1. MATH 2040 Introduction to Mathematical Finance
Instructor: Miss Liu Youmei

2. Chapter 3 Basic Annuities (年金）
Definition of an annuity
Annuity-immediate
Annuity-due
Annuity values on any date
Perpetuity-immediate &perpetuity-due
Problem with unknown time
Problem with unknown rate of interest

3. Definition of an annuity
An annuity is a series of payments made at equal intervals of time (annually or otherwise).
Payments made for certain for a fixed period of time are called an annuity-certain.
In this chapter, we deal with annuities with:
The payment frequency and the interest conversion period are equal.
The payments are leveled (all equal to each other).

4. Annuity-immediate
Payments of 1 are made at the end of every year for n years.
The present value (at t = 0) of an annuity-immediate, where the annual effective rate of interest is i, shall be denoted as and is calculated as follows:

5. Present value of annuity-immediate

6. Calculation of annuity-immediate

7. Accumulated value of annuity-immediate
The accumulated value (at t = n) of an annuity-immediate, where the annual effective rate of interest is i, shall be denoted by , and is calculated as follows:

8. Accumulated value of annuity-immediate

9. First basic relationship
The first basic relationship is:
Consider an n-year investment where 1 is invested at time 0.
Also consider the stream of payments of i at the end of each year for n years and then 1 is refunded at time t = n. 1

10. First basic relationship – cont’d
The present value of these multiple payments (at time t = 0) is
Hence we have
Note that

11. Second basic relationship
The second basic relationship is:
PV = FV v n and PV(1 + i )n = FV
If the FV (future value) at time n, , is discounted back to time 0, then we will have the PV (present value), So

12. Second basic relationship – cont’d
If the PV at time 0, , is accumulated to time n, then we will have the FV, So

13. Second basic relationship

14. Third basic relationship
The third basic relationship is:
Suppose a loan of 1 is to be paid back over n years with equal annual payments of P made at the end of each year, then or
Alternatively, suppose the same loan of 1 is to be repaid by interest payments of i made at the end of each year, and the loan amount 1 is paid back at time n.

15. Third basic relationship – cont’d
In order to produce the loan amount at time n, annual payments of D, at the end of each year for n years, will be made into an account that credits interest at an annual effective rate of interest i.
The future value of all these payment D must be equal to 1, so we have or
The total payment will therefore be

16. Third basic relationship – cont’d
Note that
Therefore, a level annual payment on a loan is the same as
making an annual interest payment each year plus
making annual deposits that will accumulate to the loan amount.

17. Third basic relationship1 1

18. Interest repayment option one
Given a loan of 1, there are 3 options in repaying the loan over the next n years.
Pay back the loan and all interest due at time n.
Total interest paid = A(n) – A(0)
= Loan  (1 + i)n  Loan
= Loan  [(1 + i)n  1]

19. Interest repayment option two
Pay at the end of each year, the interest that comes due on the loan and then pay back the loan at time n.
Annual interest payment = i  Loan
Total interest paid = i  Loan  n
= Loan  (i  n)

20. Interest repayment option three
Pay a level annual amount at the end of each year for the next n years.
Annual payment
Total payments
Total interest paid

21. Comparing interest repayment options
Option 1 and 2 is a comparison between compound v.s. simple interest.
Therefore, less interest is paid under option 2. This would make sense because if you pay off interest as it becomes due, the loan cannot grow as it does under option 1.
Option 2 and 3 is a mathematical comparison that shows less interest being paid under option 3. This can be proved by making use of the basic relationship

22. Example 3.1
Find the present value of an annuity which pays $500 at the end of each half-year for 20 years if the rate of interest is 9% convertible semi-annually.
We can view the annuity as one that pays $500 at the end of each period for 40 periods with interest rate 4.5% per period.
So the answer is:

23. Example 3.2
If a person invests $1000 at 8% p.a. convertible quarterly, how much can be withdrawn at the end of every quarter to use up the fund exactly at the end of 10 years?
Since the interest rate is 8% p.a. convertible quarterly, and withdrawals are made quarterly, we can view the problem as one with interest at 2% per period, and the fund is to be withdrawn in 40 periods.
So the answer is

24. Example 3.3
Compare the total amount of interest that would be paid on a $1000 loan over a 10-year period, if the effective rate of interest is 9% per annum, under the following three repayment methods:
(1) The entire loan plus interest is paid in one lump-sum at the end of 10 years.
(2) Interest is paid each year and the principal is repaid at the end of 10 years.
(3) The loan was repaid by level payments over the 10 year period.

