1
SOCIETY OF ACTUARIES
EXAM FM FINANCIAL MATHEMATICS
EXAM FM SAMPLE SOLUTIONS
This set of sample questions includes those published on the interest theory topic for use with
previous versions of this examination. Questions from previous versions of this document that
are not relevant for the syllabus effective with the October 2022 administration have been
deleted. The questions have been renumbered.
Some of the questions in this study note are taken from past SOA examinations.
These questions are representative of the types of questions that might be asked of candidates
sitting for the Financial Mathematics (FM) Exam. These questions are intended to represent the
depth of understanding required of candidates. The distribution of questions by topic is not
intended to represent the distribution of questions on future exams.
The following model solutions are presented for educational purposes. Alternative methods of
solution are acceptable.
In these solutions,
m
s
is the m-year spot rate and
mt
f
is the m-year forward rate, deferred t years.
Update history:
October 2022: Questions 208-275 were added
January 2023: Question 204 was deleted
June 2023 Questions 276-385 were added
Copyright 2023 by the Society of Actuaries.
2
1. Solution: C
Given the same principal invested for the same period of time yields the same accumulated
value, the two measures of interest
(2)
0.04i =
and
δ
must be equivalent, which means:
2
(2)
1
2
i
e
δ
+=
over a one-year period. Thus,
2
(2)
2
1 1.02 1.0404
2
ln(1.0404) 0.0396.
i
e
δ
δ
=+==
= =
2. Solution: E
From basic principles, the accumulated values after 20 and 40 years are
4 24
20 16 4
4
4 44
40 36 4
4
(1 ) (1 )
100[(1 ) (1 ) (1 ) ] 100
1 (1 )
(1 ) (1 )
100[(1 ) (1 ) (1 ) ] 100 .
1 (1 )
ii
ii i
i
ii
ii i
i
+ −+
+ ++ + ++ =
−+
+ −+
+ ++ + ++ =
−+
The ratio is 5, and thus (setting
4
(1 )xi= +
)
4 44 11
4 24 6
6 11
5 10
10 5
55
(1 ) (1 )
5
(1 ) (1 )
55
55 1
5 40
( 1)( 4) 0.
i i xx
i i xx
x x xx
xx
xx
xx
+ −+ −
= =
+ −+ −
−=−
−=−
− +=
− −=
Only the second root gives a positive solution. Thus
5
11
4
1.31951
1.31951 1.31951
100 6195.
1 1.31951
x
x
X
=
=
−
= =
−
3
Annuity symbols can also be used. Using the annual interest rate, the equation is
40 20
44
40 20
40 20
20
100 5(100)
(1 ) 1 (1 ) 1
5
(1 ) 5(1 ) 4 0
(1 ) 4
ss
aa
ii
ii
ii
i
=
+− +−
=
+−++=
+=
and the solution proceeds as above.
3. Solution: C
Eric’s (compound) interest in the last 6 months of the 8th year is
15
100 1
22
ii
+
.
Mike’s (simple) interest for the same period is
200
2
i
.
Thus,
15
15
100 1 200
22 2
12
2
1 1.047294
2
0.09459 9.46%.
ii i
i
i
i
+=
+=
+=
= =
4
4. Solution: C
( )
1
1
11
2
77.1
11
0.011025
0.85003 1
1.105 0.14997
ln(0.14997)
19.
ln(1.105)
n
n
n
n
n
nn
n
nn
n
n
n
nv
v Ia
i
a nv
nv
v
ii
a
nv nv
ii i
a
vv
ii
v
n
+
+
++
−
= +
−
= +
=−+
−−
= = =
= −
=
=−=
To obtain the present value without remembering the formula for an increasing annuity, consider
the payments as a perpetuity of 1 starting at time 2, a perpetuity of 1 starting at time 3, up to a
perpetuity of 1 starting at time n + 1. The present value one period before the start of each
perpetuity is 1/i. The total present value is
2
(1 / )( ) (1 / ) .
n
n
i v v v ia+++ =
5. Solution: C
The interest earned is a decreasing annuity of 6, 5.4, etc. Combined with the annual deposits of
100, the accumulated value in fund Y is
( )
( )
10 0.09 10 0.09
10
10 0.09
6( ) 100
10 1.09
6 100 15.19293
0.09
565.38 1519.29
2084.67.
Ds s
s
+
−
= +
= +
=
6. Solution: D
For the first 10 years, each payment equals 150% of interest due. The lender charges 10%,
therefore 5% of the principal outstanding will be used to reduce the principal.
At the end of 10 years, the amount outstanding is
( )
10
1000 1 0.05 598.74−=
.
Thus, the equation of value for the last 10 years using a comparison date of the end of year 10 is
10 10%
598.74 6.1446
97.44.
Xa X
X
= =
=
5
7. Solution: B
The book value at time 6 is the present value of future payments:
4
6
4 0.06
10,000 800 7920.94 2772.08 10,693.BV v a= + =+=
The interest portion is 10,693(0.06) = 641.58.
8. Solution: A
The value of the perpetuity after the fifth payment is 100/0.08 = 1250. The equation to solve is:
2 24 25
1250 ( 1.08 1.08 )
( ) (25) /1.08
50(1.08) 54.
Xv v v
Xv v v X
X
= + ++
= ++ + =
= =
9. Solution: C
Equation of value at end of 30 years:
40 40 30
40 30 40
1/40
10(1 / 4) (1.03) 20(1.03) 100
10(1 / 4) [100 20(1.03) ] /1.03 15.7738
1 / 4 1.57738 0.98867
4(1 0.98867) 0.0453 4.53%.
d
d
d
d
−
−
−
− +=
−=− =
−= =
=−==
10. Solution: E
The accumulation function is
23
0
( ) exp ( /100) exp( / 300).
t
a t s ds t
= =
∫
The accumulated value of 100 at time 3 is
3
100exp(3 / 300) 109.41743.=
The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6
minus the accumulated value at time 3. Thus
( )
109.41743 [ (6) / (3) 1]
(109.41743 )(2.0544332 /1.0941743 1)
(109.41743 )0.877613
96.026159 0.122387
784.61.
Xa a X
XX
XX
X
X
+ −=
+ −=
+=
=
=
6
11. Solution: A
5
5 9.2%
1
(1 )
167.50 10 10(1.092)
1.092
(1 ) /1.092
167.50 38.6955 6.44001
1 (1 ) /1.092
(167.50 38.6955)[1 (1 ) /1.092] 6.44001(1 ) /1.092
128.8045 135.24451(1 ) /1.092
1 1.0400
0.0400 4.0%.
t
t
k
a
k
k
kk
k
k
kK
∞
−
=
+
= +
+
= +
−+
− −+ = +
= +
+=
= ⇒=
∑
12. Solution: B
10 0.0807
Option 1: 2000
299 Total payments 2990
Option 2: Interest needs to be 2990 2000 990
990 [2000 1800 1600 200]
11,000
0.09 9.00%
Pa
P
i
i
i
=
=⇒=
−=
= ++++
=
= =
13. Solution: B
Monthly payment at time t is
1
1000(0.98)
t−
.
Because the loan amount is unknown, the outstanding balance must be calculated prospectively.
The value at time 40 months is the present value of payments from time 41 to time 60:
40 1 59 20
40
40 1 60 21
1000[0.98 0.98 ]
0.98 0.98
1000 , 1/ (
1.0075)
1 0.98
0.44238 0.25434
1000 6888.
1 0.97270
OB v v
vv
v
v
= ++
−
= =
−
−
= =
−
7
14. Solution: C
The equation of value is
( )
32
32
98 98 8000
(1 ) 1 (1 ) 1
81.63
12
81 41
81.63
10
81.63
12.25%
nn
nn
n
SS
ii
ii
i
ii
i
i
+=
+− +−
+=
+=
−−
+=
=
=
15. Solution: B
Convert 9% convertible quarterly to an effective rate of j per month:
3
0.09
(1 ) 1
4
j
+=+
or j = 0.00744.
Then
60
60 0.00744
60 0.00744
60
48.6136 38.4592
2( ) 2 2 2729.7.
0.00744 0.00744
av
Ia
−
−
= = =
16. Solution: A
Equating present values:
2 10
2 10
10
10
100 200 300 600
100 200(0.76) 300(0.76) 600
425.28 600
0.7088
0.96617
1.03501 1
0.035 3.5%.
nn
vv v
v
v
v
v
i
i
++ =
++ =
=
=
=
= +
= =
8
17. Solution: A
The accumulation function is:
( )
0
0
1
ln 8
8
8
() .
8
t
t
dr
r
r
t
at e e
+
+
+
∫
= = =
Using the equation of value at end of 10 years:
( )
10 10 10
00 0
(10) 18 / 8
20,000 8 (8 ) 18
( ) (
8 ) /
8
20,000
180 111.
180
a
k tk dt k t dt k dt
at t
kk
=+=+ =
+
= ⇒= =
∫∫∫
18. Solution: D
Let C be the redemption value and
1 / (1 )vi= +
. Then
2
2
2
2
1000
1
1000 381.50
1000(1.03125)(1 0.5889 ) 381.50
1055.11.
n
ni
n
X ra Cv
v
r
i
= +
−
= +
= −+
=
19. Solution: D
Equate net present values:
22
4000 2000 4000 2000 4000
4000 2000
6000
1.21 1.1
5460.
v v v
Xv
X
X
−++ =+−
+
= +
=
20. Solution: D
The present value of the perpetuity = X/i. Let B be the present value of Brian’s payments.
2
0.4
0.4
0.4 1 0.6
0.36 ,
n
nn
n
n
X
B Xa
i
a vv
i
X
Kv
i
X
K
i
= =
= ⇒ =−⇒=
=
=
Thus the charity’s share is 36% of the perpetuity’s present value.
9
21. Solution: D
The given information yields the following amounts of interest paid:
10
10 6%
0.12
Seth 5000 1 1 8954.24 5000 3954.24
2
Janice 5000(0.06)(10) 3000.00
5000
Lori (10) 5000 1793.40 where = 679.35
The sum is 8747.64.
PP
a
= + −= − =
= =
= −= =
22. Solution: E
For Bruce,
11 10 10
100[(1 ) (1 ) ] 100(1 i)X ii i= + −+ = +
. Similarly, for Robbie,
16
50(1 )X ii= +
.Dividing the second equation by the first gives
6
1 0.5(1 )i= +
which implies
1/6
2 1 0.122462i = −=
. Thus
10
100(1.122462) (0.122462) 38.879.X = =
23. Solution: D
Year t interest is
1
1
1
nt
nt i
ia v
−+
−+
= −
.
Year t+1 principal repaid is
1 (1 )
nt nt
vv
−−
−− =
.
1
1 1 (1 ) 1 .
nt nt nt nt
X v v v v vd
−+ − − −
=− + =+ −=+
24. Solution: B
For the first perpetuity,
36 3 3
33
3
32 10( ) 10 / (1 )
32 32 10
32 / 42.
vv v v
vv
v
= ++ = −
−=
=
For the second perpetuity,
1/3 2/3 1/3 1/3 1/9 1/9
/ (1 ) (32 / 42) / [1 (32 / 42) ] 32.599.Xv v v v= + += − = − =
25 Solution: D
Under either scenario, the company will have 822,703(0.05) = 41,135 to invest at the end of each
of the four years. Under Scenario A these payments will be invested at 4.5% and accumulate to
4 0.045
41,135 41,135(4.2782) 175,984.s = =
Adding the maturity value produces 998,687 for a loss
of 1,313. Note that only answer D has this value.
The Scenario B calculation is
4 0.055
41,135 41,135(4.3423) 178,621 822,703 1, 000,000 1,324.s = =+− =
10
26. Solution: D.
The present value is
2 2 20 20
21 21
5000[1.07 1.07 1.07 ]
1.07 1.07 1.01905 1.48622
5000 5000 122
,617.
1 1.07 1 1.01905
vv
v
vv
v
+ ++
−−
= = =
−
−
27. Solution: C.
The first cash flow of 60,000 at time 3 earns 2400 in interest for a time 4 receipt of 62,400.
Combined with the final payment, the investment returns 122,400 at time 4. The present value is
4
122,400(1.05) 100,699.
−
=
The net present value is 699.
28. Solution: B.
Using spot rates, the value of the bond is:
23
60 /1.07 60 /1.08 1060 /1.09 926.03.++ =
29. Solution: E.
Using spot rates, the value of the bond is:
23
60 /1.07 60 /1.08 1060 /1.09 926.03.++ =
The annual effective rate is the solution to
3
3
926.03 60 1000(1 )
i
ai
−
=++
. Using a calculator, the solution is 8.9%.
30. Solution: C.
Duration is the negative derivative of the price multiplied by one plus the interest rate and
divided by the price. Hence, the duration is –(–700)(1.08)/100 = 7.56.
31. Solution: C
The size of the dividend does not matter, so assume it is 1. Then the duration is
1
1
() /
1/ ( ) 1 1.1
11.
1/ 1/ 0.1
t
t
t
t
tv
Ia a i
di
a i id
v
∞
=
∞∞
∞
∞
=
= = = = = =
∑
∑
11
32. Solution: B
Duration =
11
11
1.02
() /
1
.
1/
1.02
t tt
t
jj
tt
t tt
j
t
tt
tv R tv
Ia a j
a jd
vR v
∞∞
∞∞
= =
∞∞
∞
= =
= = = =
∑∑
∑∑
The interest rate j is such that
1
(1 ) 1.02 1.02 /1.05 0.03 /1.02.jv j
−
+ = = ⇒=
Then the duration is
1/ (1 ) / (1.05 /1.02) / (0.03 /1.02) 1.05 / 0.03 35.d jj=+= = =
33. Solution: A
The outstanding balance is the present value of future payments. With only one future payment,
that payment must be 559.12(1.08) = 603.85. The amount borrowed is
4 0.08
603.85 2000.a =
The
first payment has 2000(0.08) = 160 in interest, thus the principal repaid is 603.85 – 160 =
443.85.
