ANNUITIES
Sequences and Series
Consider the following sequences of numbers:
2, 4, 6, 8, . . . .
2, 4, 8, 16, . . . .
What is the mathematical difference between the
two?
1
The first is called an arithmetic sequence since
it can be described in the following way:
2, 2+2, 2+2+2, ...
The second is a geometric sequence since it can
be described as:
2, 2×2, 2×2×2, 2×2×2×2, ...
The geometric sequence is of interest to us.
2
In general terms, the first n terms of the
sequence can be described by
a, ar , ar 2 , ar 3 , .. . . ., ar n1
(In the previous example, a=2, r=2).
3
Example 1
a = 3, r = 2
a, ar, ar 2 , ar 3 , . . . . .
3, 3  2, 3  2 , 3  2 . . . . .
2
3
3, 6, 12, 24, . . . . .
4
Example 2
a  1, r  1.5
2
3
a, ar , ar , ar , . . . . .
1, 1(1.5), 1(1.5) 2 , 1(1.5) 3 ,.....
1, 1.5, 2.25, 3.375,....
5
Example 3
$500 invested at 12% compounded annually.
500……………….$500
500(1  0.12).........$560
500(1  0.12) 2 .......$627.20
500(1  0.12) 3 ........$702.46
500(1  0.12) ........$786.76
4
a  500, r  1.12
6
An annuity is a sequence of payments made
regularly over a given time interval (eg loan
repayments).
The time interval is called the term of the
annuity. The regular places where repayments
are made are called payment periods.
7
Payments made at the end of the payment
period are called ordinary annuities.
When payments are made at the beginning
of the payment period, the process is called
an annuity due.
We first consider an ordinary annuity.
8
0
1
R
2
R
3
R
4
R
n-1
R
n
R
9
It can be shown that the present value of these
payments is given by
1  (1  r )
AR
r
n
This gives the present value A of an annuity of
$R per payment period for n periods at the rate r
per period.
10
This formulation can be rearranged so that
Ar
R
1  (1  r ) n
which gives the periodic payment R of an
annuity whose present value is A.
11
Example 4
The present value of quarterly payments of $250
for 5 years at 12% compounded quarterly is
0.12
r
 0.03
4
n  5  4  20
 1  (1  0.03) 20 
A  250 



0.03
 $3719.37
12
Example 5
$25000 is borrowed over 8 years. What will be
the monthly repayments at 18% compounded
monthly?
0.18
r
 0.015
12
n  8 12  96
25000  0.015
R
1  (1  0.015) 96
 $493.08
Monthly repayments should be $493.08.
13
If the loan is taken over 5 years, then
25000  0.015
R
1  (1  0.015) 60
 $634.84
14
Example 6
A person wishes to borrow $5000 now and
$4000 two years from now.
Both loans are to be repaid with equal monthly
payments made at the end of the month for the
next five years.
What is the monthly payment?
(Assume 10% compounded monthly.)
15
0
1
2
3
4
5
R
R
R
4000
5000
R
R
16
Bring everything back to the present value.
Loans are presently worth
 0.1 
5000  40001 

 12 
24
 5000  3277.64
 $8277.64
17
The present value of the repayments is given by
  0.1  60 
 1  1 
 
  12  
A  R

0 .1




12


18
Equating the PV of the loan and payments we
60
obtain

0.1 
 1  1  12

8277.64  R 

0.1




12
0.1
8277.64 
12  175.88
R
60
0.1
1 1
12
The monthly repayment will be $175.88.




19
Note well the discussion on
page 9-8 of the study guide
regarding avoiding rounding
errors in calculations.
20
The future value of an annuity is the value
of all payments at the end of the term.
n-3
1
2
3
4
5
R
R
R
R
R______R
n-2
R
n-1
R
n
R
21
It can be shown that the future value of an
annuity of n periodic payments of $R with a
rate r per period is
(1  r ) n  1
SR
r
22
Manipulating this equation gives a formula
for R.
Sr
R
n
(1  r )  1
R is the periodic payment that must be made to
amount to S at the end of the term.
Investing in this way to meet some future
obligation is commonly called a sinking fund.
23
Example 7
If you wish an annuity to grow to $17000 over
5 years so that you can replace your car, what
monthly deposit would be required if you
could invest at 12% compounded monthly?
0.12
r
 0.01
12
n  5  12  60
17000  0.01
R
(1  0.01) 60  1
The monthly payment
 $208.16
should be $208.16.
24
Example 8
An annuity consists of monthly repayments of
$600 made over 20 years.
(a) What is the present value of the annuity?
(b) How much money is repaid?
(c) What is the future value of the payments?
(Assume 14% compounded monthly.)
0.14
r
12
n  20  12  240
25
Question (a)
 1  (1  r ) n 
A  R

r




0.14
 1  1  12
 600 
0.14

12


240





 $48250.10
The present value of the annuity is $48250.10.
26
Question (b)
The amount repaid is
 600 12  20
 $144000
27
Question (c)
(1  r ) n  1
SR
r


240
0.14
1
12
 600
0.14
12
1
 $780 699.45
The future value of the annuity is $780 699.45.
28
An annuity where each payment is due at the
beginning of the payment period is called
an annuity due.
0
(1)
(2) R
1
2
3
4
R
R
R
R
R
R
R (ordinary)
(annuity due)
1  (1  r )
(1) A  R
r
4
1  (1  r )
(2) R  R
r
3
29
The second case, describing the annuity due,
can be thought of as an initial payment
followed by an ordinary annuity of shorter
duration (one payment period shorter).
It can be shown that the present value of an
annuity due is equal to
AD  (1  r ) AO
In a similar way
S D  (1  r ) S O
30
Example 9
If payments of $100 are received at the beginning
of each payment period for 4 years, once a year,
at a rate of 15% p.a. compounded annually, what
is the present value and the future value?
0
1
2
3
4
100
100
100
100
31
The present value is given by
100  100(1.15)-1  100(1.15)2  100(1.15)3
 100  86.96  75.61  65.75
 $328.32
Using the formula
AD  (1  r ) AO
 1  (1  0.15) 4 
 (1  0.15) 100

