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Chapter 7: Interest Rates andBond Valuation
Faculty of Business AdministrationLakehead University
Spring 2003
May 13, 2003
Outline of the Lecture
7.1 Bonds and Bond Valuation
7.2 More on Bond Features
7A On Duration
7C Callable Bonds
1
7.1 Bonds and Bond Valuation
A bond is normally an interest-only loan.
If, for example, a firm wants to borrow $1,000 for 30 years and
the actual interest rate for similar corporations is 12%, then the
firm will pay $120 in interest each year for 30 years and repay
the $1,000 loan after 30 years.
The security that guarantees these payments is called abond.
A bond may involve more than one interest payment during a
year.
2
7.1 Bonds and Bond Valuation
In the above example, interest payments could be as follows:
• one payment of $120 per year;
• two payments of $60 per year;
• four payments of $30 per year;
• any arrangement such that a total of $120 in interest is paid
each year.
3
7.1 Bonds and Bond Valuation
With a single interest payment per year, the timing of cash flows
to the lender would be as follows:
. . . -0 1 2 3 29 30
Interest
Principal
$120 $120 $120 $120 $120
$1,000
With semiannual payments, the timing would be:
. . . -0 0.5 1 1.5 2 2.5 3 29 29.5 30
Interest
Principal
$60 $60 $60 $60 $60 $60 $60 $60 $60
$1,000
4
Bond Features and Prices
A bond is basically characterized by the following items:
Face Value (F ): The amount of principal to be repaid at the
bond’s maturity date.
Coupon Rate (i): The fraction ofF paid in interest each year.
Maturity ( T ): The number of years until the face amount is
repaid.
Number of Payments (m): Number of interest payments
during a year.
5
Bond Features and Prices
Using the above parameters, we can compute the bond’s coupon
payment per period, which isi×Fm .
Bonds may have other characteristics such as
• Seniority
• Repayment
• Call Provision
• Protective covenants
6
Bond Values and Yields
The above values are set by thefirm itself, although very often
dictated by market conditions.
The bond price and/or the bond’syield to maturityare entirely
determined by market conditions.
The yield to maturity of a bond is its APR.
7
Bond Values and Yields
Consider a bond with coupon ratei, face valueF , time to
maturityT and number of paymentsm. What is the price of such
a bond?
This bond’s annual coupon payment is
C = i×F.
Hence, this bond makesmpayments of
Cm =Cm
=iFm
per year, and this for the nextT years.
8
Bond Values and Yields
Let y denote the yield to maturity, or APR, of bonds with similar
risk characteristics.
Then the APR of our bond must also bey. The period interest
rate for the bond is thenrm = y/m and there areT×m periods
until maturity.
9
Bond Values and Yields
The value of the bond, i.e. its price (P), is then
P =Cm
rm
(1−
(1
1+ rm
)Tm)
+F
(1+ rm)Tm
=iF/my/m
(1−
(1
1+y/m
)Tm)
+F
(1+y/m)Tm
=iFy
(1−
(1
1+y/m
)Tm)
+F
(1+y/m)Tm
10
Bond Values and Yields
Expressing the bond price as
P =
(iy
(1−
(1
1+y/m
)Tm)
+(
11+y/m
)Tm)×F,
We can see that
P
> F if i > y,
= F if i = y,
< F if i < y.
11
Bond Values and Yields
A bond is said to
• sell at a premium whenP > F ;
• sell at par whenP = F ;
• sell at a discount whenP < F .
12
Finding the Yield
Bond prices are often listed in the newspaper. From these prices,
we can compute a bond’s yields to maturity as long as we know
F , i, T andm. We cannot solve fory analytically, and thus a
computer or the trial-and-error method must be used to this end.
Note, however, that bond prices are determined by their yield to
maturity. Looking at the yield to maturity of bonds with
maturities may help understand the trend in interest rates in
general.
13
Interest Rate Risk
The risk arising from a change in market interest rates (i.e. a
change iny) is calledinterest rate risk.
Changes in market interest rates affect bond prices.
Let P0 denote the bond price wheny = y0, and suppose thaty
changes toy1, inducing a new bond priceP1. The bond price
sensitivity to a change in interest rate will be defined as
|P1−P0|P0
=|∆P|P0
,
i.e. the proportional increase or decrease in the bond price.
14
Interest Rate Risk
The sensitivity of a bond price to changes iny depends on the
bond’s characteristics. With respect toi, T remember the
following results:
1. All other things being equal, the loweri, the greater the
interest rate risk;
2. All other things being equal, the greaterT, the greater the
interest rate risk;
15
Interest Rate Risk
Example 1
Take, for instance, a zero coupon bond withT years to maturity.
The price of such a bond is
P =F
(1+y)T .
If y is initially y0 and then decreases toy1, the relative change in
the bond price is (note that∆P > 0)
∆PP0
=F
(1+y1)T − F(1+y0)T
F(1+y0)T
=(
1+y0
1+y1
)T
−1,
16
Interest Rate Risk
Since, in this example,y1 < y0, the term
(1+y0
1+y1
)T
−1
increases withT and thus∆PP0
increases withT.
