ANNUITY

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ANNUITIES

ANNT-F
ANNT-G
Annuities
ANNT-H
ANNT-Y
ANNT-A

ANNT-Z

ANNT-B
ANNT-C
ANNT-D
ANNT-E

ANNT-A

.................................................................. ANNT 3
Present value of an ordinary annuity .................................................... ANNT 3
Mathematics
Learning Centre
The Future Value of an ordinary annuity ............................................ ANNT 6

A
nn
ui
tie
s
Objectives ..............................
.................................................
................. ANNT 1

Annuity Due ............................................................................................ ANNT 8
Mixed Questions...................................................................................... ANNT 10
Amortisation of Loans ........................................................................... ANNT 12
Index ........................................................................................................ ANNT 16
Solutions .................................................................................................. ANNT 17

Sequences................................
.................................................
................ ANNT 2
Annuities ................................ Objectives
1. To understand the concept of an annuity.
2. To perform calculations involving annuity lump sums.
3. To perform calculations involving annuity payments.
4. To determine if an annuity is an annuity due or an ordinary annuity.
5. To do calculations on amortised loans.

ANNT 2

Annuities

ANNT-B

Mathematics Learning Centre

Sequences

A sequence is a set of numbers arranged in a set order.
For example: 5, 10, 15, 20, 25… is a sequence. The first term is 5, the second is 10, and the third
is 15 and so on.
The type of sequence used to obtain Annuity formulas is a Geometric Sequence. In a geometric
sequence any new term is found by multiplying or dividing the previous term by a constant

number called the common ratio. The sequence 2, 4, 8, 16, 32, 64… is a geometric sequence
because any new value is found by multiplying the previous value by 2. The next term in this
sequence is 128 (i.e. 64x2).
Let’s consider the Geometric Sequence 1, 3, 9, 27, 81, 243 … The second term was determined
by multiplying the first term by 3 (3=1x3). The third term was found by multiplying the second
term by 3 (9=3x3). The fourth term was found by multiplying the third term by 3 (27 = 9x3) and
so on.
In this example the common ratio is 3. The common ratio is given the pronumeral r, the first
term is given the pronumeral a, therefore any sequence can be expressed as:
a, ar, ar 2 , ar 3 , ar 4 ,.........ar n1
For the geometric sequence 100, 110, 121, 133.1, 146.41…,, the first term a is 100, the
common ratio r is110 100 1.1. It is important to note that 121 110 & 133.1 121 also
give 1.1.
For the geometric sequence where a =10 and r = 1.5, the sequence is 10, 15, 22.5, 33.75,
50.625…
When the terms of a geometric sequence are added together
ie: a ar ar 2 ar 3 ......... ar n1 , this is called a Geometric Series.
An Annuity is simply a special type of Geometric Series.

ANNT 3

Annuities

ANNT C

Mathematics Learning Centre

Annuities

An annuity is a series of regular, equally spaced, payments over a defined period of time (often
called the term) at a constant rate of interest. The payments may occur weekly, fortnightly,
monthly, quarterly or yearly. Example of annuities includes: regular payments into a savings
account or superannuation fund, loan payments and periodic payments to a person from a
retirement fund. An Ordinary annuity is an annuity where the regular payment is made at the
end of the successive time periods. Another type of annuity, called an Annuity Due, will be
covered later in the module.
ANNT D

The Present Value formula of an Ordinary Annuity.
A loan from a bank is an example of an annuity with a present value and repayments for the term
of the loan. In other words, the banks gives you the lump now (at present) and the repayments
are made in periodic payments after this.

Present
0

R

R

R

R

R

1

2

3

4

5

R - periodic payment

Each repayment must be changed to its present value (see Finance module). Doing this and then
considering the sum of the series, the sum of the series is
1 (1 r )n
r
Where $ A is the present value of an ordinary annuity, $ R is the amount of each payment, r is
the rate per period (payment) and n is the number of periods (payments). An ordinary annuity is
an annuity where the payment is made at the end of the payment period.
A R

In this module, normal rounding will be used. However, be aware that financial institutions will
use a different method.

ANNT 4

Annuities

Mathematics Learning Centre

Examples ANNT-D1
1. Mr. and Mrs. Lyons wish to have an annuity for when their daughter goes to university.
They wish to invest into an annuity that will pay their daughter $1000 per month for 4
years. What is the present value of the annuity given that current interest rates are 8%
p.a?
0.08
 0.00666667 per month, n 12 4 48 months .
12
This annuity is considered to be an ordinary annuity, that is, each payment will occur at the end
of the payment period.
The information given is: R 1000, r

A R

11 r

A 1000

n

r
1 (1 0.00666667)48

0.00666667
0.273079396
A 1000
0.00666667
A 40961.91

This means $40 961.91 invested now at 8% will provide $1000 per month for 4 years.

