# Lecture 05: Effective Interest Rates

```LEGOs Theory Continued…
● In learning about compound cash flows, we found
ways to decompose and convert the cash flow
pattern building blocks from one type to another, and
how to relocate the block patterns to different points
in time.
● The cash flow periods (time between arrows on cfd)
are like the bumps on a LEGO brick.
● The effective interest rate per period is like the holes
of the block – the bumps and holes must align for the
LEGO blocks to fit together.
● There are three major cases of effective interest rate
to consider – and four formulas to know. But first,
there are some terms to learn…
Interest Rate Terms…
● Compounding Period (cp) – the time between points
when interest is computed and added to the initial amount.
● Payment Period (pp) – the shortest time between
payments. Interest is earned on payment money once per
period (cost of money)
● Nominal Rate ( r ) – is a simplified expression of the
annual cost of money. It means nothing, unless the
compounding period is stated along with it.
● Annual Percentage Rate (APR) – is the nominal interest
rate on a yearly basis (credit cards, bank loans, …). It, too,
should have a compounding period stated.
● Effective Rate ( i ) – is the rate that is used with the table
factors or the closed form equations, and it converts the
nominal rate taking into account both the compounding
period and the payment period so that the blocks match.
Compounding Period is Equal to the
Payment Period
r = nominal annual interest rate for payments that match the
compounding period: (cp < year and pp = cp)
Examples: 12% per year compounded monthly
10% APR, compounded quarterly
i
(1)
(2)
= interest rate per compounding period = r =
m
nominal interest rate
( # of compounding periods per year)
Examples: 12% / 12 months = 1% compounded monthly
10% / 4 quarters = 2.5% compounded quarterly
(1)
(2)
Which would you rather have: 12% compounded annually
or 12% compounded monthly?
Compounding Period is More Frequent than
the Annual Payment Period
EFFECTIVE INTEREST RATE
ia
= effective interest rate per year compounded annually
= ( 1 + interest rate per cp)(# of cp per year) – 1
= 1+ r
m
m
–1
Example:
r = 12% per year compounded monthly
imonth = 12% yearly = 1 % compounded monthly
12 months
ia = (1 + .01)12 – 1 = 12.68% compounded annually
Another example…
r = 12% per year compounded semi-annually
isemi-annual
=
12% annually
2 times per year
= 6% per 6 months
ia = (1 + .06)2 – 1 = .1236
= 12.36% per year compounded annually
As the compounding period gets smaller, does the
effective interest rate increase or decrease?
r = 12% per year compounded daily
idaily
= 12%
365
= .000329
ia = (1 + .000329)365 – 1 = .12747
= 12.747% per year compounded annually
What happens if we let the compounding period
get infinitely small?
Continuous Compounding
i = e( r )(# of years) – 1
Examples:
r = 12% per year compounded continuously
ia
= e( .12 )(1) – 1 = 12.75%
What would be an effective six month interest rate
for r = 12% per year compounded continuously?
i6 month = e( .12 )(.5) – 1 = 6.184%
Compounding Period is More Frequent than
the Payment Period
EFFECTIVE INTEREST RATE
ie
= effective interest rate per payment period
= ( 1 + interest rate per cp)(# of cp per pay period) – 1
= 1+ r
m
me
–
1
Example:
r = 12% APR, compounded monthly, payments quarterly
imonth = 12% yearly = 1 % compounded monthly
12 months
ie = (1 + .01)3 – 1 = .0303 – or – 3.03% per payment
Summary of Effective Rates
An “APR” or “% per year” statement is a Nominal interest rate
– denoted r – unless there is no compounding period stated
The Effective Interest rate per period is used with tables & formulas
Formulas for Effective Interest Rate:
If continuous compounding, use i  er ( y )  1
y is length of pp, expressed in decimal years
m
 r
If cp < year, and pp = 1 year, use ia  1    1
 m
m is # compounding periods per year
If cp < year, and pp = cp, use
m is # compounding periods per year
If cp < year, and pp > cp, use
me is # cp per payment period
i
r
m
m
r e

ie  1    1
 m
CRITICAL POINT
When using the factors,
n and i must always match!
Use the effective interest rate
formulas to make sure that i
matches the period of interest
(sum any payments in-between compounding periods so
that n matches i before using formulas or tables)
Note:
Interest doesn’t start accumulating
until the money has been
invested for the full period!
Shows up here
on CFD…
Deposit
here …
(End of Period
Convention)
Returns
interest
here!
i
X
0
1
2 periods
Problem 1
The local bank branch pays interest on savings accounts at
the rate of 6% per year, compounded monthly. What is the
effective annual rate of interest paid on accounts?
GIVEN:
r = 6%/yr
m = 12mo/yr
DIAGRAM:
NONE NEEDED!
FIND ia:
m
r

ia  1    1
 m
12
 0.06 
 1 
  1  6.17%
12


Problem 2
What amount must be deposited today in an account paying
6% per year, compounded monthly in order to have \$2,000 in
the account at the end of 5 years?
