1. 4 Mathematics of Finance Compound Interest Annuities
Amortization and Sinking Funds
Arithmetic and Geometric Progressions

2. 4.1
Compound Interest

3. Simple Interest Formulas
Simple interest is the interest that is computed on the original principal only.
If I denotes the interest on a principal P (in dollars) at an interest rate of r per year for t years, then we have
I = Prt
The accumulated amount A, the sum of the principal and interest after t years is given by
A = P + I = P + Prt
= P(1 + rt)
and is a linear function of t.

4. Example
A bank pays simple interest at the rate of 8% per year for certain deposits.
If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years?
What is the interest earned in that period?
Solution
Using the accumulated amount formula with P = 1000, r = 0.08, and t = 3, we see that the total amount on deposit at the end of 3 years is given by
or $1240.

5. Example
A bank pays simple interest at the rate of 8% per year for certain deposits.
If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years?
What is the interest earned in that period?
Solution
The interest earned over the three year period is given by
or $240.

6. Applied Example: Trust Funds
An amount of $2000 is invested in a 10-year trust fund that pays 6% annual simple interest.
What is the total amount of the trust fund at the end of 10 years?
Solution
The total amount is given by
or $3200.

7. Compound Interest
Frequently, interest earned is periodically added to the principal and thereafter earns interest itself at the same rate. This is called compound interest.
Suppose $1000 (the principal) is deposited in a bank for a term of 3 years, earning interest at the rate of 8% per year compounded annually.
Using the simple interest formula we see that the accumulated amount after the first year is
or $1080.

8. Compound Interest
To find the accumulated amount A2 at the end of the second year, we use the simple interest formula again, this time with P = A1, obtaining:
or approximately $

9. Compound Interest
We can use the simple interest formula yet again to find the accumulated amount A3 at the end of the third year:
or approximately $

10. Compound Interest
Note that the accumulated amounts at the end of each year have the following form:
These observations suggest the following general rule:
If P dollars are invested over a term of t years earning interest at the rate of r per year compounded annually, then the accumulated amount is
or:

11. Compounding More Than Once a Year
The formula
was derived under the assumption that interest was compounded annually.
In practice, however, interest is usually compounded more than once a year.
The interval of time between successive interest calculations is called the conversion period.

12. Compounding More Than Once a Year
If interest at a nominal rate of r per year is compounded m times a year on a principal of P dollars, then the simple interest rate per conversion period is
For example, if the nominal interest rate is 8% per year, and interest is compounded quarterly, then
or 2% per period.

13. Compounding More Than Once a Year
To find a general formula for the accumulated amount, we apply
repeatedly with the interest rate i = r/m.
We see that the accumulated amount at the end of each period is as follows:

14. Compound Interest Formula
There are n = mt periods in t years, so the accumulated amount at the end of t years is given by
Where n = mt, and A = Accumulated amount at the end of t years P = Principal r = Nominal interest rate per year m = Number of conversion periods per year t = Term (number of years)

15. Example
Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded
Annually
Semiannually
Quarterly
Monthly
Daily

16. Example Solution Annually. Here, P = 1000, r = 0.08, and m = 1.
Thus, i = r = 0.08 and n = 3, so
or $

17. Example Solution Semiannually. Here, P = 1000, r = 0.08, and m = 2.
Thus, and n = (3)(2) = 6, so
or $

18. Example Solution Quarterly. Here, P = 1000, r = 0.08, and m = 4.
Thus, and n = (3)(4) = 12, so
or $

19. Example Solution Monthly. Here, P = 1000, r = 0.08, and m = 12.
Thus, and n = (3)(12) = 36, so
or $

20. Example Solution Daily. Here, P = 1000, r = 0.08, and m = 365.
Thus, and n = (3)(365) = 1095, so
or $

21. Continuous Compounding of Interest
One question arises on compound interest:
What happens to the accumulated amount over a fixed period of time if the interest is compounded more and more frequently?
We’ve seen that the more often interest is compounded, the larger the accumulated amount.
But does the accumulated amount approach a limit when interest is computed more and more frequently?

22. Continuous Compounding of Interest
Recall that in the compound interest formula
the number of conversion periods is m.
So, we should let m get larger and larger (approach infinity) and see what happens to the accumulated amount A.

