﻿ 无代写-CHAPTER 7: |学霸联盟

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PRACTICE PROBLEMS
CHAPTER 7: BOND MATHEMATICS

A word about BEY (i.e., Bond Equivalent Yield).
T-bills use semi-annual compounding. So, for n <
365
2
, we use simple interest.
Principle + interest is given by P(1 + y?
n
365
) = 100 at maturity.
Solving for y gives
y ? BEY =
100? P
P
?
365
n
.
For n =
365
2
, we have Principal + Interest = P(1 +
y
2
)
Thereafter, i.e., for n >
365
2
, we use simple interest again:
P(1 +
y
2
)(1 + y?
n ?
365
2
365
) = 100,
which is equivalent to the text’s equation on p. 132.
We can use the quadratic equation to solve for y:
BEY =
? ? ?
?
? ? ? ?
?
2
365 365
2 2
365
100
2
365
2 1 1
1
n n n
P
n
( ) ( )( )
.

Exercise 7.1: An investor buys a face amount \$1 million of a six-month (182 days)
U.S. Treasury bill at a discount yield of 9.25%. What is the cost of purchasing
these bills? Calculate the bond equivalent yield. Indicate clearly the formula
you used and show all the steps in your calculations. Recalculate the bond
equivalent yield if the T-bill has a maturity of 275 days.

Exercise 7.2: On November 18, 2014, a 7 78 % U.S. T-bond maturing on May 15,
2017, was quoted for settlement on November 20, 2014. The last coupon was
paid on November 15, 2014. The number of days between coupon payments is
182. The bond’s YTM is 7.91%.
(a) What is the invoice price (cash price) of the T-bond?
(b) What is the accrued interest on the T-bond?
(c) What is the quoted price of the T-bond?
(d) What is the exact duration of the T-bond?
(e) What is the approximate convexity of the T-bond?
(f) Re-do part (a) using the next coupon date; show that the prices are the same.

Exercise 7.3: What is the price of a ten-year zero-coupon bond priced to yield 10%
under each of the following assumptions?
(a) Annual yield. (b) Semi-annual yield.
(c) Monthly yield. (d) Daily yield.
What is the continuous limit?

2
Exercise 7.4: In the table below, fill in the indicated blanks. Show all the steps in
your calculations. Duration is denoted by D, coupon by C, and yield by y
(compounded semi-annually). The settlement date is February 15, 2014. Now
suppose you have a liability of \$100 million that will be paid as a lump sum in
5.5 years. Using the 9% coupon bond and the 10% bond how can you hedge
this liability? When calculating the number of bonds you need to purchase,
assume that the coupon bonds have a face value of \$100,000.

C

y
Maturity
Date
Cash
Price
Exact
D
DV01
(per \$100)
9% 9.00% 5/15/xx 102.26 4.8 xxxx
0% 9.00% 2/15/2020 xxx.xx x.x xxxx
10% 9.00% 2/15/2024 106.50 6.66 xxxx

Exercise 7.5: Supply the missing information in the following table. The settlement
date is February 15, 2014. Using this information, how would you use these
two bonds to hedge a liability with a present value of \$100 million and a
Macaulay duration of 5.0? (Zero coupon bonds have a face value of
\$1,000,000, and coupon bonds have a face value of \$100,000.)

C

y
Maturity
Date
Cash
Price

D
0% 9.12% 2/15/2018 70.00 x.x
9% 7.33% 2/15/2022 110.00 6.0

Exercise 7.6: On November 15, 2014, you bought \$10 million (face amount) of a
7.50% T-bond maturing on November 15, 2044, at a yield of 7.60%. All
coupons were reinvested at 5% yield (with simple interest for this yield only).
The bond was sold on June 20, 2015, (36 days after the last coupon payment,
and there are 184 days between coupon payments at the time that the bond is
sold) at a yield of 7.50%. Calculate the annualized rate of return from this
transaction using simple interest. (Note: there are 181 days from 15/11/2014 to
the first coupon payment on 15/05/2015.)

Exercise 7.7: Assuming that the settlement date is August 15, 2014, determine the
duration of the T-bond from the next table. From the duration formula, estimate
the price of the T-bond for a yield change of ±10 basis points. Note that the last
coupon date (LCD) was May 15, 2014 (92 days before the settlement date) and
the next coupon date is Nov. 15, 2014 (92 days after the settlement date).

