FINANCE 1 DESS Gestion des Risques/Ingéniérie Mathématique

FINANCE 1
DESS Gestion des Risques/Ingéniérie
Mathématique
Université d’EVRY VAL D’ESSONNE
EXERCICES CORRIGES
Philippe PRIAULET
1
Exercise 1 On 12/04/01 consider a fixed coupon bond whose features are the following:
· face value: $1,000
· coupon rate: 8%
· coupon frequency: semi-annual
· maturity: 05/06/04
What are the future cash-flows delivered by this bond ?
Solution 2 The coupon cash-flow is equal to $40
Coupon =
8% × $1, 000
= $40
2
It is delivered on the following future dates: 05/06/02, 11/06/02, 05/06/03, 11/06/03 and
05/06/04.
The redemption value is equal to the face value $1,000 and is delivered on maturity date
05/06/04.
Exercise 3 Consider the same bond as in the previous exercise. We are still on 12/04/01.
1- Compute the accrued interest taking into account the Actual/Actual day-count basis.
2- Same question if we are now on 09/06/02.
Solution 4 1- The last coupon has been delivered on 11/06/01. There are 28 days between
11/06/01 and 12/04/01, and 181 days between the last coupon date (11/06/01) and the next
coupon date (05/06/02). Hence the accrued interest is equal to $6.188
28
× $40 = $6.188
181
Accrued Interest =
2- The last coupon has been delivered on 05/06/02. There are 123 days between 05/06/02
and 09/06/02, and 184 days between the last coupon date (05/06/02) and the next coupon date
(11/06/02). Hence the accrued interest is equal to $26.739
Accrued Interest =
123
× $40 = $26.739
184
2
Exercise 5 An investor has a cash of $10,000,000 at disposal. He wants to invest in a bond
with $1,000 nominal value and whose dirty price is equal to 107.457%.
1- What is the number of bonds he will buy ?
2- Same question if the nominal value and the dirty price of the bond are respectively $100
and 98.453%.
Solution 6 1- The number of bonds he will buy is given by the following formula
Number of bonds bought =
Cash
No min al Value of the bond × dirty price
Here the number of bonds is equal to 9306
n=
10, 000, 000
= 9306.048
1, 000 × 107.457%
n=
10, 000, 000
= 101562.024
100 × 98.453%
2- n is equal to 101,562
Exercise 7 On 10/25/99 consider a fixed coupon bond whose features are the following:
· face value: Eur 100
· coupon rate: 10%
· coupon frequency: annual
· maturity: 04/15/08
Compute the accrued interest taking into account the four different day-count bases: Actual/Actual, Actual/365, Actual/360 and 30/360.
Solution 8 The last coupon has been delivered on 04/15/99. There are 193 days between
04/15/99 and 10/25/99, and 366 days between the last coupon date (04/15/99) and the next
coupon date (04/15/00).
· the accrued interest with the Actual/Actual day-count basis is equal to Eur 5.273
193
× 10% × Eur 100 = Eur 5.273
366
3
· the accrued interest with the Actual/365 day-count basis is equal to Eur 5.288
193
× 10% × Eur 100 = Eur 5.288
365
· the accrued interest with the Actual/360 day-count basis is equal to Eur 5.361
193
× 10% × Eur 100 = Eur 5.361
360
There are 15 days between 04/15/99 and 04/30/99, five months between May and September,
and 25 days between 09/30/99 and 10/25/99, so that there are 190 days between 04/15/99 and
10/25/99 on the 30/360 day-count basis
15 + (5 × 30) + 25 = 190
· finally the accrued interest with the 30/360 day-count basis is equal to Eur 5.278
190
× 10% × Eur 100 = Eur 5.278
360
Exercise 9 Treasury bills are quoted using the yield on a discount basis or on a money market
basis.
1- The yield on a discount basis denoted yd is computed as
yd =
F −P
B
×
F
n
where F is the face value, P the price, B the year-basis (365 or 360) and n is the number of
calendar days remaining to maturity.
Prove in this case that the price of the T-bill is obtained using the following equation
µ
n × yd
P =F 1−
B
4
¶
2- The yield on a money market basis denoted ym is computed as
ym =
B × yd
B − n × yd
Prove in this case that the price of the T-bill is obtained using the following equation
P =³
1+
F
n×ym
B
´
Solution 10 1- From the equation
yd =
F −P
B
×
F
n
we deduce
n × yd
P
−1=−
B
F
and finally we obtain
µ
n × yd
B
P =F 1−
¶
2- From the equation
ym =
B × yd
B − n × yd
we deduce
ym =
B × F F−P × B
n
F −P
B−n× F ×
B
n
Then we have
n × ym
=
B
F −P
F
P
F
=
B
× F F−P
´
= ³n
1 − F −P
F
F
F −P
= −1
P
P
Finally we obtain
P =³
1+
F
n×ym
B
´
Exercise 11 1- What is the yield on a discount basis of a bill whose face value F is 1,000,
price P is 975 and n the number of calendar days remaining to maturity is 126 ? We assume
that the year-basis is 360.
2- What is the yield on a money market basis of the same bill ?
5
Solution 12 1- The yield yd on a discount basis which satisfies the following equation is equal
to 7.143%
yd =
1, 000 − 975 360
×
= 7.143%
1, 000
126
2- The yield ym on a money market basis which satisfies the following equation is equal to
7.326%
ym =
360 × 7.143%
= 7.326%
360 − 126 × 7.143%
Exercise 13 Suppose the interest rate is 12% per year compounded continuously. What is the
effective annual interest rate ?
Solution 14 The effective annual interest rate is obtained as
R = e0.12 − 1 = 0.1274 = 12.74%.
Exercise 15 If you deposit $2,500 in a bank account which earns 8% annually on a continuously compounded basis, what will be the account balance in 7.14 years.
Solution 16 The account balance in 7.14 years will be
$2500.e8%.7.14 = $4425.98
Exercise 17 If an investment has a cumulative 63.45% rate of return over 3.78 years, what is
the annual continuously compounded rate of return?
Solution 18 The annual continuously compounded rate of return RC is such that
1.6345 = e3.78R
C
We deduce RC = ln(1.6345)/3.78 = 13%.
Exercise 19 How long does it take to double a $100 initial investment when investing at a 5%
continuously compounded interest rate?
6
Solution 20 In general, the solution is given by
CT
xeR
= 2x
or
T =
ln 2
RC
Note that this does not depend on the principle x. In this example, we obtain
T =
ln 2
= 13.86214 years
0.05
Exercise 21 A invests $1000 at 5% per annum continuously compounded. B invests $200 at
20% per annum continuously compounded. Does B ever catch up? How long does it take?
Solution 22 B catches up if the difference between them becomes zero, Alternatively if the ratio
of their amounts becomes 1. The equality condition can be stated as
1000e0.05t = 200e0.2t
equivalent to
ln(5) + 0.05t = 0.2t
or
t=
ln(5)
= 10.73
0.15
Exercise 23 1- What is the price of a 5-year bond with a nominal value of $100, a yield to
maturity of 7%, a 10% coupon rate, and an annual coupon frequency ?
2- Same question for a yield to maturity of 8%, 9% and 10%. Conclude
Solution 24 1- The price P of a bond is given by the following formula
P =
n
X
N ×c
(1 + y)n
i=1
7
+
N
(1 + y)n
which simplifies into
P =
·
¸
N ×c
1
N
1−
n +
y
(1 + y)
(1 + y)n
where N, c, y and n are respectively the nominal value, the coupon rate, the yield to maturity
and the number of years to maturity of the bond.
Here we obtain for P
"
#
1
100
10
1−
+
P =
5
7%
(1 + 7%)
(1 + 7%)5
P is then equal to 112.301% of the nominal value or $112.301. Note that we can also use the
Excel function ”Price” to obtain P.
2- Prices of the bond for different yields to maturity (YTM) are given in the following table
YTM
Price
8%
$107.985
9%
$103.890
10%
$100
Bond prices decrease with an increase in rates.
Exercise 25 1- What is the price of a 5-year bond with a nominal value of $100, a yield to
maturity of 7%, a 10% coupon rate, and semi-annual coupon payments ?
2- Same question for a yield to maturity of 8%, 9% and 10%.
Solution 26 1- The price P of this bond is given by the following formula
P =
2n
X
N × c/2
i=1
i
(1 + y/2)
+
N
(1 + y/2)2n
which simplifies into
"
#
1
N
N ×c
1−
+
P =
2n
y
(1 + y/2)
(1 + y/2)2n
where N, c, y and n are respectively the nominal value, the coupon rate, the yield to maturity
and the number of years to maturity of the bond.
8
Here we obtain for P
P =


