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The Theory of Interest  Solutions Manual
Chapter 10
1.
(
a
) We have
(
29
(
29
(
29
(
29
(
29
1
2
3
4
5
1000 1.095
1.0925
1.0875
1.08
1.07
$3976.61.





+
+
+
+
=
(
b
) The present value is greater than in Example 10.1 (1), since the lower spot rates
apply over longer periods while the higher spot rates apply over shorter periods.
2.
We have
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
2
3
4
1
2
3
4
1
2
3
4
2
3
4
1000 1
1.05 1
1.05
1
1.05
1
1.05
1
1.05
1.05
1.05
1.05
1000 1
$4786.78.
1.09
1.081
1.0729
1.06561
s
s
s
s




+
+
+
+
+
+
+
+
=
+
+
+
+
=
3.
Since
k
s
is differentiable over
0
4,
k
≤
≤
.002 .001
0
at
2
k
d
s
k
k
dk
=

=
=
which is a relative maximum or minimum. Computing values for
0,1,2,3,4
k
=
we
obtain
0
1
2
3
4
.09
.0915
.092
.0915
.09.
s
s
s
s
s
=
=
=
=
=
(
a
) Normal.
(
b
) Inverted.
4.
117
Payment at
Spot rate
Accumulated value
0
t
=
.095
(
29
5
1.095
1.57424
=
1
t
=
.0925 − .0025
(
29
4
1.09
1.41158
=
2
t
=
.0875 − .0050
(
29
3
1.0825
1.26848
=
3
t
=
.0800 − .0075
(
29
2
1.0725
1.15026
=
4
t
=
.0700 − .0100
(
29
1
1.06
1.06000
=
6.4646
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Chapter 10
5.
Adapting Section 9.4 to fit this situation we have
(
29
(
29
(
29
4
2
2
10,000 1.0925
$11,946.50.
1.05
1.04
=
6.
(
a
)
(
29
2
2 930.49
1019.31 $841.67.
B
A
P
P

=

=
(
b
)
(
29
2
2 1000.00
1000.00
$1000.00.
B
A
C
C

=

=
(
c
) We have
2
.09
s
=
and
(
29
2
841.67 1.09
1000.00 confirming the statement.
=
7.
The price of the 6% bond per 100 is
(
29
6
.06
6 .12
6
100 1.12
75.33.
P
a

=
+
=
The price of the 10% bond per 100 is
(
29
6
.10
6 .08
10
100 1.08
109.25.
P
a

=
+
=
We can adapt the technique used above in Exercise 6. If we buy 10/6 of the 6% bonds,
the coupons will exactly match those of the 10% bond. The cost will be
(
29
(
29
10
10
75.33
125.55
and will mature for
100 .
6
6
=
Thus, we have
(
29
(
29
(
29
6
6
4
125.55 109.25 1
100
6
s

+
=
and solving
6
.2645,
s
=
or 26.45%.
8.
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This document was uploaded on 12/10/2011.
 Fall '09

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