C

y
Maturity
Date
Quoted
Price

D
10% 9.00% 11/15/2026 107.3043 xx.xx

Exercise 7.8: Reconsider Exercise 7.7. You may now estimate the prices at different
yield levels by using both duration and convexity. Calculate the convexity,
(using the formula for a round number of coupons remaining) compare the
estimated prices with the actual price and explain the deviations.

3
SOLUTIONS
Exercise 7.1: We have Face value = F = \$1million, n = 182 days, discount yield = d
= 9.25%. The price is
P = F(1 ?
n d?
360
) = 1M(1 ?
182 0 0925
360
? .
) = \$953,236.11.
Since n < 182 days, BEY =
100 95 3236
953236
? .
.
?
365
182
= 9.8386%.
If n = 275 days, and the discount yield is again d = 9.25%, then the price is
P = 1M(1 ?
n d?
360
) = 1M(1 ?
275 0 0925
360
? .
) = \$929,340.28.
BEY =
? ? ?
?
? ? ? ?
?
2 275
365
275
365
2 2 275
365
100
92 934
2 275
365
2 1 1
1
( ) ( )( ). = 9.9258%.
Exercise 7.2:
(a) # days since last coupon = 5 = x ? z,
# days between coupon payments = 182 = x; coupon rate = 7.875% = c; y =
7.91%.
Now, as of the LCD there were 2 ? years remaining in the life of the bond; so, N = 5.
P = (1 +
y
2
)
x z
x
? C
y
C
y
y N
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
100
1
2
( )
= (1.03955
5
182) 5
7.8751007.875 0.0791
0.0791 (1.03955)
? ??? ?
?? ?
? ?
? ?? ?

= 100.02853.
(b) ai =
7 875
2
5
182
.
? = 0.108173.
(c) Cash price = quoted price + accrued interest, so
quoted price = Cash price – accrued interest = 100.02853 ? 0.108173
= 99.9204 (? 99 29
32
).
(d) For the duration, first we calculate the duration as of the last coupon date (LCD)
DLCD =
2
2
? y
y
?
2
2 1
2
1 2
? ? ?
? ? ?
y N c y
c
y
yN
( )
[( ) ]

=
)0791.0(2
0791.2 ?
)0791.0(2]1)
2
0791.01[()07875.0(2
)0791.007875.0(50791.2
5 ???
??
= 13.142 ? 10.824 = 2.318 years.
4
The exact duration is D = DLCD ? ?(
x z
x
?
) = 2.318 ? 0.5(
5
182
) = 2.30 years.
(e) Since the coupon rate is 7.875%, that means that C = 7.875. So, the
approximate convexity is
Cx =
?
?
2
2
P
y
1
P
,
where
?
?
2
2
P
y
=
N N
C
y
y N
( )( )
( )
? ?
? ?
1 100
4 1
2
2
?
CN
y
y N2 11
2
( )? ?
?
2
1
2
13
C
y
y N[( ) ]? ??
=
7)
2
0791.01(4
)
0791.0
875.7100)(6(5
?
?
?
62 )
2
0791.01()0791.0(
)5)(875.7(
?
? ]1)
2
0791.01[(
)0791.0(
)875.7(2 5
3 ??
?
= 2.530 – 4,986.494 – (?5,610.281) = 626.317
so, Cx = 626.317
100.02853
= 6.26.
(f) At the NCD we’ll receive a coupon of C/2 = 3.9375. Then, there will be 2 years
remaining in the life of the bond, so N = 4. Note that z = 182 – 5 = 177.
P = (1 +
y
2
)
z
x
? 100
2 (1 )
2
N
C
yC C
yy
? ??? ?
? ?? ?
? ??? ?? ?
= (1.03955
177
182)
?
4
7.8751007.875 0.07913.9375
0.0791 (1.03955)
? ??? ?
? ?? ?
? ?
? ?? ?