10 
1
100

1 − ³
´10  + ³
´10
7%
7%
1 + 7%
1
+
2
2
P is then equal to 112.475% of the nominal value or $112.475. Note that we can also use the
Excel function ”Price” to obtain P.
2- Prices of the bond for different yields to maturity (YTM) are given in the following table
YTM
Price
8%
$108.111
9%
$103.956
10%
$100
Exercise 27 We consider the following zero-coupon curve
Maturity
Zero-Coupon Rate
1 year
4.00%
2 years
4.50%
3 years
4.75%
4 years
4.90%
5 years
5.00%
1- What is the price of a 5-year bond with a $100 face value which delivers a 5% annual
coupon rate ?
2- What is the yield to maturity of this bond ?
3- We suppose that the zero-coupon curve increases instantaneously and uniformly by 0.5%.
What is the new price and the new yield to maturity of the bond ? What is the impact of this
rates increase for the bondholder ?
4- We suppose now that the zero-coupon curve will remain stable over time. You hold the
bond until maturity. What is the annual return rate of your investment ? Why is this rate
different from the yield to maturity ?
9
Solution 28 1- The price P of the bond is equal to the sum of its discounted cash-flows and
given by the following formula
P =
5
5
5
105
5
+
+
+
= $100.136
+
1 + 4% (1 + 4.5%)2 (1 + 4.75%)3 (1 + 4.9%)4 (1 + 5%)5
2- The yield to maturity R of this bond verifies the following equation
100.136 =
4
X
5
105
i +
(1 + R)5
i=1 (1 + R)
Using the Excel function ”yield”, we obtain 4.9686% for R.
3- The new price P of the bond is given by the following formula
P =
5
5
5
105
5
+
+
+
= $97.999
+
2
3
4
1 + 4.5% (1 + 5%)
(1 + 5.25%)
(1 + 5.4%)
(1 + 5.5%)5
The new yield to maturity R of this bond verifies the following equation
97.999 =
4
X
5
105
i +
(1 + R)5
i=1 (1 + R)
Using the Excel function ”yield”, we obtain 5.4682% for R.
The impact of this rates increase for the bondholder is an absolute capital loss of $2.137
Absolute Loss = 97.999 − 100.136 = −$2.137
and a relative capital loss of 2.134%
Re lative Loss =
−2.137
= −2.134%
100.136
4-Before maturity, the bondholder receives intermediate coupons that he reinvests on the
market:
- after one year, he receives $5 that he reinvests for 4 years at the 4-year zero-coupon rate
10
to obtain at the maturity date of the bond
5 × (1 + 4.9%)4 = $6.0544
- after two years, he receives $5 that he reinvests for 3 years at the 3-year zero-coupon rate
to obtain at the maturity date of the bond
5 × (1 + 4.75%)3 = $5.7469
- after three years, he receives $5 that he reinvests for 2 years at the 2-year zero-coupon rate
to obtain at the maturity date of the bond
5 × (1 + 4.5%)2 = $5.4601
- after four years, he receives $5 that he reinvests for 1 year at the 1-year zero-coupon rate
to obtain at the maturity date of the bond
5 × (1 + 4%) = $5.2
- after five years, he receives the final cash-flow equal to $105.
The bondholder finally obtains $127.4614 five years later
6.0544 + 5.7469 + 5.4601 + 5.2 + 105 = $127.4614
which corresponds to a 4.944% annual return rate
µ
127.4614
100.136
¶1/5
− 1 = 4.944%
This return rate is different from the yield to maturity of this bond (4.9686%) because the curve
is not flat at a 4.9686% level. With a flat curve at a 4.9686%, we obtain $127.6108 five years
later
6.0703 + 5.7829 + 5.5092 + 5.2484 + 105 = $127.6108
11
which corresponds exactly to a 4.9686% annual return rate.
µ
127.6108
100.136
¶1/5
− 1 = 4.9686%
Exercise 29 We consider the three zero-coupon bonds (strips) with the following features
Bond
Maturity
Price
Bond 1
1 year
96.43
Bond 2
2 years
92.47
Bond 3
3 years
87.97
Each strip delivers $100 at maturity.
1- Deduce the zero-coupon yield curve from the bond prices
2- We anticipate a rates increase in one year so the prices of strips with residual maturity 1
year, 2 years and three years are respectively 95.89, 90.97 and 84.23. What is the zero-coupon
yield curve anticipated in one year ?
Solution 30 1- The 1-year zero-coupon rate denoted R(0,1) is equal to 3.702%
R(0, 1) =
100
− 1 = 3.702%
96.43
The 2-year zero-coupon rate denoted R(0,2) is equal to 3.702%
R(0, 2) =
µ
100
92.47
¶1/2
− 1 = 3.992%
The 3-year zero-coupon rate denoted R(0,3) is equal to 3.702%
R(0, 2) =
µ
100
87.97
¶1/3
− 1 = 4.365%
2- The 1-year, 2-year and 3-year zero-coupon rates become respectively 4.286%, 4.846% and
5.887%.
12
Exercise 31 We consider the following increasing zero-coupon yield curve
Maturity
R(0,t)
Maturity
R(0,t)
1 year
5.000%
6 years
6.550%
2 years
5.500%
7 years
6.650%
3 years
5.900%
8 years
6.741%
4 years
6.200%
9 years
6.830%
5 years
6.382%
10 years
6.900%
where R(0, t) is the zero-coupon rate at date 0 with maturity t.
1- Compute the par curve.
2- Compute the forward rate curve in one year.
3- Draw the three curves in the same graph. Wwhat can you say about their relative position
?
Solution 32 1- Recall that the par yield c(n) for maturity n is given by the following formula
1−
c(n) = P
n
1
(1+R(0,n))n
1
i
i=1 (1+R(0,i))
Using this equation we obtain the following par yields
Maturity
c(n)
Maturity
c(n)
1
5.000%
6
6.466%
2
5.487%
7
6.556%
3
5.867%
8
6.636%
4
6.147%
9
6.711%
5
6.315%
10
6.770%
2- Recall that F (0, x, y − x), the forward rate as seen from date t = 0, starting at date t = x,
and with residual maturity y − x is defined as
F (0, x, y − x) ≡
·
(1 + R(0, y))y
(1 + R(0, x))x
13
¸
1
y−x
−1
Using the previous equation,we obtain the forward rate curve in one year
Maturity
F(0,1,n)
Maturity
F(0,1,n)
1
6.002%
6
6.928%
2
6.353%
7
6.992%
3
6.603%
8
7.061%
4
6.730%
9
7.113%
5
6.863%
3- The graph of the three curves shows that the forward yield curve is above the zero-coupon
yield curve, which is above the par yield curve. This is always the case when the par yield curve
is increasing.
7,5%
7,0%
Yield
6,5%
6,0%
Par Yield Curve
Zero-Coupon Yield Curve
5,5%
Forward Yield Curve
5,0%
1
2
3
4
5
6
7
8
9
10
Maturity
Exercise 33 At date t=0, we get in the market three bonds with the following features
Coupon
Maturity
Price
Bond 1
10
2 years
P01 = 108.00
Bond 2
7.5
3 years
P02 = 100.85
Bond 3
8.5
3 years
P03 = 103.50
14
Derive the zero-coupon curve until the five-year maturity.
Solution 34 Using the no-arbitrage relation, we obtain the following equations for the five bond
prices



108 = 10B(0, 1) + 110B(0, 2)



100.85 = 7.5B(0, 1) + 7.5B(0, 2) + 107.5B(0, 3)




 103.5 = 9B(0, 1) + 9B(0, 2) + 109B(0, 3)
which can be expressed in a matrix form

108



103.5



  10 110

 
 100.85  =  7.5 7.5 107.5

 
8.5
8.5
108.5





  B(0, 1) 
 

 ×  B(0, 2) 
 

B(0, 3)

Then we get the following discount factors




 B(0, 1)   0.944275 

 

 B(0, 2)  =  0.895975 

 


 

