Chapter 7 - Valuation and
Characteristics of Bonds
Characteristics of Bonds
 Bonds pay fixed coupon (interest)
payments at fixed intervals (usually
every six months) and pay the par
value at maturity.
$I
0
$I
1
$I
$I
$I
$I+$M
2
...
n
Example: AT&T 6 ½ 36
 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 2036)
 Issued by AT&T.
$32.50
0
$32.50 $32.50 $32.50 $32.50 $32.50+$1000
1
2
…
28
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 belowinvestment grade bonds; rated BB and
below. High-yield bonds.
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.
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.
 Lists restrictive provisions which are
designed to protect bondholders.
 Describes repayment provisions.
Bond Ratings
 Moody’s
 Standard & Poor’s
 Fitch Investor Services
 Involve judgment about the future
risk potential of the bond
 AAA, AA, …, BB, …, CC,…, D
Current Yield
 The ratio of the annual interest payment
to the bond’s market price
 Bond with 8 percent coupon interest rate,
a par value of 1,000, and a market price of
700, what is its current yield?
 Current Yield = (0.08 * 1,000) / 700 =
11.4%
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.
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.
Valuation
n
V =
S
t=1
$Ct
(1 + k)t
 Ct = cash flow to be received at time t.
 k = the investor’s required rate of return.
 V = the intrinsic value of the asset.
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).
Bond Valuation
S
n
Vb =
t=1
$It
(1 + kb)t
+
$M
(1 + kb)n
Vb = $It (PVIFA kb, n) + $M (PVIF kb, n)
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?
0
120
120
120
...
1000
120
1
2
3
...
20
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.
Bond Example
Mathematical Solution:
PV = PMT (PVIFA k, n ) + FV (PVIF k, n )
PV = 120 (PVIFA .12, 20 ) + 1000 (PVIF .12, 20 )
PV = PMT
1
1 - (1 + i)n
i
PV = 120 1 -
+ FV / (1 + i)n
1
(1.12 )20 + 1000/ (1.12) 20 = $1000
.12
 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?
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.
Bond Example
Mathematical Solution:
PV = PMT (PVIFA k, n ) + FV (PVIF k, n )
PV = 120 (PVIFA .10, 20 ) + 1000 (PVIF .10, 20 )
PV = PMT
PV =
1
1 - (1 + i)n
i
120 1 -
+ FV / (1 + i)n
1
(1.10 )20 + 1000/ (1.10) 20 = $1,170.27
.10
 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?
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.
Bond Example
Mathematical Solution:
PV = PMT (PVIFA k, n ) + FV (PVIF k, n )
PV = 120 (PVIFA .14, 20 ) + 1000 (PVIF .14, 20 )
PV = PMT
PV =
1
1 - (1 + i)n
i
120 1 -
+ FV / (1 + i)n
1
(1.14 )20 + 1000/ (1.14) 20 = $867.54
.14
Suppose coupons are semi-annual
Mode = end
N = 40
I%YR = 14/2=7
PMT = 60
FV = 1000
Solve PV = -$866.68
Bond Example
Mathematical Solution:
PV = PMT (PVIFA k, n ) + FV (PVIF k, n )
PV = 60 (PVIFA .14, 20 ) + 1000 (PVIF .14, 20 )
PV = PMT
PV =
1
1 - (1 + i)n
i
60 1 -
+ FV / (1 + i)n
1
(1.07 )40 + 1000 / (1.07) 40 = $866.68
.07
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.
S
n
P0 =
t=1
$It
(1 + kb)t
+
$M
(1 + kb)n
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?
YTM Example
Mode = end
N = 16
PV = -898.90
PMT = 50
FV = 1000
Solve I%YR = 6%
6%*2 = 12%
Bond Example
Mathematical Solution:
PV = PMT (PVIFA k, n ) + FV (PVIF k, n )
898.90 = 50 (PVIFA k, 16 ) + 1000 (PVIF k, 16 )
PV = PMT
1
1 - (1 + i)n
i
1
898.90 = 50 1 - (1 + i )16
i
+ FV / (1 + i)n
+ 1000 / (1 + i) 16
solve using trial and error
Bond Example
Mathematical Solution:
10%  5 %  1000
14%  7%  811.06
We know i is between 10% and 14%, because 898.9 is
between 811.06 and 1000.
