Bonds Prices and Yields
Bonds
2


Corporations and government entities can raise capital by selling bonds

Long term liability (accounting)

Debt capital (finance)
The bond has

Principal, par, or face value: F

Price: P

Yield: y (actually “yield to maturity” and the discount rate)

Maturity date, time to maturity, term, or tenor: T


Date at which the bond principal, F, is returned to investors
In the case of a coupon bond (as opposed to a zero coupon bond)

Coupon rate: c (annual, simple, nominal rate)

Annual payment frequency: m; or period Dt
 In the U.S. semiannual coupons is typical: m = 2 or Dt = .5
Zero Coupon Bonds
3



ZCBs do not pay a coupon
The return and ‘yield’ (rate) is due to the purchase price at
a discount to face value
U.S. Treasury bills (T – bills) are zero coupon bonds



Time-to-maturity at issue is 4, 13, 26, 52 weeks
Face value $100 to $5,000,000
A ZCB yield is the interest rate,
and the discount rate denoted z
F
t=0
P
t=T
Zero Coupon Bond
4

For T ≤ 1 year:
F
P
(1  z  T)
F
where z is the annual simple rate or yield

For T > 1 year
t=0
F
P
(1  z)T
P
t=T
where z is the annualized effective rate or yield
If a bond has a term of a year or less, simple interest is used, otherwise
compound annual interest is used by convention
Zero Coupon Bond Example
5

The face value is $1000, the market price is $850, and the time to
maturity is 3.5 years. What is the annualized yield ?
F
P
(1  z)T

$850 =
$1000
(1+ z)3.5
æ $1000 ö
z=ç
÷
è $850 ø
1
3.5
-1 = 4.753%
The face value is $1000, the market price is $975, and the time-tomaturity is 0.5 years. What is the annualized yield?
F
P
(1  z  T)
$1000
$975 =
(1+ 0.5×z)
æ $1000 ö
z = 2× ç
-1÷ = 5.128%
è $975
ø
Coupon Bond
6
F = face or
par value
C = coupon payment
P =
current
price
t=0.0
t=Dt
t=2∙Dt
i=0
i=1
i=2
t0=0.0
t1=Dt
t2=2Dt
t=M∙Dt=T
i=M
tM= M∙Dt =T
Coupon Payment
7


Bond coupon cash flows, C, are defined by a nominal, simple
coupon rate, c, and a compounding frequency per year, m, or
coupon period measured in years, Dt
The total cash flow at time ti, CFi, is defined as:
CFi = C
for i <M
CFM = C + F
T=num of years (floating)
N=num of years (integer)
m=periods per year
In this course, generally M=Nm
360= 30 12
C  c  F  Dt
example
c  1.625%
F  $1000
Dt  .5
C  $8.125
Effective coupon rate, y
2
 1.625% 
1 
  1  1.632%
2 

2
y

 1    1  y%
 2
Coupon Bond Yield
8

Yield to maturity is the actual yield achieved for a coupon
bond if


The bond is held to maturity, and
Each coupon payment is reinvested at a rate of return of y through
time T


The yield to maturity is the investor’s expected return on
investment and is thus the issuer’s rate cost


The risk that coupons cannot be reinvented at a rate greater than or
equal to y due to market conditions is called “reinvestment risk”
It’s the issuer’s cost of debt, kD, for the bond
The yield reflects both the time value of money and the credit
worthiness of the borrower

The expected variance in the cash flow is reflected in the yield
Bond Price
9

The discount rate y is the yield to maturity or simply the yield
on a coupon bond

It’s an internal rate of return that sets the discounted cash
flow on the right hand side to the market price of the bond, P,
on the left hand side
M
P
i1
CFi
y

1  
 m
M
i
y is the nominal annual yield
to maturity in this formula
with integer periods
CFi
P
ti
(1

y
)
i 1
y is effective annual yield to
maturity in this formula with
discrete real time periods
Fractional Initial Time Period
10