Alternatively, observe that the principal repaid in the final payment is the outstanding loan
balance at the previous payment, or 559.12. Principal repayments form a geometrically
decreasing sequence, so the principal repaid in the first payment is
3
559.12 /1.08 443.85.=
34. Solution: B
Because the yield rate equals the coupon rate, Bill paid 1000 for the bond. In return he receives
30 every six months, which accumulates to
20
30
j
s
where j is the semi-annual interest rate. The
equation of value is
10
20 20
1000(1.07) 30 1000 32.238.
jj
ss= + ⇒=
Using a calculator to solve for
the interest rate produces j = 0.0476 and so
2
1.0476 1 0.0975 9.75%.i = −= =
35. Solution: A
To receive 3000 per month at age 65 the fund must accumulate to 3,000(1,000/9.65) =
310,880.83. The equation of value is
300 0.08/12
310,880.83 957.36657 324.72.Xs X= = ⇒
36. Solution: D
(A) The left-hand side evaluates the deposits at age 0, while the right-hand side evaluates the
withdrawals at age 17.
(B) The left-hand side has 16 deposits, not 17.
(C) The left-hand side has 18 deposits, not 17.
(D) The left-hand side evaluates the deposits at age 18 and the right-hand side evaluates the
withdrawals at age 18.
(E) The left-hand side has 18 deposits, not 17 and 5 withdrawals, not 4.
12
37. Solution: D
Because only Bond II provides a cash flow at time 1, it must be considered first. The bond
provides 1025 at time 1 and thus 1000/1025 = 0.97561 units of this bond provides the required
cash. This bond then also provides 0.97561(25) = 24.39025 at time 0.5. Thus Bond I must
provide 1000 – 24.39025 = 975.60975 at time 0.5. The bond provides 1040 and thus
975.60975/1040 = 0.93809 units must be purchased.
38. Solution: C
Because only Mortgage II provides a cash flow at time two, it must be considered first. The
mortgage provides
2 0.07
/ 0.553092Ya Y=
at times one and two. Therefore, 0.553092Y = 1000
for Y = 1808.02. Mortgage I must provide 2000 – 1000 = 1000 at time one and thus X =
1000/1.06 = 943.40. The sum is 2751.42.
39. Solution: A
Bond I provides the cash flow at time one. Because 1000 is needed, one unit of the bond should
be purchased, at a cost of 1000/1.06 = 943.40.
Bond II must provide 2000 at time three. Therefore, the amount to be reinvested at time two is
2000/1.065 = 1877.93. The purchase price of the two-year bond is
2
1877.93 /1.07 1640.26=
.
The total price is 2583.66.
40. Solution: C
Given the coupon rate is greater than the yield rate, the bond sells at a premium. Thus, the
minimum yield rate for this callable bond is calculated based on a call at the earliest possible
date because that is most disadvantageous to the bond holder (earliest time at which a loss
occurs). Thus, X, the par value, which equals the redemption value because the bond is a par
value bond, must satisfy
Price =
30
0.03
30 0.03
1722.25 0.04 1.196 1440.Xa Xv X X= + = ⇒=
41. Solution: B
Because 40/1200 is greater than 0.03, for early redemption the earliest redemption should be
evaluated. If redeemed after 15 years, the price is
30
30 0.03
40 1200 /1.03 1278.40a +=
. If the bond
is redeemed at maturity, the price is
40
40 0.03
40 1100 /1.03 1261.80a +=
. The smallest value should
be selected, which is 1261.80. (When working with callable bonds, the maximum a buyer will
pay is the smallest price over the various call dates. Paying more may not earn the desired yield.)
13
42. Solution: E
Given the coupon rate is less than the yield rate, the bond sells at a discount. Thus, the minimum
yield rate for this callable bond is calculated based on a call at the latest possible date because
that is most disadvantageous to the bond holder (latest time at which a gain occurs). Thus, X, the
par value, which equals the redemption value because the bond is a par value bond, must satisfy
Price =
20
0.03
20 0.03
1021.50 0.02 0.851225 1200.Xa Xv X X= + = ⇒=
43. Solution: B
Given the price is less than the amount paid for an early call, the minimum yield rate for this
callable bond is calculated based on a call at the latest possible date. Thus, for an early call, the
effective yield rate per coupon period, j, must satisfy Price =
19
19
1021.50 22 1200
j
j
av= +
. Using
the calculator, j = 2.86%. We also must check the yield if the bond is redeemed at maturity. The
equation is
20
20
1021.50 22 1100
j
j
av= +
. The solution is j = 2.46% Thus, the yield, expressed as a
nominal annual rate of interest convertible semiannually, is twice the smaller of the two values,
or 4.92%.
44. Solution: C
First, the present value of the liability is
15 6.2%
35,000 335,530.30.PV a= =
The duration of the liability is:
2 15
35,000 2(35,000) 15(35,000) 2,312,521.95
6.89214.
335,530.30 335,530.30
t
t
t
t
tv R
vv v
d
vR
+ ++
= = = =
∑
∑
Let
X
denote the amount invested in the 5 year bond.
Then,
(5) 1 (10) 6.89214 208,556.
335,530.30 335,530.30
XX
X
+− = => =
45 . Solution: A
The present value of the first eight payments is:
89
2 78
2000 2000(1.03)
2000 2000(1.03) ... 2000(1.03) 13,136.41.
1 1.03
vv
PV v v v
v
−
= + ++ = =
−
The present value of the last eight payments is:
7 9 7 2 10 7 8 16
7 9 7 9 17
2000(1.03) 0.97 2000(1.03) (0.97) 2000(1.03) (0.97 )
2000(1.03) 0.97 2000(1.03) (0.97)
7,552.22.
1 0.97
PV v v v
vv
v
= + +
+
−
= =
−
Therefore, the total loan amount is L = 20,688.63.
14
46. Solution: E
1
2
2
3
0
2
3
3
0
0
33
3
100
2000 500exp
3
150
50
4 exp 0.5 exp 0.5ln 3
150
3
150
4 exp 0.5ln 1 1
450 450
16 1
450
18.8988
t
t
t
r
dr
r
r
r
dr
r
tt
t
t
=
+
= = +
+
= +=+
= +
=
∫
∫
47. Solution: E
Let F, C, r, and i have their usual interpretations. The discount is
()
n
Ci Fr a−
and the discount in
the coupon at time t is
1
()
nt
Ci Fr v
−+
−
. Then,
26
21
5
26
40 0.095
194.82 ( )
306.69 ( )
0.63523 0.91324 0.095
( ) 194.82(1.095) 2062.53
Discount 2062.53 21,135
Ci Fr v
Ci Fr v
vv i
Ci Fr
a
= −
= −
= ⇒ = ⇒=
−= =
= =
48. Solution: A
851
1
4
1
699.68
842.39 (annual payment)
699.68
581.14
1.0475
842.39 581.14 261.25
261.25
5500 (loan amount)
0.0475
Total interest = 842.39(8) 5500 1239.12
Pv
P
P
I
L
−+
=
=
= =
=−=
= =
−=
15
49. Solution: D
18
18
18 0.007
24 0.004
22,000(1.007) 450.30 16,337.10
16,337.10
715.27
OB s
Pa
P
= −=
=
=
50. Solution: C
If the bond has no premium or discount, it was bought at par so the yield rate equals the coupon
rate, 0.038.
( )
( )
2 14 14
2 14 14
14
14
14
14
1
1(190) 2(190) 14(190) 14(5000)
2
190 190 190 5000
95 7(5000)
190 5000
5.5554
vv v v
d
vv v v
Ia v
d
av
d
+ ++ +
=
+ ++ +
+
=
+
=
Or, taking advantage of a shortcut:
14 0.038
11.1107.da= =
This is in half years, so dividing by two,
11.1107
5.5554
2
d = =
.
51. Solution: A
[ ]
[ ]
7.959
7.425
1.072
(0.08) (0.072) 1 ( )
(0.08) 1000 1 (0.008)(7.425) 940.60
v
P P iv
P
= =
= −∆
=−=
52. Solution: E
32
3 2 12
3
3
3
2
2
2
32
12
12
(1 ) (1 ) (1 )
1
0.85892 , 0.052
(1 s )
1
0.90703 , 0.050
(1 )
1.052 1.050 (1 )
0.056
s sf
s
s
s
f
f
+=+ +
= =
+
= =
+
= +
=
16
53. Solution: C
Let
0
d
be the Macaulay duration at time 0.
0
8 0.05
10
2
7 0.05
1
2
6.7864
1 5.7864
6.0757
5.7864
0.9524
6.0757
da
dd
da
d
d
= =
= −=
= =
= =
This solution employs the fact that when a coupon bond sells at par the duration equals the
present value of an annuity-due. For the duration just before the first coupon the cash flows are
the same as for the original bond, but all occur one year sooner. Hence the duration is one year
less.
Alternatively, note that the numerators for
1
d
and
2
d
are identical. That is because they differ
only with respect to the coupon at time 1 (which is time 0 for this calculation) and so the
payment does not add anything. The denominator for
2
d
is the present value of the same bond,
but with 7 years, which is 5000. The denominator for
1
d
has the extra coupon of 250 and so is
5250. The desired ratio is then 5000/5250 = 0.9524.
54. Solution: A
Let N be the number of shares bought of the bond as indicated by the subscript.
(105) 100, 0.9524
(100) 102 0.9524(5), 0.9724
(107) 99 0.9524(5), 0.8807
CC
BB
AA
NN
NN
NN
= =
=−=
=−=
55. Solution: B
All are true except B. Immunization requires frequent rebalancing.
17
56. Solution: D
Set up the following two equations in the two unknowns:
22
13
(1.05) (1.05) 6000
2 (1.05) 2 (1.05) 0.
AB
AB
−
−
+=
−=
Solving simultaneously gives:
2721.09
3307.50
586.41.
A
B
AB
=
=
−=
57. Solution: A
Set up the following two equations in the two unknowns.
3
3
(1) 5000(1.03) (1.03) 12,000
5463.635 (1.03) 12,000 (1.03) 6536.365
(2) 3(5000)(1.03) (1.03) 0 16,390.905 6536.3650
2.5076
7039.27
2807.12
b
bb
b
B
BB
bB b
b
B
B
b
−
−−
−
+=⇒
+=⇒=
− =⇒ −=
=
=
=
58. Solution: D
29
5
3 10
6
(1 ) (1 )
95,000(1 )
2 (1 ) 9 (1 )
5(95, 000)(1 )
A
L
A
L
PAi Bi
Pi
P Ai Bi
Pi
−−
−
−−
−
=+ ++
= +
′
=− +− +
′
=−+
Set the present values and derivatives equal and solve simultaneously.
0.92456 0.70259 78, 083
1.7780 6.0801 375, 400
78,083(1.7780 / 0.92456) 375,400
47,630
0.70259(1.7780 / 0.92456) 6.0801
[78,083 0.70259(47, 630)] / 0.92456 48,259
1.0132
AB
AB
B
A
A
B
+=
−−=−
−
= =
−
=−=
=
18
59. Solution: D
Throughout the solution, let j = i/2.
For bond A, the coupon rate is (i + 0.04)/2 = j + 0.02.
For bond B, the coupon rate is (i – 0.04)/2 = j – 0.02.
The price of bond A is
20
20
10,000( 0.02) 10,000(1 )
A
j
P ja j
−
= + ++
.
The price of bond B is
20
20
10,000( 0.02) 10,000(1 )
B
j
P ja j
−
= − ++
.
Thus,
20 20
20
5,341.12 [200 ( 200)] 400
5,341.12 / 400 13.3528.
AB
jj
j
PP a a
a
− = = −− =
= =
Using the financial calculator, j = 0.042 and i =2(0.042)=0.084.
60. Solution: D
The initial level monthly payment is
15 12 0.09/12 180 0.0075
400,000 400,000
4,057.07.R
aa
×
= = =
The outstanding loan balance after the 36th
payment is
36
180 36 0.0075 144 0.0075
4,057.07 4,057.07(87.8711) 356,499.17.B Ra a
−
= = = =
The revised payment is 4,057.07 – 409.88 = 3,647.19.
Thus,
144 /12
144 /12
356,499.17 3,647.19
356,499.17 / 3,647.19 97.7463.
j
j
a
a
=
= =
Using the financial calculator, j/12 = 0.575%, for j = 6.9%.
61. Solution: D
The price of the first bond is
30 2 60
30 2 0.05/2 60 0.025
1000(0.05 / 2) 1200(1 0.05 / 2) 25 1200(1.025)
772.72 272.74 1,045.46.
aa
−× −
×
++ = +
=+=
The price of the second bond is also 1,045.46. The equation to solve is
60
60 /2
1,045.46 25 800(1 / 2) .
j
aj
−
= ++
The financial calculator can be used to solve for j/2 = 2.2% for j = 4.4%.
19
62. Solution: E
Let n = years. The equation to solve is
2 12
1000(1.03) 2(1000)(1.0025)
2 ln1.03 ln1000 12 ln1.0025 ln 2000
0.029155 0.69315
23.775.
nn
nn
n
n
=
+= +
=
=
This is 285.3 months. The next interest payment to Lucas is at a multiple of 6, which is 288
months.
63. Solution: A
Equating the accumulated values after 4 years provides an equation in K.
4
4
0
4
4
0
0
2
1
10 1 10 exp
25 0.25
11
4ln(1 0.04 ) 4 ln(K 0.25
t) 4 ln( 1) 4ln( ) 4 ln
0.25
1
1 0.04
0.04 1
5.