B(0, 3)
0.80975
we deduce the zero-coupon rates


 R(0, 1) = 5.901% 


 R(0, 2) = 5.646% 



R(0, 3) = 7.288%

Exercise 35 Suppose we know from market prices the following zero-coupon rates with matu-
15
rities inferior or equal to one year:
Maturity
Zero-Coupon Rate
1 Day
3.20%
1 Month
3.30%
2 Months
3.40%
3 Months
3.50%
6 Months
3.60%
9 Months
3.80%
1 Year
4.00%
Now we consider bonds priced by the market until the 4-year maturity:
Maturity
Coupon
Gross Price
1 Year and 3 Months
4%
102.8
1 Year and 6 Months
4.5%
102.5
2 Years
3.5%
98.3
3 Years
4%
98.7
4 Years
5%
101.6
1- Using the bootstrapping method, compute the zero-coupon rates for the following maturities 1 year and 3 months, 1 year and 6 months, 2 years, 3 years and 4 years.
2- Draw the zero-coupon yield curve using a linear interpolation
Solution 36 1- We first extract the one-year-and-three-month maturity zero-coupon rate. In
the absence of arbitrage opportunities, the price of this bond is the sum of its future discounted
cash-flows:
102.8 =
104
4
+
1/4
(1 + 3.5%)
(1 + x)1+1/4
where x is the one-year-and-three-month maturity zero-coupon rate to be determined. Solving
this equation (for example with the Excel solver) we obtain 4.16% for x. Applying the same
procedure with the one-year-and-six-month maturity and the two-year maturity bonds we obtain
respectively 4.32% and 4.41% for x. Next we have to extract the 3-year maturity zero-coupon
16
rate solving the following equation
98.7 =
104
4
4
+
+
(1 + 4%) (1 + 4.41%)2 (1 + y%)3
y is equal to 4.48% and finally we extract the 4-year maturity zero-coupon rate denoted z by
solving the following equation
101.6 =
5
105
5
5
+
+
+
3
2
(1 + 4%) (1 + 4.41%)
(1 + 4.48%)
(1 + z%)4
z is equal to 4.57%.
2- Using the linear graph option in Excel we draw the zero-coupon yield curve
4,80%
4,60%
4,40%
Zero-Coupon Rate
4,20%
4,00%
3,80%
3,60%
3,40%
3,20%
3,00%
0
1
2
3
4
Maturity
Exercise 37 From the prices of zero-coupon bonds quoted in the market, we obtain the following
17
zero-coupon curve
Maturity
Zero-Coupon Rate R(0,t)
Discount Factor B(0,t)
1 year
5.000%
0.95238
2 years
5.500%
0.89845
3 years
5.900%
0.84200
4 years
6.200%
0.78614
5 years
?
?
6 years
6.550%
0.68341
7 years
6.650%
0.63720
8 years
?
?
9 years
6.830%
0.55177
10 years
6.900%
0.51312
where R(0,t) is the zero-coupon rate at date 0 for maturity t, and B(0,t) is the discount factor
at date 0 for maturity t.
We need to know the value for the 5-year and the 8-year zero-coupon rates. We have to
estimate them, and test four different methods.
1- We use a linear interpolation with the zero-coupon rates. Deduce R(0,5), R(0,8) and the
corresponding values for B(0,5) and B(0,8).
2- We use a linear interpolation with the discount factors. Deduce B(0,5), B(0,8) and the
corresponding values for R(0,5) and R(0,8).
_
3- We postulate the following form for the zero-coupon rate function R (0, t):
_
R (0, t) = a + bt + ct2 + dt3
Estimate the coefficients a, b, c and d which best approximate the given zero-coupon rates
using the following optimization program
Min
a,b,c,d
X³
_
´2
R(0, i)− R (0, i)
i
18
where R(0, i) are the zero-coupon rates given by the market.
_
_
Deduce the value for R(0, 5) =R (0, 5), R(0, 8) =R (0, 8), and the corresponding values for
B(0,5) and B(0,8).
_
4- We postulate the following form for the discount factor function B (0, t):
_
B (0, t) = a + bt + ct2 + dt3
Estimate the coefficients a, b, c and d which best approximate the given discount factors
using the following optimization program
Min
a,b,c,d
X³
_
´2
B(0, i)− B (0, i)
i
where B(0, i) are the discount factors given by the market.
_
_
Deduce the value for B(0, 5) =B (0, 5), B(0, 8) =B (0, 8), and the corresponding values for
R(0,5) and R(0,8).
5- Conclude
Solution 38 1- Consider that we know R(0, x) and R(0, z), respectively, the x-year and the
z-year maturity zero-coupon rates and that we need R(0, y) the y-years maturity zero-coupon
rate with y ∈ [x; z]. Using the linear interpolation, R(0, y) is given by the following formula
R(0, y) =
(z − y)R(0, x) + (y − x)R(0, z)
z−x
From this equation, we deduce the value for R(0,5) and R(0,8)
R(0, 4) + R(0, 6)
(6 − 5)R(0, 4) + (5 − 4)R(0, 6)
=
= 6.375%
6−4
2
(9 − 8)R(0, 7) + (8 − 7)R(0, 9)
R(0, 7) + R(0, 9)
R(0, 8) =
=
= 6.740%
9−7
2
R(0, 5) =
Using the standard following equation which lies the zero-coupon rate R(0,t) and the discount
factor B(0,t)
B(0, t) =
1
(1 + R(0, t))t
19
we obtain 0.73418 for B(0,5) and 0.59345 for B(0,8).
2- We use the same formula as in question 1 but adapted to discount factors
B(0, y) =
(z − y)B(0, x) + (y − x)B(0, z)
z−x
we obtain 0.73478 for B(0,5) and 0.59449 for B(0,8).
Using the standard following equation which lies the zero-coupon rate R(0,t) and the discount
factor B(0,t)
R(0, t) =
µ
1
B(0, t)
¶1/t
−1
we obtain 6.358% for R(0,5) and 6.717% for R(0,8).
3- Using the Excel function ”DroiteReg”, we obtain the following values for the parameters
Parameters
Value
a
0.04351367
b
0.00720757
c
-0.000776521
d
3.11234E-05
which provides us with the following values for the zero-coupon rates and associated discount
factors
_
_
Maturity
R(0, t)
R (0, t)
B(0, t)
B (0, t)
1
5.000%
4.998%
0.95238
0.95240
2
5.500%
5.507%
0.89845
0.89833
3
5.900%
5.899%
0.84200
0.84203
4
6.200%
6.191%
0.78614
0.78641
5
?
6.403%
?
0.73322
6
6.550%
6.553%
0.68341
0.68330
7
6.650%
6.659%
0.63720
0.63681
8
?
6.741%
?
0.59339
9
6.830%
6.817%
0.55177
0.55237
10
6.900%
6.906%
0.51312
0.51283
20
4- We first note that there is a constraint in the minimization because we must have
B(0, 0) = 1
So the value for a is necessarily equal to 1.
Using the Excel function ”DroiteReg”, we obtain the following values for the parameters
Parameters
Value
a
1
b
-0.04945479
c
-0.001445358
d
0.000153698
which provides us with the following values for the discount factors and associated zero-coupon
rates
_
_
Maturity
B(0, t)
B (0, t)
R(0, t)
R (0, t)
1
0.95238
0.94925
5.000%
5.346%
2
0.89845
0.89654
5.500%
5.613%
3
0.84200
0.84278
5.900%
5.867%
4
0.78614
0.78889
6.200%
6.107%
5
?
0.73580
?
6.328%
6
0.68341
0.68444
6.550%
6.523%
7
0.63720
0.63571
6.650%
6.686%
8
?
0.59055
?
6.805%
9
0.55177
0.54988
6.830%
6.871%
10
0.51312
0.51461
6.900%
6.869%
5- The table below provides the results obtained using the four different methods of interpo-
21
lation and minimization
Rates Interpol.
DF Interpol.
Rates Min.
DF Min.
R(0,5)
6.375%
6.358%
6.403%
6.328%
R(0,8)
6.740%
6.717%
6.741%
6.805%
B(0,5)
0.73418
0.73478
0.73322
0.73580
B(0,8)
0.59345
0.59449
0.59339
0.59055
”Rates Interpol.” is for interpolation on rates (question 1). ”DF Interpol.” is for interpolation
on discount factors (question 2). ”Rates Min” is for minimization with rates (question 3). ”DF
Min.” is for minimization with discount factors (question 4).
The table shows that results are quite similar according to the two methods based on rates.
Differences appear when we compare the four methods. In particular, we can obtain a spread
of 7.5 bps for the estimation of R(0,5) between ”Rates Min.” and ”DF Min.”, and a spread
of 8.8 bps for the estimation of R(0,8) between the two methods based on discount factors. We
conclude that the zero-coupon rates and discount factors estimations are sensitive to the method
of interpolation or minimization used.
Exercise 39 We want to derive the current zero-coupon yield curve for maturities inferior to
10 years. For that goal, we use a basket of bonds quoted by the market and a discount function
modelled as a three-order polynomial spline. The features of the bonds used to derive this curve
22
are summarized in the following table
Bond
Coupon Rate
Maturity
Market Price
Bond 1
0%
7/365 year
99.92
Bond 2
0%
1/12 year
99.65
Bond 3
0%
0.25 year
98.92
Bond 4
0%
0.5 year
97.77
Bond 5
5%
1 year
100.02
Bond 6
6%
2 years
101.56
Bond 7
5%
2.5 years
101.72
Bond 8
7%
3.25 years
109.72
Bond 9
8%
4 years
108.65
Bond 10
5%
4.5 years
100.26
Bond 11
7%
5.5 years
109.89
Bond 12
7%
7 years
107.55
Bond 13
6%
8.75 years
102.75
Bond 14
7%
10 years
108.21
The coupon frequency of these bonds is annual, and the face value is Eur 100.
We model the discount function B(0,s) as a standard polynomial spline with two splines