For a fractional initial coupon period: t1 < ∆t
F = face or
par value
C = coupon payment
i=0
i=1
t0=0.0
t1
i=2
t2=t1+Dt
i=M
t M= T
For a bond with semi-annual coupons, assume that the next coupon payment
is in 3 months. The coupon payments occur at
t0=0.0, t1=0.25, t2=0.75, t3=1.25, t4 = 1.75, …
Zero Coupon Bonds Again
11

A bond dealer can split a coupon bond into ZCBs

one for the principal and
one for each coupon

This is called ‘stripping’ the bond



The advantage of a ZCB is that there is no reinvestment risk
For a ZCB, the yield, y, is the zero coupon rate denoted as z
Bond Equation Applications
12

Find the yield-to-maturity, y, from a known market price, P

Solve for y (nominal, y, or effective, y ‘bar’)
M
P
i1

y

1  
 m
i
CFi
P
ti
i1 (1  y )
Solve for the roots of a nonlinear equation


M
CFi
In this course use Excel Goal Seek
Example: Compute both the effective and nominal yield for a bond
with $1000 face value, current market price of $800, coupon rate of
7% paid semiannually, and 4.5 years to maturity.
Bond Equation Applications
13
M
P
M
CFi
P
ti
i1 (1  y )
$1,000
F
7.00%
c
nominal
13.434%
y
effective
t
CF
DF
DCF
0
$0
$0.00
0.5
$35
0.939 $32.86
1
$35
0.882 $30.85
1.5
$35
0.828 $28.97
2
$35
0.777 $27.20
2.5
$35
0.730 $25.54
3
$35
0.685 $23.98
3.5
$35
0.643 $22.51
4
$35
0.604 $21.14
4.5 $1,035
0.567 $586.94
Sum
$1,315
P $800.00
i1
13.011%
t
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Sum
y nominal
i
CF
0
$0
1
$35
2
$35
3
$35
4
$35
5
$35
6
$35
7
$35
8
$35
9 $1,035
$1,315
DF
0.939
0.882
0.828
0.777
0.730
0.685
0.643
0.604
0.567
P
CFi
y

1



m


i
DCF
$0.00
$32.86
$30.85
$28.97
$27.20
$25.54
$23.98
$22.51
$21.14
$586.94
$800.00
Bond Equation Applications
14
 Convert
the nominal yield to the effective yield
2
 13.011% 
13.434%   1 
 1
2


2
y

y  1    1
 2

Find market price from a known yield
 For
the bond in the last example, what is the price?
 Given
an effective annual yield of 12% or
 A nominal annual yield of 12%
Bond Equation Applications
15
M
P
M
CFi
P
ti
i1 (1  y )
$1,000
F
7.00%
c
nominal
12.000%
y
effective
t
CF
DF
DCF
0
$0
$0.00
0.5
$35
0.945 $33.07
1
$35
0.893 $31.25
1.5
$35
0.844 $29.53
2
$35
0.797 $27.90
2.5
$35
0.753 $26.36
3
$35
0.712 $24.91
3.5
$35
0.673 $23.54
4
$35
0.636 $22.24
4.5 $1,035
0.601 $621.53
Sum
$1,315
P $840.34
i1
12.000%
t
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Sum
y nominal
i
CF
0
$0
1
$35
2
$35
3
$35
4
$35
5
$35
6
$35
7
$35
8
$35
9 $1,035
$1,315
DF
0.943
0.890
0.840
0.792
0.747
0.705
0.665
0.627
0.592
P
CFi
y

1  
 m
DCF
$0.00
$33.02
$31.15
$29.39
$27.72
$26.15
$24.67
$23.28
$21.96
$612.61
$829.96
i
Bond Equation Applications
16
For the bond with a 12% effective yield and price $840.34
at time 0, here’s a plot of price as time progress from 0 to
4.5 years assuming a constant yield of 12%
$1,050
$1,025
$1,000
$975
Price