K
dt
Kt
K
K dt K
K
Kt K
K
K
K
K
K
+=
+
+
+ = =
+ = +− =
+
+
+=
=
=
∫
∫
Therefore,
4
10(1 5 / 25) 20.74.X =+=
64. Solution: D
The outstanding balance at time 25 is
25
25
25
100( ) 100 .
a
Da
i
−
=
The principle repaid in the 26th
payment is
25
25 25
25
2500 (100) 2500 2500 100 100 .
a
X i aa
i
−
=− =−+ =
The amount borrowed is
the present value of all 50 payments,
25
25 25
2500 100( ) .a v Da+
Interest paid in the first payment
is then
25
25 25
25 25
25
25 25 25
25
25
2500 100( )
2500(1 ) 100 (25 )
2500 2500 2500 100
2500 .
i a v Da
v va
v vv a
Xv
+
= −+ −
=−+−
= −
20
65. Solution: C
The accumulated value is
20 0.0816
1000 50,382.16.s =
This must provide a semi-annual annuity-
due of 3000. Let n be the number of payments. Then solve
0.04
3000 50,382.16
n
a =
for n = 26.47.
Therefore, there will be 26 full payments plus one final, smaller, payment. The equation is
26
26 0.04
50,382.16 3000 (1.04)aX
−
= +
with solution X = 1430. Note that the while the final
payment is the 27th payment, because this is an annuity-due, it takes place 26 periods after the
annuity begins.
66. Solution: D
For the first perpetuity,
( )
( )
2
2
1
1 7.21
11
1
11
6.21
0.0775.
i
i
i
+=
+−
=+−
=
For the second perpetuity,
( )
( )
1
3
1
1 (1.0875) 7.21
1.0775 0.01 1
1.286139 7.21(1.0875) 0.286139
1.74.
R
R
R
−
+=
+−
=
=
67. Solution: E
5
55
5
5 15 15
5
10,000 100( ) 100
0.05
10,000 1256.64 8.13273
1075
av
Ia Xv a Xv
a
X
X
−
= += +
= +
=
68. Solution: C
5
10 0.06
5000 (1.05)
5000
297.22
13.1808(1.2763)
Xs
X
=
= =
21
69. Solution: E
The monthly payment on the original loan is
180 8/12%
65,000
621.17
a
=
. After 12 payments the
outstanding balance is
168 8/12%
621.17 62,661.40a =
. The revised payment is
168 6/12%
62,661.40
552.19.
a
=
70. Solution: E
At the time of the final deposit the fund has
18 0.07
750 25, 499.27.s =
This is an immediate annuity
because the evaluation is done at the time the last payments is made (which is the end of the final
year). A tuition payment of
17
6000(1.05) 13,752.11=
is made, leaving 11,747.16. It earns 7%, so
a year later the fund has 11,747.16(1.07) = 12,569.46. Tuition has grown to 13,752.11(1.05) =
14,439.72. The amount needed is 14,439.72 – 12,569.46 = 1,870.26
71. Solution: B
The coupons are 1000(0.09)/2 = 45. The present value of the coupons and redemption value at
5% per semiannual period is
40
40 0.05
45 1200(1.05) 942.61.Pa
−
=+=
72. Solution: A
For a bond bought at discount, the minimum price will occur at the latest possible redemption
date.
20
20 0.06
50 1000(1.06) 885.30.Pa
−
=+=
(When working with callable bonds, the maximum a
buyer will pay is the smallest price over the various call dates. Paying more may not earn the
desired yield.)
73. Solution: C
5
4
1.095
1 11.5%
1.090
−=
22
74. Solution: D
The accumulated value of the first year of payments is
12 0.005
2000 24,671.12.s =
This amount
increases at 2% per year. The effective annual interest rate is
12
1.005 1 0.061678.−=
The present
value is then
25 25
1
11
26
1 1.02
24,671.12 1.02 (1.061678) 24,671.12
1.02 1.061678
0.960743 0.960743
24,187.37 374, 444.
1 0.960743
k
kk
kk
P
−−
= =
= =
−
= =
−
∑∑
This is 56 less than the lump sum amount.
75. Solution: A
The monthly interest rate is 0.072/12 = 0.006. 6500 five years from today has value
60
6500(1.006) 4539.77
−
=
. The equation of value is
2
4539.77 1700(1.006) 3400(1.006) .
nn−−
= +
Let
1.006
n
x
−
=
. Then, solve the quadratic equation
2
2
3400 1700 4539.77 0
1700 1700 4(3400)( 4539.77)
0.93225.
2(3400)
xx
x
+− =
−+ − −
= =
Then,
1.006 0.9325 ln(1.006) ln(0.93225) 11.73.
n
nn
−
= ⇒− = ⇒ =
To ensure there is 6500 in five years, the deposits must be made earlier and thus the maximum
integral value is 11.
76. Solution: C
( )
( )
( )
( )
4
4
4
2
2
12
39 1 2 38
39 39( 2) 38 38( 4)
38 1 4 39
1 /4
39 2 38 4 39 38
1/ (19.5 9.5) 0.1
1 1 / 2 .95 1.108 10.8%.
d
d
dd
d
d
d
d
id i
−
−
−
−
−
−
= ⇒ = ⇒− =−
−
−
−=−
= −=
+= − = = ⇒=
23
77. Solution: C
The monthly interest rate is 0.042/12 = 0.0035. The quarterly interest rate is
3
1.0035 1 0.0105−=
. The investor makes 41 quarterly deposits and the ending date is 124 months from the start.
Using January 1 of year y as the comparison date produces the following equation:
41
124
1
100 1.9
1.0105 1.0035
k
k
X
X
=
+=
∑
Substituting
3
1.0105 1.0035=
gives answer (C).
78. Solution: D
Convert the two annual rates, 4% and 5%, to two-year rates as
2
1.04 1 0.0816−=
and
2
1,05 1 0.1025−=
.
The accumulated value is
4
3 0.0816 2 0.1025
100 (1.05) 100 100(3.51678)(1.21551) 100(2.31801) 659.269ss+= + =
.
With only five payments, an alternative approach is to accumulate each one to time ten and add
them up.
The two-year yield rate is the solution to
5
100 659.269
i
s =
. Using the calculator, the two-year
rate is 0.093637. The annual rate is
0.5
1.093637 1 0.04577−=
which is 4.58%.
79. Solution: C
( )
1 12
15
4 8% 216 0.6434%
1.08 1 0.006434
1
25,000
1.08
25,000(3.57710)
240.38
3.17217(117.2790)
ä Xä
X
−=
=
= =
24
80. Solution: B
perp.
10
15 0.08
10
10 0.10
10
11
1
0.08 0.1
PV (15,000) 15,000
0.1 1.1
164,457.87 15,000 179, 457.87
179,458
1.10
9.244
6.759 179,458
1.10
17,384
a
Xa
X
X
−
=++
= +=
+=
+=
=
81. Solution: A
14
14
14 0.03
14 0.03
14(1.03)
1050.50 (22.50 ) 300(
1.03)
0.03
a
Xa X
−
−
−
=++ +
11.2961 9.25565
1050.50 (22.50 )11.2961 198.335 79.3111 598 7.54
0.03
X X XX
−
= + + + => = =>=
82. Solution: D
The amount of the loan is the present value of the deferred increasing annuity:
30
10 10
30 0.05
30 0.05 30 0.05 30 0.05
30(1.05)
(1.05) 500 500( ) (1.05 )(500) 64,257.
0.05 /1.05
a
a Ia a
−
−−
−
+= + =
83. Solution: C
30 30 30 30
30
29
29
1/29
(1 ) (1.03) (1 ) (1.03)
50,000 (
1 ) 5,000
(1 ) ( 0.03) 0.03
50,000 / (1 ) 5,000
(1 ) 10
10 1 0.082637
ii
i
ii i
i
i
i
+− +−
+=
+− −
+=
+=
= −=
The accumulated amount is
30 30
30
(1.082637) (1.03)
50,000 (1.082637) 797,836.82
(1.082637) (0.082637 0.03)
−
=
−
25
84. Solution: D
The first payment is 2,000, and the second payment of 2,010 is 1.005 times the first payment.
Since we are given that the series of quarterly payments is geometric, the payments multiply by
1.005 every quarter.
Based on the quarterly interest rate, the equation of value is
22 33
2,000
100,000 2,000 2,000(1.005) 2,000(1.005) 2,000(1.005)
1 1.005
1 1.005 2,000 /100,000 0.98 /1.005.
vv v
v
vv
= + + + +=
−
− = ⇒=
.
The annual effective rate is
( )
4
4
1 0.98 /1.005 1 0.10601 10.6%v
−
−
−= −= =
.
85. Solution: A
Present value for the first 10 years is
( )
( )
10
1 1.06
7.58
ln 1.06
−
−
=
Present value of the payments after 10 years is
( ) ( ) ( )
( ) ( )
10
0
0.5584
1.06 1.03 1.06 19.45
ln 1.06 ln 1.03
ss
ds
∞
−−
= =
−
∫
Total present value = 27.03
86. Solution: C
( ) ( )
( )
10
5
1
52
1
10,000 1.06 1.06 75,000
11
13,382.26 1.1236 75, 000
6
1.1236 27,526.83
24,498.78
dt
t
Xe
X
X
X
+
∫
+=
+=
=
=
87. Solution: D
The effective annual interest rate is
11
(1 ) 1 (1 0.055) 1 5.82%id
−−
= − −= − −=
The balance on the loan at time 2 is
2
15,000,000(1.0582) 16,796,809.=
The number of payments is given by
|
1, 200,000 16,796,809
n
a =
which gives n = 29.795 => 29
payments of 1,200,000. The final equation of value is
30
29
1, 200,000 (1.0582) 16,796,809
(16,796,809 16,621,012)(5.45799) 959, 490.
aX
X
−
+=
=−=
26
88. Solution: C
2 4 22
2
1 0.525(1 ) 1 0.525(1 ) 0.90476 0.95119
1 0.1427(1 ) 1 v (1 0.90476) / 0.1427 0.667414 0.332596
ln(0.332596) / ln(0.95119) 22
nn n
v v vv v
vv v
n
− = − ⇒= + ⇒ = ⇒ =
− = − ⇒− = − = ⇒ =
= =
89. Solution: C
The monthly payment is
360 0.005
200,000 / 1199.10a =
. Using the equivalent annual effective rate
of 6.17%, the present value (at time 0) of the five extra payments is 41,929.54 which reduces the
original loan amount to 200,000 – 41,929.54 = 158,070.46. The number of months required is
the solution to
0.005
158,070.46 1199.10
n
a=
. Using calculator, n = 215.78 months are needed to
pay off this amount. So there are 215 full payments plus one fractional payment at the end of the
216th month, which is December 31, 2020.
90. Solution: D
The annual effective interest rate is 0.08/(1 – 0.08) = 0.08696. The level payments are
5 0.08696
500,000 / 500,000 / 3.9205 127,535.a = =
This rounds up to 128,000. The equation of
value for X is
5
4 0.08696
128,000 (1.08696) 500,000
(500,000 417, 466.36)(1.51729) 125,227.
aX
X
−
+=
=−=
91. Solution: B
The accumulated value is the reciprocal of the price. The equation is
X[(1/0.94)+(1/0.95)+(1/0.96)+(1/0.97)+(1/0.98)+(1/0.99)] = 100,000.
X= 16,078
27
92. Solution: D
Let P be the annual payment. The fifth line is obtained by solving a quadratic equation.
10
10 6 1
10
5
10 5
5
10
1/10
10
(1 ) 3600
4871
1 3600
4871
1 0.739068
0.69656
0.485195
0.485195 1 0.075
1 3600
48,000
0.075
Pv
Pv
v
v
vv
v
v
i
v
XP
i
−+
−
−=
=
−
=
−=
=
=
= −=
−
= = =
93. Solution: A
Let j = periodic yield rate, r = periodic coupon rate, F = redemption (face) value, P = price, n =
number of time periods, and
1
1
j
v
j
=
+
. In this problem,
( )
1
2
1.0705 1 0.03465j = −=
, r = 0.035, P
= 10,000, and n = 50.
The present value equation for a bond is
n
j
nj
P Fv Fra= +
; solving for the redemption value F
yields
50
50 0.03465
10,000 10,000
9,918.
(1.03465) 0.035 0.18211 0.035(23.6044)
n
j
nj
P
F
v ra a
−
= = = =
++ +
.
94. Solution: B
Jeff’s monthly cash flows are coupons of 10,000(0.09)/12 = 75 less loan payments of
2000(0.08)/12 = 13.33 for a net income of 61.67. At the end of the ten years (in addition to the
61.67) he receives 10,000 for the bond less a 2,000 loan repayment. The equation is
(12 )
(12) 120
120 /12
(12)
12
8000 61.67 8000(1 /12)
/12 0.00770875
1.00770875 1 0.0965 9.65%.
i
ai
i
i
−
= ++
=
= −= =
28
95. Solution: B
The present value equation for a par-valued annual coupon bond is
n
i
ni
P Fv Fra= +
; solving for
the coupon rate r yields
1
nn
ii
ni ni ni
P Fv v
P
r
Fa a F a
−
= = −
.
All three bonds have the same values except for F. We can write r = x(1/F) + y. From the first
two bonds:
0.0528 = x/1000 + y and 0.0440 = x/1100 + y. Then,
0.0528 – 0.044 = x(1/1000 – 1/1100) for x = 96.8 and y = 0.0528 – 96.8/1000 = –0.044. For the
third bond, r = 96.8/1320 – 0.044 = 0.2933 = 2.93%.
96. Solution: A
The effective semi-annual yield rate is
2
(2) (2)
1.04 1 1.9804%
22
ii
=+ =>=
. Then,
2 12 12
13
12
582.53 (1.02) (1.02 ) (1.02 ) 250
1.02 (1.02 )
250 12.015 197.579 32.04.