25. Example 3.3 (1)
(1) The entire loan plus interest is paid in one lump-sum at the end of 10 years.
The accumulated value of the loan at the end of 10 years is 1000(1.09)10 =
The total amount of interest paid is therefore: $ = $

26. Example 3.3 (2)
(2) Interest is paid each year and the principal is repaid at the end of 10 years.
Interest paid each year is $1000  0.09 = $ 90
Total interest for 10 years is $90  10 = $ 900.

27. Example 3.3 (3)
(3) The loan was repaid by level payments over the 10 year period.
Let the level payment be R. Then
Which gives
So the total interest paid is  $  $1000 = $

28. Example remark
The later the principal was repaid, the more is interest paid.
(1) The entire principal plus accumulated interest was paid at the end of 10 years, so the most interest was paid for this option.
(2) Interest is paid each year, less interest was paid on this option.
(3) The loan was repaid by level payments. In each payment, some interest and some principal were paid. So the least interest amount was paid for this option.

29. Annuity due(期初年金）
For annuity immediate, payments of 1 are made at the end of every year for n years.
In another type of annuity, called annuity due, payments of 1 are made at the beginning of every year for n years.

30. Annuity due
The present value (at t = 0) of an annuity-due, where the annual effective rate of interest is i, shall be denoted as and is calculated as follows:

31. Annuity due – cont’d
The accumulated value at t = n of an annuity-due, where the annual effective rate of interest is i, shall be denoted as and is calculated as follows:

32. Basic relationships -Annuity due
Similar to annuity immediate, there are also three basic relationships on annuity-due.
(1)
(2) PV(1 + i)n = FV and PV = FV  v n
(3)
There are similar interpretation of all these relationships as those for annuity-immediate.

33. Basic relationships - Annuity due (1)
Consider an n-year investment where 1 is invested at time 0.
Suppose this investment produces
annual interest payments of d at the beginning of each year
the amount of 1 at t = n.
The present value of this is
It follows that
Note that

34. Basic relationships - Annuity due (2)
If the future value at time n, , is discounted back to time 0, then we obtained its present value,
Similarly, if the present value at time 0, , is accumulated forward to time n, then we get So

35. Basic relationships - Annuity due (3)
Consider a loan of 1, to be repaid back over n years with equal annual payments of P made at the beginning of each year.
Suppose an effective rate of interest, i, is used.
The present value of the single payment loan must be equal to the present value of the multiple payment stream, i.e.

36. Basic relationships - Annuity due (3)
Consider an alternative method of repaying the same loan of 1, where the annual interest due on the loan d, is paid at the beginning of each year for n years, and the loan amount 1 is paid at time n.
To produce the loan amount 1 at time n, annual payments, D, are deposited at the beginning of each year for n years so that it will grow with the same effective rate of interest as the loan.
The future value of the multiple deposits should equal the future value of 1 at time n. i.e.
The total annual payment is then

37. More basic relationships
In addition to the above three, there are also three additional basic relationships on annuities.
(4) and
(5)
(6)

38. Basic relationship (4)
An annuity-due starts one period ahead of an annuity-immediate, and as a result it earns interest for one more period.
That applies to both present value and accumulated value.
So we have
and

39. Basic relationship (5)
An annuity-due starts immediately. That results in an additional payment at time 0.
The remaining (n – 1) payments can be viewed as an annuity-immediate for (n – 1) years
Hence we get

40. Basic relationship (6)
The annuity due for (n – 1) years may be viewed as one that starts at the end of year one and finishes at the end of the (n – 1)-th year.
An additional payment of 1 at time n results in becoming n payments that now starts at the end of each year.
So we have

41. Example 3.4
An investor wishes to accumulate $1000 in a fund at the end of 12 years. He plans to make deposits at the end of each year, the final payment to be made at the end of the 11-th year. What should the annual deposits be if the fund earns 7% effective?
Suppose the annual deposit is $ D. Number of payments made is 11, and we want to get the accumulated value of these 11 payments at the end of 12 years, so
Solving for D, we get D = 59.21

42. Annuity values on any date
There are several other dates to valuing annuities rather than at the beginning of the term (t = 0) or at the end of the term (t = n).
Present values more than one period before the first payment date
Accumulated values more than one period after the last payment date
Current value between the first and the last payment dates.

