# Chapter 6 Interest Rates and Bond Valuation

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1 Chapter 6 Interest Rates and Bond Valuation Solutions to Problems P6-1. P6-2. LG 1: Interest Rate Fundamentals: The Real Rate of Return Basic Real rate of return = 5.5% 2.0% = 3.5% LG 1: Real Rate of Interest Supply and Demand Curve Interest Rate Required Demanders/ Supplier (%) Amount of Funds Supplied/Demanded (\$) billion Current Suppliers Demanders after new Current demanders (b) The real rate of interest creates an equilibrium between the supply of savings and the demand for funds, which is shown on the graph as the intersection of lines for current suppliers and current demanders. K 0 = 4% (c) See graph. (d) A change in the tax law causes an upward shift in the demand curve, causing the equilibrium point between the supply curve and the demand curve (the real rate of interest) to rise from k o = 4% to k 0 = 6% (intersection of lines for current suppliers and demanders after new law).

2 Chapter 6 Interest Rates and Bond Valuation 145 P6-3. LG 1: Real and Nominal Rates of Interest 4 shirts (b) \$100 + (\$ ) = \$109 (c) \$25 + (\$ ) = \$26.25 (d) The number of polo shirts in one year = \$109 \$26.25 = He can buy 3.8% more shirts ( = ). (e) The real rate of return is 9% 5% = 4%. The change in the number of shirts that can be purchased is determined by the real rate of return since the portion of the nominal return for expected inflation (5%) is available just to maintain the ability to purchase the same number of shirts. P6-4. LG 1: Yield Curve Yield Curve of U.S. Treasury Securities Yield % Time to Maturity (years) (b) The yield curve is slightly downward sloping, reflecting lower expected future rates of interest. The curve may reflect a general expectation for an economic recovery due to inflation coming under control and a stimulating impact on the economy from the lower rates. P6-5. LG 1: Nominal Interest Rates and Yield Curves Challenge k l = k * + IP + RP 1 For U.S. Treasury issues, RP = 0 R F = k * + IP 20 year bond: R F = % = 11.5% 3 month bill: R F = % = 7.5% 1 year note: R F = % = 8.5% 5 year bond: R F = % = 10.5%

3 146 Part 2 Important Financial Concepts (b) If the real rate of interest (k * ) drops to 2.0%, the nominal interest rate in each case would decrease by 0.5 percentage point. (c) Return versus Maturity Rate of Return % Years to Maturity The yield curve for U.S. Treasury issues is upward sloping, reflecting the prevailing expectation of higher future inflation rates. (d) Followers of the liquidity preference theory would state that the upward sloping shape of the curve is due to the desire by lenders to lend short-term and the desire by business to borrow long term. The dashed line in the part (c) graph shows what the curve would look like without the existence of liquidity preference, ignoring the other yield curve theories. (e) Market segmentation theorists would argue that the upward slope is due to the fact that under current economic conditions there is greater demand for long-term loans for items such as real estate than for short-term loans such as seasonal needs. P6-6. LG 1: Nominal and Real Rates and Yield Curves Challenge Real rate of interest (k * ): k i = k * + IP + RP RP = 0 for Treasury issues k * = k i IP Security Nominal Rate (k j ) IP = Real Rate of Interest (k * ) A 12.6% 9.5% = 3.1% B 11.2% 8.2% = 3.0% C 13.0% 10.0% = 3.0% D 11.0% 8.1% = 2.9% E 11.4% 8.3% = 3.1%

4 Chapter 6 Interest Rates and Bond Valuation 147 (b) The real rate of interest decreased from January to March, remained stable from March through August, and finally increased in December. Forces which may be responsible for a change in the real rate of interest include changing economic conditions such as the international trade balance, a federal government budget deficit, or changes in tax legislation. (c) Yield Curve of U.S. Treasury Securities Yield % Time to Maturity (years) (d) The yield curve is slightly downward sloping, reflecting lower expected future rates of interest. The curve may reflect a general expectation for an economic recovery due to inflation coming under control and a stimulating impact on the economy from the lower rates. P6-7. LG 1: Term Structure of Interest Rates Yield Curve of High-Quality Corporate Bonds Today Yield % years ago 5 years ago Time to Maturity (years)