2.

Roger Little borrows $20 000 to buy a car. He wishes to make monthly payments for 4
years. The interest rate he is charged is 10.5% p.a. What is the size of each monthly
payment?

The lump sum of $20 000 is given at the beginning of the loan, this indicates it is the present
0.105
12
annuity is considered to be an ordinary annuity because each payment occurs at the end of the
payment period.
A R
R
R

11 r

n

r
Ar
1 (1 r ) n
20000 0.00875
1 (1 0.00875)48

R $512.07

This means Roger will pay $512.07 per month for 4 years to repay the loan.
ANNT 5

value of an annuity: A $20000, r 0.00875 per month, n 12 4 48 months . This
Annuities

Mathematics Learning Centre

Exercise ANNT-D1
For each of the following situations, find the present value of the ordinary annuity
described.
1. $2000 per month for 5 years at a rate of 6% compounded monthly.
1
2. $5500 per quarter for 6 years at 5.6% compounded quarterly.
2
3. $150 per month for 3 years at 8% compounded semi-annually.
4. Joe wants to invest a lump sum of money now to cover a monthly commitment
of $100 over 5 years. If the lump sum is invested at 7.5% compounded monthly,
what amount will Joe invest?

5. Jane’s car insurance company charges a monthly premium of $41.50. What is
the present value of two year’s premiums if the rate of inflation is 0.3% per
month?
Exercise ANNT-D2
For each of the following situations, find the periodic payment necessary for the annuity
given the present value.
1. $25000 to fund 4 years of monthly payments at a rate of 5.25%
2. A public servant retires with $325000 in superannuation. He is offered an
annuity for the next 15 years where he will receive monthly payments. If the
interest rate is 8% compounded monthly, determine the size of the monthly
payment.
3. The executor of a will has to distribute an inheritance of $28 000 to a sole
beneficiary in equal monthly payments over 4 years while the beneficiary
undertakes university study. How much will each monthly payment be? (assume
an interest rate of 7.5%)
4. If an annuity is purchased for $50 000, how much will each quarterly payment
be over 10 years if the rate of interest is 5% compounded quarterly.

ANNT 6

Annuities

Mathematics Learning Centre

ANNT E

The Future Value formula of an Ordinary Annuity.

Where regular payments are made with a lump sum at the end, the lump sum at the end is called
the Future Value of an annuity. A good example of this is a saving scheme where regular
payments are made to build to a lump sum at the end of a period of time. In business, this is
called a sinking fund. It is used to save for the future replacement of major capital items.

Future
Value ($)
0
R

1
R

8
R

9
R

10
R

The future value of an annuity is the value of all payments at the end of the term. It is given by
(1 r )n1
S R
r
the equation:
, where S is the future value of the ordinary annuity, R is the
periodic payment, r is the interest rate per period and n is the number of payments.
Examples ANNT E1:

George deposits $150 into a bank account at the end of the month for 5 years at a rate of
1.
7% compounded monthly.
(1 r )n 1
r
R 150, r 0.07 12 0.005833&
n 512 60
S R

S 150
S 150 71.59290091
S 10738.94
The future value of the annuity is $10 738.94

ANNT 7

Annuities

Mathematics Learning Centre

2. Wally is planning for his retirement 20 years away. When he retires he wants a lump
sum of $300 000. His financial advisor suggested that 5% p.a. was a suitable interest rate to
consider. How much will he have to pay per month into his retirement fund (assume ordinary
annuity).
(1 0.005833)601
0.005833&
Given: S 300000, r 0.05 12 0.00416, n 12 20 240
S R

(1 r )n1
r

Sr
R
(1 r )n 1
300000 0.00416&
R
(1.00416)2401
R 729.87
Wally will have to pay $729.87 per month until he retires in 20 years.

Exercise ANNT-E1
1. Find the future value of an annuity where $1200 is paid at the end of each year for 15
years at a rate of 7% compounded annually.
2. The ABC concreting company set up a sinking fund to assist in buying a new truck in 5
years time. They can only afford $3000 a quarter which is paid into a savings account
with an interest rate of 8% p.a. Assuming quarterly compounding, what will be the size
&
of the sinking fund after 5 years? (Assume ordinary annuity)
3. Colin invests $500 per month, paid into a savings account for 10 years. What is the
balance of the account at the end of the period assuming an interest rate of 7%
compounded monthly?
4. In 5 years, a printing machine is to &be replaced. A new machine is expected to cost
$33000. Assuming an annual interest rate of 8% compounded monthly, what will be the
size of each monthly payment?

