate must coincide with
compounding period
•
It is important to determine if PP = CP, PP >CP, or PP<CP
Length of Time
Involves Single Amount
(P and F Only)
Involves Gradient Series
(A, G, or g)
PP = CP
P/F , F/P
P/A, P/G, P/g
F/A etc.
P/F, F/P
P/A, P/G, F/A etc.
PP > CP
PP < CP

27. Case I: When PP>CP for Single
Amount for P/F or F/P
 Step 1: Identify the number of compounding
periods (M) per year
 Step 2: Compute the effective interest rate per
payment period (i)

i = r/M
 Step 3: Determine the total number of payment
periods (n)
 Step 4: Use the SPPWF or SPCAF with i and N
above

28. Example
Determine the future value of $100 after 2 years at credit
card stated interest rate of 15% per year, compounded
monthly.
Solution:
P = $100, r = 15%, m = 12
EIR /month = 15/12 = 1.25%
n = 2 years or 24 months
Alternative Method
i = (1 + r/m)m – 1
= (1+0.15/12)12 – 1
= 16.076
F = P(F/P, i, n)
F = P(F/P, i, n)
F = P(F/P, 0.0125, 24)
F = P(F/P, 0.16076, 2)
F = 100(F/P, 0.0125, 24)
F = 100(1.3474)
F = 100(1.3474)
F = $134.74
F = 100(1.3456)
F = $134.56
The results are slightly different because of the rounding off
16.076% to 16.0%

29. Factor Values for
Untabulated i or n
There are 3 ways to find factor values for untabulated i
or n values
1. Use formula
2. Use spreadsheet function
3. Linearly interpolate in interest tables
 Formula or spreadsheet function is fast & accurate
 Interpolation is only approximate

30. Factor Values for Untabulated i
or n

Factor value
axis
f2
f
Linear
assumption
unknown
f1
X1
Required
X
X2
i or n axis
Absolute Error = 2.2215 – 2.2197
= 0.0018

31. Case II: When PP >CP for
Series for P/A or F/A
 For series cash flows, first step is to determine relationship
between PP and CP
 Determine if PP ≥ CP, or if PP < CP
 When PP ≥ CP, the only procedure (2 steps) that can be
used is as follows:
First, find effective i per PP
Example: if PP is in quarters, must find effective i/quarter
Second, determine n, the number of A values involved
Example: quarterly payments for 6 years yields n = 4×6 = 24
 You can use than the standard P = A(P/A, i , n) or F = A(F/A, i, n) etc.

32. Example: PP >CP for
Series for P/A or F/A
For the past 7 years,
Excelon Energy has paid
$500 every 6 months for a
software maintenance
contract. What is the
equivalent total amount after
the last payment, if these
funds are taken from a pool
that has been returning 8%
per year, compounded
quarterly?
Solution:
Compounding Period (CP) = Quarter & PP = 6 months
r = 8 % per year or 4% per 6 months & m=2/ quarter
PP > CP
Effective rate (i) per 6 months = (1+r/m)m -1
i= (1+0.04/2)2 – 1 => 4.04%
Since, total time is 7 years and PP is 6 months
we have total 7x2=14 payments
F = A(F/A, i, n)
F = 500(F/A, 0.0404, 14)
F = 500(18.3422) => $9171.09

33. Case III: Economic
Equivalence when PP< CP
 If a person deposits money each month into a savings account where
interest is compounded quarterly, do all the monthly deposits earn interest
before the next quarterly compounding time?
 If a person's credit card payment is due with interest on the 15th of the
month, and if the full payment is made on the 1st, does the financial
institution reduce the interest owed, based on early payment? Anyone ?
CP: 3 months = 1 quarter
0
1
│PP │
1 month
2
3
4
5
6
7
8
9
10
11
12
Months

34. Case III: Economic
Equivalence when PP< CP
Two policies:
1. Inter-period cash flows earn
no interest (most common)
2. inter-period cash flows earn
compound interest
 positive cash flows are
moved to beginning of the
interest period (no interest
earned) in which they occur
and negative cash flows are
moved to the end of the
interest period (no interest
paid)
cash flows are not moved
and equivalent P, F, and A
values are determined
using the effective interest
rate per payment period

35. Example 4.11: Example: Clean
Air Now (CAN) Company
Last year AllStar Venture Capital agreed to invest funds in 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 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-inter period-interest policy. How
much is AllStar in the “red” at the end of the year?

