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無代寫-CHAPTER 8:
時間:2021-07-12
PRACTICE PROBLEMS
CHAPTER 8: YIELD CURVE AND TERM STRUCTURE ANALYSIS

NOTE: for this problem set assume all interest rates are compounded semi-annually.

Exercise 8.1: Assume the following spot rate (zero coupon) curve for this problem.
Years 0.5 1.0 1.5 2.0 2.5
Rate 7.50 7.75 8.00 8.00 8.00
(a) If all bonds are priced consistently with this curve, what would be the price of a two-
year semi-annual, 14% coupon security?
(b) Find the par bond yield curve.
(c) Use linear interpolation to find the spot rates with maturities 10 months and 14 months.
(d) Find the forward rate curve, f0(i,i + 2
1 ), for i = 0.5, 1, 1.5, 2. Using the spot rates in part
(c), find the forward rate f0(1, 14/12).

Exercise 8.2: The data in the next table reflects the conditions on October 8, 2013.
Coupon Maturity Price YTM
9% 9/15/2015 100 332 8.96%
10 58 % 8/15/2043 100 10
5
8 %
You expect the yield curve will flatten, but you have no clue as to whether the over-all
interest rates will rise or fall. Using the two securities suggest a “spread trade” that is
consistent with your expectations.

Exercise 8.3: Assume the following spot rate (zero coupon) curve for this problem.
Years 0.5 1.0 1.5 2.0 2.5
Rate 11.00 10.50 10.00 9.50 9.00
Two-year bonds that are strippable and paying a coupon of 10.375% are selling at a yield of
10%. Is it worthwhile to strip them and sell the stripped pieces? Or is it better to sell
them as a unit? (Assume that you own them and that the strips may be sold at spot
yields.)

Exercise 8.4: Given the following data, find the zero coupon curve by using the bootstrap
method. (Coupons are paid semi-annually.)
Coupon Maturity (years) Price (decimal)
7% 0.5 101.00
4% 1.0 98.82
8% 1.5 103.68

Exercise 8.5: You’re given the following data. (Coupons are paid semi-annually.)
Coupon Maturity (years) Price (decimal)
6% 1.0 99.72
8% 1.0 101.63
5% 1.5 98.03
If A =
?
?
?
?
?
?
?
?
?
?
5.1025.25.2
01044
01033
and (from EXCEL) A?1 =
?
?
?
?
?
?
?
?
?
?
?
?
?
009756.002439.002439.0
003.004.0
003.104.1
, find
the zero coupon curve.

Exercise 8.6: Given the following data, find the zero coupon curve by using the bootstrap
method. (Coupons are paid semi-annually.)
Coupon Maturity (years) Price (decimal)
2
12% 0.5 102.17
7% 1.0 99.34
9% 1.5 101.68

Exercise 8.7: You’re given the following data. (Coupons are paid semi-annually.)
Coupon Maturity (years) Price (decimal)
10% 1.0 105.22
5% 1.0 100.39
7% 1.5 103.30
If A =
5 105 0
2.5 102.5 0
3.5 3.5 103.5
? ?
? ?
? ?
? ?? ?
and (from EXCEL) A?1 =
0.41 0.42 0
0.01 0.02 0
0.01353 0.01353 0.009662
?? ?
? ??? ?
? ??? ?
,
find the zero coupon curve.

Exercise 8.8: You have liabilities of $50 million per year for the next 3 years (with payments
made at the end of the year.) The term structure is given by y1 = 8%, y2 = 9%, and y3 =
9.5%. You want to immunize these liabilities by using a 1-year zero coupon bond (face
value = $1 million) and a 3-year 5% bond (face value = $100,000 and coupon payments
are made annually). Assuming the bonds are correctly priced, show how to immunize
the liabilities. Show that the convexity of the bond portfolio is greater than the
convexity of the liabilities.

SOLUTIONS
Exercise 8.1: We assume the interest rates are compounded semi-annually here. First, let’s
find the zero coupon bond prices. We’re using b(0,t) = (1 +
2
ty )?2?t.
Years 0.5 1.0 1.5 2.0 2.5
Rate 7.50 7.75 8.00 8.00 8.00
b(0,t) 0.9639 0.9268 0.8890 0.8548 0.8219

(a) P =
2/1
2/
1y
C
?
+ 2
2 )2/1(
2/
y
C
?
+ ? + i
iy
C
)2/1(
2/100
?
?
=
2
075.01
7
?
+
2)
2
0775.01(
7
?
+
3)
2
080.01(
7
?
+
4)
2
080.01(
107
?