1 1.02
c vcv cv v
vv
c vc c
v
= + ++ +
−
= + = + =>=
−
13
12
1.02 (1.02 )
582.53 250 12.015 197.579 32.04
1 1.02
vv
c
vc c
v
−
= + = + =>=
−
97. Solution: E
Book values are linked by BV3(1 + i) – Fr = BV4. Thus 1254.87(1.06) – Fr = 1277.38.
Therefore, the coupon is Fr = 52.7822. The prospective formula for the book value at time 3 is
( 3)
( 3)
( 3)
1 1.06
1254.87 52.7822 1890(1.06)
0.06
375.1667 1010.297(1.06)
l
n(375.1667 / 1010.297)
3 17.
ln(1.06)
n
n
n
n
−−
−−
−−
−
= +
=
−= =
−
Thus, n = 20. Note that the financial calculator can be used to solve for n – 3.
29
98. Solution: A
Book values are linked by BV3(1 + i) – Fr = BV4. Thus BV3(1.04) – 2500(0.035) = BV3 +
8.44. Therefore, BV3 = [2500(0.035) + 8.44]/0.04 = 2398.5. The prospective formula for the
book value at time 3 is, where m is the number of six-month periods.
( 3)
( 3)
( 3)
1 1.04
2398.5 2500(0.035) 2500(1.04)
0.04
211 312.5
(1.04)
ln(211/ 312.5)
3 10.
ln(1.04)
m
m
m
m
−−
−−
−−
−
= +
=
−= =
−
Thus, m = 13 and n = m/2 = 6.5. Note that the financial calculator can be used to solve for m – 3.
99. Solution: C
( )
( )
( )
( )
1 10
2
2
11 2
1
3
3
2 1/3
12 3
2
2
0.04
1
0.06 1 (
1.06)(1.04) 1 0.04995
1
1
0.08 1
[(1.08)(1.04995) ] 1 0.05987 6%.
1
sf
s
fs
s
s
fs
s
= =
+
= = − ⇒ = −=
+
+
= = − ⇒ = −= =
+
100. Solution: B
The Macaulay duration of Annuity A is
22
22
0(1) 1( ) 2( ) 2
0.93
11
v v vv
vv vv
++ +
= =
++ ++
, which leads to the
quadratic equation
2
1.07 0.07 0.93 0vv+ −=
. The unique positive solution is v = 0.9.
The Macaulay duration of Annuity B is
23
23
0(1) 1( ) 2( ) 3( )
1.369.
1
vv v
vv v
++ +
=
++ +
101. Solution: D
With v =1/1.07,
23 4
23 4
2(40,000) 3(25,000) 4(100,000)
3.314.
40,000 25,000 100,000
vv v
D
vv v
++
= =
++
30
102. Solution: C
2
0
0
1 / ( ) (1 ) / 1
30
1 / (1 ) /
n
n
n
n
nv
Ia
di i i
MacD
a d ii i
v
∞
=
∞
∞
∞
=
+
= = = = = =
+
∑
∑
and so i = 1/30.
Then,
30
29.032.
1
1
1
30
MacD
ModD
i
= = =
+
+
103. Solution: B
I) False. The yield curve structure is not relevant.
II) True.
III) False. Matching the present values is not sufficient when interest rates change.
104. Solution: A
The present value function and its derivatives are
31 4
42 5
53 6
( ) (1 ) 500(1 ) 1000(1 )
( ) 3 (1 ) 500(1 ) 4000(1 )
( ) 12 (1 ) 1000(1 ) 20,000(1 ) .
Pi X Y i i i
Pi Y i i i
Pi Y i i i
−− −
−− −
−− −
=++− +− +
′
=− ++ ++ +
′′
= +− +− +
The equations to solve for matching present values and duration (at i = 0.10) and their solution
are
(0.1) 0.7513 1137.56 0
(0.1) 2.0490 2896.91 0
2896.91/ 2.0490 1413.82
1137.56 0.7513(1413.82) 75.36.
PX Y
PY
Y
X
=+ −=
′
=−+=
= =
=−=
The second derivative is
53 6
(0.1) 12(1413.82)(1.1) 1000(1.1) 20,000(1.1) 1506.34.P
−− −
′′
= −− =−
Redington immunization requires a positive value for the second derivative, so the condition is
not satisfied.
31
105. Solution: D
This solution uses time 8 as the valuation time. The two equations to solve are
28
7
( ) 300,000(1 ) (1 ) 1,000,000 0
( ) 600,000(1 ) (8 ) (1 ) 0.
y
y
Pi i X i
Pi i yX i
−
−
= ++ + − =
′
= ++ − + =
Inserting the interest rate of 4% and solving:
28
7
28
7
7
300,000(1.04) (1.04) 1, 000,000 0
600,000(1.04) (8 ) (1.04) 0
(1.04) [1,000,000 300,000(1.04) ] /1.04 493,595.85
624,000 (8 )(1.04) (493,595.85) 0
8 624,000 / [493,595.85(1.04) ] 8.9607
X 493,5
y
y
y
X
yX
X
y
y
−
−
−
+ −=
+− =
=−=
+− =
=+=
=
8.9607
95.85(1.04) 701, 459.=
106. Solution: A
This solution uses Macaulay duration and convexity. The same conclusion would result had
modified duration and convexity been used.
The liabilities have present value
25
573 /1.07 701/1.07 1000.+=
Only portfolios A, B, and E
have a present value of 1000.
The duration of the liabilities is
25
[2(573) /1.07 5(701) /1.07 ] /1000 3.5.+=
The duration of a
zero coupon bond is its term. The portfolio duration is the weighted average of the terms. For
portfolio A the duration is [500(1) + 500(6)]/1000 = 3.5. For portfolio B it is [572(1) +
428(6)]/1000 = 3.14. For portfolio E it is 3.5. This eliminates portfolio B.
The convexity of the liabilities is
25
[4(573) /1.07 25(701) /1.07 ] /1000 14.5.+=
The convexity of
a zero-coupon bond is the square of its term. For portfolio A the convexity is [500(1) +
500(36)]/1000 = 18.5 which is greater than the convexity of the liabilities. Hence portfolio A
provides Redington immunization. As a check, the convexity of portfolio E is 12.25, which is
less than the liability convexity.
107. Solution: D
The present value of the liabilities is 1000, so that requirement is met. The duration of the
liabilities is
123
402.11[1.1 2(1.1) 3(1.1) ] /1000 1.9365.
−−−
++ =
Let X be the investment in the one-
year bond. The duration of a zero-coupon is its term. The duration of the two bonds is then [X +
(1000 – X)(3)]/1000 = 3 – 0.002X. Setting this equal to 1.9365 and solving yields X = 531.75.
32
108. Solution: A
Let x, y, and z represent the amounts invested in the 5-year, 15-year, and 20-year zero-coupon
bonds, respectively. Note that in this problem, one of these three variables is 0.
The present value, Macaulay duration, and Macaulay convexity of the assets are, respectively,
22 2
5 15 20 5 15 20
,,
x y zx y z
xyz
xyz xyz
++ + +
++
++ ++
.
We are given that the present value, Macaulay duration, and Macaulay convexity of the liabilities
are, respectively, 9697, 15.24, and 242.47.
Since present values and Macaulay durations need to match for the assets and liabilities, we have
the two equations
5 15 20
9697, 15.24
xyz
xyz
xyz
++
++= =
++
.
Note that 5 and 15 are both less than the desired Macaulay duration 15.24, so z cannot be zero.
So try either the 5-year and 20-year bonds (i.e. y = 0), or the 15-year and 20-year bonds (i.e. x =
0).
In the former case, substituting y = 0 and solving for x and z yields
(20 15.24)9697
3077.18
20 5
x
−
= =
−
and
(15.24 5)9697
6619.82
20 5
z
−
= =
−
.
We need to check if the Macaulay convexity of the assets exceeds that of the liabilities.
The Macaulay convexity of the assets is
22
5 (3077.18) 20 (6619.82)
281.00
9697
+
=
, which exceeds
the Macaulay convexity of the liabilities, 242.47. The company should invest 3077 for the 5-year
bond and 6620 for the 20-year bond.
Note that setting x = 0 produces y = 9231.54 and z = 465.46 and the convexity is 233.40, which
is less than that of the liabilities.
109. Solution: E
The correct answer is the lowest cost portfolio that provides for $11,000 at the end of year one
and provides for $12,100 at the end of year two. Let H, I, and J represent the face amount of each
purchased bond. The time one payment can be exactly matched with H + 0.12J = 11,000. The
time two payment can be matched with I + 1.12J = 12,100. The cost of the three bonds is H/1.1 +
I/1.2321 + J. This function is to be minimized under the two constraints. Substituting for H and I
gives (11,000 – 0.12J)/1.1 + (12,100 – 1.12J)/1.2321 + J = 19,820 – 0.0181J. This is minimized
by purchasing the largest possible amount of J. This is 12,100/1.12 = 10,803.57. Then, H =
11,000 – 0.12(10,803.57) = 9703.57. The cost of Bond H is 9703.57/1.1 = 8,821.43.
33
110. Solution: C
The strategy is to use the two highest yielding assets: the one-year bond and the two-year zero-
coupon bond. The cost of these bonds is
2
25,000 /1.0675 20,000 /1.05 41,560.+=
111. Solution: E
Let P be the annual interest paid. The present value of John’s payments is
0.05X
Pa
. The present
value of Karen’s payments is
0.05
(1.05) (1.05) / 0.05
XX
P aP
−−
∞
=
. Then,
0.05
(1.05) / 0.05 1.59
1.05 1 1.05
1.59
0.05 0.05
1.59 2.59(1.05)
ln1.59 ln 2.59 ln1.05
10.
X
X
XX
X
P Pa
X
X
−
−−
−
=
−
=
=
= −
=
112. Solution: A
Cheryl’s force of interest at all times is ln(1.07) = 0.06766. Gomer’s accumulation function is
from time 3 is 1 + yt and the force of interest is y/(1 + yt). To be equal at time 2, the equation is
0.06766 = y/(1 + 2y), which implies 0.06766 + 0.13532y = y for y = 0.07825. Gomer’s account
value is 1000(1 + 2x0.07825) = 1156.5.
113. Solution: D
One way to view these payments is as a sequence of level immediate perpetuities of 1 that are
deferred n-1, n, n+1,… years. The present value is then
1 1 2 2 3 22
/ / / ( / )( ) / .
n nn n n
v iv iv i v ivv v v i
− +− −
+ + += +++ =
Noting that only answers C, D, and E have this form and all have the same numerator,
22 2 2
/ /( ) / .
nn n
v ivvi vd
−
= =
114. Solution: B
The monthly interest rate is
1/12
(1.08) 1 0.643%.j = −=
Then,
4 0.08 252 0.00643
20,000 , 90,122.24 630.99 , 142.83.s Xs X X= = =
34
115. Solution: D
20 10
20 10
20 10
11
1.5 , 1.5 , 1.5 0.5 0.
ee
aa e e
δδ
δδ
δδ
−−
−−
−−
= = − +=
Let
10
.Xe
δ
−
=
We then have the
quadratic equation
2
1.5 0.5 0XX− +=
with solution X = 0.5 for
ln 0.5 / ( 10) 0.069315.
δ
= −=
Then, the accumulated value of a 7-year continuous annuity of 1 is
7(0.069315)
7
1
9.01.
0.069315
e
s
−
= =
116. Solution: B
The present value is
3 10 17 4 7
3 37 37 3
37 3
77
7
(1 ) (1 )
.
11
n
nn
n
vv v v
aa
vv v v
a
vv
−+
++
+
+ + ++
−
− − −−
= = =
−−
117. Solution: C
From the first annuity,
0.109
1.109 1
21.8 21.8 200[1.109
1].
0.109
n
n
n
Xs
−
= =⋅= −
From the second annuity,
2
1
19,208( ) 19, 208 19,208 .
1 1.109 1
n
nn
nn
v
X vv
v
= ++= =
−−
Hence,
2
1
200[1.109 1] 19, 208
1.109 1
[1.109 1] 19, 208 / 200 96.04
1.109 1 9.8
200(9.8) 1960.
n
n
n
n
X
−=
−
−= =
−=
= =
118. Solution: C
60
60
60 1%
60
45.4 33.03
2( ) 2 2 2
, 474.60.
0.01
0.01
av
Ia
−
−
= = =
35
119. Solution: E
Let j be the semi-annual interest rate. Then,
12
2
2
2
475,000 300 300 (1 ) 200( ) 300 300 / 200 /
474,700 300 200 0
300 300 4(474,700)( 200)
0.02084
2(474,700)
(1 ) 1 0.04212 4.21%.
jj
a j Ia j j
jj
j
ij
−
∞∞
= + ++ = + +
− −=
+− −
= =
= + −= =
120. Solution: B
The present value is
2
0.06 0.06
4 2( ) 4 / 0.06 2(1.06) / 0.06 655.56.a Ia
∞∞
+=+ =
121. Solution: A
The present value of the income is
0.1025
100 100 / 0.1025 975.61.a
∞
= =
The present value of the
investment is
2345
6
12345
1
1 1.05 /1.1025 (1.05 /1.1025) (1.05 /1.1025) (1.05 /1.1025) (1.05 /1.1025)
1 1.05
[1 1.05 1.05 1.05 1.05 1.05 ] 5.3295 .
1 1.05
X
X XX
−
−−−−−
−
+++++
−
=+++++ = =
−
Then 975.61=5.3295X for X = 183.06.
122. Solution: A
The present value of the ten level payments is
10 0.05
8.10782 .Xa X=
The present value of the
remaining payments is
10 10
10 11 2
1.015 1.015 /1.05
( 1.015 1.015 ) 18.69366 .
1 1.015 1 1.015 /1.05
v
Xv v X X X
v
+ += = =
−−
Then, 45,000 = 8.10782X + 18.69366X = 26.80148X for X = 1679.