 B0 (s) = 1 + c0 s + b0 s2 + a0 s3 for s ∈ [0, 3]
B (0, s) =
h
i

 B10 (s) = 1 + c0 s + b0 s2 + a0 s3 − (s − 3)3 + a1 (s − 3)3 for s ∈ [3, 10]
1- Write the theoretical price of bond 1.
2- Calculate the coefficients behind each parameter and the constant number for bond 1
3- Do the same job for bond 2 to bond 14.
4- Estimate the coefficients c0 , b0 , a0 and a1 which best approximate the market prices of
given bonds using the following optimization program
Min
c0 ,b0 ,a0 ,a1
14 ³
X
i=1
23
_
Pi − P i
´2
_
where Pi are the market prices and P i the theoretical prices. We suppose that residual are
homoscedastic so that each bond has the same weight in the minimization program.
5- Calculate
P14 ³
i=1
_
Pi − P i
´2
6- For each bond, calculate the spread between the market price and the theoretical price.
7- Draw the graph of the zero-coupon yield curve.
8- Draw the graph of the forward yield curve in one, two and three months.
_
Solution 40 1- The theoretical price P 1 of bond 1 is
_
µ
7
P i = 100.B 0,
365
¶
"
µ
7
7
+ b0 .
= 100. 1 + c0 .
365
365
¶2
µ
7
+ a0 .
365
¶3 #
2- From question 1, we deduce the coefficients behind each parameter and the constant
number for bond 1
constant number
c0
100
700
365
b0
100.
³
7
365
´2
a0
100.
³
7
365
´3
3- For each bond, we obtain the coefficients behind each parameter and the constant number
24
in the following table
Bond
constant number
c0
b0
a0
a1
Bond 1
100
1.9178
0.0368
0.0007
0
Bond 2
100
8.3333
0.6944
0.0579
0
Bond 3
100
25
6.25
1.5625
0
Bond 4
100
50
25
12.5
0
Bond 5
105
105
105
105
0
Bond 6
112
218
430
854
0
Bond 7
115
272.5
668.75
1658.125
0
Bond 8
128
374
1177
3764.9531
1.6719
Bond 9
132
480
1840
7092
108
Bond 10
125
512.5
2231.25
9523.125
355
Bond 11
142
676
3525.5
17177.625
1696.375
Bond 12
149
896
5880
32688
7100
Bond 13
154
1131.5
9234.625
56638.9688
21270.5
Bond 14
170
1385
12695
81387
39788
Note that Z, what we call the coefficients matrix; is the matrix with dimension 14 × 4 which is
represented by the four last columns of the table.
We give below the details for Bond 8:
· constant number n
n = 7 × 4 + 100 = 128
· coefficient behind c0
7 × [(3.25 − 3) + (3.25 − 2) + (3.25 − 1)] + 107 × 3.25 = 374
· coefficient behind b0
h
i
7 × (3.25 − 3)2 + (3.25 − 2)2 + (3.25 − 1)2 + 107 × 3.252 = 1177
25
· coefficient behind a0
h
i
h
i
7 × (3.25 − 3)3 + (3.25 − 2)3 + (3.25 − 1)3 + 107 × 3.253 − (3.25 − 3)3 = 3764.9531
· coefficient behind a1
107 × (3.25 − 3)3 = 1.6719
4- β = (c0 , b0 , a0 , a1 )T , the vector of parameters is the solution of the standard OLS (Ordinary Least Squared) procedure. It is the result of the following matricial calculation
³
β = ZT Z
´−1
ZT P
where Z T is the transposed matrix of Z, X−1 is the inverse matrix of X and P is the following
vector resulting for each bond of the difference between the constant number and the market
price of the bond.






















P =




















99.92 − 100







 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
=
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
99.65 − 100 

98.92 − 100
97.77 − 100
100.02 − 105
101.56 − 112
101.72 − 115
109.72 − 128
108.65 − 132
100.26 − 125
109.89 − 142
107.55 − 149
102.75 − 154
108.21 − 170
26

−0.08



−0.35 


−1.08 



−2.23 


−4.98 

−10.44
−13.28
−18.28
−23.35
−24.74
−32.11
−41.45
−51.25
−61.79




























We finally obtain for β

βT
−0.043700313




=



−0.003444022
0.000566232
−9.21747E − 05









Note that we can also use the Excel function ”Droitereg” to obtain the vector of parameters
β.
5- The sum of squared spreads
P14 ³
i=1
_
Pi − P i
´2
is equal to 0.043885606.
6- In the following table we examine for each bond the spread between the market price and
the theoretical price
Bond
Market Price
Theoretical Price
Spread
Bond 1
99.92
99.9161
0.0039
Bond 2
99.65
99.6335
0.0165
Bond 3
98.92
98.8869
0.0331
Bond 4
97.77
97.7360
0.0340
Bond 5
100.02
100.1093
-0.0893
Bond 6
101.56
101.4760
0.0840
Bond 7
101.72
101.7274
-0.0074
Bond 8
109.72
109.7342
-0.0142
Bond 9
108.65
108.6926
-0.0426
Bond 10
100.26
100.2787
-0.0187
Bond 11
109.89
109.8869
0.0031
Bond 12
107.55
107.4482
0.1018
Bond 13
102.75
102.8591
-0.1091
Bond 14
108.21
108.1697
0.0403
27
7- We draw below the graph of the zero-coupon yield curve
6,25%
Zero-Coupon Rate
5,75%
5,25%
4,75%
4,25%
0
1
2
3
4
5
6
7
8
9
10
Maturity
We draw below the graph of the three forward yields curves beginining in one, two and three
months. Recall first that F (0, x, y − x), the forward rate as seen from date t = 0, starting at
date t = x, and with residual maturity y − x is defined as
·
(1 + R(0, y))y
F (0, x, y − x) ≡
(1 + R(0, x))x
28
¸
1
y−x
−1
We give successively the values 1/12, 2/12 and 3/12 to x.
6,10%
5,90%
Forward Yield
5,70%
5,50%
5,30%
Forward Yield Curve in 3 Months
Forward Yield Curve in 2 Months
Forward Yield Curve in 1 Month
5,10%
4,90%
4,70%
4,50%
0
20
40
60
80
100
120
Maturity in Months
Because the zero-coupon curve is increasing, the forward yield curve in three months is above
the forward yield curve in two months, which is above the forward yield curve in one month.
Exercise 41 We consider three bonds with the following features:
Bond
Maturity
Coupon Rate
YTM
Bond 1
2 years
5%
5%
Bond 2
10 years
6%
5.5%
Bond 3
30 years
7%
6%
YTM is for yield to maturity. Coupon frequency is annual.
1- Compute the dirty price and the modified duration of each bond
2- a) The YTM of each of these bonds decreases instantaneously by 0.2%. For each bond,
compute the new exact price given by discounting its future cash-flows, the price approximation
given using the first order Taylor expansion, and the difference between these two prices.
b) Same question if the YTM of each of these bonds decreases instantaneously by 1%. Conclude
29
c) For bond 3, draw the difference between the two prices (the new exact price given by
discounting its future cash-flows and the price given using the first order Taylor expansion)
depending on the YTM change.
3- Compute the convexity of each bond
4- We suppose that the YTM of each of these bonds decreases instantaneously by 1%. Compute the price approximation given using the second order Taylor expansion. Compare it to the
exact price given by discounting its future cash-flows.
Solution 42 1- The dirty price and the modified duration of each of these bonds is given in
the following table:
Bond
Price
Modified Duration
Bond 1
100
1.859
Bond 2
103.769
7.438
Bond 3
113.765
13.394
2- a) When the YTM of each of these bonds decreases by 0.2%, we obtain the following
results
Bond
New Exact Price
FOTE Price
Spread
Bond 1
100.373
100.372
0.001
Bond 2
105.327
105.312
0.015
Bond 3
116.877
116.812
0.065
Recall that the FOTE price is given by the following formula
F OT E Pr ice = P + −MD × ∆y × P
where P is the original price, MD the modified duration and ∆y, the YTM change.
b) When the YTM of each of these bonds decreases by 1%, we obtain the following results
Bond
New Exact Price
FOTE Price
Spread
Bond 1
101.886
101.859
0.027
Bond 2
111.869
111.487
0.382
Bond 3
130.745
129.003
1.742
30
When the YTM change is high, the spread between the two prices is not negligible. We have
to use the second order approximation.
c) For bond 3, we draw below the difference between the new exact price given by discounting
its future cash-flows and the price given using the first order Taylor expansion depending on the
YTM change.
15
10
5
0
3.
0%
3.
3%
3.
6%
3.
9%
4.
2%
4.
5%
4.
8%
5.
1%
5.
4%
5.
7%
6.
0%
6.
3%
6.
6%
6.
9%
7.
2%
7.
5%
7.
8%
8.
1%
8.
4%
8.
7%
9.
0%
Difference between the Two Prices
20
-5
YTM Level
3- The convexity of each of these bonds is given in the following table
Bond
Convexity
Bond 1
5.27
Bond 2
70.95
Bond 3
280.97
4- Recall that the SOTE (second order Taylor expansion) price is given by the following
formula
SOT E Pr ice = P + −M D × P × ∆y + RC × P × (∆y)2 /2
where RC is the (relative) convexity.
31
When the YTM of each of these bonds decreases by 1%, we obtain the following results
Bond
Exact Price
SOTE Price
Spread
Bond 1
101.886
101.886
0.000
Bond 2
111.869
111.855
0.014
Bond 3
130.745
130.601
0.144
Exercise 43 Today is 1/1/98. On 6/30/99 we will have to make a payment of $100. We can
only invest in a riskfree pure discount bond (nominal $100) that matures on 12/31/98 and in a
riskfree coupon bond, nominal $100 that pays an annual interest (on 12/31) of 8% and matures
on 12/31/00. Assume a flat term structure of 7%. How many units of each of the bonds should
we buy in order to be perfectly immunized?
Solution 44 We first have to compute the present value P V of the debt, which is the amount
we will have to deposit
PV =
100
= 90.35
(1.07)1.5
We also compute the price P1 of the one-year pure discount bond
P1 =
100
= 93.46
1.07
Similarly, the price P3 of the three year coupon bond is
P3 =
8
8
108
+
+
= 102.6
2
1.07 (1.07)
(1.07)3
The duration of the one-year pure discount bond is obviously one. The duration D3 of the
three-year coupon bond is
D3 = 1 ×
8
1.07
102.6
+2×
8
(1.07)2
102.6
+3×
8
(1.07)3
102.6
= 2.786
We now compute the number of units of the one-year and the tree-year bonds (q1 and q3 respectively), so as to achieve a dollar duration equal to that of debt, and also a present value
of the portfolio equal to that of debt. We know that the duration of the debt we are trying to
32
immnunize is 1.5, therefore q1 and q3 are given as the unique solution to the following system
of equations