ANNT 8

Annuities

Mathematics Learning Centre

ANNT F Annuity Due
So far, all calculations have been performed with ordinary annuities, where periodic payments
are made at the end of the period. An annuity due differs from an ordinary annuity since
periodic payments are made at the beginning of the period. Some payments such as house rents
and insurance premiums are paid at the beginning of the period before the service is provided. If
the payment is made at the beginning of the period, each payment will be subject to an extra
month’s interest when compared to ordinary annuities. A simple modification to the ordinary
annuity formulae give:
PRESENT VALUE OF AN ANNUITY DUE
AD1 r AO
AD R1 r

11 r

n

r

FUTURE VALUE OF AN ANNUITY DUE
S D1 r SO
S D R1 r

1 r

1

Examples ANNT F1
1. A company wishes to deposit an amount of money into an account at the beginning of
each year for the next 5 years to purchase a new machine costing $50000. How much will each
yearly payment be if the current interest rate is 7.2%p.a?
This is a sinking fund scenario. It is an annuity due.
Given: n 5, r 0.072, S 50000
S D R1 r
SD r

1 r

1

R

n

⎣⎦
50000 0.072

R

⎣⎦
3600

R
0.445639816
R 8078.27

$8 078.27 is required at the beginning of the year for 5 years at 7.2% to have $50 000 in the
sinking fund at the end of 5 years.

ANNT 9

Annuities

Mathematics Learning Centre

2. Joe pays $250 rent per week at the beginning of each week. He is considering paying a
whole year’s rent in advance; given the interest rate is 5.2% p.a. How much is this amount?
This is a present value calculation. This is an annuity due situation.
Given: R 250, r 0.052 52 0.001, n 52

1rr
⎡1 r−1⎤1
1

AD R1 r

n

r

11.001
1 1.072
A⎡D1.072
 25051.001
0.001
AD 12674.28
Joe would have to pay $12 674.28 now to cover his rent for the year.
52

Exercise ANNT – F1
1. Find the future value of an annuity due if $800 is paid into an account at the beginning of
each month for 5 years at a rate of interest of 5% p.a. compounded monthly.
2. Find the present value of an annuity if the periodic amount is $450 per quarter for 20
years at the rate of 4.5% p.a. compounded quarterly.

3. A company leases office space for a period of 12 months. The monthly rent of $2500 is
paid at the beginning of each month. If the company is to cover all rents with a single
lump sum at the beginning of the year and invests this at 6.3% p.a. how much will the
lump sum be?

ANNT 10

Annuities

ANNT G

Mathematics Learning Centre

Mixed Questions

For the following questions, follow these steps.
1. Read the question thoroughly
2. Decide on the type of annuity: either ordinary annuity, where periodic payments are
made at the end of the period OR annuity due, where periodic payments are made at the
beginning of the period.
3. Decide if the annuity contains a lump sum in the present, the lump sum is at the
beginning of the term of the annuity, OR future, the lump sum is at the end of the
annuity.
4. Select the appropriate formula

Example ANNT G1
Kevin Smith won $820 000 in a lotto game. With this lump sum he purchases an annuity to give
him a monthly income for the next twenty years. The first payment occurs a month after the start
of the annuity. If the interest rate is 6.8% p.a. compounded monthly, calculate the amount of
each payment.
This is an ordinary annuity (The periodic payments occur at the end of the period.)
This annuity involves a present value (The lump sum is at the beginning of the term of the
annuity.)
The formula required is A R

1 (1 r ) n
r

ANNT 11

Presnt

uVealA

Ordinary Annuity
Annuities

n

1 (1 r )
A R
r

Annuity Due
n

11 r
Mathematics
Centre
AD R1 Learning
r
r

Futre

uVealS

n
n
0.068
 0.005666& n 2012 240 1 r1
Information given is A 820000
(1 r r
)1
12
S R
S D R1 r
r
r
Rearranging the formula to make R the subject:
1 (1 r ) n
A R
r
Ar
R
1 (1 r ) n

820000 0.005666&
1 (1.005666)240
R $6259.38
Kevin will receive $6259.38 per month for 20 years.
R

Exercise ANNT G1
1. Sally borrows $10 000 to buy a car. If the interest rate charged is 10.5% p.a. calculate the
monthly repayment over the term of the loan, 5 years. The first payment is made a month
after the loan lump sum is advanced.
2. Brett wishes to set aside all his rent for one year. This money will be put into an account
paying 6.6% p.a. compounded monthly and his rent is $1040 per month. Rent is always
paid in advance. Calculate the amount Brett must deposit.

