Fundamentals of Financial Arythmetics Lecture 3 Dr Wioletta Nowak

Fundamentals of Financial Arythmetics Lecture 3

Dr Wioletta Nowak

• Period-certain annuity (fixed period

annuity) – a type of annuity that guarantees

benefit payments for a designated period of

time (5-year certain=annuity payments for 5

years).

Annuity – compound interest

Annuity-immediate

Annuity due

�_�=� ∙ (1+� )^� −1

� �_�=� ∙(1+� ) ∙ (1+� )^� −1

�

Period-certain annuity

� ∙ ( 1 +� ) ^� = � _�

� ∙ (1+� )^� =� ∙ (1+� )^� − 1

�

Annuity-immediate

Annuity due

� ∙ (1+� )^�=� ∙ (1+� )∙ (1+� )^� −1

�

Period-certain annuity

Annuity-immediate Annuity due

 

 





 



r r

K a

) 1

(

1 1 



 





 



 

r r

r K a

) 1

( 1 1 ) 1

(

r

r

a K





 

) 1

(

1 ^{a  1  1 r}  ^{1  ( K 1   r r )}



Period-certain annuity

Annuity-immediate Annuity due

) 1

ln(

1 ln

r a

K r

N 

 

 



 





 ln( 1 )

) 1

1 ( ln

r

a r

K r

N 

 

 









 





Perpetuity

• A perpetuity immediate • A perpetuity due

r a r

r

K a

N

 



 





 



( 1 )

1 1 lim

r r K a  ( 1  ) r 

K  a

Example 1 – Annuity-immediate

N a

1 15.0 15.6 4.1 11.5

2 11.5 11.9 4.1 7.8

3 7.8 8.1 4.1 4.0

4 4.0 4.1 4.1 0

r

r

a K





 

) 1

( 1

 4

N K  15 r  4 %

1

K

( 1  r ) K

a K

r

K

 ( 1  )



K

Example 2– Annuity-due

N a

1 15.0 4.0 11.0

2 11.0 11.5 4.0 7.5

3 7.5 7.8 4.0 3.8

4 3.8 4.0 4.0 0

r

r K

a r







 

) 1

( 1 1

1

 4

N K  15 r  4 %

1

K

( 1  r ) K

a K

r

K

 ( 1  )



K

Example 3 Annuity-immediate

• An investment of 100 PLN is to be used to

make payments of 15 PLN at the end of every

year for as long as possible. If the fund earns

an annual rate of interest 1%, find how many

regular payments can be made.

Example 3 Annuity-immediate

) 1

( ln

1 ln

r a

K r

N 

 

 



  



 93

.

 6 N

 100

K a  15 r  0 . 01

Example 3 – Additional annuity payment

� ∙ (1 +� )⁷= �₆ ∙ (1 + � )+ �

� ∙ (1 +� )⁶= �₆+ � 1 +�

� = �₆

( 1 +� )⁶ + �

( 1 +� )⁷

 

r a r

A 1

1

6



 



Example 3 – Additional annuity payment

� ∙ (1 +� )⁷= �₆ ∙ (1 + � )+ �

�₆=15 ∙ (1+0.01)⁶ −1

0.01 =92.28

� = � ∙ (1 +� )⁷ − �₆ ∙ (1 +� )

�=100 ∙(1.01)⁷− 92.28 ∙ ( 1.01)=14.01

Example 3 – Enlargement of one of the payments

� ∙ ( 1 +� ) ⁶= �₆ + �

x=100=13.87

87 .

 13

x

Example 3 – New annuity payments

� =7

86 . ) 14

01 . 0 1

( 1

01 . 0 100

7







 

a

r

r

a K





 

) 1

(

1