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Chapter 7 - Valuation and Characteristics of Bonds.

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1 Chapter 7 - Valuation and Characteristics of Bonds

2 Characteristics of Bonds  Bonds pay fixed coupon (interest) payments at fixed intervals (usually every six months) and pay the par value at maturity.

3 Characteristics of Bonds  Bonds pay fixed coupon (interest) payments at fixed intervals (usually every six months) and pay the par value at maturity. 0 12...n $I $I $I $I $I $I+$M

4 Example: AT&T 6 ½ 32  Par value = $1,000  Coupon = 6.5% or par value per year, or $65 per year ($32.50 every six months).  Maturity = 28 years (matures in 2032).  Issued by AT&T.

5 Example: AT&T 6 ½ 32  Par value = $1,000  Coupon = 6.5% or par value per year, or $65 per year ($32.50 every six months).  Maturity = 28 years (matures in 2032).  Issued by AT&T. 0 12… 28 $32.50 $32.50 $32.50 $32.50 $32.50 $32.50+$1000

6 Types of Bonds  Debentures - unsecured bonds.  Subordinated debentures - unsecured “junior” debt.  Mortgage bonds - secured bonds.  Zeros - bonds that pay only par value at maturity; no coupons.  Junk bonds - speculative or below- investment grade bonds; rated BB and below. High-yield bonds.

7 Types of Bonds  Eurobonds - bonds denominated in one currency and sold in another country. (Borrowing overseas.)  example - suppose Disney decides to sell $1,000 bonds in France. These are U.S. denominated bonds trading in a foreign country. Why do this?

8 Types of Bonds  Eurobonds - bonds denominated in one currency and sold in another country. (Borrowing overseas.)  example - suppose Disney decides to sell $1,000 bonds in France. These are U.S. denominated bonds trading in a foreign country. Why do this?  If borrowing rates are lower in France.

9 Types of Bonds  Eurobonds - bonds denominated in one currency and sold in another country. (Borrowing overseas).  example - suppose Disney decides to sell $1,000 bonds in France. These are U.S. denominated bonds trading in a foreign country. Why do this?  If borrowing rates are lower in France.  To avoid SEC regulations.

10 The Bond Indenture  The bond contract between the firm and the trustee representing the bondholders.  Lists all of the bond’s features: coupon, par value, maturity, etc. coupon, par value, maturity, etc.  Lists restrictive provisions which are designed to protect bondholders.  Describes repayment provisions.

11 Value  Book value: value of an asset as shown on a firm’s balance sheet; historical cost.  Liquidation value: amount that could be received if an asset were sold individually.  Market value: observed value of an asset in the marketplace; determined by supply and demand.  Intrinsic value: economic or fair value of an asset; the present value of the asset’s expected future cash flows.

12 Security Valuation  In general, the intrinsic value of an asset = the present value of the stream of expected cash flows discounted at an appropriate required rate of return.  Can the intrinsic value of an asset differ from its market value?

13 Valuation  C t = cash flow to be received at time t.  k = the investor’s required rate of return.  V = the intrinsic value of the asset. V = t = 1 n $C t (1 + k) t

14 Bond Valuation  Discount the bond’s cash flows at the investor’s required rate of return.

15 Bond Valuation  Discount the bond’s cash flows at the investor’s required rate of return.  The coupon payment stream (an annuity).

16 Bond Valuation  Discount the bond’s cash flows at the investor’s required rate of return.  The coupon payment stream (an annuity).  The par value payment (a single sum).

17 Bond Valuation V b = $I t (PVIFA k b, n ) + $M (PVIF k b, n ) $I t $M (1 + k b ) t (1 + k b ) n V b = + n t = 1 

18 Bond Example  Suppose our firm decides to issue 20-year bonds with a par value of $1,000 and annual coupon payments. The return on other corporate bonds of similar risk is currently 12%, so we decide to offer a 12% coupon interest rate.  What would be a fair price for these bonds?

19 0 1 2 3...20 1000 120 120 120... 120 P/YR = 1 N = 20 I%YR = 12 FV = 1,000 PMT = 120 Solve PV = -$1,000 Note: If the coupon rate = discount rate, the bond will sell for par value.

20 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 )

21 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i

22 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 ) PV = 120 (PVIFA.12, 20 ) + 1000 (PVIF.12, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 PV = 120 1 - (1.12 ) 20 +1000/ (1.12) 20 = $1000.12.12

23  Suppose interest rates fall immediately after we issue the bonds. The required return on bonds of similar risk drops to 10%.  What would happen to the bond’s intrinsic value?

24 P/YR = 1 Mode = end N = 20 I%YR = 10 PMT = 120 FV = 1000 Solve PV = -$1,170.27

25 P/YR = 1 Mode = end N = 20 I%YR = 10 PMT = 120 FV = 1000 Solve PV = -$1,170.27 Note: If the coupon rate > discount rate, the bond will sell for a premium.