3. From the time Jane and John’s daughter was born, they decided to save for her university
education. Jane and John assume their daughter will require $1000 per month for her
four years of study, payments being made at the beginning of each month. If Jane and
John save for 18 years, calculate the amount they must save at the beginning of each
month. Assume 6% p.a. interest is compounded monthly. Hint: there are two annuities
here, calculate the lump sum required to fund the $1000 p.m. allowance first, then the
monthly amount the parents must save.
4. The PS Transport Company decide that they must start saving for a new vehicle in 5
years time. In an account that pays 5.4% p.a. compounded monthly they deposit a one
off payment of $20 000 and $500 at the end of each month. How much will they have at
the end of 5 years? (Hint: treat the two amounts separately, one is an annuity and the
other compound growth)
&

ANNT 12

Annuities

ANNT H

Mathematics Learning Centre

Amortization of Loans

The term amortization refers to loans where each payment is made up of principal and interest
components. An example of where amortization is commonly used is housing mortgage loans
which usually have a term of many years. Early in the term of the loan, the amount of principal
outstanding is large, hence the interest component of the loan is high and the amount paid off the
loan is small. As the loan progresses, the amount of principal paid each repayment increases and
the amount of interest decreases. The operation of the loan can be shown using an amortization
schedule.
Example ANNT H1
For this example, a loan of $100 000 is being considered over a term of 10 years at an interest
rate of 9%p.a. with monthly repayments. Repayments on loans are made at the end of the month
so this is an ordinary annuity. The lump sum is at the beginning of the annuity so the present
value formula is used.
The first step is to calculate the repayment amount of the loan.
0.09
12
A R
R
R

11 r

n

r
Ar
1 (1 r ) n
100000 0.0075

1 (1 0.0075)120
R $1266.76

The monthly repayment for this loan (annuity) is $1266.76 per month.
The amount of interest for the first month is $100000 0.0075 $750 , so the amount of principal
paid off the loan during the first month is $1266.76 - $750 = $516.76. So the balance of the loan
for the second month is $100000 - $516.76 = $99483.24, and this continues until the loan is paid
off after 10 years. The amortization schedule for the first six months of the loan is shown below.

ANNT 13

Annuities

Mathematics Learning Centre

A $100000, r 0.0075, n 1012 120

The total amount of interest paid over the 10 years of the loan is calculated by calculating the
total paid in repayments 1201266.76 $152011.20 subtract the amount borrowed $100 000 to
give $52 011.20
The amount of principal outstanding at the beginning of the 25th month: At the beginning of
the 25th month, 24 payments have been made, resulting in 96 months left on the loan. Using the
present value of an annuity, the amount of principal outstanding is given by:
11 r
A R
r

n

11 .0075
A 1266.76
0.0075
A 86467.06
The amount of principal outstanding at the beginning of the 25th month (which is also the
balance at the end of the 24th month) is $86 467.06
96

Because this is the amount of principal outstanding at the beginning of the 25th month, the

amount of interest for the 25th month is 0.0075 $86467.06 $648.50
Consequentially, the amount of principal paid in the 25th month will be $1266.76-$648.50 =
$618.26
Month
Principal
Interest per
Payment at
Principal paid Principal
Example ANNTOutstanding
H2
month
the end of the off
Outstanding
at the
month
outstanding
at the end of
beginning
principal
A person amortizes
a loan of $30000 for a new car by obtaining a 10
year loan at a the
ratemonth
of 12%
the
month
with monthly payments.
1
$100 payment
000
$750
$1266.76
$516.76
$99 483.24
Find
(a) the monthly
2
$99 483.24
$746.12
$1266.76
$520.64
$98 962.60
(b) the total interest charged
3
$98 962.60
$742.22
$1266.76
$524.54
$98 438.06
(c) the principal remaining after 4 years
4
$98 438.06
$738.29
$1266.76
$528.47
$97 909.59
(d) the interest
paid and the
principal paid $1266.76
during the 49th month
5
$97 909.59
$734.32
$532.44
$97 377.15
6

