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Chapter 5

Chapter 5 Mathematics of Finance Compound Interest In the last class session, we covered the basics regarding compound interest Tonight, we build on these results to get some help for working with financial investments.

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Chapter 5

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  1. Chapter 5 Mathematics of Finance

  2. Compound Interest • In the last class session, we covered the basics regarding compound interest • Tonight, we build on these results to get some help for working with financial investments. • Let P = principal invested, let S = the compound amount (i.e., the amount of money we have after n time periods) and let r = interest rate.

  3. Compound Interest (cont.) • Then S = P(1 + r)n • Example: Suppose that we purchase a certificate of deposit for $1000 to mature in 4 years with an interest rate of 4.5% • S = $1000(1 + 0.045)n =1000(1.1925) • S = $1,192.50 • Note that in this problem, the idea is to find S, given P, r, and n

  4. Quarterly Compounding • Suppose now compounding is quarterly. How long does it take for $1500 to become $2100? • $2100 = $1500(1 + 0.0125)n • ln(2100/1500)/ln(1.0125) = n • So, n = 27.088 quarters or 6.77 years • Shows that quarterly compounding decreases the length of time it takes for $1500 to become $2100 at 5% interest.

  5. Power of Compound Interest • Suppose a person at age 23 puts $1000 aside at 6% interest for retirement. How much money will be available at age 65? • S = $1000(1 + 0.06)42 = $11,560 • Suppose a person at age 60 puts $1000 aside at 6% interest for retirement. How much money will be available at age 65? • S = $1000(1 + 0.06)5 = $1,340

  6. Power of Compound Interest • Suppose that you have outstanding credit card debt of $4700. The interest rate on the card is 23% per year. How much would be owed at the end of one year if there was no need to make the minimum payments? • S = 4700(1 + 0.24) = $5828.00 • After 5 years it would be: • S = 4700(1 + 0.24)5 = $13,778.63 • After 5 years with monthly compounding: • S = 4700(1 + 0.02)60 = $15,420.84

  7. Two Variations on Compound Interest • Suppose we know P, S, and n and want to find r. • Example: Assume that we buy a CD for $600 and after 5 years it is worth $900. What interest rate was earned? • Then, $900 = $600(1 + r)5 • So, (900/600)1/5 = (1 + r) • Or, 1.0845 = 1 + r • This means that r = 0.0845

  8. Two Variations (cont.) • Another use for the compound interest formula comes up if we know S, P, and r and want to find n. • Suppose we know that a CD purchased for $1500 will eventually be valued at $2100. At 5% interest, how long will it take for this to happen. • $2100 = $1500(1.05)n • ln(2100/1500) = n(ln(1.05)) • n = ln(2100/1500)/(ln(1.05)) = 6.897 years

  9. Doubling Time of Money • Given that S = P(1+ r)n then the principal invested, P, doubles when S = 2P • Example: How long does it take for $1500 to double at 5% • $3000 = $1500(1 +0.05)n • ln(3000/1500) = ln(2) = n(ln(1.05)) • n = ln(2)/ln(1.05) = 14.21 • n = 14.21 years

  10. Doubling Time of Money • With quarterly compounding, how long does it take for the $1500 to double? • $3000 = $1500(1 + 0.0125)n • n = ln(2)/ln(1.0125) • Daily compounding? • n = ln(2)/ln(1.0001) • Continuous compounding: Use rule of 70, so at 5% interest it takes 14 years(i.e., 5x14 = 70

  11. Effective Rate of Interest • Definition: The effective rate re that is equivalent to a nominal (annual) rate of interest r compounded n times per year is • re = (1 + (r/n))n – 1 • Example: suppose that compounding occurs 4 times per year • re = (1 + (r/4))4 - 1

  12. Effective Interest Rates Suppose compounding is quarterly and we want to know the equivalent annual rate of interest • Again, let P = $1500, let the given (nominal) interest rate equal 5% and suppose money is invested for one year • Annual compounding: S = $1500(1.05) = $1575 • Quarterly compounding: S = $1500(1.0125)4 = $1500(1.0509) = $1576.41

  13. EffectiveRate (cont.) • So, the effective interest rate calculation asks: What annual rate would bring $1500 to $1576.41? This would be re = 0.0509 or 5.09% • Another way to say this is that 5% interest compounded quarterly is the same as 5.09% compounded annually.

  14. Another example • What effective rate is equivalent to a nominal rate of 6% compounded monthly • re = (1 + (0.06/12))12 = 1.0617 • What if compounding occurs daily? • re = (1 + (0.06/365))365 = 1.0618 • What if compounding occurs continuously? • In this case, as we saw in Chapter 4, the effective rate of interest is e0.06 = 1.0618

  15. #27 p. 222 • A zero-coupon bond is a bond that is sold for less than its face value and has no periodic interest payments; instead the bond is redeemed at face value at maturity. Suppose that such a bond sells today for $420 and can be redeemed in 14 years for $1000. The bond earns what nominal rate compounded semiannually?