23. Continuous Compounding of Interest
If we let u = m/r so that m = ru, then the above formula becomes
The table shows us that when u gets larger and larger the expression
approaches (rounding to five decimal places).
It can be shown that as u gets larger and larger, the value of the expression approaches the irrational number … which we denote by e. u 10
100
1000
10,000
100,000
1,000,000

24. Continuous Compounding of Interest
Continuous Compound Interest Formula
A = Pert
where
P = Principal
r = Annual interest rate compounded
continuously
t = Time in years
A = Accumulated amount at the end
of t years

25. Examples
Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded (a) daily, and (b) continuously.
Solution
Using the compound interest formula with P = 1000, r = 0.08, m = 365, and t = 3, we find
Using the continuous compound interest formula with P = 1000, r = 0.08, and t = 3, we find
A = Pert = 1000e(0.08)(3) ≈
Note that the two solutions are very close to each other.

26. Effective Rate of Interest
The last example demonstrates that the interest actually earned on an investment depends on the frequency with which the interest is compounded.
For clarity when comparing interest rates, we can use what is called the effective rate (also called the annual percentage yield):
This is the simple interest rate that would produce the same accumulated amount in 1 year as the nominal rate compounded m times a year.
We want to derive a relation between the nominal compounded rate and the effective rate.

27. Effective Rate of Interest
The accumulated amount after 1 year at a simple interest rate R per year is
The accumulated amount after 1 year at a nominal interest rate r per year compounded m times a year is
Equating the two expressions gives

28. Effective Rate of Interest Formula
Solving the last equation for R we obtain the formula for computing the effective rate of interest:
where
reff = Effective rate of interest
r = Nominal interest rate per year
m = Number of conversion periods per year

29. Example
Find the effective rate of interest corresponding to a nominal rate of 8% per year compounded
Annually
Semiannually
Quarterly
Monthly
Daily

30. Example
Solution
Annually.
Let r = 0.08 and m = 1. Then
or 8%.

31. Example
Solution
Semiannually.
Let r = 0.08 and m = 2. Then
or 8.16%.

32. Example
Solution
Quarterly.
Let r = 0.08 and m = 4. Then
or 8.243%.

33. Example
Solution
Monthly.
Let r = 0.08 and m = 12. Then
or 8.300%.

34. Example
Solution
Daily.
Let r = 0.08 and m = 365. Then
or 8.328%.

35. Effective Rate Over Several Years
If the effective rate of interest reff is known, then the accumulated amount after t years on an investment of P dollars can be more readily computed by using the formula

36. Present Value Consider the compound interest formula:
The principal P is often referred to as the present value, and the accumulated value A is called the future value, since it is realized at a future date.
On occasion, investors may wish to determine how much money they should invest now, at a fixed rate of interest, so that they will realize a certain sum at some future date.
This problem may be solved by expressing P in terms of A.

37. Present Value
Present value formula for compound interest
Where and

38. Examples
How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000?
Solution
Here, A = 20,000, r = 0.06, m = 12, and t = 3.
Using the present value formula we get

39. Examples
Find the present value of $49, due in 5 years at an interest rate of 10% per year compounded quarterly.
Solution
Here, A = 49,158.60, r = 0.1, m = 4, and t = 5.
Using the present value formula we get

40. Present Value with Continuously Compounded Interest
If we solve the continuous compound interest formula
A = Pert
for P, we get
P = Ae–rt
This formula gives the present value in terms of the future (accumulated) value for the case of continuous compounding.

41. Applied Example: Real Estate Investment
Blakely Investment Company owns an office building located in the commercial district of a city.
As a result of the continued success of an urban renewal program, local business is enjoying a mini-boom.
The market value of Blakely’s property is
where V(t) is measured in dollars and t is the time in years from the present.
If the expected rate of appreciation is 9% compounded continuously for the next 10 years, find an expression for the present value P(t) of the market price of the property that will be valid for the next 10 years.
Compute P(7), P(8), and P(9), and then interpret your results.

42. Applied Example: Real Estate Investment
Solution
Using the present value formula for continuous compounding
P = Ae–rt
with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is
Letting t = 7, we find that
or $599,837.

43. Applied Example: Real Estate Investment
Solution
Using the present value formula for continuous compounding
P = Ae–rt
with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is
Letting t = 8, we find that
or $600,640.

44. Applied Example: Real Estate Investment
Solution
Using the present value formula for continuous compounding
P = Ae–rt
with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is
Letting t = 9, we find that
or $598,115.