$97377.15

Annuities

$730.33

$1266.76

ANNT 14

$536.43

$96 840.72

Mathematics Learning Centre

Exercise ANN-H1
1. Construct an amortization schedule for the following scenario: A loan of $1200 is repaid

by 6 quarterly payments, interest is 10% compounded quarterly.
2. John and Joanne take out a mortgage housing loan for $250 000 over 20 years at an
interest rate of 8% with monthly payments. Calculate the size of each repayment.
(a) The monthly repayment:
(b) The total interest charged:
3.
John
and
Joanne’s
interest
rate
(from
question
2) n
hasR
just
A 30000 r 0.12 12 0.01 n 1012 120
I
A been increased to 9%, by how
much will their
repayments
increase
by
so
they
can
still
repay the loan after 20 years?
n
I 120 430.41 30000
11 r
A R
I $21649.20
r mortgage loan for 25 years for home
4. A $45 000
additions is obtained at a rate of 7.75%
rearranging
repaid in monthly repayments. Find
The total interest charged is $ 21 649.20
Ar
a. The monthly repayment
R n
1 (1 r )
ANNT 15
30000 0.01
R120
1 (1.01)
R $430.41
Annuities
The monthly payment is

$430.41 per month.

Mathematics Learning Centre

(c) The principal remaining
after 4 years
or 48 at the
(d) The
interestofpaid
and
the principal paid
outstanding
beginning
the 36
th month.
b. The principal
th
months.
during the 49 month.
c. The interest in the 36th payment.
After 48 months, there
are
120
48
72
months
d. The principal in the 36th payment.
th
remaining.
The start amount for the 49 month is $22
The
total
interest
paid.
e.
n
015.64 (see (c)).
11 r
Interest,
A R
r
5. Marilyn and Murray take out a home mortgageIloan
for $150 000 over 20 years at an
r 22015.64
72
1rate
(1.01)
interest
of 6% with monthly payments. I 0.01 22015
A 430.41
0.01
a. Calculate the amount of each monthly
repayment.
I $220.16
A $22015.64
th being made
After 10 years Marilyn receives a lump sumThe
payout
of $30
interest
for000
the after
49 month
is $220.16
At the end
of
the
48th
month
(or
the
beginning
redundant from her university position. She decides to pay this off the principal of the
th
of the 49th
thedecides
amounttoofshorten
principal
loan.month)
She also
thestill
term ofThe
the principal
loan to just
5 more
years
(15month
in total).
paid
during
the 49
is
outstanding in $22 015.64
b. What will the size of the new repayment
R I 49th monthbe?
 430.41 220.16 $210.15

ANNT 16

Annuities

TRIG-Y

Mathematics Learning Centre

Index

Topic
Amortization of Loans
Amortization Schedule
Annuities
Annuity Due
Future Value of an Ordinary Annuity
Future Value of an Annuity Due
Geometric Sequence
Geometric Series
Interest Paid
Ordinary Annuity
Periodic Payment
Present Value of an Ordinary Annuity
Present Value of an Annuity Due
Principal Outstanding
Sequence
Term

Page

ANNT
12
ANNT
12
ANNT
3
ANNT
3,
ANNT
8

ANNT 2
ANNT 2
ANNT 13
ANNT 2
ANNT 3
ANNT 3
ANNT 8
ANNT 13

ANNT
6
ANNT 2
ANNT
8
ANNT 3

ANNT 17

Exercise ANNT-D1 Solutions
1. $2000 per month for 5 years at a rate of 6%
compounded monthly.
Both n & i must be expressed in months.

4. Joe wants to invest a lump sum of money now to
cover a monthly commitment of $100 over 5 years. If
the lump-sum is invested at 7.5% compounded
monthly, what amount will Joe invest?

R 2000, n 512 60, i 0.06 12 0.005

R100, n 512 60, i 0.075 12 0.00625
n

1(1 r)
A R
r
60
1(10.00625)
A100
0.00625
A10049.90530818
A 4990.53

n

1 (1 r)
A R
r
60
1 (1 0.005)
A 2000
0.005
A 2000 51.72556075
A 103451.12

If Joe invests $4 990.53 now he will meet his monthly
commitment of $100 per month for 5 years.

The present value of this ordinary annuity is
$103 451.12

5. Jane’s car insurance company charges a monthly
premium of $41.50. What is the present value of two
year’s premiums if the rate of inflation is 0.3% per
month?

1
6 years at 5.6%
2
compounded quarterly. Both n & i must be
2. $5500 per quarter for
expressed in quarters.