5 148 Part 2 Important Financial Concepts (b) and (c) Five years ago, the yield curve was relatively flat, reflecting expectations of stable interest rates and stable inflation. Two years ago, the yield curve was downward sloping, reflecting lower expected interest rates due to a decline in the expected level of inflation. Today, the yield curve is upward sloping, reflecting higher expected inflation and higher future rates of interest. P6-8. LG 1: Risk-Free Rate and Risk Premiums Basic Risk-free rate: R F = k * + IP Security K * + IP = R F A 3% + 6% = 9% B 3% + 9% = 12% C 3% + 8% = 11% D 3% + 5% = 8% E 3% + 11% = 14% (b) Since the expected inflation rates differ, it is probable that the maturity of each security differs. (c) Nominal rate: k = k * + IP + RP Security k * + IP + RP = k A 3% + 6% + 3% = 12% B 3% + 9% + 2% = 14% C 3% + 8% + 2% = 13% D 3% + 5% + 4% = 12% E 3% + 11% + 1% = 15% P6-9. LG 1: Risk Premiums R Ft = k * + IP t Security A: R F3 = 2% + 9% = 11% Security B: R F15 = 2% + 7% = 9% (b) Risk premium: RP = default risk + interest rate risk + liquidity risk + other risk Security A: RP = 1% + 0.5% + 1% + 0.5% = 3% Security B: RP = 2% + 1.5% + 1% + 1.5% = 6% (c) k i = k * + IP + RP or k 1 = R F + Risk premium Security A: k 1 = 11% + 3% = 14% Security B: k 1 = 9% + 6% = 15%

6 Chapter 6 Interest Rates and Bond Valuation 149 Security A has a higher risk-free rate of return than Security B due to expectations of higher nearterm inflation rates. The issue characteristics of Security A in comparison to Security B indicate that Security A is less risky. P6-10. LG 2: Bond Interest Payments Before and After Taxes Yearly interest = (\$1, ) = \$70.00 (b) Total interest expense = \$70.00 per bond 2,500 bonds = \$175,000 (c) Total before tax interest \$175,000 Interest expense tax savings (0.35 \$175,000) 61,250 Net after-tax interest expense \$113,750 P6-11. LG 4: Bond Quotation Basic Tuesday, November 7 (b) \$1,000 = \$ (c) May 15, 2013 (d) \$47,807,000 (e) 5.7% (f) last yield = 6.06%. This yield represents the expected compounded rate of return the investor would earn if the bond is purchased at the price quoted and the bond is held until the maturity date. (g) The spread of this FM bond over a similar time to maturity U.S. Treasury bond is 129 basic points, or 1.29%. P6-12. LG 4: Valuation Fundamentals Basic Cash Flows: CF 1 5 \$1,200 CF 5 \$5,000 Required return: 6% (b) 5 0 = (1 k) (1 + k) V = \$1,200 + \$1,200 + \$1,200 + \$1,200 + \$6,200 ( ) ( ) ( ) ( ) ( ) V 0 = \$8,791 Using PVIF formula: V 0 = [(CF 1 PVIF 6%,l ) + (CF 2 PVIF 6%, 2 )... (CF 5 PVIF 6%,5 )] V 0 = [(\$1, ) + (\$1, ) + (\$1, ) + (\$1, ) + (\$6, )] V 0 = \$1, \$1, \$1,008 + \$ \$4, V 0 = \$8, Calculator solution: \$8, The maximum price you should be willing to pay for the car is \$8,789, since if you paid more than that amount, you would be receiving less than your required 6% return. 5

7 150 Part 2 Important Financial Concepts P6-13. LG 4: Valuation of Assets Basic PVIF or Present Value of Asset End of Year Amount PVIFA k%,n Cash Flow A 1 \$ \$ \$5000 \$10, Calculator solution: \$10, B 1 \$ \$2,000 C \$35, \$16, Calculator solution: \$16, D 1 5 \$1, \$5, , , \$9, Calculator solution: \$9, E 1 \$2, \$1, , , , , , , , , , \$14, P6-14. LG 4: Asset Valuation and Risk Calculator solution: \$14, % Low Risk 15% Average Risk 22% High Risk PVIFA PV of CF PVIFA PV of CF PVIFA PV of CF CF 1 4 \$3, \$9, \$8, \$7,482 CF 5 15, , , ,550 Present Value of CF: \$18,825 \$16,020 \$13,032 Calculator solutions: \$18, \$16, \$13,030.91

