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```					Annuities: Math 360 Chapter 2

Lecture 2

Annuities
Introduction
• An annuity is a series of periodic
payments
• For us, the payments are contingent only
on the passage of time, not on certain
events (i.e. all annuities are annuity certain)
• We will need a simple, but very useful,
result from algebra:
Geometric Series
1  x  x  .... x  X
2       n

and      so

(1  x) X  1  x  1  x  x  .... x     2              n

                             n
 
 1  x  x  .... x  x 1  x  x  .... x
2                                       2               n

 1  x  x    2
 .... x   n
  x  x   2
 .... x  x  n   n 1

n 1
 1 x
n 1
1 x
 X  k  0 x 
n      k

1 x
Example 2.1
• The federal gov’t sends Smith a family
allowance payment of 30 every month for
Smith’s child. Smith deposits the payments in a
bank account on the last day of each month. The
account earns interest at the annual rate of 9%
compounded monthly and payable on the last
day of each month, on the minimum monthly
balance. If the first payment is deposited on May
31, 1998, what is the account balance on
December 31, 2009, including the payment just
Ex 2.1 Solution
140 total       deposits
first deposit  1
imonthly  i    .09     .0075
12   12
value at time of            last   deposit  30(1.0075)139
TotalValue  30  30(1.0075)  30(1.0075)  ... 30(1.0075)
2                 139

 1  1.0075140 
 1  1.0075   7385.91
 30               
               
Level Payment Annuities
• Number of payments in series of
payments is called the term of the annuity
• Time between the successive payments is
called payment period, or frequency
• A series of payments whose value is found
at the time of the final payment is known
as an accumulated annuity immediate
More Notation

1  (1  i )
n
(1  i )  1
n
sn|i  k 0 (1  i ) 
n 1       k

1  (1  i )         i
or
(1  i )  1  i  sn|i
n
Example 2.2
• What level amount must be deposited on
May 1 and Nov 1 each year from 1998 to
2005, inclusive, to accumulate to 7000 on
November 1, 2005 if the nominal annual
rate of interest compounded semi-annually
is 9% ?
Ex 2.2 Solution
16 total deposits
i   ( 2)
 0.09
X  level amount                deposited / year
EoV :

X (1.045)  (1.045)  ..  (1.045)  1 
15         14

X  s16|0.045  22.719337 X  7000
 X  308.11
Example 2.3
• Suppose that in Example 2.1, Smith’s child is
born in April 1998 and the first payment is
received in May (and deposited at the end of
May.) The payments continue and the deposits
are made at the end of each month until (and
including the month of) the child’s 16th birthday.
The payments stop after the 16th birthday, but
the balance continues to accumulate with
interest until the end of the month of the child’s
21st birthday. What is the balance in the account
at that time?
Ex 2.3 solution

X  (1.0075)  30  s192|.0075  12792.31
60
Some Arithmetic

Value@ time _ n  growth                from n  n  k
(1  i ) n  1
 sn|i  (1  i )                    1  i 
k                          k

i
(1  i ) n  k  (1  i ) k (1  i ) n  k  1 (1  i ) k  1
                                               
i                        i             i
 sn  k |i  sk |i
or
sn  k |i  sn|i  sk |i (1  i ) n
Example 2.4
• Suppose that in Example 2.1, the nominal
annual interest rate earned on the account
changes to 7.5 %, still compounded
monthly, as of January 1 2004. What is
the accumulated value of the account on
December 31 2009?
Ex 2.4 Solution


Value  30  s68|.0075  (1.00625)  s72|.00625
72

Example 2.5
• Suppose 10 monthly payments of 50 each are
followed by 14 monthly payments of 75 each. If
interest is at an effective monthly rate of 1%,
what is the accumulated value of the series at
the time of the final payment?