= 100.02853.
Exercise 7.3:
(a) Annual compounding: P =
100
11 10( . )
= 38.5543.
(b) Semi-annual compounding: P =
100
105 20( . )
= 37.6889.
(c) Monthly compounding: P =
100
1
010
12
120(
.
)?
= 36.9407.
(d) Daily compounding: P =
100
1
010
365
3650(
.
)?
= 36.7930.
Continuous compounding: P = 100e?(10?0.10) = 36.7879.
5
Note that the PV or price decreases. As the number of compounding periods
increases, the effective annual rate increases, so the PV decreases.
Exercise 7.4:
Bond 1: Since D = 4.8, we check years 2019, 2020, etc. By trial and error, using the
formula for round # of coupons, N remaining:
D =
2
2
? y
y
?
2
2 1
2
1 2
? ? ?
? ? ?
y N c y
c
y
yN
( )
[( ) ]
,
or the exact formula
D =
2
2
? y
y
?
2
2 1
2
1 2
? ? ?
? ? ?
y N c y
c
y
yN
( )
[( ) ]
?
1
2
(1 ?
z
x
),
where z = # of days to next coupon, and x = # of days between coupon payments.
Using 2019, the approximate formula gives (y = 9%, N = 11)
D =
2 0 09
2 0 09
? .
( . )
?
2 0 09 11 0 09 0 09
2 0 09 1
0 09
2
1 2 0 0911
? ? ?
? ? ?
. ( . . )
( . )[(
.
) ] ( . )

= 11.61 ? 7.15 = 4.45,
while the exact formula gives D = 4.46 ?
1
2
(1 ?
89
181
) = 4.20.
Using 2020, the formula gives (y = 9%, N = 13)
D =
2 0 09
2 0 09
? .
( . )
?
2 0 09 13 0 09 0 09
2 0 09 1
0 09
2
1 2 0 0913
? ? ?
? ? ?
. ( . . )
( . )[(
.
) ] ( . )
?
1
2
(1 ?
89
181
) = 4.8. (For 2021,
the approximate D = 5.6, the exact D = 5.4—definitely too large.)
Value of 0.01 = DV01(per 100% of par) = ?p?
1
100
, where
?p = ?
?
?
P
y
?
1
100
, and D = ?
?
?
P
y
?
( )1
2
?
y
P
. Combining these formulas gives DV01(per
100% of par) =
D P
y
?
?1
2
?
1
10 000,
;
Bond 1: DV01(per 100% of par) =
2
09.01
26.1028.4
?
? ?
1
10 000,
= 0.047.
Bond 2: For a zero coupon bond,
6
D = time to maturity = 6 years.
P =
100
1 0 045 12( . )?
= 58.97.
DV01(per 100% of par) =
6 59
1
0 09
2
?
?
. ?
1
10 000,
= 0.034.
Bond 3: DV01(per 100% of par) =
6 66 1065
1
0 09
2
. .
.
?
?
?
1
10 000,
= 0.0679.
To hedge the liability, we set
x1 + x2 = 1
and Dp = x1D1 + x2D2,
or 5.5 = x14.8 + x26.66,
which gives x1(4.8 – 6.66) = 5.5 – 6.66, so x1 =
8.466.6
5.566.6
?
? = 0.6237,
x2 = 0.3763. The present value of the liability is 25.5)045.01(
000,000,100
??
= \$61,619,874.
So, we invest 61,619,874?0.6237 = 38,429,599 in Bond 1, and 23,190,275 in Bond 2.
The number of bonds required are n1 = 38,429,599/102,260 = 376, and n2 =
23,190,275/106,500 = 218. [I stored x1 and x2 into my calculator’s memory, so
that I avoided rounding off until the final step.]