Therefore,the greaterT, the greater the interest rate risk of a
zero-coupon bond.
Does this result hold with coupon bonds?
17
Interest Rate Risk
Example 2
Consider two bonds, bond 1 and bond 2, withF = $1,000,
i = 10%andm= 1.
Bond 1 has 5 years to maturity while bond 2 has 10 years to
maturity.
If, initialls y = 10%, then the price of each bond is $1,000.
18
Interest Rate Risk
Suppose thaty suddenly decreases to 5%.
Then the price of Bond 1 becomes $1,216, an increase of
1,216−1,0001,000
= 21.6%,
and the price of Bond 2 becomes $1,386, an increase of
1,386−1,0001,000
= 38.6%.
The increase is proportionally greater for the longer-term bond.
19
Interest Rate Risk
Suppose now thaty increases to 15%.
Then the price of Bond 1 becomes $832, a decrease of
1,000−8321,000
= 16.8%,
and the price of Bond 2 becomes $749, a decrease of
1,000−7491,000
= 25.1%.
Again, the decrease is proportionally greater for the longer-term
bond.
20
Interest Rate Risk
Example 3
Consider now two bonds with the same characteristics except for
the coupon rates. More specifically,F = $1,000, T = 10and
m= 1 for both bonds, but Bond 1 has a 5% coupon rate while
Bond 2 has a 10% coupon rate.
If, initialls y = 10%, then
• the price of Bond 1 is $693;
• the price of Bond 2 is $1,000.
21
Interest Rate Risk
Suppose thaty decreases to 5%.
Then the price of Bond 1 becomes $1,000, an increase of
1,000−693693
= 44.3%,
and the price of Bond 2 becomes $1,386, an increase of
1,386−1,0001,000
= 38.6%.
The increase is proportionally greater for the bond with a lower
coupon rate.
22
Interest Rate Risk
Suppose now thaty increases to 15%.
Then the price of Bond 1 becomes $498, a decrease of
693−498693
= 28.1%,
and the price of Bond 2 becomes $749, a decrease of
1,000−7491,000
= 25.1%.
Again, the decrease is proportionally greater for the bond with a
lower coupon rate.
23
7A Duration
How to compare bonds with different coupon ratesanddifferent
time to maturity?
Starting point: It is easy to compare zero-coupon bonds, in
which case interest rate risk increases with time to maturity.
If it were possible to find the “equivalent” zero-coupon bond to
any coupon bond, it would then be easy to compare different
bonds: We would just have to compare the time to maturity of
their equivalent zero-coupon counterparts.
24
7A Duration
The time to maturity of a bond’s equivalent zero-coupon
counterpart is calledduration.
How to calculate duration?
Note that a coupon bond can be viewed as a set ofTm
zero-coupon bonds.
1. A zero-coupon bond payingiFm at time 1m (in years);
2. A zero-coupon bond payingiFm at time 2m;
...
Tm. A zero-coupon bond payingF + iFm at timeT.
25
7A Duration
Take a 3-year bond with a 10% coupon rate,F = $1,000and
m= 1. This bond can be viewed as a set of three zero-coupon
bonds:
Bond z1; A zero-coupon bond paying $100 in one year;
Bond z2; A zero-coupon bond paying $100 in two years;
Bond z3; A zero-coupon bond paying $1,100 in three years.
26
7A Duration
If y = 8%, then the price of the coupon bond is
P =1001.08︸︷︷︸Pz1
+100
(1.08)2︸ ︷︷ ︸
Pz2
+1,100(1.08)3︸ ︷︷ ︸
Pz3
= $1,052,
i.e. the sum of its zero-coupon bond prices.
27
7A Duration
Duration,D, is a weighted average of the time to maturity of thezero-coupon bonds composing a bond:
D =Pz1
P×1 +
Pz2
P×2 +
Pz3
P×3
=100/1.08
1,052×1 +
100/(1.08)2
1,052×2 +
1,100/(1.08)3
1,052×3
= 2.74years.
28
7A Duration
More generally, duration is calculated as follows
D =
1m
Tm
∑n=1
nXn
(1+y/m)n
P,
whereXn is the cash flow in periodn (not necessarily a year
here), including the principal whenn = Tm.
The term1m in front of
Tm
∑n=1
allows to have a measure in years.
Durationapproximatesinterest rate risk: The greater duration,
the greatershouldinterest rate risk be.
29
7A Duration
D =Tm
∑n=1
[Xn
P(1+y/m)n ×nm
]
where
Xn
P(1+y/m)n ≡ weight on the “nth zero-coupon bond”,
nm≡ time to maturity (in years) of the “nth zero-coupon bond”.
30
7A Duration
Note that
Tm
∑n=1
Xn
P(1+y/m)n =
Tm
∑n=1
Xn
(1+y/m)n
P
=
Tm
∑n=1
Xn
(1+y/m)n
Tm
∑n=1
Xn
(1+y/m)n
= 1.
31
7A Duration
Note that the duration of a zero-coupon bond is
D =
1m
Tm
∑n=1
nXn
(1+y/m)n
P=
TF
(1+y/m)T
P= T.