45. Applied Example: Real Estate Investment
Solution
From these results, we see that the present value of the property’s market price seems to decrease after a certain period of growth.
This suggests that there is an optimal time for the owners to sell.
You can show that the highest present value of the property’s market value is $600,779, and that it occurs at time t ≈ 7.72 years, by sketching the graph of the function P.

46. Example: Using Logarithms in Financial Problems
How long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly?
Solution
Using the compound interest formula
with A = 15,000, P = 10,000, r = 0.12, and m = 4, we obtain

47. Example: Using Logarithms in Financial Problems
How long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly?
Solution
We’ve got
Taking logarithms on both sides
logbmn = nlogbm
So, it will take about 3.4 years for the investment to grow from $10,000 to $15,000.

48. 4.2
Annuities

49. Annuities
An annuity is a sequence of payments made at regular time intervals.
The time period in which these payments are made is called the term of the annuity
Examples:
Regular deposits to a savings account
Monthly home mortgage payments
Monthly insurance payments

50. Annuities
1. The terms are given by fixed time intervals(annuity certain). 2. The period payments are equal in size. 3. The payments are made at the end of the payment period(ordinary annuity) 4. The payment periods coincide with the interest conversion period (simple annuity).

51. Future Value of an Annuity
The future value S of an annuity of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period, is

52. Example
Find the amount of an ordinary annuity consisting of 12 monthly payments of $100 that earn interest at 12% per year compounded monthly.
Solution
Since i is the interest rate per period and since interest is compounded monthly in this case, we have
Using the future value of an annuity formula, with R = 100, n = 12, and i = 0.01, we have
or $

53. Present Value of an Annuity
The present value P of an annuity consisting of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period, is

54. Example
Find the present value of an ordinary annuity consisting of 24 monthly payments of $100 each and earning interest of 9% per year compounded monthly.
Solution
Here, R = 100, i = r/m = 0.09/12 = , and n = 24, so
or $

55. Applied Example: Saving for a College Education
As a savings program towards Alberto’s college education, his parents decide to deposit $100 at the end of every month into a bank account paying interest at the rate of 6% per year compounded monthly.
If the savings program began when Alberto was 6 years old, how much money would have accumulated by the time he turns 18?

56. Applied Example: Saving for a College Education
Solution
By the time the child turns 18, the parents would have made
deposits into the account, so n = 144.
Furthermore, we have R = 100, r = 0.06, and m = 12, so
Using the future value of an annuity formula, we get
or $21,015.

57. Applied Example: Financing a Car
After making a down payment of $4000 for an automobile, Murphy paid $400 per month for 36 months with interest charged at 12% per year compounded monthly on the unpaid balance.
What was the original cost of the car?
What portion of Murphy’s total car payments went toward interest charges?

58. Applied Example: Financing a Car
Solution
The loan taken up by Murphy is given by the present value of the annuity formula
or $12,043.
Therefore, the original cost of the automobile is $16,043 ($12,043 plus the $4000 down payment).
The interest charges paid by Murphy are given by(36)(400) – 12,043 = 2,357or $2,357.

59. Amortization and Sinking Funds
4.3
Amortization and Sinking Funds

60. Amortization Formula
The periodic payment R on a loan of P dollars to be amortized over n periods with interest charged at the rate of i per period is

61. Applied Example: Home Mortgage Payment
The Blakelys borrowed $120,000 from a bank to help finance the purchase of a house.
The bank charges interest at a rate of 9% per year on the unpaid balance, with interest computations made at the end of each month.
The Blakelys have agreed to repay the loan in equal monthly installments over 30 years.
How much should each payment be if the loan is to be amortized at the end of term?

62. Applied Example: Home Mortgage Payment
Solution
Here, P = 120,000, i = r/m = 0.09/12 = , and
n = (30)(12) = 360.
Using the amortization formula we find that the size of each monthly installment required is given by
or $

63. Applied Example: Home Affordability
The Jacksons have determined that, after making a down payment, they could afford at most $2000 for a monthly house payment.
The bank charges interest at a rate of 7.2% per year on the unpaid balance, with interest computations made at the end of each month.
If the loan is to be amortized in equal monthly installments over 30 years, what is the maximum amount that the Jacksons can borrow from the bank?

64. Applied Example: Home Affordability
Solution
We are required to find P, given i = r/m = 0.072/12 = 0.006, n = (30)(12) = 360, and R = 2000.
We first solve for P in the amortization formula
Substituting the numerical values for R, n, and i we obtain
Therefore, the Jacksons can borrow at most $294,643.