  16. Solution • 1000 = 420(1 + (r/2))28 • Let x = r/2 • 2.381 = (1 + x)28 • ln(2.381) = 28ln(1 + x) • Ln(1 + x ) = ln(2.381)/28 = 0.031 • So, (1 + x) = e0.031 = 1.0315 • This means that x = 0.0315; r = 0.0630

  17. #28, p. 222 • Suppose that $1000 is hidden under a mattress for safekeeping. Each year the purchasing power of money is 96.5% of what it was the previous year. After 6 years, what is the purchasing power of the $1000? • S = $1000(1 + (-0.035))6 • S = $1000(1 – 0.035)6 = $807.54

  18. Homework • Pp. 221-222: 7,9,11,15,21,25

  19. Present Value • We know that the future sum, S, that is obtained by investing the principal, P, at 100r% for n years is given by • S = P(1 + r)n • So, if we want to find the value today of a future sum to be paid in n years when the interest rate is 100r%, we would calculate • P = S/(1 + r)n • Or, P = S(1 + r)-n

  20. Example #1 • Find the present value of $1000 due after 3 years if the interest rate is 9% • P = 1000(1.09)-3 = 1000/1.295=772.18 • Same problem but with monthly compounding • P = 1000(1.0075)-36 =1000/1.309=764.15

  21. Example #2 • Suppose that a trust fund for a child’s education is set up by a single payment today so that at the end of 15 years $50,000 will be available for college tuition payments. If the interest rate is 7% compounded semiannually, how much should be paid into the fund today? • P = 50,000(1.035)-30 = $17,813.92

  22. Equations of Value • Suppose that Mr. Smith owes Mr. Jones two sums of money: $600 due in 5 years and $1000 due in 2 years. The interest rate is 8% compounded quarterly. How much should Mr. Smith pay today to retire both of the debts? • x = 1000(1.02)-8 + 600(1.02)-20 • x = 1000/1.17 + 600/1.49 = 854.70 + 403.78 = $1,258.48

  23. Equations of Value • A debt of $3000 due in 6 years is to be paid off by three payments, $500 now, $1500 in 3 years and a final payment after 5 years. How large should the final payment be? The interest rate is 6% compounded annually. • 3000 = 500(1.06)6 + 1500(1.06)3 + x(1.06) • Divide both sides by (1.06) • x = 3000(1.06)-1 – 500(1.06)5 – 1500(1.060)2 • x = 2830.19 – 669.11 – 1685.40 = $475.68

  24. Net Present Value • Suppose that you can invest $20,000 in a business that will guarantee you a cash flow of $10,000 in year 2, $8,000 in year 3, and $6,000 in year 4. the interest rate 7% compounded annually. What is the net present value of the business venture? This is the value today of the cash flow less investment cost. • NPV = 10,000(1.07)-2 + 8,000(1.07)-3 + 6,000(1.07)-4 – 20,000 = -$471.31 • With a negative NPV, the investment should not be undertaken.

  25. Homework • P. 226-227: 1, 13, 15, 19, 21

  26. Geometric Sequence • Definition: The sequence of n numbers a, ar, ar2, ar3, … ,arn-1 where a ≠ 0 is called a geometric sequence with first term a and common ratio r Example: Suppose $100 is invested at 6% for 4 years. Then the compound amounts at the end of each year is a geometric sequence and can be written as: 100(1.06), 100(1.06)2, 100(1.06)3, 100(1.06)4

  27. Geometric Series • Definition: A geometric series is defined as the sum of terms in a geometric sequence. • Example: 100(1.06) + 100(1.06)2 + 100(1.06)3 + 100(1.06)4 • In general, a geometric series can be written as: a + ar + ar2 + ar3 + …. + arn-1

  28. Geometric Series • Suppose we wish to compute the sum of terms in a geometric series • s = a + ar + ar2 + ar3 + …. + arn-1 • rs = ar + ar2 + ar3 + ar4 + …. + arn • s – rs = a – arn • s(1 – r) = a(1 – rn) • s = a(1 – rn)/(1 – r)

  29. Examples • Suppose a = 1, r = ½, and n = 6. Find s. • s = a(1 – rn)/(1 – r) • s = (1 – 0.57)/(1 – 0.5) = .9922/0.5 = 1.9844 ≈ 127/64 (see p. 230). • Suppose we wish to evaluate 35 + 36 + 37 + 38 + 39 + 310 + 311 • a = 35 = 243, r = 3, n = 7 • s = 243(1 – 37)/(1 – 3) = 243(-2,186)/(-2) = 265,599