0.15


 $328.32
32
0
1
2
3
100
100
100
100
4
33
The future value is given by
100(1.15)1  100(1.15) 2  100(1.15)3
 100(1.15)
4
 115  132.25  152.09  174.90
 $574.24
34
Using the formula
S D  (1  r ) SO
 1.15 4  1
 1.15100

0.15 

 $574.24
In summary, treat all annuity problems as
ordinary and then make the correction (multiply
by 1+r) for annuities due if necessary.
35
Example 10
A company wishes to lease temporary office space
for a period of 6 months.
The rental fee is $500 a month payable in advance.
Suppose that the company wants to make a lumpsum payment, at the beginning of the rental period,
to cover all rental fees due over the 6-month period.
If money is worth 9% compounded monthly, how
much should the payment be?
36
0.09
r
 0.0075, n  6, R  500
12
This is an annuity due. However, treating the
problem as an ordinary annuity, the present
value is given by
1  (1  r )  n
AR
r
1  (1.0075) 6
 500
0.0075
 2922.80
37
Correcting for an annuity due
A  1.0075 2922.80
 $2944.72
A lump sum payment of $2944.72 should be
made to cover the 6 month rental.
38
Example 11
A owes B the sum of $5000 and agrees to pay B
the sum of $1000 at the end of each year for 5
years and a final payment at the end of the sixth
year.
How much should the final payment be if
interest is at 8% compounded annually?
39
r  0.08, n  5, R  1000
This is an ordinary annuity.
The future value of the annuity is given by
(1  r )  1
SR
r
 1.085  1
 1000

 0.08 
n
 $5866.60
40
The value of this money at the end of the sixth
year is 5866.60 (1  0.08)
 $6335.93
The value of the debt $5000 six years into the
future is equal to 5000(1  0.08) 6
 $7934.37
The final payment is then
7934.37  6335.93  $1598.44
41
0
1
2
3
4
5
6
7934.37
5000
Debt:
S  5000 1  0.08 
6
 7934.37
42
0
1
2
3
4
5
6
7934.37
5000
1000
1000
1000
1000
1000
5866.60
The value of the repayments after the 5th year is
 1.085  1 
S  1000 

 0.08 
 5866.60
43
0
1
2
3
4
5
6
7934.37
5000
1000
1000
1000
1000
1000
5866.60
6335.93
At the end of the next year the money has the
value of S  5866.60 (1  0.08)1
 6335.93
44
The equation of value at the end of the sixth
year is
6335.93  x  7934.37
x  1598.44
The final payment should be $1598.44.
45
Example 12
In 10 years a $40000 machine will have a
salvage value of $4000. A new machine at that
time is expected to sell for $52000. In order to
provide funds for the difference between the
replacement cost and the salvage value, a sinking
fund is set up into which equal payments are
placed at the end of each year. If the fund earns
7% compounded annually, how much should
each payment be?
46
r  .07
n  10
0
10
40 000
4 000
52 000
-R - R
48 000
47
 (1  r ) n  1 

S  R
r


 1.0710  1 

48 000  R
 .07 
48 000  R 13.82
48 000
R
13.82
 3474.12
Sinking fund repayments should be $3474.12.
48
Example 13
A paper company is considering the purchase of
a forest that is estimated to yield an annual return
of $50000 for 10 years, after which the forest
will have no value. The company wants to earn
8% on its investment and also set up a sinking
fund to replace the purchase price. If money is
placed in the fund at the end of each year and
earns 6% compounded annually, find the price
the company should pay for the forest. Give
your answer to the nearest hundred dollars.
49
50 000
8% return
sinking fund repayment
Let the purchase price be x.
50 000  0.08 x  R
(*)
50
To recoup the purchase price, the repayments R
must amount to x in 10 years.
r  0.06
n  10
 1.0610  1 

x  R
 .06 
x  13.18R
51
x
R
13.18
(**)
Substitute (**) into (*).
x
50 000  0.08 x 
13.18
x  320 784.34
The purchase price for the forest should be
$320 800 (nearest hundred).
52
Example 14
In order to replace a machine in the future, a
company is placing equal payments into a sinking
fund at the end of each year so that after 10 years the
amount in the fund is $25000. The fund earns 6%
compounded annually. After 6 years, the interest
rate increases and the fund pays 7% compounded
annually. Because of the higher interest rate, the
company decreases the amount of the remaining
payments. Find the amount of the new payment.
Give your answer to the nearest dollar.
53
.06
0
.07
6
10
25 000
 1.0610  1 

25 000  R
 .06 
R  $1896.70
54
.06
0
.07
6
10
25 000
R
13 230.08
After 6 years the repayments amount to
 1.06 6  1 

S  1896.70
 .06 
 $13 230.08
55
.06
0
.07
6
10
25 000
R
13 230.08
17 341.94
The value of this money after a further 4 years
at 7% is
S  13 230.08 (1.07)4
 $17 341.94
56
Amount to be raised
 25 000  17 341.94
 7 658.06
This is to be done with 4 payments.
 1.07  1
7658.06  R

 .07 
4
R  1724.81
The reduced repayment should be $1725 (to
the nearest dollar).
57