8 Chapter 6 Interest Rates and Bond Valuation 151 (b) The maximum price Laura should pay is \$13,032. Unable to assess the risk, Laura would use the most conservative price, therefore assuming the highest risk. (c) By increasing the risk of receiving cash flow from an asset, the required rate of return increases, which reduces the value of the asset. P6-15. LG 5: Basic Bond Valuation B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) B o = 120 (PVIFA 10%,16 ) + M (PVIF 10%,16 ) B o = \$120 (7.824) + \$1,000 (0.218) B o = \$ \$218 B o = \$1, Calculator solution: \$1, (b) Since Complex Systems bonds were issued, there may have been a shift in the supplydemand relationship for money or a change in the risk of the firm. (c) B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) B o = 120 (PVIFA 12%,16 ) + M (PVIF 12%,16 ) B o = \$120 (6.974) + \$1,000 (0.163) B o = \$ \$163 B o = \$ Calculator solution: \$1,000 When the required return is equal to the coupon rate, the bond value is equal to the par value. In contrast to above, if the required return is less than the coupon rate, the bond will sell at a premium (its value will be greater than par). P6-16. LG 5: Bond Valuation Annual Interest Basic B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) Bond Table Values Calculator Solution A B o = \$140 (7.469) + \$1,000 (0.104) = \$1, \$1, B B o = \$80 (8.851) + \$1,000 (0.292) = \$1, \$1, C B o = \$10 (4.799) + \$100 (0.376) = \$85.59 \$85.60 D B o = \$80 (4.910) + \$500 (0.116) = \$ \$ E B o = \$120 (6.145) + \$1,000 (0.386) = \$1, \$1,122.89

9 152 Part 2 Important Financial Concepts P6-17. LG 5: Bond Value and Changing Required Returns B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) Bond Table Values Calculator Solution (b) (1) B o = \$110 (6.492) + \$1,000 (0.286) = \$1, \$1, (2) B o = \$110 (5.421) + \$1,000 (0.187) = \$ \$ (3) B o = \$110 (7.536) + \$1,000 (0.397) = \$1, \$1, Bond Value versus Required Return 1,300 1,200 1,100 Bond Value (\$) 1, % 9% 10% 11% 12% 13% 14% 15% Required Return (%) (c) When the required return is less than the coupon rate, the market value is greater than the par value and the bond sells at a premium. When the required return is greater than the coupon rate, the market value is less than the par value; the bond therefore sells at a discount. (d) The required return on the bond is likely to differ from the coupon interest rate because either (1) economic conditions have changed, causing a shift in the basic cost of long-term funds, or (2) the firm s risk has changed.

10 Chapter 6 Interest Rates and Bond Valuation 153 P6-18. LG 5: Bond Value and Time Constant Required Returns B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) Bond Table Values Calculator Solution (b) (1) B o = \$120 (6.142) + \$1,000 (0.140) = \$ \$ (2) B o = \$120 (5.660) + \$1,000 (0.208) = \$ \$ (3) B o = \$120 (4.946) + \$1,000 (0.308) = \$ \$ (4) B o = \$120 (3.889) + \$1,000 (0.456) = \$ \$ (5) B o = \$120 (2.322) + \$1,000 (0.675) = \$ \$ (6) B o = \$120 (0.877) + \$1,000 (0.877) = \$ \$ Bond Value versus Years to Maturity Bond Value (\$) (c) The bond value approaches the par value Years to Maturity P6-19. LG 5: Bond Value and Time Changing Required Returns Challenge B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) Bond Table Values Calculator Solution (1) B 0 = \$110 (3.993) + \$1,000 (0.681) = \$1, \$1, (2) B 0 = \$110 (3.696) + \$1,000 (0.593) = \$1, \$1, (3) B 0 = \$110 (3.433) + \$1,000 (0.519) = \$ \$897.01

11 154 Part 2 Important Financial Concepts (b) (c) Bond Table Values Calculator Solution (1) B 0 = \$110 (8.560) + \$1,000 (0.315) = \$1, \$1, (2) B 0 = \$110 (7.191) + \$1,000 (0.209) = \$1, \$1, (3) B 0 = \$110 (6.142) + \$1,000 (0.140) = \$ \$ Value Required Return Bond A Bond B 8% \$1, \$1, % 1, , % The greater the length of time to maturity, the more responsive the market value of the bond to changing required returns, and vice versa. (d) If Lynn wants to minimize interest rate risk in the future, she would choose Bond A with the shorter maturity. Any change in interest rates will impact the market value of Bond A less than if she held Bond B. P6-20. LG 6: Yield to Maturity Basic Bond A is selling at a discount to par. Bond B is selling at par value. Bond C is selling at a premium to par. Bond D is selling at a discount to par. Bond E is selling at a premium to par. P6-21. LG 6: Yield to Maturity Using a financial calculator the YTM is %. The correctness of this number is proven by putting the YTM in the bond valuation model. This proof is as follows: B o = 120 (PVIFA %,15 ) + 1,000 (PVIF %,15 ) B o = \$120 (6.569) + \$1,000 (0.167) B o = \$ B o = \$ Since B o is \$ and the market value of the bond is \$955, the YTM is equal to the rate derived on the financial calculator. (b) The market value of the bond approaches its par value as the time to maturity declines. The yield to maturity approaches the coupon interest rate as the time to maturity declines.