R 41.50, n 212 24, i 0.03

1
R 5500, n 6 4 26,
2
i 0.056 4 0.014

n

1(1 r)
A R
r
24
1(1 0.003)
A 41.50
0.003
A 41.5023.122934
A 959.60

n

1 (1 r )
A R
r
26
1 (1 0.014)
A 5500
0.014
A 5500 21.66802573
A 119174.14

The present value of Jane’s car insurance premiums is
$959.60

The present value of this ordinary annuity is $119
174.14
3. $150 per month for 3 years at 8% compounded
semi-annually. Semi-annually means half yearly.
R, n & i must be expressed in half years.

R 150 6 900, n 3 3 6,
i 0.08 2 0.04
n

1 (1 r)
r
6
1 (1 0.04)
A 900
0.04
A 900 5.242136857
A 4717.92
A R

The present value of this ordinary annuity is $4 717.92

ANNT 17

Exercise ANNT-D2 Solutions
1. $25000 to fund 4 years of monthly payments at a
rate of 5.25%

A 25000, n 412 48,
r 0.0525 12 0.004375
n

A R

1 (1 r)
r
Ar

R n
1 (1 r)
25000 0.004375
R48
1 (1 0.004375)
109.375
R
0.189044389
R 578.57
The monthly payment will be $578.57

3. The executor of a will has to distribute an
inheritance of $28000 to a sole beneficiary in equal
monthly payments over 4 years while the beneficiary
undertakes university study. How much will each
monthly payment be?

A 28000, n 412 48,
r 0.075 12 0.00625
Ar
R n
1 (1 r)
28000 0.00625
R48
1 (1 0.00625)
175
R
0.258489819
R 677.01
The student will receive a monthly payment of
$677.01

2. A public servant retires with $325000 in
superannuation. He is offered an annuity for the next
15 years where he will receive monthly payments. If
the interest rate is 8% compounded monthly,
determine the size of the monthly payment.

A 325000, n 1512 180,
r 0.08 12 0.006666&
Ar
R n
1 (1 r)
325000 0.0066&
R180
1 (1 0.0066)&
2166.666
R
0.697603911
R 3105.87

4. If an annuity is purchased for $50 000, how much
will each quarterly payment be over 10 years if the
rate of interest is 5% compounded quarterly.

A 50000, n 10 4 40,
r 0.05 4 0.0125
Ar
R n
1 (1 r)
50000 0.0125
R40
1 (1 0.0125)
625
R
0.391586664
R 1596.07
The quarterly payment will be $1596.07.

The monthly payment will be $3105.87

ANNT 18

Exercise ANNT-E1 Solutions
1. Find the future value of an annuity where $1200 is
paid at the end of each year for 15 years at a rate of
7% compounded annually.

3. Colin invests $500 per month, paid into a savings
account for 10 years. What is the balance of the
account at the end of the period assuming an interest
rate of 7% compounded monthly?

R 1200, r 0.07, n 15.

R 500, r 0.07 12 0.00583,&
n 1012 120.

n

(1 r)1
r
15
(1.07)1
S 1200
0.07
S $30154.83
S R

n

(1 r)1
S R
r
120
(1.00583)&1
S 500
0.00583&
S $86542.40

The future of this ordinary annuity is $30154.83

Colin will have a savings account balance of
$86542.40
4. In 5 years, a printing machine is to be replaced. A
new machine is expected to cost $33000. Assuming
an annual interest rate of 8% compounded monthly,
what will be the size of each monthly payment?

2. The ABC concreting company set up a sinking
fund to assist in buying a new truck in 5 years time.
They can only afford $3000 a quarter which is paid
into a savings account with an interest rate of 8%p.a.
Assuming quarterly compounding, what will be the
size of the sinking fund after 5 years?

S 33000, r 0.08 12 0.006,&
n 512 60.

R 3000, r 0.08 4 0.02, n 5 4 20.

n

(1 r )1
S R
r

n

(1 r)1
S R
r
20
(1.02)1
S 3000
0.02
S $72892.11
The company will have saved $72892.11 towards the
new truck.

Sr

n R
(1 r)1
220
R
0.489845649
R $449.12

A payment of $449.12 per month will be required to
raise $33000 for new machinery in 5 years time.