Value  50s24|.01  25s14|.01  1722.36
Present Value of an Annuity
Immediate
• Make a lump sum payment X now to receive
periodic payments C , starting one period from
today. If the market bears a constant interest of i,
then Present Value of this Annuity Immediate is
calculated as
C         C              C
X                   ...
1  i (1  i ) 2
(1  i ) n
1  n
n

 C k 1 k  C   1    ..   n 1       C  
1 
1  n
C         Can|i
i
Loan Repayment – Ex 2.7
• Brown has bought a new car and requires a loan
of 12000 to pay for it. The car dealer offers
Brown two alternatives on the loan:
• A.) Monthly payments for 3 years, starting one
month after purchase, with an annual interest
rate of 12% compounded monthly, or
• B.) Monthly payments for 4 years, also starting
one month after purchase, with annual interest
rate 15% compounded monthly.
• Find Brown’s monthly payment and the total
amount paid over the course of the repayment
period under each of the two options
Ex 2.17 Solution

a.)12000  P  a36|.01  P  398.57
1             1

b.)12000  P2  a48|.0125  P2  333.97
TotalValue( A)  36  P  14348.52
1

TotalValue( B)  48  P2  16030.56
Present Value of an Annuity Some
Time Before Payments Begin
• Example 2.8:
• Suppose that in Example 2.7 Brown can repay
the loan, still with 36 payments under option (a)
or 48 payments under option (b), with the first
payment made 9 months after the car is
purchased in either case. Assuming interest
accrues from the time of the car purchase, find
the payments required under options (a) and (b).
• This is known as a deferred annuity.
Ex 2.8 Solution

1
'
   9
12000  P  1.01  1.01  10             44
 ..  1.01       P 
1
'
1
8
 a36|.01
so
12000
P 
'
8
 431.60
1.01  a36|.01
1

12000
P 
'
8
 368.86
1.0125  a48|.0125
2
Duality of Total Value an Present
Value for Annuities

sn|i  1  i  an|i
n

an|i   sn|i
n
Perpetuities
• What happens as the term approaches infinity?

1 n
1
a|i    lim an|i  lim        
n        n     i   i
Ex. 2.10
• A perpetuity immediate pays X per year.
Brian receives the first n payments,
Colleen receives the next n payments, and
Brian’s share of the present value of the
original perpetuity is 40%, and Jeff’s share
is K. Calculate K.
Ex. 2.10 Solution
 1  n 
PV ( Brian)  X  an|i  X  
 i       0.4 X  a|i
        
X
 0.4   1  n  0.4
i
X           X
PV (Colleen)    X  an|i  0.4  0.6   0.24 
n

i           i
X
 PV  PV ( Brian)  PV ( Brian)  PV ( Jeff )
i
X          X
 K  PV ( Jeff )  (1  0.4  0.24)  0.36
i           i
Annuity Immediate
n 1     (1  i )  1
n
S _  1  (1  i )  ...  (1  i )          
n|i                                                i
1 v           n
a _  v  v  ...  v 
2           n
n|i                      i
S _  Payment Accumulation at Final Time
n|i

a _  Present Value one period before
n|i

first payment
Annuity Due


S n|i  (1  i )  ... (1  i ) n  (1  i ) S n|i
_                                           _

 value 1 period after final payment

a n|i  (1  i )a _  Present Value at time of first payment
_

n|i

0          1        2         ...      n -1           n
                                                           
                                                 
_                                         _        _
a_         a n|i                                      S n|i   S n|i
n|i
Level Payment Annuities-Some
Generalizations
• What happens when compounding interest
period and the annuity payment period do
not coincide ?
Ex 2.11
• Smith deposits quarterly of 1000 each
over 16 years. Find balance after last
payment.
a) Interest quoted at 9% nominal annual
rate compounded monthly
b) Interest quoted at effective annual rate 10%
Compounded interest paid for fraction of years.
Ex 2.11 Soln
64 total deposit. Suppose effective quarter-
rate is j.                  (1  j )  1
64
_
value = 1000 S 64| j =1000           j
a)               0.09 3
1  j  (1      )  (1  0.0075)     3

12
so j  0.02266917.
b)
1  j  (1  0.01)1/ 4 so j  0.02411369 .

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