Compound Interest Annuities Amortization and Sinking Funds Arithmetic and Geometric Progressions

# Compound Interest Annuities Amortization and Sinking Funds Arithmetic and Geometric Progressions

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## Compound Interest Annuities Amortization and Sinking Funds Arithmetic and Geometric Progressions

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

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 principalP (in dollars) at an interest rate of r per year for t years, then we have I = Prt • The accumulated amountA, the sum of the principal and interest after tyears 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 10years? 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 A2at the end of the second year, we use the simple interest formula again, this time with P =A1, obtaining: or approximately \$1166.40.

9. Compound Interest • We can use the simple interest formulayet again to find the accumulated amount A3 at the end of the third year: or approximately \$1259.71.

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 tyears 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 compoundedannually. • In practice, however, interest is usually compoundedmore 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 compoundedm times a year on a principal of P dollars, then the simple interest rate perconversion 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 oftyears 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 \$1259.71.

17. Example Solution • Semiannually. Here, P = 1000, r = 0.08, and m = 2. Thus, and n = (3)(2) = 6, so or \$1265.32.

18. Example Solution • Quarterly. Here, P = 1000, r = 0.08, and m = 4. Thus, and n = (3)(4) = 12, so or \$1268.24.

19. Example Solution • Monthly. Here, P = 1000, r = 0.08, and m = 12. Thus, and n = (3)(12) = 36, so or \$1270.24.

20. Example Solution • Daily. Here, P = 1000, r = 0.08, and m = 365. Thus, and n = (3)(365) = 1095, so or \$1271.22.

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 compoundedmore and more frequently? • We’ve seen that the more often interest is compounded, the larger the accumulated amount. • But does the accumulated amountapproach alimit 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 amountA.

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 approaches2.71828 (rounding to five decimal places). • It can be shown that as u gets larger and larger, the value of the expression approaches the irrational number2.71828… which we denote by e.

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)≈ 1271.25 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 compoundedmtimes 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 interestrateR per year is • The accumulated amount after 1 year at a nominal interest rater 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 reffis 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 principalP is often referred to as the present value, and the accumulated valueA 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 expressingPin terms ofA.

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 \$49,158.60due 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 continuouscompound 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 valueP(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 tyears from now is • Lettingt = 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 tyears from now is • Lettingt = 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 tyears from now is • Lettingt = 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 timet≈ 7.72 years, by sketching the graph of the functionP.

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. Future Value of an Annuity • The future valueS of an annuity of npayments 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

50. Example • Find the amount of an ordinaryannuity consisting of 12monthlypayments 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 \$1268.25.