 93.46 × 1 × q1 + 102.6 × 2.786 × q3 = 90.35 × 1.5


=⇒
93.46 × q1 + 102.6 × q3 = 90.35


 q1 = 0.696117


q3 = 0.2465
In terms of dollar amount 93.46×q1 = $65.0591invested in 1 year maturity bond and 102.6×q3 =
25.2939 invested in 3 years maturity bond.
Exercise 45 An investor holds 100,000 units of a bond whose features are summarized in the
following table. He wishes to be hedged against a rise in interest rates.
Maturity
Coupon Rate
YTM
Duration
Price
18 Years
9.5%
8%
9.5055
$114,181
Characteristics of the hedging instrument which is here a bond are the following:
Maturity
Coupon Rate
YTM
Duration
Price
20 Years
10%
8%
9.8703
$119.792
Coupon frequency is semi-annual. YTM is for yield to maturity. The YTM curve is flat at a
8% level.
1- What is the quantity φ of the hedging instrument that the investor has to sell ?
2- We suppose that the YTM curve increases instantaneously by 0.1%.
a) What happens if the bond portfolio has not been hedged ?
b) And if it has been hedged?
3- Same question as the previous one when the YTM curve increases instantaneously by 2%.
4- Conclude
Solution 46 1- The quantity φ of the hedging instrument is obtained by resolving the following
equation (??)
φ=−
11, 418, 100 × 9.5055
= −91, 793
119.792 × 9.8703
The investor has to sell 91,793 units of the hedging instrument.
33
2- Prices of bonds with maturity 18 years and 20 years becomes respectively $113.145 and
$118.664.
a) If the bond portfolio has not been hedged, the investor loses money. The loss incurred is
given by the following formula (exactly -$103,657 if we take all the decimals into account)
Loss = $100, 000 × (113.145 − 114.181) = −$103, 600
b) If the bond portfolio has been hedged, the investor is quasi-neutral to an increase (and a
decrease) of the YTM curve. The P&L of the position is given by the following formula
P &L = −$103, 600 + $91, 793 × (119.792 − 118.664) = −$57
3- Prices of bonds with maturity 18 years and 20 years becomes respectively $95.863 and
$100.
a) If the bond portfolio has not been hedged, the loss incurred is given by the following
formula
Loss = $100, 000 × (95.863 − 114.181) = −$1, 831, 800
b) If the bond portfolio has been hedged, the P&L of the position is given by the following
formula
P &L = −$1, 831, 800 + $91, 793 × (119.792 − 100) = −$15, 032
4- For a small move of the YTM curve, the quality of the hedge is good. For a large move
of the YTM curve, we see that the hedge is not perfect because of the convexity term which is
no more negligible.
Exercise 47 Same exercise as the previous one except that we now consider a non flat YTM
curve. The YTM of the bond with maturity 18 years is 7.5%, and the YTM of the bond with
maturity 20 years is 8%.
Solution 48 We compute the modified duration and the price of the bond to hedge.
Maturity
Coupon Rate
YTM
Modified Duration
Price
18 Years
9.5%
7.5%
9.6928
$119.581
34
The modified duration of the hedging instrument is 9.4906.
1- The quantity φ of the hedging instrument is obtained by resolving the following equation
(??)
φ=−
11, 958, 100 × 9.6928
= −98, 264
119.792 × 9.4906
The investor has to sell 98,264 units of the hedging instrument. With a non flat curve, note
that we use the modified duration instead of the duration in the previous equation.
2- Prices of bonds with maturity 18 years and 20 years becomes respectively $118.471 and
$118.664.
a) If the bond portfolio has not been hedged, the investor loses money. The loss incurred is
given by the following formula
Loss = $100, 000 × (118.471 − 119.581) = −$111, 000
b) If the bond portfolio has been hedged, the investor is quasi-neutral to an increase (and a
decrease) of the YTM curve. The P&L of the position is given by the following formula
P &L = −$111, 000 + $98, 264 × (119.792 − 118.664) = −$158
3- Prices of bonds with maturity 18 years and 20 years becomes respectively $100 and $100.
a) If the bond portfolio has not been hedged, the loss incurred is given by the following
formula
Loss = $100, 000 × (100 − 119.581) = −$1, 958, 100
b) If the bond portfolio has been hedged, the P&L of the position is given by the following
formula
P &L = −$1, 958, 100 + $98, 264 × (119.792 − 100) = −$13, 258
Exercise 49 Riding the Yield Curve
At date t = 0, we consider the following zero-coupon curve for the short-term segment which
35
is upward sloping.
Maturity
Zero-Coupon Rate
Zero-Coupon Bond Price
92 days
4.00%
$99.016
183 days
4.50%
$97.817
274 days
4.80%
$96.542
An investor has funds to invest for six months. He has two different opportunities:
· buying a 183-days T-Bill and holding it until the maturity.
· or riding down the yield curve by buying a 274-day T-bill and selling it six months after.
1- Calculate the total return rate of these two strategies assuming that the zero-coupon curve
remains stable.
2- If the zero-coupon yield curve rises by 2% after date t=0 and stay stable at this level,
what would be the total return rate of riding down the yield curve ?
Solution 50 1- From the table above, the price of the 183-days T-Bill is 99.016$ so that buying
and holding it will provide a total return rate of
100 − 99.016
= 2.231%
99.016
From the table above, the price of the 274-day T-Bill is $96.542. The 274-day T-Bill becomes
a 92-day T-Bill at our horizon date. If we assume no change in the zero-coupon yield curve,
riding down the yield curve would bring a total return rate of
99.016 − 96.542
= 2.563%
96.542
that is to say a 0.332% surplus of total return rate compared to the buy and hold strategy.
2- But if the zero-coupon yield curve rise by 2%, the price of the 92-day T-Bill is $98.542
so riding down the yield curve only brings a total return rate of
98.542 − 96.542
= 2.072%
96.542
In this case riding down the yield curve is worse than the buy-and-hold strategy.
36
Exercise 51 Rollover Strategy
An investor has funds to invest for one year. He anticipates an 1% increase of the curve in
six months. Six-month and one-year zero-coupon rates are respectively 3% and 3.2%. He has
two different opportunities:
· he can buy the 1-year zero-coupon T-bond and hold it until maturity.
· or he can applicate the rollover strategy by buying the 6-month T-bill, holding it until
maturity, and buying a new six-month T-bill in six month and holding it until maturity
1- Calculate the annualized total return rate of these two strategies assuming that the investor’s anticipation is correct.
2- Same question when rates decrease by 1% in six months ?
Solution 52 1- The annualized total return rate of the first strategy is of course 3.2%
100
(1+3.2%)
100
(1+3.2%)
100 −
=
100 − 96.899
= 3.2%
96.899
By doing a rollover, the investor will invest at date t = 0 $98.533 to obtain $100 six months
later. Note that $98.533 is obtained as follows
$100
6
(1 + 3%) 12
= $98.533
Six months later he will then buy a quantity α = 100/98.058 of a six-month T-Bill which pays
$100 at maturity so that the annualized total return rate of the second strategy is 3.5%
100 × α − 98.533
= 3.5%
98.533
2- The annualized total return rate of the first strategy is still 3.2% as it is only 2.5% for
the rollover strategy
100 × β − 98.533
= 2.5%
98.533
where β = 100/99.015.
In this case the rollover strategy is worse than the simple buy-and-hold strategy.
Exercise 53 Bond versus Strips Arbitrage
37
On 05/15/02, we suppose that the price of strips with maturity 05/15/03, 05/15/04, 05/15/05
and 05/15/06 are respectively 96.05, 91.23, 86.5 and 81.1. The principal amount of strips is
$100. At the same time the price of the bond with maturity 05/15/06, coupon rate 5% and
principal amount $1,000 is 98.75.
1- Compute the price of the reconstructed bond
2- A trader wants to buy 10,000 bonds. Is there an arbitrage he can benefit ?
Solution 54 1- The price of the reconstructed bond using strips is 98.844
5 × 0.9605 + 5 × 0.9123 + 5 × 0.865 + 105 × 0.811 = 98.844
2- For the trader, the arbitrage consists in buying the bond and selling the strips. Then he
buys a quantity of 10,000 bonds, sells a quantity of 5,000 strips with maturity 05/15/03, 5,000
strips with maturity 05/15/04, 5,000 strips with maturity 05/15/05 and 105,000 strips with
maturity 05/15/06. The gain of the trader is
$1, 000 × 10, 000 × (98.844 − 98.75) % = $9, 400
Exercise 55 Butterfly
We consider three bonds with short, medium and long maturities whose features are summarized in the following table
Maturity
Coupon Rate
YTM
Bond Price
$Duration
Quantity
2 Years
6%
6%
100
183.34
qs
10 Years
6%
6%
100
736.01
−1, 000
30 Years
6%
6%
100
1, 376.48
ql
Face value of bonds is $100, YTM stands for yield-to-maturity, bond prices are dirty prices
and we assume a flat yield-to-maturity curve in the exercise. We structure a butterfly in the
following way:
· we sell 10, 000 10-year maturity bonds
· we buy qs 2-year maturity bonds and ql 30-year maturity bonds
38
1- Determine the quantities qs and ql so that the butterfly is cash and $duration neutral.
2- What is the P&L of the butterfly if the yield to maturity curve still flat goes up to a 7%
level ? And down to a 5% level ?
3- Draw the P&L of the butterfly depending on the value of the yield to maturity
Solution 56 1- The quantities qs and ql , which are determined so that the butterfly is cash and
$duration neutral, satisfy the following system