36
123. Solution: D
The equation of value is
0.06
2 32
0.06
10,000 ( 0.996 0.996 ) 15.189 .
1 0.996
1 0.996
ve
Xv v v X X X
v
e
−
−
= + + += = =
−
−
The
solution is X = 10,000/15.189 = 658.37.
124. Solution: D
Discounting at 10%, the net present values are 4.59, –2.36, and –9.54 for Projects A, B, and C
respectively. Hence, only Project A should be funded. Note that Project C’s net present value
need not be calculated. Its cash flows are the same as Project B except being 50 less at time 2
and 50 more at time 4. This indicates Project C must have a lower net present value and therefore
be negative.
125. Solution: D
The loan balance after 10 years is still 100,000. For the next 10 payments, the interest paid is
10% of the outstanding balance and therefore the principal repaid is 5% of the outstanding
balance. After 10 years the oustanding balance is
10
100,000(0.95) 59,874.=
Then,
10 0.1
59,874 / 59,874 / 6.14457 9,744.Xa= = =
126. Solution: B
First determine number of regular payments:
44
0.06 0.06
4000 600 , (4000 / 600)1.06 8.4165.
nn
va a= = =
Using the calculator, n = 12.07 and
thus there are 11 regular payments. The equation for the balloon payment, X, is:
4 16
11 0.06
4000 600 3748.29 0.39365 , 639.43.v a Xv X X= += + =
127. Solution: C
( )
5
5 0.11 5 0.12
20,000 1.11 (3.69590 3.60478 /1.68506) 5.83516X
X 20,000 / 5.83516 3427.50.
Xa a X
−
=+=+ =
= =
37
128. Solution: A
The principal repaid in the first payment is 100 – iL. The outstanding principal is L – 100 + iL =
L + 25. Hence, iL = 125. Also,
16 8
16 8
8 16
16 8
2
8
300(1 ) 200(1 )
300 200
125 100 200 300
300 200 25 0
200 200 4(300)(25)
200 100
0.5.
600 600
vv
La a
i
iL v v
vv
v
−− −
= −=
==+−
− +=
±−
±
= = =
The larger of the two values is used due to the value being known to exceed 0.3. The outstanding
valance at time eight is the present value of the remaining payments:
8
1/8
1 0.5
300 300 1657.
21
a
−
= =
−
129. Solution: E
Let j be the monthly rate and X be the level monthly payment. The principal repaid in the first
payment is 1400 = X – 60,000j. The principal repaid in the second payment is 1414 = X –
(60,000 – 1400)j. Substituting X = 1400 + 60,000j from the first equation gives 1414 = 1400 +
60,000j – 58,600j or 14 = 1400j and thus j = 0.01 and X = 2000. Let n be the number of
payments. Then
0.01
60,000 2000
n
a=
and the calculator (or algebra) gives n = 35.8455. The
equation for the drop payment, P, is
36
35 0.01
60,000 2000 58,817.16 0.698925a Pv P= += +
for P
= 1692.
130. Solution: C
The accumulated value is
( )
24
24 0.06/12 24 0.08/12
1000 (1 0.08 /12) 1000(25.4320(1.1729) 25.9332) 55,762.ss+ + = +=
131. Solution: C
Each month the principal paid increases by
1/12
1.1
. Thus, the amount of principal paid increases to
1/12 30 6 2
500(1.1 ) 500(1.1) 605.
−
= =
38
132. Solution: C
20 10 10 20
11
20 10
10
21
10
10 20 10
11 21
20 10 10
10
21
Int 900 300 900(1 ) 300(1 ) 1200 300 900
Int 900 900(1 )
Int 2Int 1200 300 900 1800 1800
9 15 6 0 2 / 3
Int 900(1 ) 300
ii
i
ia a v v vv
ia v
vv v
vv v
v
=⋅ ⋅+ = −+ −= − −
= ⋅= −
= ⇒− − =−
⇒ − +=⇒ =
= −=
133. Solution: C
The original monthly payment is
240 0.005
85,000 / 85,000 /139.5808 608.97a = =
. On July 1, 2009
there has been 4 years of payments, hence 16x12 = 192 remaining payments. The outstanding
balance is
192 0.005
608.97 608.97(123.2380) 75,048.24a = =
. The number of remaining payments
after refinancing is determined as
0.0045
1 1.0045
75,048.24 500 500
0.0045
0.67543 1 1.0045
ln(0.32457) / ln(1.0045) 250.62.
n
n
n
a
n
−
−
−
= =
= −
=−=
Thus the final payment will be 251 months from June 30, 2009. This is 20 years and 11 months
and so the final payment is May 31, 2030.
134. Solution: B
Just prior to the extra payment at time 5, the outstand balance is
20 0.07
1300 1300(10.5940) 13,772.20a = =
. After the extra payment it is 11,172.20. Paying this off
in 15 years requires annual payments of
15 0.07
11,172.20 / 11,172.20 / 9.1079 1226.65a = =
.
39
135. Solution: C
During the first redemption period the modified coupon rate is 1000(0.035)/1250 = 2.80% which
is larger than the desired yield rate. If redeemed during this period, bond sells at a premium and
so the worst case for the buyer is the earliest redemption. The price if called at that time is
20
20 0.025
35 1250(1.025) 35(15.5892) 762.84 1308.46a
−
+ = +=
. During the second redemption
period the modified coupon rate is 1000(0.035)/1125 = 3.11% which is also larger than the
desired yield rate and the worst case for the buyer is again the earliest redemption. The price if
called at that time is
40
40 0.025
35 1125(1.025) 35(25.1028) 418.98 1297.58a
−
+ = +=
. Finally, if the
bond is not called, its value is
60
60 0.025
35 1000(1.025) 35(30.9087) 227.28 1309.08a
−
+ = +=
.
The appropriate price is the lowest of these three, which relates to the bond being called after the
40th coupon is paid.
136. Solution: B
Because the yield is less than the coupon rate, the bond sells at a premium and the worst case for
the buyer is an early call. Hence the price should be calculated based on the bond being called at
time 16. The price is
16
16 0.05
100 1000(1.05) 100(10.0378) 458.11 1542a
−
+ = +=
. (When working
with callable bonds, the maximum a buyer will pay is the smallest price over the various call
dates. Paying more may not earn the desired yield.)
137. Solution: A
All calculations are in millions. For the ten-year bond, at time ten it is redeemed for
10
2(1.08) 4.31785=
. After being reinvested at 12% it matures at time twenty for
10
4.31785(1.12) 13.4106=
. The thirty-year bond has a redemption value of
30
4(1.08) 40.2506=
.
For the buyer to earn 10%, it is sold for
10
40.2506(1.1) 15.5184
−
=
. The gain is 13.4106 +
15.5184 – 6 = 22.9290.
138. Solution: A
The book value after the third coupon is
37
37 0.0265
7500(0.037) (1.0265) 6493.05 0.379943aC C
−
+=+
and after the fourth coupon it is
36
36 0.0265
7500(0.037) (1.0265) 6387.61 0.390012aC C
−
+=+
. Then,
6493.05 0.379943 (6387.61 0.390012 ) 28.31
105.44 0.010069 28.31
7660.15.
CC
C
C
+ −+ =
−=
=
40
139. Solution: C
The semiannual yield rate is
1/2
1.1 1 0.0488−=
. Assuming the bond is called for 2900 after four
years, the purchase price is
8
8 0.0488
150 2900(1.0488) 150(6.4947) 1980.87 2955.08a
−
+ = +=
. With
a call after the first coupon, the equation to solve for the semi-annual yield rate (j) and then the
annual effective rate (i) is
2
2955.08 (150 2960) / (1 )
1 1.05242
1.05242 1 0.10759.
j
j
i
=++
+=
= −=
140. Solution: C
The book value after the sixth coupon is
34
34 0.036
1000( / 2) 1000(1.036) 9716.01 300.45ra r
−
+=+
.
After the seventh coupon it is
33
33 0.036
1000( / 2) 1000(1.036) 9565.79 311.26ra r
−
+=+
. Then,
4.36 9565.79 311.26 (9716.01 300.45) 10.81 150.22
(10.81 4.36) /150.22 0.0429.
rr r
r
= +− + =−
=−=
141. Solution: B
The two equations are:
5
5 0.04
5
5 0.04
(10,000 ) 9,000(1.04) 44,518.22 7,397.34
1.2 [10,000( 0.01)] 11, 000(1.04) 44,518.22 9,486.38.
P ra r
P ra r
−
−
= +=+
=++ = +
Subtracting the first equation from the second gives 0.2P = 2089.04 for P = 10,445.20. Inserting
this in the first equation gives r = (10,445.20 – 7,397.34)/44,518.22 = 0.0685.
142. Solution: C
When the yield is 6.8% < 8%, the bond is sold at a premium and hence an early call is most
disadvantageous. Therefore,
10
10 0.034
40 1000(1.034) 1050.15Pa
−
=+=
. When the yield is 8.8% >
8%, the bond is sold at discount. Hence, Q < 1000 < P. and thus Q = 1050.15 – 123.36 = 926.79.
Also, because the bond is sold at a discount, the latest call is the most disadvantageous. Thus,
22 2
2 0.044
2
40 40
926.79 40 1000(1.044) (1.044) 1000 909.09 90.90(1.044)
0.044 .044
17.70 90.90(1.044)
2
ln(17.70 / 90.90) / ln(1.044) 38
19.
nn n
n
n
a
n
n
−− −
−
= + =+ −= +
=
=−=
=
41
143. Solution: B
The fund will have
4
4 0.05
500(1.05) 100 176.74s−=
after four years. After returning 75% to the
insured, the insurer receives 0.25(176.74) = 44.19. So the insurer’s cash flows are to pay 100 at
time 0, receive 125 at time 2, and receive 44.19 at time four. The equation of value and the
solution are:
42
2
2
100(1 ) 125(1 ) 44.19 0
125 ( 125) 4(100)( 44.19)
(1 ) 1.5374
200
1 1.2399
24%.
ii
i
i
i
+− +− =
±− − −
+= =
+=
=
144. Solution: B
The Macaulay duration of the perpetuity is
2
1
1
()
(1 ) / 1
1 1/ 17.6.
1/
n
n
n
n
nv
Ia
ii i
i
a ii
v
∞
=
∞
∞
∞
=
++
= = = =+=
∑
∑
This implies that i = 1/16.6. With i = 2i = 2/16.6, the duration is 1 + 16.6/2 = 9.3.
145. Solution: A
Because the interest rate is greater than zero, the Macaulay duration of each bond is greater than
its modified duration. Therefore, the bond with a Macaulay duration of c must be the bond with a
modified duration of a and a = c/(1 + i) which implies 1 + i = c/a. The Macaulay duration of the
other bond is b(1 + i) =bc/a.
146. Solution: B
11
1.10
(0.1025) (0.10) 0.97534 (0.10).
1.1025
PP P
≈=
Therefore, the approximate percentage price
change is 100(0.97534 – 1) = –2.47%.
147. Solution: B
Cash-flow matching limits the number of investment choices available to the portfolio manager
to a subset of the choices available for immunization.
42
148. Solution: C
Options for full immunization are:
2J (cost is 3000), K+2L (cost is 2500), and M (cost is 4000). The lowest possible cost is 2500.
Another way to view this is that the prices divided by total cash flows are 0.6, 0.5, 0.5, and 0.8.
The cheapest option will be to use K and L, if possible.
149. Solution: B
The present value of the assets is 15,000 + 45,000 = 60,000 which is also the present value of the
liability. The modified duration of the assets is the weighted average, or 0.25(1.80) + 0.75Dmod.
The modified duration of the liability is 3/1.1 and so Dmod = (3/1.1 – 0.45)/0.75 = 3.04.
150. Solution: C
Let A be the redemption value of the zero-coupon bonds purchased and B the number of two-
year bonds purchased. The total present value is:
2
1783.76 /1.05 (100 /1.06 1100 /1.06 ) 0.95238 1073.3357 .AB A B=++ =+
To exactly match the cash flow at time one, A + 100B = 1000. Substituting B = 10 – 0.01A in the
first equation gives 1783.76 = 0.95238A + 10733.357 – 10.733357A for A = 8949.597/9.780977
= 915. The amount invested is then 915/1.05 = 871.
151. Solution: B
The company must purchase 4000 in one-year bonds and 6000 in two-year bonds. The total
purchase price is
2
4000 /1.08 6000 /1.11 8573.+=
152. Solution: C
The modified duration is 11/1.10 = 10. Then,
(0.1025) (0.10)[1 (0.1025 0.10)10] 0.975 (0.10).PP P≈ −− =
Therefore, the approximate
percentage price change is 100(0.975 – 1) = –2.50%.
153. Solution: B
7.959
1.072
(0.08) 1000 942.54.
1.08
P
≈=
43
154. Solution: E
Modified duration = (Macaulay duration)/(1 + i) and so Macaulay duration = 8(1.064) = 8.512.
8.512
1.064
112,955 107, 676
1.07
MAC
E
= =
and
112,955[1 (0.07 0.064)(8)] 107,533.
MOD
E = −− =
Then,
107,676 107,533 143.
MAC MOD
EE−= − =
155. Solution: C
The Macaulay duration of the portfolio is
35,000(7.28) 65,000(12.74)
10.829.
35,000 65,000
+
=
+
Then,
1/10.829
10.829
1.0432 1.0432 105, 000
105,000 100,000 1.004516 0.0385.
1 1 100,000
i
ii
= ⇒ = = ⇒=
++
156. Solution: A
1.05 ln(121, 212 /123,000)
121, 212 123,000 3.8512.
1.054 ln(1.05 /1.054)
MAC
D
MAC
D
= ⇒= =
Then,
3.8512 /1.05 3.67.
MOD
D = =
157. Solution: A
I provides cash flows that exactly matches the liabilities. II only has PV(A) = PV(B), which is
not sufficient for exact matching. III describes Reddington immunization, not exact matching.