$950
$925
$900
$875
$850
$825
0.0
0.5
1.0
1.5
2.0
2.5
Time
3.0
3.5
4.0
4.5
Corporate Credit Rating
17
AAA companies
From Investopedia
Reinvestment Risk
18

Consider a $1000 bond with a coupon rate of 10% paid annually for 10 years.
Initially, the yield is 11%, the price is $941.11, and the yield curve is flat. Prior to
the payment of the next coupon, we consider three scenarios
1.
the yield curve shifts parallel down to 9%
2.
the yield curve remains flat at 11%
3.
the yield curve shifts parallel up to 12%
What are the actual yields?
Bond Price Calculation
$1,000 F
10.00% c nominal Year
CF
DF
DCF
11.00% y nominal
1 $ 100 0.9009 $ 90.09
2 $ 100 0.8116 $ 81.16
3 $ 100 0.7312 $ 73.12
4 $ 100 0.6587 $ 65.87
5 $ 100 0.5935 $ 59.35
6 $ 100 0.5346 $ 53.46
7 $ 100 0.4817 $ 48.17
8 $ 100 0.4339 $ 43.39
9 $ 100 0.3909 $ 39.09
10 $ 1,100 0.3522 $ 387.40
Sum
$ 941.11
Yield To Maturity
Future Value of Coupon Reinvestment
9%
11%
12%
$
217.19 $ 255.80 $ 277.31
$
199.26 $ 230.45 $ 247.60
$
182.80 $ 207.62 $ 221.07
$
167.71 $ 187.04 $ 197.38
$
153.86 $ 168.51 $ 176.23
$
141.16 $ 151.81 $ 157.35
$
129.50 $ 136.76 $ 140.49
$
118.81 $ 123.21 $ 125.44
$
109.00 $ 111.00 $ 112.00
$
1,100.00 $ 1,100.00 $ 1,100.00
$
2,519.29 $ 2,672.20 $ 2,754.87
10.35%
11.00%
11.34%
Plot price v. yields to maturity
19
Bond “price – yield” or P-y curve
$1,300
$1,200
Each point
represents a
DCF calculation
$1,100
Price
F=$1000
c=7% semiannual
T=4.5 yrs
M
$1,000
P
CFi
(1  y)ti
14%
16%
i 1
$900
$800
$700
0%
2%
4%
6%
8%
10%
12%
Yield
Illustrates how price changes as yield-to-maturity changes for a
particular bond ( c, m, M, and F are constant)
Home Mortgage Calculation
20

Given the nominal interest rate, m=12, P, and N,
what is the monthly payment, C?
M

C : monthly payment

P
Includes principal repayment and interest –
there is no return of principal “F”
i 1
Ci
y

1



m



N : number of years

m : number of compounding periods per year (12 for home loans)

y : nominal fixed interest rate for the loan

P : loan principal (the mortgage amount)

Solve for C using Excel Goal Seek

Find the value of C that equates the left and right hand sides
i
Mortgage Example
21




You wish to borrow $300,000 at 6.5% fixed for 30 years.
The following excel table shows the calculations for the first
12 months and the last 5 months.
The monthly payment of $1896 is determined using goal seek
to force the sum of the last column to $300,000.
Note that you will pay out $682,633 in principal and interest


$300,000 in principal
$382,633 in interest
Mortgage Example
22
$300,000
6.500%
12
6.697%
0.542%
M
P
y nominal
m
y annual effective
y monthly effective
P
i 1
Ci
y