= 7(0.9639) + 7(0.9268) + 7(0.8890) + 107(0.8548)
= 110.9215.
(b) For each maturity, we want to find C so that
100 =
2
C [b(0,0.5) + b(0,1) + ... + b(0,t)] + 100b(0,t)
t = 0.5: C = 7.50;
t = 1:
2
C = 100[1 ? b(0,1)]/{b(0,0.5) + b(0,1)}
C = 200[1 ? 0.9268]/{0.9639 + 0.9268} = 7.745
3
t = 1.5: C = 200[1 ? b(0,1.5)]/{b(0,0.5) + b(0,1) + b(0,1.5)}; C = 7.987
t = 2: C = 200[1 ? b(0,2)]/{b(0,0.5) + b(0,1) + b(0,1.5) + b(0,2)}; C = 7.990
t = 2.5: C = 200[1 ? b(0,2.5)]/{b(0,0.5) + b(0,1) + b(0,1.5) + b(0,2) + b(0,2.5)}; C = 7.992.
(c) From p. 19 of Ch. 5 notes, y2 = y1 + [y3 ? y1]
13
12
TT
TT
?
? . For T2 = 10/12,
y10/12 = y0.5 + (y1 – y0.5)
)5.01(
12
6
12
10
?
?
= 7.50 – (7.75 – 7.50)
3
2 = 7.667%.
For T2 = 14/12,
y14/12 = y1 + (y1.5 – y1)
)15.1(
12
12
12
14
?
?
= 7.75 – (8.00 – 7.75)
3
1 = 7.833%.
(d) With semi-annual compounding, we have
(1 +
2
5.0y )(1 +
2
)1,5.0(0f ) = (1 +
2
1y )2; 1 +
2
)1,5.0(0f =
15.0
21
)
2
1(
)
2
1(
y
y
?
?
;
f0(0.5,1) = 2[ 1
2
)0375.1(
)03875.1( ? 1] = 8.00%.
(1 +
2
1y )2(1 +
2
)5.1,1(0f ) = (1 +
2
5.1y )3; 1 +
2
)5.1,1(0f =
21
35.1
)
2
1(
)
2
1(
y
y
?
?
;
f0(1,1.5) = 2[ 2
3
)03875.1(
)04.1( ? 1] = 8.50%.
(1 +
2
5.1y )3(1 +
2
)2,5.1(0f ) = (1 +
2
2y )4; 1 +
2
)2,5.1(0f =
35.1
42
)
2
1(
)
2
1(
y
y
?
?
;
f0(1.5,2) = 2[ 3
4
)04.1(
)04.1( ? 1] = 8.00%; f0(2,2.5) = 2[ 4
5
)04.1(
)04.1( ? 1] = 8.00%.
For the rate f0(1, 14/12), (1 +
2
1y )2
12/22
0
2
)12/14,1(
1
?
?
?
??
?
? ? f =
12/142
12/14
2
1
?
?
?
??
?
? ? y ;
3/1
0
2
)12/14,1(
1 ?
?
??
?
? ? f =
21
3/712/14
)
2
1(
)
2
1(
y
y
?
?
; f0(1,14/12) = 2[ 6
7
)03875.1(
)03917.1( ? 1] = 8.33%.
Exercise 8.2: We have a 2-year bond and a 30-year bond. Currently, the yield curve is
upward sloping. If the yield curve is to flatten, that means that y2 is expected to rise
4
and y30 is expected to fall (relatively speaking). If y2 ? then the bond price, P2 ?; if
y30?, then the 30-year bond price, P30 ?. So, sell the two-year bond, and buy the 30
year bond.

Exercise 8.3: Note that coupon payments are 10.375/2 = 5.1875.
Price calculated from the yield curve is
P1 =
2
11.01
5.1875
?
+
2)
2
1050.01(
5.1875
?
+
3)
2
10.01(
5.1875
?
+
4)
2
095.01(
5.187510
?
= 101.4483.
Price calculated from the YTM is
P2 =
05.1
5.1875 + 2)05.1(
5.1875 + 3)05.1(
5.1875 + 4)05.1(
5.187510 = 100.6649.
So, buy the coupon bond, strip the bond and sell the stripped pieces.

Exercise 8.4:
101.00 = 103.5b(0,0.5); b(0,0.5) =
5.103
00.101 = 0.9758
98.82 = 2b(0,0.5) + 102b(0,1); b(0, 1) =
)102(
)9758.0(282.98 ? = 0.9497;
103.68 = 4[b(0,0.5) + b(0,1)] + 104b(0,1.5);
b(0, 1.5) =
)104(
)9497.09758.0(468.103 ?? = 0.9229;
b(0, 0.5) = (1 +
y0 5
2
. )?1 = (0.9758); y0.5 = 2[(0.9758)?1 ? 1] = 4.95%;
b(0, 1) = (1 +
y1
2
)?2 = (0.9497); y1 = 2[(0.9497)?1/2 ? 1] = 5.23%;
b(0, 1.5) = (1 +
y1 5
2
. )?3 = (0.9229); y1 = 2[(0.9229)?1/3 ? 1] = 5.42%.