36. Example 4.11: Example: Clean
Air Now (CAN) Company
The venture fund manager generated the cash flow diagram in $1000 units from
AllStar’s perspective as given below. 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-inter period-interest policy. How much is AllStar in the “red” at the end of the
year?

37. Example: Clean Air Now (CAN)
Company
Given cash flows
Positive Cash flows (inflows) at the start
of CP period
Negative Cash flows (outflows) at the
end of CP period

38. Example: Clean Air Now (CAN)
Company
Solution:
Effective rate per quarter = 12/4 = 3%
Now
F = 1000[-150(F/P, 3%, 4)-200(F/P, 3%, 3) +(180-175 )(F/P, 3%, 2)+ 165(F/P, 3%, 1)-50]
F = $ (-262111) Investment after one year

39. Continuous Compounding
 If compounding is allowed to occur more and more frequently,
the compounding period becomes shorter and shorter, and m
(the number of compounding periods per payments) increases
 Continuous compounding is present when the duration of
CP, the compounding period, becomes infinitely small and m ,
the number of times interest is compounded per period,
becomes infinite.
 Businesses with large numbers of cash flows each day consider
the interest to be continuously compounded for all transactions.

40. Continuous Compounding
 We have effective interest rate generalize formula
as follows: i = (1 + r/m)m – 1
 Taking “m” limits tends to infinity… and simplifying
the equation we get the following expression for
continuous effective interest rate
i = er – 1

41. Example: Continuous
Compounding
Example: If a person deposits $500 into an account every 3 months
at an interest rate of 6% per year, compounded continuously, how
much 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%
Continuous Effective rate per 3 months: i = e0.015 – 1 = 1.51%
F = 500(F/A,1.51%,20) = $11,573
Practice:
Example 4.12 & 4.13

42. Varying Interest Rates
 Interest rate does not remain constant full life time of a project
 In order to do incorporate varying interest rates in our
calculations, normally, engineering studies do consider average
values that do care of these variations.
 But sometimes variation can be large and having significant
effects on Present or future values calculated via using average
values
 Mathematically, varying interest rates can be accommodated in
engineering studies

43. Varying Interest Rates
When interest rates vary over time, use the interest rates
associated with their respective time periods to find P
 The general formula for varying interest rate is given as:
P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + …..
+ Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)
 For single F or P only the last term of the equation can be used.
 For uniform series replace “F” with “A”

44. Example: Varying Interest
Rates
Given below the cash flow calculate the Present value.
$70,000
i=7%
$35,000
i=7%
i=9%
0
P=?
1
2
$25,000
i=10%
3
4
Year
P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + …..
+ Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)
P = [ 70,000(P/A, 7%, 2) + 35,000 (P/F, 9%, 1) (P/F, 7%, 2) + 25000(P/F, 10%, 1)
(P/F, 7%, 2) (P/F, 9%, 1)
= 70,000 (1.8080) + 35,000 (0.9174)(0.8734) + 25,000 (0.9091)(0.8734)(0.0.9174)
= $172,816

45. Problem 4.57: Varying Interest Rates
Calculate (a) the Present value (b) the
uniform Annual worth A of the following
Cash flow series
P=?
i=10%
1
2
3
4
5
6
7
i=14%
8
Year
0
$100 $100 $100 $100 $100
$160 $160
$160
P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + …..
+ Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)
P = 100(P/A, 10%, 5) + 160 (P/A, 14%, 3) (P/F, 10%, 5)
= 100(3.7908) + 160(2.3216)(0.6209)
= $609.72

46. Problem 4.57: Varying Interest Rates
(b) the uniform Annual worth A of the
following Cash flow series
P = 609.72
i=10%
1
2
3
4
5
6
7
i=14%
8
Year
0
A=?
P = 100(P/A, 10%, 5) + 150 (P/A, 14%, 3) (P/F, 10%, 5)
= 100(3.7908) + 160(2.3216)(0.6209)
= $609.72
609.72 = A(3.7908) + A(2.3216)(0.6209)
A = 609.72 / 5.2323
A = $ 116.53 per year

47. THANK YOU