1



 m
i
t
0.000
0.083
0.167
0.250
0.333
0.417
0.500
0.583
0.667
0.750
0.833
0.917
1.000
i
0
1
2
3
4
5
6
7
8
9
10
11
12
29.667
29.750
29.833
29.917
30.000
Sum
356
357
358
359
360
CF
$
$
$
$
$
$
$
$
$
$
$
$
$
DF
DCF
1,886
1,876
1,866
1,856
1,846
1,836
1,826
1,816
1,806
1,796
1,787
1,777
1,896
1,896
1,896
1,896
1,896
1,896
1,896
1,896
1,896
1,896
1,896
1,896
0.995
0.989
0.984
0.979
0.973
0.968
0.963
0.958
0.953
0.947
0.942
0.937
$
$
$
$
$
$
$
$
$
$
$
$
$
$ 1,896
$ 1,896
$ 1,896
$ 1,896
$ 1,896
$ 682,633
0.146
0.145
0.145
0.144
0.143
P
$
277
$
276
$
274
$
273
$
271
$ 300,000
Perpetuity
23
M
P=å
i=1
Example: How much money do you need to invest, P, to pay out $1 per
year forever if the pay out rate is 10% (effective) per year?
Ci
(1+y)i
P=
Now in the case that
M=∞
C is constant
C
y
M
P=å
∞
1
i
i=1 (1+y)
P=C× å
i=1
Ci
(1+y)i
and of course y < 1
C
C
P=
y
This is a perpetuity
P
i
If a nominal annual rate, y,
is used then
P
C
y
 
m
Annuity
24
Now how much money do you need to invest at
10% to receive a $1 / year payout for M years ?
C
i
M M+1
That’s an annuity (a perpetuity would pay out
forever)
P
C
1
P= ×
y 1+y
PM =
( )
C
= × (1+y )
y
C
y
Annuity: Payout
M
-M
Annuity: Present Value
( )
-M
C C
P= - × 1+y
y y
-M ö
C æ
= × ç1 - 1+y ÷
ø
y è
( )
C=
P×y
( )
P × y × (1+y )
=
(1+y ) -1
1- 1+y
-M
M
M
Annuity
25
Now how much money do you need to invest at 10% to receive a $1 / year payout for M years ?
That’s an annuity (a perpetuity would pay out forever)
-M ù
é
C×m ê æ y ö ú
P=
× 1- ç1+ ÷
y ê è mø ú
ë
û
æyö æ yö
P× ç ÷ × ç1+ ÷
èmø è mø
C=
M
æ yö
ç1+ ÷ -1
è mø
M
M=20 years
C=$1
Y=10%
P=$8.51
Annuities
26
Closed Form Formulas
27

Annuity

Home mortgage annuity formula example
 $300,000  0.542%  (1  0.542%)360 
  $1896.20
C  
360
(1  0.542%)  1



Bonds
 Annuity for coupon payment plus the discounted face
value




1
1
F

P  C  

M
M
 y   y 
y  
y
     1     1  
  m   m  m    m 
Closed Form Formulas
28

Bonds
 Example
of bond w/ F=$1000, c=7% semi-annual,
T=4.5yrs, y annual nominal = 13.011%


1
1

P  $35  

 13.011%   13.011%  13.011%  9
 
 
 1 

2


2
2




 Bond


 $1000
 $800.00
9

  1  y 
2
 
with fractional initial period
 



 

1
1
F
1



P  C   1 


M
M
e

y y 
y  
y   
   m      1     1     1  y  d
     m   m    m    m 







Closed Form Formulas
29
Clean and Dirty Price example (p. 7.10) using closed form
last coupon
next coupon
.825
8/15/08
.175
8/15/09
8/15/10
8/15/11
8/15/12
8/15/13
8/15/14
6/12/09


1
1
$100 
1


  $108.70


P

$
5

1





e=64 days d = 365 days
64
5
5 

4% 4%(1  4%)  (1  4%)  (1  4%) 365 


e/d=.175
$110
$109
$108
$107
F=$100
y=4% annual
c=5% annual
y & c are effective & nominal
Price
$106
$105
$104
$103
$102
$101
$100
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Time
3.5
4.0
4.5
5.0
5.5
6.0