Homework Assignment #1
Economics 333
Money and Banking
Assigned: September 21, 2003
Due: September 28, 2003
1. A discount bond with a face value of 100, and an initial maturity of 2 years is
sold with a yield to maturity of 6%. After 1 year, the initial investor sells the
bond at a price of 98. What is the gross return earned by the initial investor
over the course of the year? What is the yield earned by the investor who buys
the bond at 98 and holds it for the remaining year?
100
 88.999644 . The gross return
(1.06) 2
of a discount bond after 1 year is the resale price relative to the initial purchase
98
 1.101128 . The yield after 1 year is
price. Rt 1 
88.999644
100
1  itYTM
 1.02040816
1 
98
The bond is sold at a price equal to Pt ,2B 
2. A new company issues a coupon bond that will pay a face value of 100 in 10
years. However, the company does not expect to earn income for the first 4
years of its life. As such, the bond will not pay a coupon in the first 4 years of
its life. In the 5th, 6th, 7th, 8th, 9th, and 10th year it will pay a coupon of 20.
Using a discount factor of 10% (i = .10), calculate the present value of the
bond. This can be done in 3 steps A) Calculate the present value of a standard
coupon bond with a maturity of T = 10, a coupon of 20 paid every year, and a
face value of 100; B) Calculate the present value of 4 annual payments of 20
using the formula for a fixed payment loan; and C) subtract B) from A).
Steps A.The present value of a 10 year coupon bond with a face value of 100 and a
T
10
C
FACE 20
100
1
20 coupon is PV  [1   11 i  ] 
 [1   1.1
 ]  10  161.4456711.
T
i
(1  i)
.1
(1.1)
B The present value of a 4 year fixed payment loan with a payment of 20 is equal
T
4
C
20
1
PVB  [1   11 i  ]  [1   1.1
 ]  63.39730893 . The difference is 98.04836
to
i
.1
3. You take a simple loan of $100,000 to buy a car. The auto-finance company
tells you that you can either pay back $150,000 in 5 years or $180,000 in 7
years. What is the annual interest rate on each of these loans? If you took a 5
year fixed payment loan, what size of annual payment would you have to
make if the interest rate were equal to that on the 5 year simple loan
mentioned above?
The interest rate on a simple loan of LOAN and a repayment of REPAY in T
REPAY
periods is the T-th root of the ratio of the ratio of repayment to loan T
.
LOAN
We can solve this as
5 Year Loan
5
7 Year Loan
7
150000 5
 1.5
100000
1.084472
180000 7
 1.8
100000
1.087596
If the bank required an interest rate of 1.084472 on a loan of $100,000 repaid in
constant annual payments over 5 years, the bank would apply the formula
C  i  LOAN
1 5
)
1 i
 . 084472 100000
 8,447.20 1
1
1- 1.5
5
1 (
)
1.084472
 3*8,447.20=25341.53136
1 (
4. The Malaysian national oil company Petroleum Nasional (Petronas) has an
outstanding coupon bond with a maturity date in 23 years, coupon rate of
7.625% and a yield to maturity of 7.083%. What is the current yield of the
bond?
First you need to solve for the price of the bond. The yield to maturity is the
discount factor that would set the price equal to the present value. We can thus
calculate the price as
C
C
C
FACE C 
1  FACE
P

 ... 

 1 

. The price
2
23
23
1  i 1  i 
i  (1  i ) 23  (1  i) 23
1  i  1  i 

7.625 
1
100
1

 106.3608 . Note that the price is

23 
.07083  (1.07083)  (1.07083) 23
greater than the face value because the coupon rate is larger than the yield to
maturity. The current yield of 7.169% is the ratio of the coupon to the price.
7.625
 0.07169 which is greater than the yield to maturity but smaller than
106.3608
the coupon rate.
is P 
5. Exactly one year ago, the government of the Philippines sold three coupon
bonds of maturity dates in 5, 10, and 20 years. The 5 year bond had a coupon
rate of 8.875%, the 10 year bond had a coupon rate of 9.875%, and the 20 year
bond had a coupon rate of 10.625%. Currently the yield to maturity on these
bonds is 6.425%, 7.313%, and 9.006%. Calculate the return on each of these
bonds if you had held them for one year and sold them in secondary markets
after you had collected a coupon.
6. Events. A hurricane hits the United States damaging many people’s homes.
Assume that many people cash in their savings to pay for the rebuilding.
Further, the government borrows money to finance further rebuilding of
infrastructure. Draw pictures of the bond market from the bond market &
loanable funds perspective, showing the effects of these events on bond prices
and yields.
P
Bond Market
SB
i
Loanable Funds Market
SLF
DB
QB
7. Uncovered Parity. On the same day, you find 1 year yields on comparable
bonds issued in Euros, Yen, British Pounds, and Swiss Francs s well as the
current spot exchange rate with the Hong Kong dollar, treating Hong Kong as
the domestic interest rate. Assume that the uncovered interest parity holds
true. Use domestic and foreign interest rates, and the current spot rate to
calculate the market’s expectation of the exchange rate 1 year from now.
Euros
Yen
Pounds
US Dollars
Yield
Foreign
HK
9/10/2005
2.32
0.15
5.00167
2.09
1.42
1.42
1.42
1.42
According to uncovered interest parity,
Exchange Rate
9/10/2005
9/10/2006
0.105607773
0.106544935
14.08450704
13.90813824
0.07203573
0.074579688
0.128369705
0.12921774
EX t 1 
1 iF
 EX t
1 i
8. Term Structure You open up the financial page on September 10, 2005 and
find the following yield curve for Exchange Fund notes. Assume that the
expectations theory of the term structure is true. Use the yield curve to
calculate the market’s expectation of some future interest rates.
a. Calculate the market’s expectation of the 1 year yield to maturity on
September 10, 2006 (1 year from now), September 10, 2007 (2 years
from now), and September 10, 2008 (3 years from now).
DLF
b. Calculate the market’s expectation of the yield to maturity on a 2 year
note on September 10, 2010 (5 years from now).
c. Calculate the markets expectations of the average yield over the period
September 10, 2010 (5 years from now) to September 10, 2015 (10
years from now).
Yield Curve: September 10, 2005
Years
2
1.975
1
1.42
Yield
Maturity
4
2.833
To
3
2.493
5
3.154
7
3.611
10
3.981
a. Under the expectations theory, the 2 year yield is the geometric
average of the 1 year yield and expectations of the 1 year yield in 1
1  i 1  i   1  i 
1  i2,t 
year
e
1,t 1
1,t
1  i 