 (qs × 183.34) + (ql × 1, 376.48) = 10, 000 × 736.01


which solution is


(qs × 100) + (ql × 100) = 10, 000 × 100

−1
 qs   183.34 1, 376.48 

=

ql
100
100




 7, 360, 100   5, 368 
=

×
1, 000, 000
4, 632
2- If the yield to maturity curve goes up to a 7% level or goes down to a 5% level, bond
prices become
Maturity
Price if YTM = 7%
Price if YTM = 5%
2 Years
98.19
101.86
10 Years
92.98
107.72
30 Years
87.59
115.37
P&Ls are respectively $3,051 and $3,969 when the yield to maturity curve goes up to a 7% level
or goes down to a 5% level.
3- We draw below the profile of the P&L’s butterfly depending on the value of the yield to
39
maturity
100000
90000
80000
Total Return in $
70000
60000
50000
40000
30000
20000
10000
0
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
Yield to Maturity
The butterfly has a positive convexity. Whatever the value of the yield to maturity the strategy
always generates a gain. This gain is all the more substantial as the yield-to-maturity reaches
a level further away from 6%.
Exercise 57 We consider two firms A and B which have the same financial needs in terms
of maturity and principal. The two firms can borrow money in the market at the following
conditions:
· Firm A: 11% at a fixed rate or Libor + 2% for a $10 million loan and a five-year maturity.
· Firm B: 9% at a fixed rate or Libor + 0.25% for a $10 million loan and a five-year
maturity.
1- We suppose that firm B prefers a floating rate debt as firm A prefers a fixed rate debt.
What is the swap they will structure to optimize their financial conditions ?
2- If firm B prefers a fixed rate debt as firm A prefers a floating rate debt, is there a swap
to structure so that the two firms optimize their financial conditions ? Conclude
Solution 58 1- Firm B has 2% better conditions at a fixed rate and 1.75% better conditions at
a floating rate than firm A. The spread between the conditions obtained by firm A and firm B at
40
a fixed rate and the spread between the conditions obtained by firm A and firm B at a floating
rate is different from 0.25%.
To optimize their financial conditions:
Firm B borrows money at a 9% fixed rate, firm A contracts a loan at a Libor + 2% floating
rate and they structure the following swap. Firm B pays Libor + 0.75% and receives the fixed 9%
as firm A receives Libor + 0.75% and pays the fixed 9%. The financing operation is summarized
in the following table
Firm A
Firm B
Initial Financing
(Libor + 2%)
(9%)
Swap A to B
(9%)
9%
Swap B to A
Libor + 0.125%
(Libor + 0.125%)
Financing Cost
(10.875%)
(Libor + 0.125%)
Financing Cost without Swap
(11%)
(Libor + 0.25%)
Gain
0.125%
0.125%
By structuring a swap, firm A and firm B have optimized their financial conditions and each
firm has gained 0.125%.
2- There is no swap to structure between the two firms so that they ameliorate their financial
conditions.
Exercise 59 We consider at date T0 , a 6-month LIBOR standard swap contract with maturity
6 years with the following cash-flow schedule
−F
T0
−F
−F
−F
−F
−F
T1
T2
T3
T4
T5
T6
T7
T8
T9
T10
T11
T12
V1
V2
V3
V4
V5
V6
V7
V8
V9
V10
V11
V12
Note that Ti+1 − Ti = 6 months, ∀i ∈ {0, 1, 2, ..., 11}. We suppose that the swap face value is
$1 million, and that the rate F of the fixed leg is 6%.
At date T0 , zero-coupon rates and discount factors for maturities T1 , T2 , ...T12 are given in
41
the following table
Maturity
ZC rates
Maturity
ZC rates
T1
4.005%
T7
5.785%
T2
4.575%
T8
5.896%
T3
4.925%
T9
6.001%
T4
5.134%
T10
6.069%
T5
5.412%
T11
6.121%
T6
5.599%
T12
6.148%
1- What is the pricing formula for this plain vanilla swap using the zero-coupon method ?
2- Compute the discount factors for maturities T1 , T2 , ...T12 .
3- Give the price of this swap ?
4- What is the swap rate such that the price of this swap is zero ?
5- An investor has bought 100,000 5-year bond with a 7.2% annual coupon rate and a nominal
amount of $1,000. What is the price, the yield to maturity and the modified duration of this
bond.
6- The investor fears a rates increase. How many swaps must he sell to protect its bond
portfolio ?
7- The yield to maturity curve increases uniformly by 0.3%. What is his new position with
and without the hedge ?
Solution 60 1- The price of this swap at date T0 is given by
SW APT0 = $1, 000, 000 ·
Ãi=6
X
i=1
!
F.B(T0 , T2i ) − 1 + B(T0 , T12 )
2- Discount factors are given in the following table using the following formula
B(0, t) =
1
(1 + R(0, t))t
where, at date 0, B(0,t) is the discount factor for maturity t, and R(0,t) the zero-coupon rate
42
for maturity t.
Maturity
ZC rates
Discount Factors
T1
4.005%
0.98056
T2
4.575%
0.95625
T3
4.925%
0.93043
T4
5.134%
0.90472
T5
5.412%
0.87655
T6
5.599%
0.84922
T7
5.785%
0.82133
T8
5.896%
0.79521
T9
6.001%
0.76932
T10
6.069%
0.74483
T11
6.121%
0.72126
T12
6.148%
0.69908
3- The price of the swap at date T0 is simply −$3957.5.





 0.95625 + 0.90472 + 0.84922 

 − 1 + 0.69908 = −$3, 957.5
SW APT0 = $1, 000, 000×6% 
+0.79521 + 0.74483 + 0.69908
4- In the previous formula, we have to replace 6% with 6.07996% to obtain a swap price
equal to zero.
5- The price P of the bond is obtained by discounting its future cash-flows according to
P = 7.2 × (0.95625 + 0.90472 + 0.84922 + 0.79521) + 107.2 × 0.74483 = 105.0847
The yield to maturity R verifies the following formula
105.0847 =
4
X
i=1
7
107
i +
(1 + R)5
(1 + R)
Using for example the Excel function ”yield”, we obtain 5.9931% for R.
The modified duration of this bond is given again by using the Excel function ”MDuration”.
43
We obtain 4.14094.
6- The solution of the third question shows that the swap is the sum of two assets:
· a 6-year bond with a 6% annual coupon rate and $1,000,000 face value
· minus a $1,000,000 cash value.
The price of the bond contained in the swap is given by the following formula
6 × 0.95625 + 0.90472 + 0.84922 + 0.79521 + 0.74483) + 106 × 0.69908 = 99.6043
In the same way as in question 5, we deduce the yield to maturity and the modified duration of
this bond which are respectively equal to 6.08068% and 4.9119.
Assuming that the yield to maturity move of the bond to hedge is equal to the yield to
maturity of the bond contained in the swap, we deduce the number n of swaps that the investor
has to sell according to the following equation
QB × NB × $DurationB = n × NS × $DurationS
where QB is the quantity of bonds bought by the investor, NB the nominal amount of the bond,
NS the nominal amount of the swap. $DurationB is the $Duration of the bond to hedge as
$DurationS is the $Duration of the bond contained in the swap.
We finally obtain
n=
100, 000 × 1, 000 × 105.0847 × 4.14094
= 88.94
1, 000, 000 × 99.6043 × 4.9119
so that the investor has to sell 89 swaps.
7- The new price of the bond to hedge is 103.78982. the new price of the bond contained in
the swap is 98.1502 so that the new swap price is
SW AP = $1, 000, 000 × (0.981502 − 1) = −$18, 498
· Without the hedge, the investor’s loss is
Loss = 100, 000 × $1, 000 × (103.78982 − 105.0847) % = −$1, 294, 880
44
· With the hedge, the investor loses -$1,294,880 in the bond market and gains $1,294,105
in the swap market
Gain = 89 × (−3, 957.5 − 18, 498) = $1, 294, 105
so that the net position is very near from zero (exactly -$775).
Exercise 61 Deriving the current interbank zero-coupon yield curve with the Nelson and Siegel
model we obtain the following parameter values
β0
β1
β2
τ
6.9%
−3.5%
−1%
3
The goal of this exercise is to obtain the Nelson and Siegel level, slope and curvature $Durations of some plain vanilla swaps, and to hedge a bond portfolio against a change of these
parameters.
1- Compute the price and the Nelson and Siegel level, slope and curvature $durations of the
three following swaps:
a- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity
of 2 years, semi-annual payments on the fixed leg and a fixed rate equal to 4%.
b- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity
of 5 years, semi-annual payments on the fixed leg and a fixed rate equal to 5%.
c- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity
of 10 years, semi-annual payments on the fixed leg and a fixed rate equal to 5.50%.
2- An investor wants to hedge a bond portfolio whose level, slope and curvature $ durations
are respectively -789,456,145, -142,256,548 and -97,897,254. He decides to use the three swaps
to hedge his position. How many of these three swaps must he buy or sell ?
3- Compute the financing cost of the hedge by assuming that the investor hedges his position
during 14 days with no rebalancing. The Libor 2 weeks is equal to 3.5%.
4- Explain why hedging with swaps is better than hedging with bonds.
Solution 62 1- Recall that the discount factor B(0, θ) is given by the following formula using
45
the Nelson and Siegel model
B(0, θ) = e−θ.R
C (0,θ)
µ
−θ. β 0 +β 1
=e
·
θ
1−exp − τ
θ
τ
(
)
¸
+β 2
·
θ
1−exp − τ
θ
τ
(
) −exp − θ
( τ)
¸¶
and that the level, slope and curvature $durations of a bond delivering cash-flows Ci at dates
θi , which are denoted respectively S0 , S1 and S2 , are given by