158. Solution: D
Let F be the face amount of Bond X. Then,
15 15
15 15
2695.39 200 and 3490.78 200a Fv a Fv=+=+
.
Subtract the first equation from the second to obtain
15.
795.39 Fv=
Then for bond X,
15 15
2695.39 200 795.39 (2695.39 795.39) / 200 9.5aa= + ⇒= − =
. This
implies i =0.0634. Then
15 15
9.5 (1 ) / 0.0634 1 0.0634(9.5) 0.3977vv=− ⇒=− =
and
795.39 / 0.3977 2000.F = =
The coupon rate is 200/2000 = 10.0%.
159. Solution: D
The value of 1 invested with bank P after three years is
3
1.04 0.02 1.144864.+=
The yield from
Bank Q satisfies
3 1/3
1.144864 (1 ) 1.144864 1 0.04613 4.6%.ii= + ⇒= −= =
44
160. Solution: D
With a continuously compounded annual interest rate of 6%,
0.06
.ve
−
=
The value of the first
annuity is
1.2
1.2
20
0.06
20
0.06
20
1
20
20
1
600,000 ( ) 102.614 .
1
e
e
av
e
X Ia X X X
de
−
−
−
−
−
−
−
−
= = = =
−
Hence,
X = 600,000/102.614 = 5847.155. Then the value of the second annuity is
1.5
1.5
25
0.06
25
0.06
1
25
25
1
5847.155 5847.155 779,366.
1
e
e
av
e
de
−
−
−
−
−
−
−
−
= =
−
161. Solution: B
The amount of principal repaid at payment 15 is (where R is the quarterly payment)
40 15 1 26 26
10,030.27 (1.03) 10,030.27(1.03) 21, 631.19.Rv R R
−+ −
= = ⇒= =
The amount of interest in payment 25 is
40 25 1 16
21,631.19(1 ) 21, 631.19(1 1.03 ) 8,151.35.v
−+ −
−= −=
162. Solution: E
The present value of the payments (4000 at month 36 plus the payments of X) must match the
present value of the present value of the amounts borrowed (4000 at month 0 plus the payments
of 800).
The quarterly interest rate is 0.264/4 and all payment times should be in quarters of a year. On
that time scale, the 4000 at month 36 is at time 12. The payments of 4000 are at times 1/6, 3/6,
5/6, …, 71/6 and there are 36 such payments. One way to write the present value of these
payments is
36
12 0.5
1
3
4000
.
0.264
0.264
1
1
4
4
n
n
X
−
=
+
+
+
∑
The payments of 800 are at times 1, 3, 5, 7, 9, and 11,, in quarters. One way to write the present
value of these payments plus the initial debt of 4000 is
6
21
1
800
4000 .
0.264
1
12
n
n
−
=
+
+
∑
These are the two sides of equation in answer choice E.
45
163. Solution: B
Let r be the coupon rate for Bond A. The coupon rate for Bond B is then r + 0.01. Then,
20 20
20 0.1 20 0.1
20
20 0.1 20 0.1
11
1600 1000 ( 0.01)
(1.1) (1.1)
2
1.6 2 0.01 0.29729 17.02713 0.08514
(1.1)
1.6 0.29729 0.08514
0.0715 7.15%.
17.02713
ra r a
ra a r
r
= + + ++
=++ = + +
−−
= = =
164. Solution: E
Let n be the number of payments and let j be the interest rate per half-year. Because the given
values are n – 1 half-years apart,
1
7,968.89(1 ) 19,549.25.
n
j
−
+=
Also,
1
1
1 1 7,968.89 /19,549.25
7,968.89 1,000 1,000( 1) 1,000 1 1, 000 1
n
nn
v
aa
jj
−
−
−−
= = += + = +
.
Then,
1 7,968.89 /19,549.25
0.085
7,968.89
1
1,000
j
−
= =
−
for
2
(1.085) 1 0.1772 17.7%.i = −= =
165. Solution: B
The denominator of the duration is the present value of the annuity:
20
20 0.02 30 0.02
4 78.1729 .Xa Xv a X+=
The numerator is the time-weighted present value of the annuity. In units of X we need the
present value of 0, 1, …, 19, 80, 84, …, 196. One way to view this is as four times a 49-year
increasing immediate annuity (so payments of 4, 8, …, 76, 80, 84, …, 196) less three times a 19-
year increasing immediate annuity (so payments of 3, 6, …, 57). The present value is:
49 0.02 19 0.02
4 ( ) 3 ( ) [4(655.2078) 3(147.4923)] 2,178.3542 .X Ia X Ia X X−=− =
The duration is the ratio, 2,178.3542/78.1729 = 27.87.
46
166. Solution: A
( )
22 2 8 2
14
1/14
100 1 1 1 2 100 1 1 1
2 2 22
21
2
(2 1)2 0.1015 10.15%.
i i ii
i
i
+ +−= + +−
= +
= −= =
167. Solution: D
Let C be the amount of the semiannual coupon for bond B.
10
10 0.03
10
10 0.035
40 1000(1.03) 1085.30
1085.30 1000(1.035) 8.3166 708.9188
(1085.30 708.9188) / 8.3166 45.2566
45.2566 2
0.0905 9.05%.
1000
Xa
X Ca C
C
y
−
−
=+=
==+=+
=−=
×
= = =
168. Solution: C
Let X be the original loan value. From the original loan terms,
15
50Xa=
. Under the revised
repayment plan,
5
10 5
50 30X a va= +
. Equating the two gives
5
15 10 5
50 50 30a a va= +
which
does not match answer A. All the other choices use s. Multiplying both sides by
10
(1 )i+
gives
5
15 10 5
50 50 30vs s s= +
, which is answer C. This can also be obtained by equating the values of
the two payment streams at time 10 rather than time 0.
169. Solution: A
The effective monthly rate is
1/12
1.065 1 0.0052617−=
. The accumulated value is
180 0.0052617 180 0.0052617
180 0.0052617
1097 5( )
180
1097(298.733) 5
0.0052617
300.3049 180
327,710 5 442,031.
0.0052617
s Is
s
+
−
= +
−
=
+=
47
170. Solution D
Fund K receives 1000 at the end of each year and also receives interest payments of 1300, 1235,
1170, …, 65. The accumulated value is
20 0.0825 20 0.0825
20
20 0.0825
1000 65( )
20(1.0825)
1000(47.0491) 65
0.0825
97.6311 47.0491
47,049.1 65 86
,902.
0.0825
s Ds
s
+
−
= +
−
=
+=
171. Solution: D
For Q the accumulated value is
25 0.09
84.7009Xs X=
. For R the accumulated value is
24 0.08
24 0.08
24
72.1059 24
100(25) 9( ) 2500 9 2500 9 7911.91.
0.08 0.08
s
Is
−
−
+=+
=+ =
Then
X = 7911.91/84.7009 = 93.41.
172. Solution: E
The accumulated values for Funds X and Y are
10
1000 1
2
k
+
and
10
921.90 1
2
k
−
−
respectively. Equating them and solving for k:
10 10
10
10
2
2
2
1000 1 921.90 1
22
0.9219 1 1 1
22 4
1 0.9919
4
0.0324
0.18.
kk
kk k
k
k
k
−
+= −
=+− =−
−=
=
=
Then
10
0.18
1000 1 2367.36.
2
P
=+=
48
173. Solution: A
The 3-year interest rate is
3
1.07 1 0.225043−=
. Then,
0.225043
1.225043
735 5.443595
0.225043
Xa X X
∞
= = =
and X = 735/5.443595 = 135.02.
174. Solution: B
0.06 0.06
2
2600 9( )
1 1 1.06
9
0.06 1.06
0.06
2500
0.06
(2600 2500)(0.06) 6.
Pa v Ia
P
P
P
∞∞
= +
= +
= +
=−=
175. Solution: B
( )
( ) ( )
10
10 5
10 5 10
10 5 10
10 5
2
5
1/5
475 400
11
475 400
475 1 400 1
875 400 75 0
400 400 4(875)(75)
0.6
2(875)
(1/ 0.6) 1 0.1076 10.76%
ii
a a va
v vv
ii
v vv
vv
v
i
∞
= +
− −+
=
− = −+
− −=
±+
= =
= −= =
176. Solution: E
Based on the effective yield rate,
2
100 /1.1 2 /1.1 39.03XX X= + ⇒=
. After one year, the
outstanding loan balance is 100 + 8 – 39.03 = 68.97. For the balance to be zero after two years,
68.97(1 ) 2(39.03) 0 78.06 / 68.97 1 0.1318 13.2%.ii+ − = ⇒= −= =
177. Solution: D
The amount borrowed is
5
5 0.1 5 0.1
1000 2000 8498.35.a va+=
The outstanding balance after five
years is
5 0.1
2000 7581.57a =
. The principal repaid is 8498.35 – 7581.57 = 916.78. The interest
paid is 5000 – 916.78 = 4-83.22.
49
178. Solution: B
The accumulation is
10 0.08
10 0.08
10
15.6455 10
10 0.12 ( ) 10 0.12 10 0.12 18.4683 .
0.08 0.08
s
XXIs XX XX X
−
−
+=+ =+ =
Then, X = 10,000/18.4683 = 541.47
179. Solution: C
The equation of value is
30
30 0.05
30 0.05 30 0.05 30 0.05
30
10, 000 ( 5) 5( ) ( 5) 5
0.05
16.14107 6.94132
( 5)(15.37245) 5 15.37245
843.11275
0.05
(10
,000 843.11275) /15.37245 705.36.
av
Xa Ia Xa
XX
X
−
=+− =+−
−
=+− =−
=+=
180. Solution: A
2 345
150 100 150 50 150 150 150
600 221.94
1.15
1.15 1.15 1.15 1.15
NPV
−−
=−+ + + + + =−
181. Solution: A
I is true.
II is false, the price sensitivity of assets and liabilities must be equal.
III is false, the convexity of assets should be greater than the convexity of liabilities.
182. Solution: D
The effective annual rate of interest is
12
(1.005) 1 0.06168−=
. The present value of the tuition
payments six months before the first payment is
6
4 0.06168
25,000(1.005) 24, 262.95(3.66473) 88,917.16a
−
= =
. The accumulated value of the
deposits at that time is
0.005
1000
n
s
. Equating the two amounts:
1.005 1
88,917.16 1000
0.005
1.44459 1.005
ln(1.44459) / ln(1.005) 73.75.
n
n
n
−
=
=
= =
Therefore, at least 74 payments will be required.
50
183. Solution: E
Let x be the annual payment amount. Macaulay duration is
23 7
23 7
12 3 7
17.6315
1.1 1.1 1.1 1.1
3.62
4.8684
1.1 1.1 1.1 1.1
xxx x
xx x x
+ + ++
= =
+ + ++
.
Alternatively, the duration can be calculated as
7 0.1 7 0.1
() / .Ia a
184. Solution: E
PV of liabilities is
23
402.11(1/1.1 1/1.1 1/1.1 ) 1000.++ =
Duration of liabilities is
23
402.11(1/1.1 2 /1.1 3 /1.1 ) /1000 1.93653.++ =
Let X be the investment in one-year bonds. To
match duration, since zero-coupon bonds have duration = maturity, 1.93653 = [X + 3(1000 –
X)]/1000. Then, 2X = 3000 – 1936.53 = 1063.47 and X = 532.
185. Solution: E
A change in face value multiplies all cash flows by the same amount. Therefore, there is no
change in the duration. If the coupon rate increases, the coupons become larger, but the
redemption value stays the same. This causes payments prior to redemption to receive more
weight relative to the payment at redemption and thus the duration will decrease.
186. Solution: A
We are given
(4)
8%i =
and want to determine
(12)
/12i
. The equation that links the two and its
solution is:
12 4
4
(12) (4)
4/12
(12)
1/3
(12)
8%
1 11
12 4 4
8%
11
12 4
8%
1 1.
12 4
ii
i
i
+ =+=+
+=+
=+−
51
187. Solution: E
Let m be the monthly payment and i be the monthly interest rate. The interest in the first payment
is 125,000i and the principal repaid is 125,000 – 124,750 = 250. Thus m = 125,000i + 250.
Similarly, for the second payment, m = 124,750i +252. Thus, 250i = 2 for i = 2/250 = 0.008 and
then m = 1250. To obtain the number of payments, the equation to solve is
0.008
125,000 1250
1 1.008
100
0.008
0.2 1.008
ln(0.2) / ln(1.008) 202.
n
n
n
a
n
−
−
=
−
=
=
=−=
188. Solution: C
Let x and y be the amount invested in the five and twenty year bonds respectively. To match the
present values:
0.07(10) 0.07(15)
500,000 500,000 423, 262.xy e e
−−
+= + =
To match the durations,
noting that the denominators of the durations for assets and liabilities are the same,
0.07(10) 0.07(15)
5 20 500,000(10) 500,000(15) 5,107, 460.xy e e
−−
+= + =
Subtracting five times the
first equation from the second one gives 15y = 2,991,150 for y = 199,410 and x = 423,262 –
199,410 = 223,852.
189. Solution: C
Let n be the term of the bond in half-years. We know that
601 1080
n
v=
and thus
601/1080
n
v =
. Then
0.05
1 1 601/1080
8.87037.
0.05 0.05
n
n
v
a
−−
= = =
The purchase price of the bond is
0.05
40 1080 40(8.87037) 610 956.
n
n
av+ = +=
190. Solution: B
Principal repaid in the first payment is 1000 – 10,000(0.04) = 600. Therefore, the principle
repaid in the tenth payment is
9
600(1.04) 854=
and the interest paid is 1000 – 854 = 146.
52
191. Solution: C
After one year the outstanding balance is
48 0.025
500 13,886.58a =
. This must match the present
value of the revised payments:
6 12
6 0.025 36 0.025
13,886.58 500 4.74964 8,757.69
(13,886.58 8,757.69) / 4.74964 1, 079.85.