43. Values before first payment (1)
Consider a series of payments of 1 that are made at time t = 3, 4, 5, 6, 7, 8 and 9.
At t = 2, there exists 7 future payments whose present value is represented by If this value is discounted back to time t = 0, then the value of this series of payments is (2 periods before the first end-of-the-year payment) is
Alternatively, at t = 3, there exists 7 future payments whose present value is represented by If this is discounted back to time t = 0, then the value of this series of payments (3 periods before the first beginning-of-the-year payment) is

44. Values before first payment (2)
Another way to examine this situation is to pretend there are 9 end of year payments. This can be done by adding 2 more payments to the existing 7. In this case, let the 2 additional payments be made at t = 1 and 2.
At time t = 0, the value of the 9 payments and the two added-on payments are and respectively.
Therefore the present value at t = 0 is
Note that this results in the formula:
A more general formula is: For positive integers m and n

45. Values before first payment (3)
With the annuity due version, suppose 3 more payments are added to the existing 7 payments. In this case, let the 3 additional payments be made at t = 0, 1 and 2.
At time t = 0, the value of the 10 payments and the three added-on payments are and respectively.
Therefore the present value at t = 0 is
Note that this results in the formula:
A more general formula is: For positive integers m and n

46. Values before first payment (4)
This type of annuity is often called a deferred annuity, since payments commence only after a deferred period.
A symbol for an annuity-immediate deferred for m periods with a term of n periods after the deferral period is
And for annuity-due is
Thus the above annuity could be labeled or

47. Values after last payment (1)
At t = 9, there exists 7 past end-of-year payments whose accumulated value is represented by If this value is accumulated forward to time t = 12, then the value of this series of payments (3 periods after the last end-of-year payments) is:
Alternatively, at t = 10, there exists 7 past beginning-of-year payments whose accumulated value is represented by If this value is accumulated forward to time t = 12, then the value of this series of payments (2 periods after the last beginning-of-year payments) is:

48. Values after last payment (2)
Another way to examine this situation is to add three more payments to the existing 7 so that now there are 10 end-of-year payments. Let the 3 additional payments be made at t = 10, 11 and 12 (denoted as ).
At t = 12, there now exists 10 end-of-year payments whose accumulated value is represented by This value would then be reduced by the value of the three added-on payments, i.e
Therefore the accumulated value at t = 12 is
This results in
The general formula is:

49. Values after last payment (3)
To use the annuity due approach, we may add just 2 more payments to the existing 7 so that now there are 9 beginning-of-year payments. Let the 2 additional payments be made at t = 10 and 11 (denoted as ).
At t = 12, there now exists 9 beginning-of-year payments whose accumulated value is represented by This value would then be reduced by the value of the two added-on payments, i.e
Therefore the accumulated value at t = 12 is
This results in
The general formula is

50. Values between first and last dates (1)
The 7 payments can be represented by an annuity-immediate or by an annuity-due depending on the time that they are evaluated
For example, at t = 2, the present value of the 7 end-of-year payments is At t = 9, the future value of these payments is
At a point between t = 2 and t = 9, the present value and future value can be accumulated to or discounted back respectively.
Take t = 6, the present value would be accumulated forward 4 years, the accumulated value would be discounted back 3 years.
So we have
The general formula is

51. Values between first and last dates (2)
Alternatively, at t = 3, we can view the payments as beginning-of-year payments and its present value is At t = 10, the future value of these payments would be
At t = 6, for example, the present value would be accumulated forward 3 years and the accumulated would be discounted back 4 years.
So we have
The general formula is

52. Values between first and last dates (3)
At any time during the payments, there will exist a series of past payments and a series of future payments.
We can view all payments as end-of-year payments. At t = 6, for example, we have 4 past payments whose accumulated value is , and 3 future payments whose present value is
So the value at t = 6 of all 7 payments is
Alternatively, all the payments can be viewed as beginning-of-year payments. Then at t = 6, we have 3 past payments and 4 future payments with accumulated values and present values of and respectively.
This results in
The general formula is

53. Values of annuities at other dates
In general, the present value of an n-period annuity where the first payment is m years from now is
The accumulated value of an n-period annuity m periods after the last payment date is
The current value of an n-period annuity immediately after the first m ( ≤ n) payments have been made is

54. Perpetuity-immediate（永续年金） (1)
A perpetuity-immediate makes payment of 1 at the end of every year forever, i.e. n = .
The present value of a perpetuity-immediate, where the annual effective rate of interest is i, shall be denoted by , and is calculated as follows:

55. Perpetuity-immediate (2)
We can also derive the above formula by taking the limit as n tends to infinity in the original present value formula:
Note that suppose an initial amount of is invested at t = 0. Then the interest payments, payable at the end of each year, produced by this investment is exactly 1.
is not defined because it would equal .