12 Chapter 6 Interest Rates and Bond Valuation 155 P6-22. LG 6: Yield to Maturity Trial-and-Error Calculator Bond Approximate YTM YTM Approach Error (%) Solution A = \$90 + [(\$1,000 \$820) 8] [(\$1,000 + \$820) 2] = 12.36% 12.71% % B = 12.00% 12.00% % C = D = E = \$60 + [(\$500 \$560) 12] [(\$500 + \$560) 2] = 10.38% 10.22% % \$150 + [(\$1, 000 \$120) 10] [(\$1, \$1,120 2] = 13.02% 12.81% % \$50 + [(\$1, 000 \$900) 3] [(\$1, \$900) 2] = 8.77% 8.94% % (b) The market value of the bond approaches its par value as the time to maturity declines. The yield-to-maturity approaches the coupon interest rate as the time to maturity declines. P6-23. LG 2, 5, 6: Bond Valuation and Yield to Maturity Challenge B A = \$60(PVIFA 12%,5 ) + \$1,000(PVIF 12%,5 ) B A = \$60(3.605) + \$1,000(0.567) B A = \$ B A = \$ B B = \$140(PVIFA 12%,5 ) + \$1,000(PVIF 12%,5 ) B B = \$140(3.605) + \$1,000(0.567) B B = \$ B B = \$1,071.70

13 156 Part 2 Important Financial Concepts (b) Number of bonds = \$20,000 = of bond A \$ Number of bonds = \$20,000 = of bond B \$1, (c) Interest income of A = bonds \$60 = \$1, Interest income of B = bonds \$140 = \$2, (d) At the end of the 5 years both bonds mature and will sell for par of \$1,000. FV A = \$60(FVIFA 10%,5 ) + \$1,000 FV A = \$60(6.105) + \$1,000 FV A = \$ \$1,000 = \$1, FV B = \$140(FVIFA 10%,5 ) + \$1,000 FV B = \$140(6.105) + \$1,000 FV B = \$ \$1,000 = \$1, (e) The difference is due to the differences in interest payments received each year. The principal payments at maturity will be the same for both bonds. Using the calculator, the yield to maturity of bond A is 11.77% and the yield to maturity of bond B is 11.59% with the 10% reinvestment rate for the interest payments. Mark would be better off investing in bond A. The reasoning behind this result is that for both bonds the principal is priced to yield the YTM of 12%. However, bond B is more dependent upon the reinvestment of the large coupon payment at the YTM to earn the 12% than is the lower coupon payment of A. P6-24. LG 6: Bond Valuation Semiannual Interest B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) B o = \$50 (PVIFA 7%,12 ) + M (PVIF 7%,12 ) B o = \$50 (7.943) + \$1,000 (0.444) B o = \$ \$444 B o = \$ Calculator solution: \$ P6-25. LG 6: Bond Valuation Semiannual Interest B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) Bond Table Values Calculator Solution A B o = \$50 (15.247) + \$1,000 (0.390) = \$1, \$1, B B o = \$60 (15.046) + \$1,000 (0.097) = \$1, \$1, C B o = \$30 (7.024) + \$500 (0.508) = \$ \$ D B o = \$70 (12.462) + \$1,000 (0.377) = \$1, \$1, E B o = \$3 (5.971) + \$100 (0.582) = \$76.11 \$76.11

14 Chapter 6 Interest Rates and Bond Valuation 157 P6-26. LG 6: Bond Valuation Quarterly Interest Challenge B o = I (PVIFA kd%,n ) + M (PVIF kd%,n ) B o = \$125 (PVIFA 3%,40 ) + \$5,000 (PVIF 3%,40 ) B o = \$125 (23.115) + \$5,000 (0.307) B o = \$2, \$1,535 B o = \$4, Calculator solution: \$4, P6-27. Ethics Problem Absolutely not if anything, they cast raters in an even worse light. The primary ethical issue is, are investors being provided an accurate, timely, and unbiased reading on a company s bond issues creditworthiness? Rating agencies should be expected to invest heavily in statistical models and management meetings, in order to get the most accurate data and rating methods possible. The fact they have done all this and still not carried out their responsibility to investors at a high degree of proficiency is concerning.

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