26 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 )

27 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i

28 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 ) PV = 120 (PVIFA.10, 20 ) + 1000 (PVIF.10, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 PV = 120 1 - (1.10 ) 20 + 1000/ (1.10) 20 = $1,170.27.10.10

29  Suppose interest rates rise immediately after we issue the bonds. The required return on bonds of similar risk rises to 14%.  What would happen to the bond’s intrinsic value?

30 P/YR = 1 Mode = end N = 20 I%YR = 14 PMT = 120 FV = 1000 Solve PV = -$867.54

31 P/YR = 1 Mode = end N = 20 I%YR = 14 PMT = 120 FV = 1000 Solve PV = -$867.54 Note: If the coupon rate < discount rate, the bond will sell for a discount.

32 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 )

33 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i

34 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 120 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 PV = 120 1 - (1.14 ) 20 + 1000/ (1.14) 20 = $867.54.14.14

35 Suppose coupons are semi-annual P/YR = 2 Mode = end N = 40 I%YR = 14 PMT = 60 FV = 1000 Solve PV = -$866.68

36 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 )

37 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i

38 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) PV = 60 (PVIFA.14, 20 ) + 1000 (PVIF.14, 20 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 PV = 60 1 - (1.07 ) 40 + 1000 / (1.07) 40 = $866.68.07.07

39 Yield To Maturity  The expected rate of return on a bond.  The rate of return investors earn on a bond if they hold it to maturity.

40 Yield To Maturity  The expected rate of return on a bond.  The rate of return investors earn on a bond if they hold it to maturity. $I t $M (1 + k b ) t (1 + k b ) n P 0 = + n t = 1 

41 YTM Example  Suppose we paid $898.90 for a $1,000 par 10% coupon bond with 8 years to maturity and semi-annual coupon payments.  What is our yield to maturity?

42 P/YR = 2 Mode = end N = 16 PV = -898.90 PMT = 50 FV = 1000 Solve I%YR = 12% YTM Example

43 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 )

44 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i

45 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 898.90 = 50 1 - (1 + i ) 16 +1000 / (1 + i) 16 i i

46 Bond Example Mathematical Solution: PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) PV = PMT (PVIFA k, n ) + FV (PVIF k, n ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 ) 1 1 PV = PMT 1 - (1 + i) n + FV / (1 + i) n i 1 898.90 = 50 1 - (1 + i ) 16 +1000 / (1 + i) 16 i solve using trial and error i solve using trial and error

47 Zero Coupon Bonds  No coupon interest payments.  The bond holder’s return is determined entirely by the price discount.

48 Zero Example  Suppose you pay $508 for a zero coupon bond that has 10 years left to maturity.  What is your yield to maturity?

49 Zero Example  Suppose you pay $508 for a zero coupon bond that has 10 years left to maturity.  What is your yield to maturity? 0 10 -$508 $1000

50 Zero Example P/YR = 1 Mode = End N = 10 PV = -508 FV = 1000 Solve: I%YR = 7%

51 Mathematical Solution: PV = FV (PVIF i, n ) PV = FV (PVIF i, n ) 508 = 1000 (PVIF i, 10 ) 508 = 1000 (PVIF i, 10 ).508 = (PVIF i, 10 ) [use PVIF table].508 = (PVIF i, 10 ) [use PVIF table] PV = FV /(1 + i) 10 PV = FV /(1 + i) 10 508 = 1000 /(1 + i) 10 508 = 1000 /(1 + i) 10 1.9685 = (1 + i) 10 1.9685 = (1 + i) 10 i = 7% i = 7% Zero Example 0 10 PV = -508 FV = 1000

52 The Financial Pages: Corporate Bonds Cur Net Yld Vol Close Chg Polaroid 11 1 / 2 06 19.3 395 59 3 / 4...  What is the yield to maturity for this bond? P/YR = 2, N = 10, FV = 1000, PV = $-597.50, PMT = 57.50  Solve: I/YR = 26.48%

53 The Financial Pages: Corporate Bonds Cur Net Yld Vol Close Chg HewlPkd zr 17... 20 51 1 / 2 +1  What is the yield to maturity for this bond? P/YR = 1, N = 16, FV = 1000, PV = $-515, PMT = 0  Solve: I/YR = 4.24%

54 The Financial Pages: Treasury Bonds Maturity Ask Maturity Ask Rate Mo/Yr Bid Asked Chg Yld 9 Nov 18139:14 139:20 -34 5.46 9 Nov 18139:14 139:20 -34 5.46  What is the yield to maturity for this Treasury bond? (assume 35 half years) P/YR = 2, N = 35, FV = 1000, PMT = 45, PV = - 1,396.25 (139.625% of par)  Solve: I/YR = 5.457%

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