Then we try 11% and 13%....
Finally we try 12% to get the correct answer.
Zero Coupon Bonds
 No coupon interest payments.
 The bond holder’s return is
determined entirely by the
price discount.
Zero Example
 Suppose you pay $508 for a zero
coupon bond that has 10 years
left to maturity.
 What is your yield to maturity?
-$508
0
$1000
10
Zero Example
Mode = End
N = 10
PMT = 0
PV = -508
FV = 1000
Solve: I%YR = 7%
PV = -508
0
Zero Example
FV = 1000
Mathematical Solution:
PV = FV (PVIF i, n )
508 = 1000 (PVIF i, 10 )
.508 = (PVIF i, 10 ) [use PVIF table]
PV = FV /(1 + i) 10
508 = 1000 /(1 + i)10
1.9685 = (1 + i)10
i = 7%
10
The Financial Pages: Corporate Bonds
Polaroid 11 1/2 13
Cur
Yld
Vol
19.3
395 59 3/4
Close
Net
Chg
...
 What is the yield to maturity for this bond?
N = 10, FV = 1000,
PV = $-597.50,
PMT = 57.50
 Solve: I/YR = 13.24% *2 = 26.48%
The Financial Pages: Corporate Bonds
HewlPkd zr 24
Cur
Yld
Vol
...
20
Close
51 1/2
Net
Chg
+1
 What is the yield to maturity for this bond?
N = 16, FV = 1000,
PV = $-515,
PMT = 0
 Solve: I/YR = 4.24%
The Financial Pages: Treasury Bonds
Maturity
Rate Mo/Yr
9
Nov 25
Bid
Asked
139:14 139:20
Close
139 5/8
 What is the yield to maturity for this
Treasury bond? (assume 35 half years)
N = 35, FV = 1000,
PMT = 45,
PV = - 1,396.25 (139.625% of par)
 Solve: I/YR = 2.728%*2 = 5.457%
Ask
Yld
5.46
Five Relationships
 Value of bond is inversely related to
changes in the investor’s present
required rate of return
 Interest rate increases, bond value
decreases
Five Relationships
 Market value of bond will be less than
the par value if investor’s required rate
is above the coupon interest rate
Five Relationships
 As maturity date approaches, the
market value of bond approaches its
par value
Five Relationships
 Long-term bonds have greater interest
rate risk than do short-term bonds
Five Relationships
 The sensitivity of bond’s value to
changing interest rate depends not
only on length of time to maturity, but
also on the pattern of cash flows
provided by the bond
Five Relationships
 Duration: weighted average of time to
maturity in which the weight attached
to each year is the present value of the
cash flow for that year
Five Relationships
 A, B mature in 3 years, present value
$1000, par value $1000, market
interest is 10%
 A: coupon $100 per year
 B: pay whole interest $331 at the end
of year 3
 The duration of the two bonds?
Five Relationships
 Duration for A:
 [(1) * 100 /(1.1) + (2) * 100 /(1.1)^2 +
(3) * 1,100 / (1.1)^3] / 1000 = 2.74
 Duration for B:
 [(1) * 0 /(1.1) + (2) * 0/(1.1)^2 + (3) *
1,331 / (1.1)^3] / 1000 = 3
Five Relationships
 If interest rate falls to 6% from 10%
 Value of A = $1,106
 Value of B = $1,117
 B’s cash flows are received in more
distant future, and change in interest
rate has a greater impact on the
present value of later cash flow
Five Relationships
 Conclusion:
-- the longer the duration, the greater
the relative percentage change in bond
price in response to a given percentage
change in the interest rate
-- duration is more proper than term
to maturity in measuring bond’s
sensitivity to interest rate change