Exercise 7.5:
Bond 1: (zero coupon bond) D = 4 years.
Bond 2: All information is given.
For the \$100 million portfolio, we know that
Dp = x1D1 + x2D2,
where x1 = proportion of our wealth invested in Bond 1 (the zero coupon bond), and
x2 is the proportion invested in Bond 2. Here we require that
5 = x1(4) + x2(6), and 1 = x1 + x2.
Solving, we get x1(4 – 6) = 5 – 6, or x1 =
46
56
?
? = 0.50 = x2. So, we invest \$50
million in the zero coupon bond and \$50 million in the 9% bond to construct a
\$100 million portfolio with a duration of five years. The number of bonds we
need to buy are n1 = 50M/700,000 = 71, and n2 = 50M/110,000 = 455.
7
Exercise 7.6: Since we’re given the bonds’ yields, we can calculate the cash prices
directly using the PLCD (the price on the last coupon date) and the formula
P = (1 +
y
2
)
x z
x
?
?PLCD = (1 +
y
2
)
x z
x
?
?[
C
y
+
100
1
2
?
?
C
y
y N( )
].
We first calculate the purchase price of the bond. This bond had a round number of
periods remaining; note that at date t = 0, y = 7.6%, x = 181, and x – z = 0:
P0 = [
7 5
0 076
.
.
+
100
7 5
0 076
1038 60
?
.
.
( . )
] = 98.8246.
Price at sale date (y = 7.5%, x = 184 days, z = 148 days, x ? z = 36 days, N = 59)
P1 = (1.0375 )
36
184 ?[
7 5
0 075
.
.
+
100
7 5
0 075
10375 59
?
.
.
( )
] = 100.7229.
Coupon interest = 7.5/2 = 3.75; Interest on coupon = 3.75?(5%)?36/365 = 0.0185.
So, for an initial investment of 98.8246, we receive in 217 days (i.e., 181 + 36 days),
100.7229 + 3.75 + 0.0185 = 104.4914, for an annualized return of
104 4914 988246
988246
365
217
. .
.
?
? = 9.645%.
Exercise 7.7: N = 25.
# days since last coupon = 92,
# days between coupon payments = 184. Coupon rate = 10%. ai =
10
2
92
184
? = 2.5.
Cash price = quoted price + accrued interest
= 107.3043 + 2.5 = 109.8043.
D =
2
2
? y
y
?
2
2 1
2
1 2
? ? ?
? ? ?
y N c y
c
y
yN
( )
[( ) ]
?
1
2
(1 ?
z
x
)
=
2 09
2 0 09
.
( . )
?
2 09 25 010 0 09
2 010 1
0 09
2
1 2 0 0925
. ( . . )
( . )[(
.
) ] ( . )
? ?
? ? ?
?
1
2
(1 ?
92
184
),
= 11.61 ? 4.03 ? 0.25 = 7.3342.
Now,??P ?
?
?
P
y
?y = ?D?
P
y( / )1 2? ??y.
Since ?y = 10 basis points = 0.001,
8
? ? ?P ? ?7.3342?
109 8043
1045
.
.
?y = ?0.7706.
So, for a ten basis point increase, (y = 9.1%) our estimated price is
109.8043 ? 0.7706 = 109.0337,
while, for a ten basis point decrease, (y = 8.9%) the estimated price is
109.8043 + 0.7706 = 110.5749.
(The actual prices would be 109.0375, and 110.5788.)
Exercise 7.8: The convexity (for round number of periods) is Cx =
?
?
2
2
P
y
1
P
,

?
?
2
2
P
y
=
N N
C
y
y N
( )( )
( )
? ?
? ?
1 100
4 1
2
2
?
CN
y
y N2 11
2
( )? ?
?
2
1
2
13
C
y
y N[( ) ]? ?? ,
where C is the cash flow per year, in this case; y = rate per year (with semi-
annual compounding) and N = #of coupon payments remaining. (C = 10,
N = 25, y = 9%)
So,
?
?
2
2
P
y
= 27)045.1(4
)
09.0
10100)(26(25 ?
? 262 )045.1()09.0(
)25)(10( ? ]1)045.1[(
)09.0(
)10(2 25
3 ?
?
= ?550.137 – 9,827.237 + 18,306.431 = 7,929.056.
(Cx = 72.21)
So, for ?y = +0.001,
?P ?
?
?
P
y
??y +
1
2
?
?
2
2
P
y
??y2 = ?0.7706 +
1
2
?7,929.056?(0.001)2
= ?0.7706 + 0.00396 = ?0.7666.
For ?y = ?0.001, ?P = +0.7706 + 0.00396 = 0.7746.
Note that the convexity effect is small.
So, the estimates for P are 109.0377 (at y = 9.1%) and 110.5789 (at y = 8.9%).
(Again, the actual prices would be 109.0375, and 110.5788—very close.)

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