Annuity

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What is meant by Annuity?
Suppose, we want to make payment of Tk.5000 each year for 10 years to an account
which earns 6 percent interest, compounded annually, and we want to calculate the total
amount which will be in the account after 10 payments.
This series of periodic payments is called an annuity.
( )
1
]
1

¸

− +
·
i
i
A M
formula the by given is annuity ordinary of sum The
n
1 1
,
M = Sum of the annuity
A = Payment per period
i = Interest rate per period
n = Number of peroids
What are various types of annuity?
An ordinary annuity is a series of equal payments, each payment of which is made at the
end of a period. Different annuities may be stated as under:
Annuity Certain
(Payments are to be made for a
certain or fixed number of years)
Annuity Contingent
(Payments are to be made till the
happening of some contingent
event such as the death of a person,
marriage of a girl ete.)
Annuity due
(Where the first payment
falls due at the beginning
of the ist interval, and so
on)
Annuity immediate
(Where the first payment
falls due at the end of the
first interval, and so on)
Annuity
Calculation of different types of annuities:
1. Formula for the present value of an Annuity due
( )
( )
years of Number
interest of Rate
Annuity
due Annuity an of value Present
where
·
·
·
·
¹
;
¹
¹
'
¹
+
+
− + ·
n
i
A
V
i
i
i
i
A
V
n
,
1
1
1 1
2. Formula for the present value of an Immediate Annuity
( )
1
]
1

¸

+
− ·
n
i i
A
V
1
1
1
3. Formula for the amount of an Immediate Annuity
( ) { } 1 1 − + ·
n
i
i
A
M
4. Formula for the amount of Annuity due
( ) ( ) { } 1 1 1 − + + ·
n
i i
i
A
M
5. Present Value of an Annuity
The present value of an Annuity is the sum of the present values of its instalments.
Example: 1
2
Sum on Annuity
1. How much must be deposited each year at 5 percent compounded
annually to accumulate Tk.1000 after 10 years from now?

( )
( )
( )
50 . 79
50 . 79
57788 . 12
1000
57788 . 12
05 . 0
628894 . 0
05 . 0
1 628894 . 1
05 . 0
1 05 . 1
05 . 0
1 05 . 1
1000
?
1000 .
10
% 5 ,
,
1 1
10
isTk. year each deposited be to required amount, The
years
amount Total
Period
payment each of Rate
interest of Rate
where know, We

· · ∴
× ·
× ·
1
]
1

¸

·
1
]
1

¸


·
1
]
1

¸

− +
· ∴
·
·
·
·
·
·
·
·
1
]
1

¸

− +
·
A
A
A
A
A
A
A
Tk M
n
i Here
M
n
A
i
i
i
A M
n
n
Example: 2
A man borrows Tk. 6000 at 6% and promises to pay off the loan in 20 annual
payments beginning at the end of the first year. What is the annual payment
necessary?
3
Log (1.05)
10
=10 log 1.05
= 10× .021189299
= 0.21189299
Antilog 0.21189299
=1.628894

Log (1.05)
10
= 1.628894
( )
1
]
1

¸

+
− ·
n
i i
A
V
1
1
1
know, We
, where
V = Present value
A = Annuity (annual payment)
i = Rate of interest
n = Number of years
( )
( )
19 . 523
19 . 523
206 . 2
206 . 3 06 . 0 6000
206 . 3
206 . 2
06 . 0
206 . 3
1 206 .. 3
06 . 0
206 . 3
1
1
06 . 0
06 . 1
1
1
06 . 0
06 . 0 1
1
1
06 . 0
6000
?
20
06 . 0
100
6
000 , 6 ,
20
20
Tk. is necessary payment annual The
Here

·
× ×
· ∴
× ·

× ·
1
]
1

¸

− ·
1
]
1

¸

− ·
1
]
1

¸

+
− · ∴
·
·
· ·
·
A
A
A
A
A
A
A
n
i
V
Example: 3
How long will it take for a sum of money to increase by 25 percent at
compound interest of 5 percent per year? [ Mathematics- Bowen]
Answer : The sum will increase by 25%
This means if the principal amount is Tk. 100, the increased
compounded amount will be Tk (100+25) =Tk.125
4
Log (1.06)
20

= 20 log 1.06
=20×0.0253
=0.506
∴Antilog 0.506
=3.206
∴ (1.06)
20
=3.206
∴ A = Tk.125
P = Tk.100

( )
( )
( )
( )
( )
( )
years take will It
have We
57 . 4
57 . 4
0212 .
0969 .
05 . 1 log
25 . 1 log
05 . 1 log 25 . 1 log
05 . 1 log 25 . 1 log
05 . 1 25 . 1
05 . 1
4
5
05 . 1
100
125
1
1
05 .
100
5