Xv a v a X
X
=+=+
=−=
Alternatively, each missing payment is being replaced with a larger payment six months later.
The larger payment should be the payment due plus the missed payment with interest, or
6
500 500(1.025) 1,079.85.+=
192. Solution: A
Only Bond III can match the liability at time 3. The bond must mature for 1000. Only Bond II
can match the liability at time 2. The face value and coupon must total 1000. If X is the face
value, then X + 0.02X = 1000 and thus X = 980.39. Only Answer A has these to values. To
check, Bond II also provides a coupon of 0.02(980.39) = 19.61 at time 1. Therefore, Bond I must
provide the remaining 980.39 from its coupon and redemption value. If Y is the face value, then
Y + 0.01Y = 980.39 for Y = 970.68.
193. Solution: A
Cash flows (in thousands) are 12, 12, 12, 12, and 162. The first bond provides payments of 10,
10, 10, 10, and 110. Therefore, the second bond must provide 2, 2, 2, 2, and 52. This implies a
coupon rate of 2/50 = 4% and a face amount of 50. Only Answers A and B provide these. At an
8% yield, the price of this bond is 42.015 (or 42,015).
194. Solution: C
Let i be the yield rate. Then,
30
30
30 30
30
1/30
3609.29 2000(2 ) 2250(1 )
4000[1 (1 ) ] 2250(1 )
(1 ) (4000 3609.29) / (4000 2250) 0.22326
0.22326 1 0.051251.
i
ia i
ii
i
i
−
−−
−
−
= ++
= −+ + +
+= − − =
= −=
Modified duration is Macaulay duration divided by one plus the yield rate: 14.14/1.051251 =
13.71.
53
195. Solution: A
The amount of the dividends does not matter, so they will be assumed to be 1. First, calculate the
Macaulay duration. The present value of the dividends is
44
1.1 (1/ 0.1) 6.83013.va
−
∞
= =
The
numerator is the present value of “payments” of 5, 6, 7, … starting five years from now. This can
be decomposed as a level of annuity of 4 and an increasing annuity of 1, 2, 3, … . The present
value is
44
24 2
4 1 1 4 1.1
[4 ( ) ] 102.452.
0.1
1.1 0.1
i
v a Ia v
i
i
∞∞
+
+ =+= +=
The Macaulay duration is
102.452/6.83013 = 15. The modified duration is 15/1.1 = 13.64.
196. Solution: A
Let r be the semiannual coupon rate. For the original bond,
6
6 0.05
1000 1000 5075.692 746.215.P ra v r= += +
For the modified bond,
12
12 0.05
49 1000 1000 8863.252 556.837.P ra v r−= + = +
Subtracting the second equation from the
first gives 49 = –3787.56r + 189.378. The solution is r = 0.037 and the coupon is 37.
197. Solution: A
Let h(i) be the present value of the cash flows. For Redington immunization, the value of the
function and its first derivative at 25% must be zero and the second derivative must be positive.
X is immunized because:
3
24
35
(0.25) 102, 400 192,000 /1.25 100,000 /1.25 0
(0.25) 192,000 /1.25 100,000(3) /1.25 0
(0.25) 192,000(2) / 1.25 100,000(3)(4) /1.25 196,608 0
h
h
h
=−+ =
′
=−=
′′
=−+
=>
Y is not immunized because:
23
234
(0.25) 158, 400 342,000 / 1.25 100,000 / 1.25 100,000 /1.25 0
(0.25) 342,000 /1.25 100,000(2) /1.25 100,000(3) /1.25 6,400 0
h
h
=−+ + =
′
= − − =−≠
Z is not immunized because
23
(0.25) 89,600 288,000 /1.25 100,000 /1.25 300,000 /1.25 51, 200 0h =−+ + − = ≠
54
198. Solution: E
The bond sells at a premium, so the worst-case scenario is redemption at time six. Then,
6
6
66
6
1/6
1023 1000 1000(1 )
0.96
1000
[1 (1 ) ] 1000(1 )
0.96
(1 ) 0.448
0.448 1 0.1432
14.32%.
i
i
ai
ii
i
i
−
−−
−
−
= ++
= −+ + +
+=
= −=
199. Solution: E
Answer E is false because the convexity of the assets must be greater than the convexity of the
liabilities.
200. Solution: E
The accumulated value to time 4 is
3
0.42 0.32 1.26 0.42
33
0.5 0.08(4 ) 0.32 0.42 0.32
11
1
100 ( )
100 100 100 657.
0.42 0.42
t
tt t
e ee e
e e dt e e dt e
−
−
= = = =
∫∫
201. Solution: E
The value at time 17 of the payments beginning at time 18 is
2
2
1
1 (1 ) 1
1.035
2500 2500
2500
.
1
1.035 0.035
0.035
1
1.035
k
kk k
k
k
+
++ +
+ += =
+
−
−
The total present value is
2 17
15 0.035
1
115,000 2500(1.035 ) 2500
0.035
46(0.035 ) 10.7516(0.035 ) 0.55720(1 )
1.61 0.37631 0.55720
0.01889 1.89%.
46 10.7516 0.55720
k
av
k
k kk
k
−
+
= +
−
−= −+ +
−−
= =
=
−+
55
202. Solution: B
The initial payment, X is
19 20 21
2 20
1 1.02 1.02 1/1.03 1.02 /1.03
200,000 17.7267
1.03 1 1.02 / 1.03
1.03
1.03
11, 282.42.
XX
X
−
= + ++ = =
−
=
The final payment is
19
11, 282.42(1.02) 16, 436.36.=
203. Solution: E
The annual payment is
10 0.1
10,000 / 10,000 / 6.14457 1627.45.a = =
The balance at time 3 is
7 0.1
1627.45 1627.45(4.8684) 7923.08.a = =
With one-half year at simple interest, the balance at
time 3.5 is 7923.08(1.05) = 8319.23.
204.
DELETED
205. Solution: E
Let x be the amount invested in Bond A and y the amount invested in Bond B. Then 2y is
invested in Bond C. To match the present value of the assets and liabilities:
20.5
2 190,000(1.07)
3 47, 466.39.
xy y
xy
−
++ =
+=
To match the Macauley durations,
10 15 30(2 )
20.5
47, 466.39
xy y++
=
. Then,
20.5(47, 466.39) 10(47, 466.39 3 ) 75
20.5(47,366.39) 10(47, 466.39)
11,075.49
75 30
yy
y
= −+
−
= =
−
and
X = 47,466.39 – 3(11,075.49) = 14,239.92.
56
206. Solution: B
Let X be the price of Bond X. Then, for the two bonds:
2
2 0.035
2
2 0.035
10,000(0.03) (1.035)
969.52 10,000(0.025) ( 50)(1.035) .
n
n
n
n
X ac
X ac
−
−
= +
− = ++
Subtracting the second equation from the first gives
2
2 0.035
22
2
969.52 50 50(1.035)
50
969.52 [1 (1.035) ] 50(1.035)
0.035
1.035 459.05 /1478.57 0.310469
(0.5)ln(0.310469) / ln(1.035) 17.
n
n
nn
n
a
n
−
−−
−
= −
=−−
= =
=−=
207. Solution: B
With simple interest, the deposit in Bank X earns 1000(0.07) = 70 in year 8.
With compound interest, the earning in Bank Y in year 8 is
32 28
1000(1.0125) 1000(1.0125) 72.14.−=
The absolute difference is 2.14.
208. Solution: E
Split this into two perpetuities. One starts at time 0.5 at 500 increasing by 10 every year. The
other starts at time 1 at 500 with payments increasing by 10 every year. The semiannual interest
rate is 1.075
0.5
-1=0.0368221. The present value of an increasing perpetuity immediate is found
using the formula:
2
PQ
i
i
+
, where P is the initial amount and Q is the increase amount.
The first perpetuity, valued at time 0:
2
500 10
(1.0368221) 8755.39
0.075
0.075
+=
The second perpetuity, valued at time 0:
2
500 10
8444.44
0.075
0.075
+=
The total is 8755.39 + 8444.44 = 17,199.83.
209. Solution: B
10 0.05
10 0.05
5000(10) 5000 ( ) 100,000
10
5000 50,000
0.05
13.206787 10
10
0.05
0.15592
i Is
s
i
i
i
+=
−
=
−
=
=
57
210. Solution: C
( )
10 0.08
10 0.10
Payment equals: 10,000 / 1, 490.29
Accumulated total equals: 1, 490.29 23, 751.46
23, 751.46 10,000 13,751.46
a
s
=
=
−=
211. Solution: D
120
120 0.01
Value of fund after 20 years: 500 (1.01) 383, 404.42
383, 404.42 (1 ) (1.01)
0.01
3796.08
s
XX X
i
di
X
=
= = +=
=
212. Solution: D
12
12
12 0.10
Using the retrospective method, OB 12,000(1.10) 1000
37,661.14 1000(21.38428) 16, 276.86
s= −
=−=
213. Solution: E
Using the BAII Plus calculator:
n = 12
PV = 911.37
PMT = -40
FV = -1000
CPT I/Y and you get 5.0% is the half-year rate:
(2)
2
. (1 ) (1 0.05) ; 0.1025.
2
i
ii+=+ =
214. Solution: B
( ) ( )
4
0.08
0.12
100 1 200
4
4 ln 0.97 ln(200 /100) 0.08
0.0418 ln 2
16.57
n
n
e
nn
n
n
−
−=
−= +
=
=
58
215. Solution: D
Equating present values gives
10
10
1.092
1
5000
1.04
.04 .092
1.04
3377.82 (12.094127)
279.29
X
X
X
−
=
−
=
=
216. Solution: E
20 0.05
()
962.92 (0.04 0.05 )
962.92 ( 0.01 )12.46221
962.92 0.875378
1100
n
P C Fr Ci a
C C Ca
CC
C
C
=+−
=+−
= +−
=
=
The discount is 1100 – 962.92 = 137.08.
217. Solution: D
Equate the accumulated value of the deposits to the present value of the perpetuity:
( )
10
10
10
10
10
10
10 10
20 (using the BAII Plus) 12.3%
s
Is
ii
s
si
−
= =
−=
= ⇒=
The PV of the perpetuity is 10/0.123 = 81.30.
218. Solution: C
20 0.08 10 0.08
1300 4000
12,763.59 4000 (6.71008) 1306.03
a Xa
X
−=
−= =
219. Solution: C
32
32
32
32
2 3 1 2 6.16078
2.6452
2.32908
21
3 2 4.01370
2.6233
1.53001
2.6452 2.6233 0.0219
L
A
vv
d
vv
Xv Xv
d
Xv Xv
+
= = =
+
+
= = =
+
−=
59
220. Solution: D
The bank’s accumulated value at the end of 30 years is:
( )
30 0.04
30 0.05
30
100,000
364,841
100,000 1 364,841
0.044
s
a
i
i
=
+=
=
221. Solution: E
999.35 x 1.06 = 1059.31 will be available to make the first payment of 1000, leaving 59.31 to be
reinvested at X%.
817.65 x 1.07
2
= 936.13 will be available from the second bond to make the second payment of
1000, leaving 63.87 to come from the reinvestment of 59.31.
X = 100(63.87 / 59.31 – 1) = 7.69.
222. Solution: E
45
4 14 5
45
14
14
(1 ) (1 ) (1 )
(1.09) (1 ) (1.095)
0.1152
sfs
f
f
+ +=+
+=
=
223. Solution: C
1
1
mod
(1 )
(1 )
(1 )
(1 ) (1 )
n
n
nn
d
i
ni
di
D ni
ii
−
−−
−
−−
+
+
=−==+
++
224. Solution: E
The amount invested in three-year bond is equal to the PV of the third year’s payout,
( )
3
1000 1.1 751.31=
The amount invested in one-year bond is equal to the PV of the first year’s payout,
1000 1.08 925.93=
925.93 751.31 174.61−=
60
225. Solution: B
PV =
04.1
10
+
( )
2
045.1
12
+
( )
3
055.1
15
+
( )
4
07.1
20
= 9.615 + 10.989 + 12.774 + 15.258 = 48.64
226. Solution: D
Let i = yield rate, r = coupon rate (if any), F = face value, P = price, n = # of years.
For the first bond:
36
36
0.8
0.8
0.006218
P F Fv
v
i
= =
=
=
For the second bond:
0.006218
0.006218
4
0.8 (0.006218)
9
0.8 (0.0027634)
n
n
n
n
P F Fv Fa
va
= = +
= +
Using the BAII Plus, where PV=0.8, I/Y=.6218, PMT=0.0027634, FV=1
CPT N results in n=72.
227. Solution: B
Since the bond has no coupons, the Macaulay duration is the same as the amount of time until
maturity, namely 4 years.
Thus, the effective annual yield rate, y, is
1/4
1200
1 0.046635
1000
−=
.
The modified duration equals the Macaulay duration divided by (1 + y). Thus the modified
duration is
4
3.82177
1.046635
=
years.
228. Solution: C
Using the general Macaulay duration formula:
t
t
t
t
Rvt
Rv
∑
∑
where R is the cashflow:
Period Cashflow PV at 8% Period ×PV
1 10 9.26 9.26
2 12 10.29 20.58
3 15 11.91 35.73
4 20 14.70 58.80
5 30 20.42 102.10
Total 66.58 226.47
Macaulay duration = 226.47/66.58 = 3.401472 years
61
229. Solution: C
When a company’s position is Redington immunized, its position is definitely protected from
sufficiently small changes in yield rate, in either direction. However, its position may or may not
be protected from large changes in yield rate.