ANNT 19

Exercise ANNT-F1 Solutions
1. Find the future value of an annuity due if $800 is paid
into an account at the beginning of each month for 5
years at a rate of interest of 5% p.a. compounded monthly

0.05
R 800, r 0.0041666,& n 512 60
12

n

(1 r)1
S R(1 r )
r
60

(1.0041666)&1
S 8001.0041666&
0.0041666&
S 54631.55
The future value of this annuity due is $54 631.55

2. Find the present value of an annuity if the periodic
amount is $450 per quarter for 20 years at a rate of 4.5%
p.a. compounded quarterly

0.045
R 450, r 0.01125, n 20 4 80
4

The present value of this annuity due is $23 921.41

3. A company leases office space for a period of 12
months. The monthly rent of $2500 is paid at the
beginning of each month. If the company is to cover all
rents with a single lump sum at the beginning of the year
and invests this at 6.3% p.a. how much will the lump sum
be?
This is the present value of an annuity due.

0.063
R 2500, r 0.00525, n 112 12
12

n

1 (1 r )
A R(1 r )
r
80
1 (1.01125)
A 4501.01125
0.01125
A 23921.41

n

1 (1 r )
A R(1 r )
r
A 25001.00525

1 (1.00525)
0.00525

12

A 29153.10
The company must invest $29 153.10 to cover rent
for the entire year.

ANNT 20

Exercise ANNT G1 Solutions
1. Sally borrows $10 000 to buy a car. If the interest
rate charged is 10.5%p.a. calculate the monthly
repayment over the term of the loan, 5 years. The first
payment is made one month after the loan lump sum is
advanced.
This involves the present value of an ordinary annuity.
R 10000, i 0.105 12 0.00875, n 512 60
n

A R

3. From the time Jane and John’s daughter was born,
they decided to save for her university education. Jane
and John assume their daughter will require $1000 per
month for her four years of study, payments being
made at the beginning of each month. If Jane and John
save for 18 years, calculate the amount they must save
at the beginning of each month. Assume 6% p.a.
interest is compounded monthly.
Amount required to give $1000 per month for 48
months is: present value of an annuity due.

1 (1 r )
r
Ar

R 1000, r 0.06 12 0.005, n 412 48

R n
1 (1 r )
10000 0.00875
R60
1 (1.00875)
R 214.94
Sally will pay $214.94 per month.

n

1 (1 r )

AD R(1 r )

r
1 (1.005)

AD 10001.005

48

0.005

AD 42793.22

The monthly deposit for 18 years is: future value of an
annuity due.

S 42793.22, r 0.005, n 1812 216
n

(1 r)1
r

S R(1 r)
Sr
n

(1 r)(1
r ) 1

R

R

42793.22 0.005
216

1.005[(1.005)1]
R 109.93

Jane and John should deposit $109.93 at the beginning
of each month for 18 years.
2. Brett wishes to set aside all his rent for one year.
This money will be put into an account paying 6.6%
p.a. compounded monthly and his rent is $1040 per
month. Rent is always paid in advance. Calculate the
amount Brett must deposit.

4. The PS Transport Company decide that they must
start saving for a new vehicle in 5 years time. In an
account that pays 5.4% p.a. compounded monthly,
they deposit a one off payment of $20 000 and $500 at
the end of each month. How much will they have at
the end of 5 years?

This involves the present value of an annuity due.

R1040, r 0.066 12 0.0055, n11212
n

P 20000, i 0.0045, n 512 60

1 (1 r )
AD R(1 r )
r
AD 10401.0055

1 (1.0055)
0.0055

AD 12111.31
Brett must deposit $12 111.31

Using the compound interest (growth) formula, the
$20000 will grow to:
n

60

A P(1 i) 20000(1.0045) 26183.43
12

$500 per month is used to calculate the future value of
an ordinary annuity (sinking fund).

R 500, r 0.0045, n 60
n

(1 r ) 1
r
60
(1.0045) 1
S 500
0.0045
S 34352.36
Altogether, there will be $60 535.79 available to
purchase a new vehicle.
S R

ANNT 21

Exercise ANN H1
1. Construct an amortization schedule for the following scenario: a loan of $1200 is repaid by 6 quarterly
payments, interest is 10% compounded quarterly.
To do this a repayment figure is required. This scenario involves the present value of an ordinary annuity.
R 1200,
A R
R
R

i 0.10 4 0.025, n 6

1 (1 r ) n
r
Ar
1 (1 r ) n
1200 0.025
1 (1.025)6

R 217.86
Quarter
1
2
3
4
5
6

Beginning Principal
1200
1012.14
819.58
622.21
419.91
212.55

Interest
0.025x 1200=30
0.025x1012.14=25..30
0.025x819.58=20.49
0.025x622.21=15.56
0.025x419.91=10.50
0.025x212.55=5.31