· · · ⇒
· ⇒
· ⇒
· ⇒
· ⇒
+ · ⇒
+ · ⇒
+ ·
· ·
n
n
r
P
A
r P A
r
n
n
n
n
n
n
Example :4
How long will it take for sum of money to double itself at 6 percent,
compounded annually? [Bowen]
Answer : The compound amount will be double. Hence, if the principal amount
is x, the compounded amount will be 2x.
We know,
A= P(1+r)
n,
where
A= Compounded amount
5
P = Principal amount
r = rate of interest
n = period
Here, A = 2x

( )
( )
( )
years 11.9 take will it ∴
·
· · ⇒
· ⇒
· ⇒
· ⇒
· ⇒
+ · ∴
· ·
·
9 . 11
0253 . 0
3010 . 0
06 . 1 log
2 log
06 . 1 log
2 log
06 . 1 log 2 log
06 . 1 log 2 log
06 . 1 2
06 . 1 2
06 . 0
100
6
,
n
n
n
x x
r
x P
n
n
n

Example: 5
a) Find the compound interest on Tk. 1,00,000 for 4 years at 5% per annum.
b) What will be the simple interest in the above case?
Answer: a)
We know,
A= P (1+r)
n
, where
A=compounded amount
P = Principal amount
r= Rate interest
n = years
6

( )
( )
( )
121500
5.08476 A
5.08476
0.08476 5
0.02119 4 1 5
1.05 log 4 10 log 5
(1.05) log 10 log
} (1.05) log{10 A log
05 . 1 10
05 . 1 000 , 00 , 1
05 . 0 1 000 , 00 , 1
05 . 0
100
5
4
000 , 00 , 1
4 5
4 5
4 5
·
· ∴
·
+ ·
× + × ·
+ ·
+ ·
× ·
× ·
× ·
+ · ∴
·
·
·
·
Antilog
P Here,
n
n
A
r
n
Now compound Interest = A - P
=Tk.1,21,500 -1,00,000
= Tk.21,500/-
b) What will be the simple interest in the above case?
Simple Interest = P n r
= 1,00,000×4×0.05
= Tk. 20,000/-
Example: 6
Find the compound interest on 25,800 for 5 years, if the rate of interest be 2% in
the ist year,
%
2
1
2
interest second year, 3% in the 3
rd
year and there after at 4%
p.a.
We know,
A = P( 1+r)
n
, where
A = Compounded amount
r = Rate of interest per annum
n = Period

Rate of interest in the 2
nd

7
( )
. , 316 , 26
02 . 1 800 , 25
02 . 1 800 , 25
1
02 .
100
2
800 , 25
1
year 2nd the for principal the is which
A
P Here,
·
× ·
+ · ∴
·
· ·
·
n
r
year =
%
2
1
2
= 2.5%
( ) . , 9 973 , 26 025 . 0 1 316 , 26
025 . 0
100
5 . 2
1
year 3rd the for principal the is which ⋅ · + · ∴
·
·
A
Rate of interest in the 3
rd
year
· · ·
100
3
% 3
.03
Rate of interest after 3
rd
year
· · ·
100
4
% 4
0.04
( )
( )
4250
800 , 25 . 050 , 30 .
0816 . 1 12 . 783 , 27
04 . 1 12 . 783 , 27
04 . 1 12 . 783 , 27
2
2
Tk.
Tk Tk Interst Compound
·
− · ∴
× ·
× ·
+ · ∴ A
Example :7
Find the compound interest on Tk. 6,950 for 3 years, if interest is payable half
yearly, rate for the first two years being 6%, and for the third year 9% p.a.
P = 6950
n= 2,
· ·
100
6
r
.06
8
( )
years rest the for princepal the is which , 12 . 783 , 27
03 . 1 9 . 973 , 26
03 . 0 1 9 . 973 , 26
1
·
× ·
+ · ∴ A
Let x = (1.04)
2
∴log x = log (1.04)
2
=2 log 1.04
=2×0.0170
= 0.034
∴x = Antilog 0.034
=1.0816
∴(1.04)2=1.0816
( ) ( )
( )
( ) ( )
( )
Tk.1589
Tk6950 - Tk.8539 Interest Compound
8539 3.9314
A Now,
years 3rd the for principal the is which 7820 3.8932 Antilog A
·
· ∴
· · ∴
·
+ ·
× + · + ·
+ · ∴
× · + ·
1
]
1