65. Sinking Fund Payment
The periodic payment R required to accumulate a sum of S dollars over n periods with interest charged at the rate of i per period is

66. Applied Example: Sinking Fund
The proprietor of Carson Hardware has decided to set up a sinking fund for the purpose of purchasing a truck in 2 years’ time.
It is expected that the truck will cost $30,000.
If the fund earns 10% interest per year compounded quarterly, determine the size of each (equal) quarterly installment the proprietor should pay into the fund.

67. Applied Example: Sinking Fund
Solution
Here, S = 30,000, i = r/m = 0.1/4 = 0.025, and n = (2)(4) = 8.
Using the sinking fund payment formula we find that the required size of each quarterly payment is given by
or $

68. Arithmetic and Geometric Progressions
4.4
Arithmetic and Geometric Progressions

69. Arithmetic Progressions
An arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant d to the preceding term.
The constant d is called the common difference.
An arithmetic progression is completely determined if the first term and the common difference are known.

70. nth Term of an Arithmetic Progression
The nth term of an arithmetic progression with first term a and common difference d is given by
an = a + (n – 1)d

71. Example Find the twelfth term of the arithmetic progression
2, 7, 12, 17, 22, …
Solution
The first term of the arithmetic progression is a1 = a = 2, and the common difference is d = 5.
So, upon setting n = 12 in the arithmetic progression formula, we find
an = a + (n – 1)d
a12 = 2 + (12 – 1)5
= 57

72. Sum of Terms in an Arithmetic Progression
The sum of the first n terms of an arithmetic progression with first term a and common difference d is given by

73. Example
Find the sum of the first 20 terms of the arithmetic progression
2, 7, 12, 17, 22, …
Solution
Letting a = 2, d = 5, and n = 20 in the sum of terms formula, we get

74. Applied Example: Company Sales
Madison Electric Company had sales of $200,000 in its first year of operation.
If the sales increased by $30,000 per year thereafter, find Madison’s sales in the fifth year and its total sales over the first 5 years of operation.
Solution
Madison’s yearly sales follow an arithmetic progression, with a = 200,000 and d = 30,000.
The sales in the fifth year are found by using the nth term formula with n = 5.
Thus,
or $320,000.
an = a + (n – 1)d
a5 = 200,000 + (5 – 1)30,000
= 320,000

75. Applied Example: Company Sales
Madison Electric Company had sales of $200,000 in its first year of operation.
If the sales increased by $30,000 per year thereafter, find Madison’s sales in the fifth year and its total sales over the first 5 years of operation.
Solution
Madison’s total sales over the first 5 years of operation are found by using the sum of terms formula.
Thus,
or $1,300,000.

76. Geometric Progressions
A geometric progression is a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by a constant.
The constant is called the common ratio.
A geometric progression is completely determined if the first term a and the common ratio r are given.

77. nth Term of a Geometric Progression
The nth term of a geometric progression with first term a and common ratio r is given by

78. Example
Find the eighth term of a geometric progression whose first five terms are 162, 54, 18, 6, and 2.
Solution
The common ratio is found by taking the ratio of any term other than the first to the preceding term.
Taking the ratio of the fourth term to the third term, for example, gives
Using the nth term formula to find the eighth term gives

79. Sum of Terms in a Geometric Progression
The sum of the first n terms of a geometric progression with first term a and common ratio r is given by

80. Example
Find the sum of the first six terms of the geometric progression 3, 6, 12, 24, …
Solution
Using the sum of terms formula
with a = 3, r = 6/3 = 2, and n = 6, gives

81. Applied Example: Company Sales
Michaelson Land Development Company had sales of $1 million in its first year of operation.
If sales increased by 10% per year thereafter, find Michaelson’s sales in the fifth year and its total sales over the first 5 years of operation.
Solution
The sales follow a geometric progression, with first term a = 1,000,000 and common ratio r = 1.1.
The sales in the fifth year are thus
or $1, 464,100.

82. Applied Example: Company Sales
Michaelson Land Development Company had sales of $1 million in its first year of operation.
If sales increased by 10% per year thereafter, find Michaelson’s sales in the fifth year and its total sales over the first 5 years of operation.
Solution
The sales follow a geometric progression, with first term a = 100,000,000 and common ratio r = 1.1.
The total sales over the first 5 years of operation are
or $6,105,100.

83. End of Chapter