  30. Annuities • An annuity pays a fixed amount of money per period (year, quarter, month) for a fixed number of years. • Suppose that the payment per period is $R, a period is one year long, the interest rate is r, and there are n payments. In general, how much would we pay today for this income stream? • A = R(1 + r)-1 + R(1 + r)-2 + … + R(1 + r)-n • A = R(1 + r)-1 [1 - (1 + r)-n]/[1 - (1 + r)-1]

  31. Annuities (cont.) • With a little algebra, The annuity value can be written as • A = R[1 – (1 + r)-n]/r • This formula gives the present value A of an annuity of R dollars per payment for n periods at the interest rate of r per period. • Example: Find the present value of an annuity of $100 per month for 3.5 years at interest rate 6% compounded monthly • Using Appendix B , p. 954, this is $100(37.7983) = $3779.83

  32. Annuities (cont.) • Or, this value can be found approximately using a calculator. Let anr = [1 – (1 + r)-n]/r • For r = 0.005 and n = 42, anR = 37.8, so A = $3780 • Value of a consol (promise to pay in perpetuity). In this case, anr = 1/r. So, a promise to pay $100 each year in perpetuity would be valued at $2000 if the interest rate is 5%. If the interest rate is 10%, A= $1000

  33. Example 6, p. 231 • Given an interest rate of 5% compounded annually, find the present value of an annuity of $2000 due at the end of each year for three years and $5000 due at thereafter at the end of each year for four years. • A = $5000(5.786) - $3000(2.723) = $2885.91 • where a7,0.05 = 5.786 and a3,0.05 = 2.723

  34. Example 7, p. 232 • If $10,000 is used to purchase an annuity consisting of equal annual payments for four years and the interest rate is 6% compounded annually, find the amount of each payment. • Know that A = Ranr so, R = A/anr • R = $10,000/a4,0.06 = $10,000/3.465 = $2885.91

  35. Another Example • Suppose that a home mortgage of $150,000 is to be retired by making 48 monthly payments. The interest rate is 6%. Determine the monthly payment. • The way to look at this is that the bank is purchasing an annuity valued today at $150,000 with monthly payment for 4 years at 6% compounded monthly. • R = A/anr = 150,000/a48,0.005 = 150,000/42.580 = $3522.78

  36. Bond Values • Suppose that you buy a 20-year corporate bond for $10,000. The coupon rate is 5% paid semiannually. How much would you pay for the bond if the interest rate is now 6%? • S = Ranr + 10,000/(1 + r)40 • R = $250, n = 40, r = 0.03, anr = 23.114772 • R = $5778.69 S = $5778.69 + $3065.57 = $8844.26

  37. Bond Prices (cont.) • Suppose now that interest rates are 4%. What would you pay for the same bond? • S = Ranr + 10,000/(1 + r)40 • R = $250, n = 40, r = 0.02, anr=27.355479 • S = $6838.87 + 4528.90 = $11,367.77 • Two conclusions • When interest rates are above (below) the coupon rate, bonds trade at a discount (premium) • Bond prices vary inversely with the interest rate

  38. Amount of an annuity • Thus far, we have considered the value of an annuity. This is what should be paid today for a stream of future periodic payments • Now we wish to consider the amount of an annuity: The amount of money available at the end of the period of a stream of payments compounded forward • The formula for the amount of an annuity is • S = R[(1 + r)n – 1]/r = Rsnr

  39. Amount of an Annuity • Find the amount of an annuity consisting of payments of $50 at the end of every 3 months for three years at 6% compounded quarterly. Also, find the compound interest that would accrue. • S = Rsnr • S = 50s12,0.015 = 50(13.041) = $652.06 • Compound interest accruals are $52.06. • $52.06 = $652.06 - $600

  40. Sinking Fund • A sinking fund is a fund into which periodic payments are made to satisfy a future obligation. Suppose that a machine costing $7000 is to be replaced at the end of 8 years, when it will have a salvage value of $700. How much must be set aside each quarter to replace the machine if the interest rate is 8% compounded quarterly?

  41. Solution • The amount needed after 8 years is $7000 - $700 = $6300 • The required quarterly payment is • R = S/snr • R = $6300/s32,0.02 = 6300/44.227 = $142.45 • So, a quarterly payment of $142.45 will accumulate to $6300 after 8 years at 8% interest compounded quarterly

  42. Retirement Planning • A woman at age 25 has $17,000 in her retirement plan and wishes to contribute $2000 each year until she retires at age 65. Assume an interest rate of 6% compounded annually. How much will she have available for retirement • S = Rsnr + 17000(1 + r)40 • S = 2000(154.762) + 17000(10.29) • S = 309,254 + 174,857.21 = $484,381.21

  43. Homework • P. 236-238: 1, 5, 7, 13, 19, 23, 25, 27, 31,33, 41, 43

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