56. Perpetuity-immediate (3)
The present value formula for an annuity-immediate can be expressed as the difference of two perpetuity-immediate:
In this case, a perpetuity-immediate that is payable forever is reduced by perpetuity-immediate payments that start after n years. The present value at t = 0, results end-of-year payments remaining only for the first n years.

57. Perpetuity-due (1)
A perpetuity-due makes payment of 1 at the beginning of every year forever, i.e. n = .
The present value of a perpetuity-due, where the annual effective rate of interest is i, shall be denoted by , and is calculated as follows:

58. Perpetuity-due (2)
We can also derive the above formula by taking the limit as n tends to infinity in the original present value formula:
Suppose an initial amount of is invested at t = 0. The annual interest, payable at the beginning of the year, produced by this investment would be 1.

59. Example 3.5
A leaves an estate of $100,000. Interest on the estate is paid to beneficiary B for the first 10 years, to beneficiary C for the second 10 years, and to charity D thereafter. All payments are made at the end of year. Find the relative shares of B, C, and D in the estate, if it is assumed the estate will earn a 7% annual effective rate of interest.

60. Example 3.5….cont’d The present value of B’s share is
The present value of C’s share is
The present value of D’s share is
From the above, we can find that the relative shares of B, C, and D is approximately 49%, 25%,26%.

61. Example 3.5….cont’d
Note that the sum of the shares of B,C, and D is equal to $100,000 as expected.
Also note that the present value of the estate at the end of 20 years is , which equals D’s share. This confirms the fact that charity D continuing to receive the interest into perpetuity or receiving the estate value in a lump-sum at the end of 20 years are equivalent in value.

62. Problem with unknown time
When solving a problem in annuity with unknown time, the solution we often get is not an integer.
An adjustment to the payments can be made so that n does become an integer.

63. Example
How long will it take to payoff a loan of $1000 if $100 is paid at the end of every year and the annual effective rate of interest is 5%?
We need to solve the equation: Or

64. Example ….cont’d
By inspection of the interest tables, we find the value of n lying between 14 and 15.
The equation of value at t = 14 is:
The equation of value at t = 15 (after only 14 payments are made) is:
Solving, we have for R1 = and R2=21.07
We should note that R1 (1.05) = R2.

65. Example ….cont’d In our case, the equation of value is: Or i.e.
we can determine that k =
So the final payment is 100(1.05k-1)/0.05 = 20.27, or estimated value of 100k=20.67 to be made at the time

66. Unknown time
From above argument, we see that it is very inconvenient to determine the final time of payment, which is t = , with a payment of $20.67.
In practice, either a final payment at time t = 14 of $(100 + R1 = ) is made, or
a final payment at t = 15 of $ R2 = is made.

67. Unknown rate of interest
Assuming that we do not have a calculator
There are three approaches to solving for an unknown rate of interest when
They are:
Algebraic Techniques
Linear Interpolation
Successive Approximation (Iteration)

68. Algebraic Techniques
Note that is an n degree polynomial and can be easily solved if n is small.
When n gets too big, we can use a series expansion of , or even better, of Then we can solve for a low degree polynomial equation, considering only the first few terms

69. Algebraic Techniques
For a series expansion of , if considering the first two terms, we can derive an approximation for i.

70. Example 3.8
At what rate of interest, convertible quarterly, is $16,000 the present value of $1000 paid at the end of every quarter for five years?
Use the approximate formula to obtain the answer.
We have n=20 and k=16, so
That is, the interest rate is approximately 2.38% per quarter.

71. Linear Interpolation
In this method, we need to find the value of at two different interest rates i1 and i2 such that and
Then we can obtain by linear interpolation:

72. Example on Linear Interpolation
At what interest rate, convertible semi-annually, is $32,000 the present value of $2,000 payable every six month for 10 years?
Let j = i(2)/2, so the equation of value is
Define
We need to find the value of j for which f (j) = 0.

73. Example on Linear Interpolation – cont’d
By looking up the tables, we found that f (0.0200) = – 16 =
and f (0.0250) = – 16 = –
Now performing a linear interpolation, we get:
Which gives i(2) = 2j = 4.46%.