S0 = −
P
S1 = −
P
C (0,θ
θi Ci e−θi R
i)
´
 i ³
θ
1−exp − τ i
θi 
1
θi
τ1
 Ci e−θi RC (0,θi )
i


³
´




θ

³
´
1−exp − τ i

P
C

1


S = − θi 
− exp − τθ1i  Ci e−θi R (0,θi )
θi

 2
i
τ
1
a- The pricing formula for such a swap using the zero-coupon method is
SW AP = 10, 000, 000.
à 4
X
i=1
!
2%.B(0, i/2) − 1 + B(0, 2) = −$3, 103
As we can see in the previous formula, a standard swap is equivalent to:
· a bond with the nominal amount and the fixed rate of the swap, and the coupon frequency
of the fixed leg
· minus the nominal amount of the swap.
We deduce that the level, slope and curvature $durations of a swap are equivalent of those
of the bond contained in the swap. For this swap we obtain









S0 = −106 ×
S =
−106
×
1


µ 4



P

6

 S2 = −10 ×
i=1
µ
i
2
4
P
·i=1
i
2
·
4
P
i
2 2%B(0, i/2)
i=1 ¸
i/2
τ
i/2
τ
1−e−
i/2
τ
i/2
τ
1−e−
µ
−
−e
2%B(0, i/2) + 2
i/2
τ
¸
·
1−e−
2%B(0, i/2) + 2
46
¶
+ 2B(0, 2) = −1, 935, 651
·
i/2
τ
2
τ
¸
B(0, 2) = −1, 418, 594
2
1−e− τ
2
τ
¶
−e
− τ2
¸
¶
B(0, 2) = −414, 983
b- Using the same method we obtain for the second swap



SW AP = $4, 072





 S0 = −4, 499, 361


S1 = −2, 261, 944





 S = −1, 317, 652
2
c- Using the same method we obtain for the third swap



SW AP = −$5, 291





 S0 = −7, 723, 248


S1 = −2, 537, 714





 S = −1, 997, 192
2
2- This conducts to search for the quantity q1 , q2 and q3 to invest respectively in the three
swaps with maturity 2, 5 and 10 years that verify the following linear system






q1 × 1, 935, 651 + q2 × 1, 418, 594 + q3 × 414, 983 = −789, 456, 145
q1 × 4, 499, 361 + q2 × 2, 261, 944 + q3 × 1, 317, 652 = −142, 256, 548




 q × 7, 723, 248 + q × 2, 537, 714 + q × 1, 997, 192 = −97, 897, 254
1
2
3
which provides
The investor will:






q1 = 527.42
q2 = 667.11




 q = −2, 838.21
3
· buy 527 swaps with 2-year maturity
· buy 667 swaps with 5-year maturity
· sell 2,838 swaps with 10-year maturity
to protect his bond portfolio against a change of the level, slope and curvature factors.
3- The financing cost of the hedge is equal to
[(527 × −3, 103) + (667 × 4, 072) − (2, 838 × −5, 291)] ×
47
14
× 3.5% = $21, 909
360
4- It is preferable to use swaps instead of bonds to hedge a position because the hedging cost
is usually lower with these instruments.
Exercise 63 The Cheapest to Deliver on the Repartition Date
We are now on the repartition date. Suppose a futures contract with Eur100,000 contract
size whose price is 95 and three bonds denoted A,B and C with the following features. What is
the bond that the seller of the futures contract will choose to deliver.
Clean Price
Conversion Factor
Bond A
112.67
119.96%
Bond B
111.54
118.66%
Bond C
111.47
119.78%
Solution 64 The seller of the futures contract chooses to deliver the bond that maximises the
difference between the invoice price IP and the cost of purchasing the bond CP , which is called
the cheapest to deliver. The quantity IP -CP is given by the following formula
IP − CP = futures price × CF − clean price
where CF is the conversion factor. Here we obtain the following results
IP-CP
Bond A
1.292
Bond B
1.187
Bond C
2.321
The seller of the futures contract will choose to deliver Bond C.
Exercise 65 A treasurer of a firm knows today that he will have to invest Eur 50,000,000 in
a particular bond A in one month. Today, this bond quotes 116.414 (the accrued interest is
included in the price). He fears a decrease in rates and then uses the futures market to hedge
its interest rate risk. The price of the ten year futures contract which expires in one month is
98.55. Its nominal amount is Eur 100,000. The conversion factor for bond A is 1.18125.
48
1- Has the treasurer to buy or sell futures contracts to hedge its interest rate risk ?
2- What is the position he has to take on the futures market ?
3- One month later, the price of bond A is 121.137 as the price of the futures contract is
102.55. What is the result of the hedge transaction for the treasurer ?
Solution 66 1- The tresurer fears a decrease in rates and then an increase in price of bond A.
To hedge this interest rate risk, he has to buy futures contract because prices of futures contracts
move in the same way as bond prices when interest rates decrease or increase.
2- The number n of futures contract is given by the following hedging ratio
n=
50, 000, 000
MDA .PA .NA
× 1.18125 = 590.775
.CF =
MDA .PA .NF
100, 000
where MDA is the modified duration of bond A, PA its gross price, NA the amount the treasurer
will invest in bond A, and NF the nominal amount of the futures contract. Note here that bond
A is the cheapest to deliver which simplifies the previous formula.
Here, the treasurer buys 591 futures contracts.
3- The result is the following for the treasurer:
· gain on the futures market:
591 × (102.55 − 98.55) % × Eur 100, 000 = Eur 2,364,000
· opportunity loss on the bond market:
(116.414 − 121.137) % × Eur 50, 000, 000 = Eur 2.361,500
which represents a net gain of Eur 2,500. The net position is not exactly equal to zero because
the treasurer buys 591 futures contract and not 590.775 as given by the hedge ratio.
Exercise 67 Speculation with Futures
Today, the gross price of a 10-year maturity bond with $1,000 principal amount is 116.277.
At the same moment the price of the ten year futures contract which expires in two month is
98.03. Its nominal amount is $100,000, and the deposit margin is $1,000. One month later the
price of the bond is 120.815 as the price of the futures is 102.24.
49
1- What is the leverage effect on this futures contract ?
2- An investor anticipates that rates will decrease in a short-term period. His cash at disposal
is $100,000.
a) What is the position he can take on the market using the bond ? What is his absolute
gain after one month ? What is the return rate of its investment ?
b) Same question as the previous one using the futures contract ?
3- Conclude
Solution 68 1- The leverage effect of the futures contract is 100
Leverage Effect =
100, 000
No min al Amount
=
= 100
Deposit Margin
1, 000
2a) The investor anticipates a decrease in rates so he will buy bonds. His cash at disposal is
$100,000. then he buys 86 bonds
Number of bonds bought =
$100, 000
= 86
$1, 000 × 116.277%
. His absolute gain over the period is $4,723
Absolute Gain = 86 × $1, 000 × [120.815 − 116.277] % = $3, 902.68
The return rate of its investment over the period is 3.903%
Re turn Rate =
$3, 902.68
= 3.903%
$100, 000
b) Futures contracts move in the same way as bonds when interest rates change, so the
investor will buy futures contracts. His cash at disposal is $100,000. then he buys 102 futures
contracts
Number of futures contracts bought =
$100, 000
$100, 000
=
= 102
deposit margin × futures price
$1, 000 × 98.03%
50
His absolute gain over the period is
Absolute Gain = 102 × $100, 000 × [102.24 − 98.03] % = $429, 420
The return rate of its investment over the period is 429.42%
Re turn Rate =
$429, 420
= 429.42%
$100, 000
3- The difference of performance between the two investments is explained by the leverage
effect of the futures contract.
Exercise 69 Let us consider a caplet contracted at date t = 05/13/02 with nominal amount
³
$10, 000, 000, exercise rate E = 5%, based upon the six-month Libor RL t, 12
following schedule
´
t = 05/13/02
T0 = 06/03/02
T1 = 12/03/02
When the caplet is contracted
When the caplet starts
Pay-off payment
and with the
At date T1 = 12/03/02, the cap holder receives the cash-flow C
·
µ
µ
C = $10, 000, 000 × Max 0; δ · RL T0 ,
¶
¶¸
1
− 5%
2
where δ = T1 − T0 (expressed in fraction of years using the Exact/360 day-count basis).
1- Draw the P&L of this caplet considering that the premium paid by the buyer is equal to
0.1% of the nominal amount.
2- What is the price formula of this caplet using the Black model ?
3- Assuming that the 6-month Libor forward is 5.17% at date t, its volatility 15% and that
the zero-coupon rate RC (t, T1 − t) is equal to 5.25%, give the price of this caplet.
4- Deduce the margin taken by the seller of this caplet.
We know want to compute the sensitivities of the caplet
5- Delta
a- Compute the delta of the caplet.
b- Recalculate the price of the caplet if the 6-month forward goes from 5.17% to 5.18%.
51
c- Compare the price difference with the quantity ”delta×change of the forward rate”.
6- Gamma
a- Compute the gamma of the caplet.
b- The 6-month forward goes from 5.17% to 5.18%. Compare the price difference of the
caplet with the quantity ”delta×change of the forward rate+gamma×(change of the forward
rate)2 /2”.
7- Vega
a- Compute the vega of the caplet.
b- Recalculate the price of the caplet if the volatility goes from 15% to 16%.
c- Compare the price difference with the quantity ”vega×change of volatility”.
8- Rho
a- Compute the rho of the caplet.
b- Recalculate the price of the caplet if the interest rate RC (t, T1 − t) goes from 5.25% to
5.35%.
c- Compare the price difference with the quantity ”rho×change of rate”.
9- Theta
a- Compute the theta of the caplet.
b- Recalculate the price of the caplet one day later on 05/14/02..
1
”.
c- Compare the price difference with the quantity ”theta× 365
Solution 70 1- The P&L of the caplet depends on the value of the 6-month Libor at date
T0 , and is given by the following formula (considering the buyer position; of course the seller
position is the opposite one)
½
·
µ
µ
P &L = $10, 000, 000 × Max 0; δ · RL T0 ,
52
¶
¶¸
1
− 5%
2
¾
− 0.1%
It appears on the followng graph
100000
90000
80000
70000
P&L in $
60000
50000
40000
30000
20000
10000
0
3,00%
-10000
3,50%
4,00%
4,50%
5,00%
5,50%
6,00%
6,50%
7,00%
-20000
Value of the 6-month Libor on 06/03/02
2- The caplet price at date t in Black (1976) model is given by
h
³
p
´i
Caplett = N · δ · B(t, T1 ) · F (t, T0 , T1 )Φ(d) − EΦ d − σ T0 − t
where:
N is the nominal amount.
B(t, T1 ) is the discount factor equal to
C (t,T −t)
1
B(t, T1 ) = e−(T1 −t).R
Φ is the cumulative distribution function of the standard Gaussian distribution.
F (t, T0 , T1 ) is the 6-month Libor forward as seen from date t, starting at date T0 and finishing
at date T1 . Note in particular that
F (T0 , T0 , T1 ) = R
53
L
µ
1
T0 ,
2
¶
d is given by
ln
d=
³
F (t,T0 ,T1 )
E
´
+ 0.5σ2 (T0 − t)
√
σ T0 − t
σ is the volatility of the underlying rate F (t, T0 , T1 ), which is usually referred to as the caplet
volatility.
3- N = $10, 000, 000 and δ =
183
360 .
The discount factor is equal to
204
B(t, T1 ) = e− 365 .5.25% = 0.97108384
√
We obtain the following values for d and d − σ T0 − t
d = 0.94100172
p
d − σ T0 − t = 0.90477328
√
so that Φ(d) and Φ(d − σ T0 − t) are equal to
Φ(d) = 0.82664803
p
Φ(d − σ T0 − t) = 0.81720728
Note that Φ the cumulative distribution function of the standard Gaussian law is already preprogrammed in Excel (see Excel functions).
We finally obtain the caplet price
Caplet = $9, 267 = 0.09267% of the nominal amount
4- The seller of this caplet takes a margin equal to 7.91%
10, 000
− 1 = 7.91%
9, 267
5- Delta
a- The delta ∆, which is the first derivative of the caplet price with respect to the underlying
54
rate F (t, Tj−1 , Tj ), is given by
∆ = N · δ · B(t, T1 )Φ(d) = 4, 080, 618
b- If the 6-month Libor forward goes from 5.17% to 5.18%, the caplet price becomes $9,678.
c- The price difference is equal to $411
Pr ice Difference = $9, 678 − $9, 267 = $411
very near from the quantity ”delta×change of the forward rate” equal to $408
delta × change of the forward rate = 4, 080, 618 × 0.01% = $408
6- Gamma
a- The gamma γ, which is the second derivative of the caplet price with respect to the
underlying rate F (t, Tj−1 , Tj ), is given by
B(t, T1 )
0
Φ (d1 ) = 675, 296, 853
γ =N ×δ× √
σ T0 − tF (t, T0 , T1 )
where Φ0 is the density function of the standard Gaussian law
1
2
Φ0 (x) = √ e−x /2
2π
b- If the 6-month Libor forward goes from 5.17% to 5.18%, the caplet price becomes $9,678
so that the price difference of the caplet is equal to $411 (see 5-c) and very well explained by
the quantity ”delta×change of the forward rate+gamma×(change of the forward rate)2 /2”
delta×change of the forward rate+gamma×(change of the f orward rate)2 /2 = $408+$3 = $411
7- Vega
a- The vega ν, which is the first derivative of the caplet price with respect to the volatility
55
parameter σ, is given by
p
0
υ = N × δ × B(t, T1 ) T0 − tF (t, T0 , T1 )Φ (d1 ) = 15, 794
b- If the volatility goes from 15% to 16%, the caplet price becomes $9,429.
c- The price difference is equal to $162
Pr ice Difference = $9, 429 − $9, 267 = $162
very near from the quantity ”vega×change of volatility” equal to $157.94
delta × change of rate = 15, 794 × 1% = $157.94
8- Rho
a- The rho ρ, which is the first derivative of the caplet price with respect to the interest rate
RC (t, T1 − t) , is given by
ρ = −(T1 − t) × Caplet = −5, 251
b- If the interest rate RC (t, T1 − t) goes from 5.25% to 5.35%, the caplet price becomes
$9,262.
c- The price difference is equal to -$7
Pr ice Difference = $9, 262 − $9, 267 = −$5
very near from the quantity ”rho×change of rate” equal to -$5.25
delta × change of rate = −5, 251 × 0.1% = −$5.25
9- Theta
a- The theta θ, which is the first derivative of the caplet price with respect to time, is given
56
by
µ
θ = N × δ × RC (t, T1 − t) ×
¶
Caplet B(t, T1 )σF (t, T0 , T1 ) 0
√
Φ (d1 ) = −19, 820
−
N ×δ
2 T0 − t
where r is the short term rate.
b- One day later on 05/14/02, the caplet price becomes $9,212.
c- The price difference is equal to -$55
Pr ice Difference = $9, 212 − $9, 267 = −$55
1
” equal to -$54.3
very near from the quantity ”theta× 365
theta ×
1
1
= −19, 820 ×
= −$54.3
365
365
Exercise 71 On 05/13/02 a firm buys a barrier caplet up and in whose features are the following:
· notional amount: $ 10,000,000
· starting date: 06/03/02
· reference rate: 3-month Libor
· barrier: 6%
· strike rate: 5%
· day-count: Actual/360
1- What is the pay-off of this option for the buyer ?
2- Draw the P&L of this caplet considering that the premium paid by the buyer is equal to
0.08% of the nominal amount.
3- What is the advantage and the drawback of this option compared to a classical caplet ?
Solution 72 1- The pay-off of this option for the buyer on 09/03/02 is
Pay-Off = $10, 000, 000 ×
³
where R 06/03/02, 14
´
·
µ
¶
¸
1
92
− 5% × 1R(06/03/02, 1 )≥6%
× Max 0; R 06/03/02,
4
360
4
is the 3-month Libor rate observed on 06/03/02, 92 is the number of
days between the 06/03/02 and the 09/03/02, and 1A = 1 if event A occurs and 0 otherwise.
2- The P&L is given by the following formula
P &L = Pay-Off − $10, 000, 000 × 0.08%
57
It appears in the following graph.
80000
70000
60000
50000
P&L in $
40000
30000
20000
10000
0
3,0%
-10000
3,5%
4,0%
4,5%
5,0%
5,5%
6,0%
6,5%
7,0%
7,5%
8,0%
-20000
Value of the 3-month Libor on 06/03/02
3- The advantage of this option compared to a classical caplet is that the buyer will pay a
lower premium. The drawback is that he will gain only if the reference rate is equal or above
the barrier.
58