5
Exercise 8.5: Now, p = Ab, and
b =
?
?
?
?
?
?
?
?
?
?
)5.1,0(
)1,0(
)5.0,0(
b
b
b
= A?1p =
?
?
?
?
?
?
?
?
?
?
?
?
?
009756.002439.002439.0
003.004.0
003.104.1 99.72
101.63
98.03
? ?
? ?
? ?
? ?? ?
=
0.9701
0.9399
0.9098
? ?
? ?
? ?
? ?? ?
;
so,
b(0,0.5) = (1 +
y0 5
2
. )?1 = (0.9701); y0.5 = 2[(0.9701)?1 ? 1] = 6.16%;
b(0,1) = (1 +
y1
2
)?2 = (0.9399); y1 = 2[(0.9399)?1/2 ? 1] = 6.30%;
b(0,1.5) = (1 +
y1 5
2
. )?3 = (0.9098); y1.5 = 2[(0.9098)?1/3 ? 1] = 6.40%.
Exercise 8.6:
102.17 = 106b(0,0.5); b(0,0.5) = 102.17
106
= 0.9639
99.34 = 3.5b(0,0.5) + 103.5b(0,1); b(0, 1) = 99.34 3.5(0.9639)
(103.5)
? = 0.9272;
101.68 = 4.5[b(0,0.5) + b(0,1)] + 104.5b(0,1.5);
b(0, 1.5) = 101.68 4.5(0.9639 0.9272)
(104.5)
? ? = 0.8916;
(1 +
y0 5
2
. )?1 = (0.9639); y0.5 = 2[(0.9639)?1 ? 1] = 7.50%;
(1 +
y1
2
)?2 = (0.9272); y1 = 2[(0.9272)?1/2 ? 1] = 7.70%;
(1 +
y1 5
2
. )?3 = (0.8916); y1 = 2[(0.8916)?1/3 ? 1] = 7.80%.
Exercise 8.7: Now, p = Ab, and
b =
?
?
?
?
?
?
?
?
?
?
)5.1,0(
)1,0(
)5.0,0(
b
b
b
= A?1p =
0.41 0.42 0
0.01 0.02 0
0.01353 0.01353 0.009662
?? ?
? ??? ?
? ??? ?
105.22
100.39
103.30
? ?
? ?
? ?
? ?? ?
=
0.9764
0.9556
0.9327
? ?
? ?
? ?
? ?? ?
;
so,
b(0,0.5) = (1 +
y0 5
2
. )?1 = (0.9764); y0.5 = 2[(0.9764)?1 ? 1] = 4.83%;
b(0,1) = (1 +
y1
2
)?2 = (0.9556); y1 = 2[(0.9553)?1/2 ? 1] = 4.59%;
b(0,1.5) = (1 +
y1 5
2
. )?3 = (0.9327); y1.5 = 2[(0.9327)?1/3 ? 1] = 4.70 %.

6
Solution 8.8: Using the yields, the present values of the liabilities and assets are
Vp = 50( 2 3
1 1 1
1.08 (1.09) (1.095)
? ? ) = 126.463 (million);
P1 =
100
1.08
= 92.5926, or, $925,926 (with F = $1 million);
P2 = 2 3
5 5 105
1.08 (1.09) (1.095)
? ? = 88.8117, or $88,811.70 (with F = $100,000).
The Fisher-Weil durations are given by
DFW(p) =
1
pV
[ 2 3
1 50 2 50 3 50
1.08 (1.09) (1.095)
? ? ?? ? ] = 1.9351,
DFW(1) =
1
1
P
[ 1 100
1.08
? ] = 1.0,
DFW(2) =
2
1
P
[ 2 3
1 5 2 5 3 105
1.08 (1.09) (1.095)
? ? ?? ? ] = 2.8484.
The (Fisher-Weil-type) convexities are given by
Cx(p) = 1
pV
[ 1 2 2 2 3 2
1(1 1)50 2(2 1)50 3(3 1)50
(1.08) (1.09) (1.095)? ? ?
? ? ?? ? ] = 5.322,
Cx(1) =
1
1
P
[ 1 2
1(1 1)100
(1.08) ?
? ] = 1.714,
Cx(2) =
2
1
P
[ 1 2 2 2 3 2
1(1 1)5 2(2 1)5 3(3 1)105
(1.08) (1.09) (1.095)? ? ?
? ? ?? ? ] = 9.340.
To immunize our portfolio of liabilities, we need (ignoring the FW subscripts):
x1 + x2 = 1, (or x2 = 1 ? x1) and
x1D1 + x2D2 = Dp.
Solving gives x1 = 2
1 2
pD D
D D
?
?
= 1.9351 2.8484
1 2.8484
?
?
= 0.4941,
x2 = 1 ? 0.4941 = 0.5059.
Since xi = niPi/Vp, we have ni = xiVp/Pi, so,
n1 = (0.4941)(126.463M)/925,926 = 67.48 ? 67.
n2 = (0.5059)(126.463M)/88,811.7 = 720.4 ? 720.
Now, let’s check that our portfolio of assets has a higher convexity than our portfolio of
liabilities:
x1Cx(1) + x2Cx(2) = 0.4941?1.714+ 0.5059?9.340 = 5.572, which is indeed larger than Cx(p)
= 5.322.


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