Learning objectives : future and present value of annuity

2.4 Annuity
MAHANI FSKM

Learning
objectives
Student should be able to:
• find the future and present value of annuity
• solve for annuity payment, R
no. of payments, n
interest rate, k
• identify the problems of present and future value

2.4.1
Introduction to
Annuity and Types
of Annuities

Introduction
• Annuity is a series of equal payments made at equal intervals of time
• Equal payments are deposited or paid periodically over time (at the end
of payment period)-ordinary annuity certain
• Ex: shop rental, insurance policy premiums, annual dividends received
and instalment payments
• Annuity enable people to plan for future investments, savings and
retirement.
• Use by financial consultants to advise people on investments
• Use by property agent to estimates the repayments of housing and car
loans

2.4.2
Future Value of
Ordinary Annuity
Certain

Future value
• sum of all the future values of the periodic payments.
• 𝑆 = 𝑅 1+𝑖 𝑛−1
𝑖 or 𝑺 = 𝑹 𝟏+𝒌
𝒎
𝒎𝒕
−𝟏
𝒌
𝒎
• where
• R=equal payments/periodic payments
• i=interest rate per interest period
• n =term of investment
• S=future value
• 𝑰 = 𝑺 − 𝒏𝑹

Example 1
Solution:
RM100 is deposited every month for 2 years 7 months at 12% compounded monthly. What is the future
and how much interest is earned?
𝑅 = 100, 𝑡 = 2
7
12 = 31
12, 𝑘 = 0.12, 𝑚 = 12
𝑆 = 𝑅
1 + 𝑘
𝑚
𝑚𝑡
− 1
𝑘
𝑚
=100 1+0.12
12
12(31
12)
−1
0.12
12
=𝑅𝑀3613.27
𝐼 = 𝑆 − 𝑛𝑅
=3613.27 − 31 100
=𝑅𝑀513.27

Example 2
Solution:
RM150 was deposited every month in an account for 15 years at 3.6% compounded monthly. Find the
amount in the account at the end of 15 years and the interest earned.
𝑅 = 150, 𝑡 = 15, 𝑘 = 0.036, 𝑚 = 12
𝑆 = 𝑅
1 + 𝑘
𝑚
𝑚𝑡
− 1
𝑘
𝑚
=150 1+0.036
12
12(15)
−1
0.036
12
=𝑅𝑀35731.01
𝐼 = 𝑆 − 𝑛𝑅
=35731.01 − 12(15) 150
=𝑅𝑀8731.01

Example 3: Forborne annuity-annuity earns interest for some periods after the last payment
Solution:
RM100 was invested every month in an account for 2 years at 12% compounded monthly. After two yea
no more deposit was made. Find the amount of the account at the end of the five years and the interest
𝑅 = 100, 𝑡 = 2, 𝑘 = 0.12, 𝑚 = 12 𝑆 = 𝑅
1 + 𝑘
𝑚
𝑚𝑡
− 1
𝑘
𝑚
=100 1+0.12
12
12(2)
−1
0.12
12
=𝑅𝑀2697.35
𝑆 = 𝑃 1 +𝑘
𝑚
𝑚𝑡
=2697.35 1 +
0.12
12
12(3)
=𝑅𝑀3859.28
𝐼 = 𝑆 − 𝑛𝑅
=3859.28 − 24 100
=𝑅𝑀1459.28
no deposit 50 2annuity

2.4.3
Present Value of
Ordinary Annuity
Certain

Present value (discounted value)
• sum of all present values of the periodic payments.
• 𝐴 = 𝑅1− 1+𝑖 −𝑛
𝑖 or 𝑨 = 𝑹𝟏− 𝟏+ 𝒌
𝒎
−𝒎𝒕
𝒌
𝒎
• where
• R=equal payments/periodic payments
• i=interest rate per interest period
• n =term of investment
• A=present value
• 𝑰 = 𝒏𝑹 − 𝑨

Example 4
Solution:
Raymond has to pay RM300 every month for 24 months to settle a loan at 12% compounded monthly.
What is the original value of the loan and the total interest that he has to pay?
𝑅 = 300, 𝑡 = 2, 𝑘 = 0.12, 𝑚 = 12
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
=300 1− 1+0.12
12
−12(2)
0.12
12
=𝑅𝑀6373.02
𝐼 = 𝑛𝑅 − 𝐴
=24 300 − 6373.02
=𝑅𝑀826.98

