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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