32
Duration: Example 1F = $1,000, i = 4%, T = 3, m= 2, y = 8%.
(1) (2) (3) (4) (5)
Year Cash Flow PV of Flow (PV of Flow)/P (1)×(4)
0.5 20 19.23 0.0215 0.0107
1.0 20 18.49 0.0207 0.0207
1.5 20 17.78 0.0199 0.0298
2.0 20 17.10 0.0191 0.0382
2.5 20 16.44 0.0184 0.0459
3.0 1,020 806.12 0.9005 2.7016
895.16 2.8469
33
Duration: Example 2F = $1,000, i = 6%, T = 3, m= 2, y = 8%.
(1) (2) (3) (4) (5)
Year Cash Flow PV of Flow (PV of Flow)/P (1)×(4)
0.5 30 28.85 0.0304 0.0152
1.0 30 27.74 0.0293 0.0293
1.5 30 26.67 0.0281 0.0422
2.0 30 25.64 0.0271 0.0541
2.5 30 24.66 0.0260 0.0651
3.0 1,030 814.02 0.8591 2.5772
947.58 2.7831
34
Duration Hedging
A firm can hedge against interest rate risk by matching liabilities
with assets.
What do we match? Either
Duration ofassets
= Duration ofliabilities
orDuration
ofassets
×Market value
ofassets
=Duration
ofliabilities
×Market value
ofliabilities
Using this procedure, a firm canimmunizeitself against interest
rate risk.
35
Duration Hedging
The IWG BankMarket Value Balance Sheet (in millions of $)
Assets Liabilities and Owners’ Equity
Value Duration Value Duration
Overnight money 35 0 C&S accounts 400 0
A/R-backed loans 500 3 months CD 300 1 year
Inventory loans 275 6 months LT financing 200 10 years
Industrial loans 40 2 years Equity 100
Mortgages 150 14.8 years
Total 1,000 Total 1,000
36
Duration Hedging
Duration of IWG’s assets:
35×0+500× 14 +275× 1
2 +40×2+150×14.8
1,000= 2.56years.
Duration of IWG’s liabilities:
400×0+300×1+200×10900
= 2.56years.
37
Duration Hedging
Is IWG immunized against interest rate risk?
2.56×1,000 > 2.56×900.
Both durations being equal, the percentage change in the value of
assets following a small change in interest rates is close to the
percentage change in the value of liabilities but, since there are
more assets than liabilities, the overall change in assets will be
greater than the overall change in liabilities.
38
Duration Hedging
What should the firm do?
1. Increase the duration of liabilities without changing the
duration of assets:
Duration ofliabilities
= 2.56× 1,000900
= 2.84years.
2. Decrease the duration of assets without changing the
duration of liabilities:
Duration ofassets
= 2.56× 9001,000
= 2.30years.
39
7C Callable Bonds
The process of replacing all or part of an issue of outstanding
bonds is calledbond refunding.
It is possible to do so when the issue iscallable.
Thecall priceis the price at which the firm can repurchase the
bond.
Why would a firm want to refund its debt issue? To take
advantage of lower interest rates.
40
7C Callable Bonds
Consider a bond withT years to maturity, a face valueF and a
coupon ratei. The bond makes annual coupon payments.
The actual interest rate isy0, but is expected to change one year
from now. No further changes iny are expected untilT.
More specifically, the YTM one year from now will be
y1 =
y with probabilityπ,
y with probability1−π,
wherey > y and0≤ π≤ 1.
41
7C Callable BondsLet P1 denote the value of the bond at the beginnig of year 1.Then
P1 =
Pl = iFy
(1−
(1
1+y
)T−1)
+ F(
11+y
)T−1if y = y,
Ph = iFy
(1−
(1
1+y
)T−1)
+ F(
11+y
)T−1if y = y.
Note thatPh > Pl
42
7C Callable Bonds
If the bond is not callable, its price as of time 0 is
Pn =iF +πPl +(1−π)Ph
1+y0.
43
7C Callable Bonds
Suppose now that the firm can call the bond in period 1 only and
at a at a call priceCP.
The firm will call the bond only if
CP < Ph,
and thus there is no point in settingCP> Ph.
44
7C Callable Bonds
The value as of time 0 of a callable bond with call priceCP is
Pc =iF +πPl +(1−π)CP
1+y0.
45
7C Callable Bonds
Let’s compare two bonds, one callable and one non-callable,
where
in≡ coupon rate of the non-callable bond;
ic≡ coupon rate of the callable bond;
CP≡ call price;
Pn≡ price of the non-callable bond;
Pc≡ price of the callable bond;
y = 10%, y = 6%, andy0 = 8%.
46
7C Callable Bonds
SupposeF = $100for the two bonds,T = 20, π = 0.5 and
CP= $110. Then
Pn = $100 whenin≈ 7.75%;
Pc = $100 whenic≈ 8.65%;
Pl = $88.71 < $110;Ph = $129.57 > $110;
Thus a8.65−7.75= 0.9%premium is needed for the callable
bond to be issued at par.
47