Payment
217.86
217.86
217.86
217.86
217.86
217.86

Principal paid off
217.86-30 = 187.86
217.86-25.30=192.56
217.86-20.49=197.37
217.86-15.56=202.30
217.86-10.50=207.36
217.86-5.31-212.55

End Principal
1200-187.86=1012.14
1012.14-192.56=819.58
819.58-197.37=622.21
622.21-202.30=419.91
419.91-207.36=212.55
212.55-212.55=0

2. John and Joanne take out a mortgage housing loan for $250 000 over 20 years at an interest rate of 8% p.a.
with monthly payments. Calculate the size of each payment.
A loan like this is an example of an ordinary annuity with a present value (the amount borrowed).

A 250000, r
A R
R
R

0.08
 0.00666,& n 2012 240
12

1 (1 r ) n
r
Ar
1 (1 r ) n

A monthly repayment of $2091.10 is required to service the loan.

250000 0.00666&

1 (1.00666)&240
R 2091.10
3. John and Joanne’s interest rate (from question 2) has just been increased to 9%, by how much will their
payments increase by so they can still repay the loan after 20 years?

A 250000, r
A R
R
R

0.09
 0.0075, n 2012 240
12

1 (1 r ) n
r
Ar
1 (1 r ) n
250000 0.0075
1 (1.0075)240

R 2249.31
The new monthly repayment is $2249.31, an increase of $158.21

ANNT 22

4. A $45 000 mortgage loan for 25 years for home additions is obtained at a rate of 7.75% repaid in monthly
repayments. Find
(a) The monthly repayment: A loan like this is an example of an ordinary annuity with a present value (the
amount borrowed).

A 45000, r
A R
R
R

0.0775
12

 0.0064583,& n 2512 300

1 (1 r ) n
r
Ar
1 (1 r ) n

The monthly repayment is $290.62

45000 0.0064583&
1 (1.0064583)300

R 290.62
(b) The principal outstanding at the beginning of the 36th month.
At the beginning of the 36th month,35 payments have been made, so there are 300-35=265 months remaining.
1 (1 r ) n

A R

r
1 (1 0.0064583)265 The principal outstanding at the beginning of

A

the 36th month is $36 827.42


2
9
0.
6
2

A 36827.42
(c) The interest in the 36th payment.
Interest for the 36th payment is

0.0064583& 36827.42 237.84

(d) The principal in the 36th payment.
Principal = $290.62-$237.84 = $52.78
In the 36th month, the principal outstanding will reduce by $52.78
(e) The total interest paid.
Total paid off the loan is $290.62 x 300=$87 186, total interest paid = $87 186 - $45 000 = $42 186
5. Marilyn and Murray take out a home mortgage loan for $150 000 over 20 years at an interest rate of 6% with
monthly payments.
(a) Calculate the amount of each monthly payment.

A150000, r
A R
R
R

0.06
12

 0.005, n 2012 240

1 (1 r ) n

r
Ar
1 (1 r ) n
150000 0.005
1 (1.005)240

R 1074.65
Each repayment will be $1074.65

ANNT 23

&

&

(b) After 10 years, Marilyn receives a lump sum payout of $30 000 after being made redundant from her
university position. She decides to pay this off the principal of the loan. She also decides to shorten the term of
the loan to just 5 more years (15 in total).
What will the size of the new repayments be?
After 10 years (120 months) the $30 000 is paid off the principal of the loan, therefore the lump sum is paid off
during the 121st payment. The outstanding principal at the end of the 121st month, which is the same as the
outstanding principal at the beginning of the 122nd month. After the 121st month, there are 240-121 = 119
months remaining
1 (1 r ) n

A R

r

A 1074.65

1 (1 0.005)119

0.005
A 96206.77
The balance of the loan at the end of the 121st month (beginning of the 122nd month) is $96 206.77 .
Because $30 000 is paid off the loan during the 121st month, the new principal outstanding will be $96206.77 $30 000 = $66 206.77 .
If the loan is now shortened to 4 years 11 months at the beginning of 122 nd month, a new repayment figure can
be calculated.

A 66206.77, r
A R
R
R

0.06
 0.005, n 59
12

1 (1 r ) n

r
Ar
1 (1 r ) n
66206.77 0.005
1 (1.005)59

R 1298.57
The new repayment figure is $1298.57

ANNT 24

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