¸

· · ·
,
_

¸
¸
+ ·
· · ∴
·
× + ·
+ · + · ∴
× · + ·
,
_

¸
¸
+ ·
,
_

¸
¸
+ ·
×
×
Antilog A
A
r
A
r
P A
n
9314 . 3
0382 . 0 8932 . 3
019 . 0 2 8932 . 3 045 . 1 log 2 7820 log
045 . 1 log 7820 log log
045 . 1 7820 045 . 1 7820
09 . 0
100
9
% 9
2
09 .
1 7820
8932 . 3
0128 . 4 8420 . 3
03 . 1 log 4 6950 log 03 . 1 log 6950 log log
03 . 1 6950 03 . 1 6950
2
06 .
1 6950
2
1
2
2 2
1 2
4
4 4
2 2 2

Example: 8
Depreciation
A machine depreciates at the rate of 10% of its value at the beginning of a year.
The machine was purchased for Tk. 5850 and the scrap value realized when sold
was Tk. 2250.Find the number of years that the machine was used.
We know,
SV = PV (1-d)
n
, where
SV = Scrap value
PV = Purchase value
d = depreciation
9
n = period
Here, SV = Tk. 2250
PV = Tk. 5810
( )
( )
( )
( ) ( )
( )
9 year of number The
n
d
· ∴

·


· · · ⇒
· ⇒
· ⇒
· ⇒
· ⇒
− · ∴
· · ·
9
9956 . 8
0458 . 0
4120 .
9542 . 1
5880 . 1
90 . 0 log
3873 . 0 log
90 . 0 log 3873 . 0 log
90 . 0 log 3873 . 0 log
90 . 0 3873 . 0
90 . 0
5850
2250
10 . 0 1 5850 2250
10 . 0
100
10
% 10
o
n
n
n
n
n
Example: 9
A company buys a machine for TK 1,00,000. Its estimated life is 12 years and scrap
value is TK. 5000. What amount is to be retained every year from the profit and allowed
to accumulate at 5 % C.I. for buying a new machine at the same price after 12 years.
Or
The cost of a machine is TK. 1,00,000 and its effective life is 12 years. If the scrap
realises only TK. 5,000, what amount should be retained out of profits at the end of each
year to accumulate at 5 % C.I. p.a.
Answer. Cost of the machine = TK. 1,00,000
Its scrap value = TK 5,000
Amount of the annuity = Cost – Scrap value
10
= TK 1,00,000 – TK 5,000
= TK.95,000
n = 12, i = 5%=
100
5
= 0.05
Annuity , A = ?
( )
Annuity A
rate Interest i
Period n
where, Amount know, We
·
·
·
¹
;
¹
¹
'
¹
− +
·
i
i
A
n
1 1


( )
( )
41 . 5964
79639 . 0
05 . 0 000 , 95
05 . 0
79639 . 0
05 . 0
1 79639 . 1
05 . 0
1 05 . 1
000 , 95
05 . 0
1 05 . 0 1
000 , 95
12
12
·
×
· ∴
×
·
¹
;
¹
¹
'
¹ −
·
¹
;
¹
¹
'
¹

·
¹
;
¹
¹
'
¹
− +
· ∴
A
A
A
A
A

Example :10
A machine is depreciated in a such way that the value of the machine at the end of any
year is 90 % of the value at the beginning of the year. The cost of the machine was TK.
12,000 and it was sold eventually as waste material for Tk 200. Find out the number of
years during which the machine was in use.
We know
SV = PV (1- d )
n
, where
SV= Scrap value
PV= Purchase value
d = depreciation
n = period
Here,
Selling price i.e. Scrap value = TK. 200
PV = TK. 12,000
d =
100
10
% 10 % 90 % 100 · · −
11
Let X = ( 1.05)
12
∴log X = log (1.05)
12
= 12 log 1.05
= 12×0.0212
= 0.2544
∴ X = Antilog 0.2544
= 1.79639
∴(1.05)
12
= 1.79639
years
n
n
n
n
n
n
n
39
83 . 38
0458 . 0
7782 . 1
1 9542 . 0
7782 . 1 0
10 log 9 log
60 log 1 log
10
9
log
60
1
log
10
9
log
12000
200
log
12000
200
log
10
9
log
1200
200
log
10
9
log
1200
200
10
9
10
9
12000 200
10
1
1 12000 200
100
10
1 000 , 12 200
·
·


·


·


· · · ⇒
·
,
_

¸
¸

·
,
_

¸
¸

·
,
_

¸
¸


,
_

¸
¸
· ⇒

,
_

¸
¸
− · ⇒

,
_

¸
¸
− · ∴

the number of years during which the machine was in use = 39 years.
12

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