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Compound Interest Annuities Amortization and Sinking Funds Arithmetic and Geometric Progressions

4. Compound Interest Annuities Amortization and Sinking Funds Arithmetic and Geometric Progressions. Mathematics of Finance. 4.1. Compound Interest. Simple Interest Formulas. Simple interest is the interest that is computed on the original principal only.

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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.

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