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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