GIVEN:
F5 = \$2 000
r = 6%/yr
m = 12 mo/yr
FIND P:
n  5 yrs
DIAGRAM:
m
\$2 000
0
1
2
5 yrs
P?
12
r 
0.06 


ia   1    1   1 
  1  6.17% / yr
m
12




P  \$2000( P | F , 6.17%, 5)
 \$2000(1  0.0617 )5  \$2000(.74129)
 \$1482.59
Problem 2 – Alternate Soln
What amount must be deposited today in an account paying
6% per year, compounded monthly in order to have \$2,000 in
the account at the end of 5 years?
GIVEN:
F5 = \$2 000
r = 6%/yr
m = 12 mo/yr
FIND P:
DIAGRAM:
\$2 000
0
1
2
60mos
P?
 12 mos 
( 5 yrs )  60 mos
n  ( m )(# yrs )  
 yr 
r  0.06  1yr 

i

  0.5% / mo
m  yr  12 mo 
P  \$2000( P | F ,0.5%,60 )  \$2000( 0.7414 )
 \$1482.80
Problem 3
A loan of \$5,000 is to be repaid in equal monthly payments
over the next 2 years. The first payment is to be made 1
month from now. Determine the payment amount if interest
is charged at a nominal interest rate of 12% per year,
compounded monthly.
GIVEN:
P = \$5 000
r = 12%/yr
m = 12 mo/yr
FIND A:
DIAGRAM:
\$5 000
1
2 yrs
0
A?
Problem 4
You have decided to begin a savings plan in order to make a
down payment on a new house. You will deposit \$1000
every 3 months for 4 years into an account that pays interest
at the rate of 8% per year, compounded monthly. The first
deposit will be made in 3 months. How much will be in the
account in 4 years?
DIAGRAM:
0
1
F?
2
3
4 yrs
\$1 000
Problem 5
Determine the total amount accumulated in an account
paying interest at the rate of 10% per year, compounded
continuously if deposits of \$1,000 are made at the end of
each of the next 5 years.
DIAGRAM:
0
1
2
F?
3
5 yrs
\$1 000
Problem 6
A firm pays back a \$10 000 loan with quarterly payments
over the next 5 years. The \$10 000 returns 4% APR
compounded monthly. What is the quarterly payment
amount?
DIAGRAM:
\$10 000
1
2
3
5 yrs = 20 qtrs
0
\$A
Problem 7
Anita Plass-Tuwurk, who owns an engineering consulting firm,
bought an old house to use as her business office. She found that
the ceiling was poorly insulated and that the heat loss could be cut
significantly if 6 inches of foam insulation were installed. She
estimated that with the insulation she could cut the heating bill by
\$40 per month and the air conditioning cost by \$25 per month.
Assuming that the summer season is 3 months (June, July,
August) of the year and the winter season is another 3 months
(December, January, and February) of the year, how much can she
spend on insulation if she expects to keep the property for 5 years?
Assume that neither heating nor air conditioning would be required
during the fall and spring seasons.
She is making this decision in April, about whether to install the
insulation in May. If the insulation is installed, it will be paid for at
the end of May. Anita’s interest rate is 9%, compounded monthly.
Problem 7
GIVEN:
SAVINGS = \$40/MO (DEC,JAN, FEB); \$25/MO (JUN, JUL, AUG)
r = 9%/YR, CPD MONTHLY
FIND P(SAVINGS OVER 5 YEARS):
1ST YR DIAGRAM:
PA ?
\$25
5 YR DIAGRAM:
PA
\$40
0
0 1
2 3 4 5 6 7 8 9 10 11 12 MO
1 2 3 4 5 YRS
P?
i = r = 0.09 = 0.75% / MO
m
12
PA = Pα + Pβ(PPβ) = Aα(P|A,i,nα) + Aβ(P|A,i,nβ)(P|F,i,6)
= \$25(P|A,0.75%,3) + \$40(P|A,0.75%,3)(P|F,0.75%,6)
= \$25(2.9556) + \$40(2.9556)(0.9562) = \$186.94 at the start of each year…
Problem 7 cont.
GIVEN:
SAVINGS = \$40/MO (DEC,JAN, FEB); \$25/MO (JUN, JUL, AUG)
r = 9%/YR, CPD MONTHLY
FIND P(SAVINGS OVER 5 YEARS):
5 YR DIAGRAM:
\$186.94
0
1 2 3 4 5 YRS
P?
m
r

ia   1    1
m

12
 0.09 
 1 
1

12 

 9.38%
P  P0  PA  A  A(P | A, i,N)
 (1  i)N  1
 A  A
N 
i
(
1

i
)


 \$186.94  \$186.94(P | A,9.38%,4)
 (1  0.0938)4  1 
 \$186.94  \$186.94 
4
0
.
0938
(
1

0
.
0938
)


 \$186.94  \$186.943.21288  \$787.56
```