Example 5
Solution:
John won an annuity that pays RM1000 every three months for three years. What is the present value of
annuity if money is worth 16% compounded quarterly?
𝑅 = 1000, 𝑡 = 3, 𝑘 = 0.16, 𝑚 = 4
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
=1000 1− 1+0.16
4
−4(3)
0.16
4
=𝑅𝑀9385.07

Example 6
Solution:
James intends to give a scholarship worth RM5000 every year for six years. How much must he deposit n
into an account that pays 7% per annum to provide the scholarship?
𝑅 = 5000, 𝑡 = 6, 𝑘 = 0.07, 𝑚 = 1
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
=5000 1− 1+0.07
1
−1(6)
0.07
1
=𝑅𝑀23832.70

Example 7: Deferred annuity-the first payment at some later date
Solution:
Find the present value of an annuity of RM500 every year for 5 years if the first payment is made in 2 ye
Assume that money is worth 6% compounded annually.
𝑅 = 500, 𝑡 = 5, 𝑘 = 0.06, 𝑚 = 1
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
=500 1− 1+0.06
1
−1(5)
0.06
1
=𝑅𝑀2106.18
𝑆 = 𝑃 1 +𝑘
𝑚
𝑚𝑡
2106.18=𝑃 1 + 0.06
1
1(1)
𝑃 = 𝑅𝑀1986.96
0 1 2 3 4 5 6
annuityno deposit

Example 8: Given S, find R
Solution:
Find the amount to be invested every three months at 10% compounded quarterly to accumulate RM10,0
in three years. Find the interest earned.
S= 10000, 𝑡 = 3, 𝑘 = 0.1, 𝑚 = 4
𝑆 = 𝑅
1 + 𝑘
𝑚
𝑚𝑡
− 1
𝑘
𝑚
10000 = 𝑅 1+0.1
4
4(3)
−1
0.1
4
𝑅 = 𝑅𝑀724.90
𝐼 = 𝑆 − 𝑛𝑅
=10000 − 4(3) 724.90
=𝑅𝑀1301.20

Example 9: Given A, find R
Solution:
Mariam invests RM12000 in an account that pays 6% compounded monthly. She intends to withdraw an
amount every months for two years and when she makes her last withdrawal, her account will have zero
Find the size of these withdrawals.
𝐴 = 12000, 𝑡 = 2, 𝑘 = 0.06, 𝑚 = 12
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
12000 =𝑅 1− 1+0.06
12
−12(2)
0.06
12
R=𝑅𝑀531.85

Example 10
Solution:
Rosalind borrowed RM80000 at 12% compounded monthly for three years.
a) Find her monthly payment
b) If she had not paid her first 5 monthly payments, how much should she pay on her sixth payment
to settle all outstanding arrears?
c) If she wanted to settle all of the loan immediately after paying the first 5 monthly payments, how mu
additional payment did she have to make?
𝐴 = 80000, 𝑡 = 3, 𝑘 = 0.12, 𝑚 = 12
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
80000 = 𝑅1− 1+0.12
12
−12(3)
0.12
12
R=𝑅𝑀2657.14
𝑎) 𝑏) 𝑐) n=12(3)-5=31
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑛
𝑘
𝑚
= 2657.141 − 1 + 0.12
12
−31
0.12
12
R=𝑅𝑀70526.57
𝑆 = 𝑅
1 + 𝑘
𝑚
𝑛
− 1
𝑘
𝑚
= 2657.14 1 +0.12
12
6
− 1
0.12
12
= 𝑅𝑀16346.77

Example 11: Find n
Solution:
Jimmy has to pay RM443.21 every month to settle a loan of RM10,000 at 6% compounded monthly.
Find the number of payments he has to make.
A=10000, R= 443.21, 𝑘 = 0.06, 𝑚 = 12
𝐴 = 𝑅
1 − 1 + 𝑘
𝑚
−𝑚𝑡
𝑘
𝑚
10000 =443.21 1− 1+0.06
12
−𝑛
0.06
12
0.112813 = 1 − 1.005−𝑛
1.005−𝑛 = 0.887187
-n=
lg(0.887187 )
lg(1.005)
-n=-24
n=24

Q & A