1  i 
e
1,t 1
2
2,t
1,t

1.019752
 1.025330371
1.0142
. We can approximate this
relationship by saying that the net 2 year yield is equal to the
arithmetic average of the net yield of a 1 year bond and the markets
expectations of the net yield of a 1 year bond in 1 year
i1,t  i1,e t 1
i2,t 
 i1,e t 1  2i2,t  i1,t  0.0253 . Notice the approximation
2
is very close. The yield on the three year bond is the geometric average
of the expected interest rates on 1 year bonds over their life.
1  i3,t 
3
1  i 1  i 1  i 
1  i 
1  i 
  1 i 1 i 
 
 1  i 
e
1,t 1
1,t
e
1,t  2
3

 1 i
3,t
e
1,t  2
1,t
3

3,t
e
1,t 1
2
2,t
1.02433
 1.035369072
1.019752
. We could also write the net 3 year yield as the arithmetic average of
i1,t  i1,e t 1  i1,et  2
i3,t 
 i1,e t  2  3i3,t  i1,t  i1,et 1 
3
three net interest rates.  3i3,t  2i2,t
i1,e t  2  3  .02493  2  .01975  0.03529
i1,e t T  (T  1)  iT 1,t  (T )  iT ,t
In general, we can write,
1  i 
)
1  i 
T 1
(1  i1,t T
. If we want to find
T 1,t
T
T ,t
the expectation of the1 year interest rate, 3 years from now, we would set
i1,e t 3  (4)  i4,t  (3)  i3,t  4  .02833  3  .02493  0.03853
T = 3, and solve
(1  i1,t T
1  i 
)
1  i 
4
4,t
3,t
3
1.02833

3
1.02493
4
 1.038598
b. The yields on the5 & 7 year bond are the geometric average of the 1
year interest rates over the life of the bond.
1  i5,5 t  1  i1,t 1  i1,e t 1 1  i1,e t  2 1  i1,e t 3 1  i1,e t  4
 




1  i  1  i 1  i 1  i 1  i 1  i 1  i 1  i 
1  i 
 1  i 1  i  
1  i 
7
7,t
e
1,t 1
1,t
e
1,t  2
e
1,t  3
e
1,t  4
e
1,t  5
e
1,t  6
7
e
1,t  5
3,t
e
1,t  6
5
2,t
But in 5 years, the yield on the 2 year bond will be equal to the geometric average
of the 1 year interest rates over those 2 years.
1  i   1  i 1  i
e
2,t  5
2
e
1,t  5
e
1,t  6
1  i   1.03611

1  i  1.03154
7
7
5
5
7,t
=1.097515572
5,t
1  i2,e t 5  1.097515572  1.047623774
Similarly, we can calculate the expectation of the average interest rates between 7 and
10 years from now using the yields on 7 and 10 year bonds. This would be equal to
the expectation of the 3 year yield in 7 years.
1  i 

1  i 
10
(1  i

)  1 i
e
3
3,t  7
e
1,t  7
1  i 1  i
e
1,t 8
e
1,t  9
10,t
7,t
1  i3,e t  7  3 1.152653679  1.048494798
7
1.03981

7
1.03611
10
=1.152653679
9.
10. ddsd
The total, gross return is written as the price after 1 period plus the coupon
divided by the initial price. In the primary market, the price equals the face value,
so Pt = 100. The formula for the price after 1 year is
C
1  FACE
where T is the initial maturity minus 1. The formula
P1  1 

i  (1  i )T  (1  i )T
P C
 1.
for the net return in percentage terms is 1
P0
Bond
Initial
Maturity Coupon Initial
Yield
Price
Net
Maturity After 1
(C)
Price/
after 1
After 1
Return
year (T)
Face
year
year
Value
A)
5
4
8.875
100
6.425
108.4075
17.28%
B)
10
9
9.875
100
7.313
116.472
26.35%
C)
20
19
10.625
100
9.006
114.4842
25.11%