230. Solution: D
Amount of loan = L
10 10%
6
10 10% 4 10% 6 7%
Initial expected yield rate 10.00%
Annual payment /
Accumulated value at time 10 ( / )( 1.07 )
La
La s s
=
=
= +
1/10
1/10
6
4 10% 6
7%
10 10%
1/1
0
Accum Value
Yield rate = 1
1.07
1
4.6410(1.5007) 7.1533
1
6.1446
8.67%
L
ss
a
−
+
= −
+
= −
=
231. Solution: E
Let L = the loan amount. Note that
5
1 (1 )ij+= +
. The equation of value is
5
120
ki kj
Pa L a⋅==⋅
so that
5
5
5
120
1 (1 )
120
1 (1 )
1 (1 )
120
1 (1 )
120
(1 ) 1
120
kj
ki
k
k
k
k
a
P
a
ji
ji
ii
ji
i
j
j
j
−
−
−
−
=
−+
=
−+
−+
=
−+
=
+−
=
Next, using the fact that 0 < j < 0.04, we get
5
(1 ) 1
5 5.41633
j
j
+−
<<
by plugging in a small value like 0.000000001 and 0.04 resulting in P
equaling more than 600 but less than 650.
62
232. Solution: B
Using the retrospective method:
6
6 0.05
4000(1.05) 250
5360.38 1700.48 3659.90
s−
−=
233. Solution: D
Let t represent the number of years since the beginning of year 1. Since the annual effective
interest rate is 3% in each of years 1 through 10, and 2% each year thereafter, the present value
of an amount is calculated by multiplying it by a discounting factor of
1
(1.03)
t
if
0 10t≤≤
, and
10 10
1
(1.03) (1.02)
t−
if
10t >
.
The balance is initially 0 (the account is new before the first deposit). Deposits of X are made at
times t = 0, 1, 2, 3,…, 25, or equivalently at time
1tk= −
for each whole number k from 1 to 26
inclusive.
For the final balance to become 0, a withdrawal of 100,000 at time t = 25 would be needed. Since
the net present value of the cash flows (withdrawals minus deposits) must be zero, in a time
period from a zero balance to another zero balance, we have
11 26
10 25 10 1 10 1 10
1 12
100,000 1 1
0
(1.03) (1.02) (1.03) (1.03) (1.02)
kk
kk
XX
− − −−
= =
−− =
∑∑
11 26
10 15 1 10 11
1 12
100,000 1 1
(1.03) (1.02) (1.03) (1.03) (1.02)
kk
kk
XX
−−
= =
= +
∑∑
.
234. Solution: A
For the first bond:
20
20 0.065
(0.076)(6000) 6000P av= +
P = 6727.22.
For the second bond:
20
20 0.065
6727.22 ( )(7500) 7500ra v= +
7500r = 417.37, so r = 0.0556.
63
235. Solution: E
40 40
40 40
40
40 |
40
40
40
40|
75
257.18 60 ( )
257.18 15
15 257.18
(0.252572) 320.33 257.18
250.01
X
YX
XY
P Fra Cv a Cv
P P a C Kv
P P a Kv
Kv
a
K
K
= +=
+
=− = ++
−= = −
= −
= −
=
236. Solution: C
The PV of the first twenty payments is:
20
1.04
1
1.10
14,000 157,
337.48
0.10 0.04
−
=
−
The PV of the remaining payments starting at time 21 is:
19 20
1
14,000(1.04) (1.01) (1.10) 49, 202.44
0.10
0.01
−
=
−
Total equals 206,539.92.
237. Solution: B
1
16,000 1,000
0.057 ( /100)
1
16
0.057 /100
1
0.057 /100
16
0.55
r
r
r
r
=
−−
=
+
+=
=
238. Solution: E
Let j equal the five-year interest rate.
5
2
(1 ) (1.09)
0.538624
2 10
38.18
0.538624
(0.538624)
j
j
PV
+=
=
=+=
64
239. Solution: A
From the first bond:
8
8 0.03
25P a Cv= +
From the second bond:
4
4 0.03
0.93 25P a Cv= +
Multiply the first equation by 0.93 and plug into the second equation:
4
8 0.03 4 0.03
0.93(25) 0.93 25
163.2078 0.73415 92.9275 0.88849
70.2804 0.15434
455.37
a C a Cv
CC
C
C
+= +
+=+
=
=
240. Solution: D
The PV of the liability is
2
600,000
548,387.92
1.046
=
and its Macaulay duration is 2.
Then, equating present values:
92.387,548
046.1
046.1
4
=+
yx
And equating durations:
4
( /1.046) ( /1.046 )
(1) ( 4) 2
548,387.92 548,387.92
xy
+=
Solving the system of equations results in x = 382,409
241. Solution: E
Solution:
32
4000(1.04 ) 1400[1 (1 ) (1 ) ],ii= ++++
or i solves the quadratic equation:
i
2
+ 3i − 0.2139 = 0.
Thus, i = 6.97% because the other root is negative.
242. Solution: B
4 10
0.01 0.01
6 0.01
100,000
0.05
100,000 (5.79548)(0.96098) (20)(0.90529)
100,000 (23.6751)
4223.85
X
X
a v v
XX
X
X
= +
= +
=
=
65
243. Solution: D
Using BA II Plus:
180 0.075/12
0.075/12
60,000
556.21
49,893 556.21
132
m
Xa
X
a
m
=
=
=
=
So, (180 – 132) = 48 payments have been made so far:
0.6/12
49,893 556.21
119.3
n
a
n
=
=
Use 120 future payments including the smaller one.
48+120=168.
244. Solution: C
20
20 0.10
20 0.10
20
2000 110
0.10
2000 110(8.51356) (55.40691)
1063.51 (55.40691)
19.1945
av
aX
X
X
X
−
= +
= +
=
=
245. Solution: E
10
1.15
1
1.05
100 1483.62
0.05 0.15
PV
−
= =
−
246. Solution: D
( )
1
1
41
5
4
5
4
4
1
1144.5
865
865
0.75579
1144.5
0.75579, 0.93240, 0.07251
(0.07251) 797.50
10,999.02
n
nn
n
OB Ra Rv
P Rv Rv
Rv
v
Rv
vvi
IX
X
−
−−+
−
= = =
= = =
= = =
= = =
= =
=
66
247. Solution: C
Given that the problem states that the inequality is true for all interest rates from 0% to 10% and
all values of Y, it is sufficient to determine it for one set of values. Select i = 7% and Y = 121.
Then,
3
3
121/ (1 3(0.07)) 100
121/ (1.07) 98 / 77
121(1 0.07(3)) 95.59
121(0.93) 97.33
Q
R
S
T
=+=
= =
=−=
= =
Hence,
S < T < R < Q
248. Solution: C
The yield rate on Kate’s bond is
(2)
10
10
2
(2)
(1000 100) 25 1000
0.0371551
2
i
av
i
−= +
=
The discount on Wallace’s bond is
8
8 0.05
(1000 ) 25 1000
1000 838.42, 161.58
Da v
DD
−= +
−= =
The book value of Kate’s bond at time 1 is
8
8 0.0371551
25 1000
917.19
Ba v
B
= +
=
The difference is B – D =917.19 – 161.58 = 755.61
67
249. Solution: B
24
35
13
24
1000(1 ) 300(1 )
1153.84
2000(1 ) 1200(1 )
2667.91
(1 ) (1 )
1153.84 0.95238 0.86384
(1 ) 3 (1 )
2667.91 0.90703 2.46811
L
L
L
L
A
L
Pi i
P
Pii
P
PX i Y i
XY
P Xi Yi
XY
−−
−−
−−
−−
= ++ +
=
′
=− +− +
′
= −
= +++
= +
′
=−+− +
−=− −
So, we have two equations and two unknowns. Solving simultaneously, we get:
953.57, 346.61, 2.75.
Y
YX
X
= = =
250. Solution: C
20 0.03
1000
67.22
67.22(20) 1344.31
1344.31 1000 (1000 950 900 50)
344.31 (10,500)
0.03279
Pa
P
i
i
i
=
=
=
= + ++++
=
=
251. Solution: B
(12)
1/12
120 6 1
120 24 1
24
1.08 1 0.006434
12
600 (1/1.006434)
1254.47
1254.47 673.42
i
P
P
Pv
−+
−+
= −=
=
=
= =
68
252. Solution: B
20
20
0
0
2
ln(1 2 )
12
20
5
41
41 (1 )
0.204035
(1 0.204035) 2.53
dt
t
t
ee
i
i
+
+
∫
= =
= +
=
+=
253. Solution: B
( )
( )
10
20
20
10
20 10
20 10
10
600 600
1000
13
51
55 33
05 3 2
3 9 4(5)(2)
37
Let 0.4 9.59582%
2(5) 10
600
1 0.4 8753.8
0.0959582
i
v
a
ii
v
v
ii
vv
vv
xv i
X
⋅= +
−
= +
−=+
=+−
−± +
−±
= = =
= ⇒=
= +=
254. Solution: B
2
2
2
150 10
5000
5000 150 10 0
150 ( 150) 4(5000)( 10)
10,000
0.06217
ii
ii
i
i
= +
− −=
±− − −
=
=
69
255. Solution: D
If paid in one lump sum, the total interest paid is
20
(1.05 1) 1.65330 .XX−=
With level payments for 10 years, the total interest paid is
10 0.05
10 0.29505 .
X
XX
a
−=
Then,
1.65330 1000 0.29505
736.24.
XX
X
= +
=
256. Solution: B
20 1 20 1
20 1
21
21
(1 )
21
0.5
1.05 2
(21 )ln(1.05) ln 2
7
tt
tt
t
t
t
PI
vv
v
v
t
t
−+ −+
−+
−
−
=
= −
=
=
=
−=
=
257. Solution: B
2
2
10,000 10,815 20,800
20,800 20,800 4(10,815)(10,000)
21,630
0.970873 or 0.952381
0.03 or 0.05
0.03 0.05 0.02
vv
v
v
i
+=
±−
=
=
=
−=
70
258. Solution: A
30
30 0.042
4
4 0.021
45 1200
1108.85
1108.85 20 1376.69
A
A
n
n
Pa v
P
av
= +
=
= +
Using the BA II Plus:
PV = 1108.85
PMT = 20
FV = 1376.69
I/Y = 2.1
Solve for 4n and get 4n = 48, n = 12.
Or solve the equation to get
4
0.36876
n
v =
and then solve for n.
259. Solution: C
Statement I should have an
60
s
on the left.
Statement II has an annuity-due rather than an annuity-immediate on the right.
Statement III is correct.
260. Solution: D
15(961.54)(1) 20(966.14)(2) 30(878.41)(3)
2.198495
15(961.54) 20(966.14) 30(878.41)
++
=
++
261. Solution: B
17
18 0.10
1,000,000(1.10) 1,000,000
5,054, 470.285 (45.59917) 1,000,000
4,054, 470.285 (45.59917)
88,915.43
Ps
P
P
P
−=
−=
=
=
262. Solution: D
Using time 5 as the first reference point, then bringing that value back to time 0:
( )
5
5
5
500 500
i
i
v a Ia
+
This combines a five-year level annuity-due of 500 plus an increasing annuity-due starting with
500 and increasing by 500.
71
263. Solution: B
Let face amount equal 1.
( )
18
18
18 18
18
1
1.61 2.25
1.61 2.25(1 )
1.25 0.64
0.96342
1
1.45 2.25
1.45 2.25 1
1.25 0.8
0.64
ln 0.963492 ln 0.64
12
n
n
nn
n
n
v
iv
i
vv
v
v
v
iv
i
vv
v
v
n
n
−
= +
= −+
=
=
−
= +
= −+
=
=
=
=
264. Solution: C
10
10 0.10
2
20
20 0.048881
450 1000
Using BA II Plus calculator:
10.50
1.10 (1 ) , 0.048881
5.25 1000
451.64
Xa v
X
jj
Pa v
P
= +
=
=+=
= +
=
72
265. Solution: E
Let I be the amount of interest in the first month.
0 1 10
, ()PmIPIP Pm− += = − −
In the first month, the interest
10
()P Pm−−
was charged on a principal of
0
P
, so the effective
monthly interest rate (expressed as a decimal) of the first loan is
1 0 10
00
()P Pm PPm
PP
− − −+
=
.
The nominal annual interest rate (expressed as a decimal) for both loans is therefore
10
0
12
PPm
P
−+
, so the effective daily rate (expressed as a decimal) for the second loan is
10
0
12
365
PPm
P
−+
.
Finally, the effective monthly rate (expressed as a decimal) for the second loan is
365/12
10
0
12
11
365
PPm
P
−+
+−
.
266. Solution: D.
(0, ) (1 )
(0, ) (1 )
(1 )
(1 )
(1 )
(0, )
(0, )
m
n
m
mn
n
Pm i
Pn i
i
Xi
i
Pm
X
Pn
−
−
−
−+
−
= +
= +
+
= = +
+
=
267. Solution: A
180
120 0.005
180 0.005
360 0.005
2000 180,146.91
180,146.91
1520.18
1520.18 253,553.61
OB a
Pa
P
La
= =
=
=
= =
268. Solution: E
The conditions
1) assets and liabilities have equal present values and equal modified durations, and
2) the convexity of its assets exceeds the convexity of its liabilities
are precisely what is required for Redington immunization.
73
269. Solution: E
6
6
6
6
90.17 4
132.47 4 1.6
j
j
a Xv
a Xv
= +
= +
Multiply the first equation by 1.6:
6
6
144.272 6.4 1.6
j
a Xv= +
Subtract the second equation:
6
11.802=2.4
j
a
Use annuity calculation on BA II Plus:
(2)
(2)
6% , 12%
2
i
ji= = =
270. Solution: B
2 0.05
2
9297
5000
5000
4535.12
1.05
Pa
P
=
=
=
271. Solution: B
Using the BA II Plus calculator:
2
(2)
(2)
10
10 0.019804
5
5
1.04= 1
2
0.019804
2
25 1000
1046.72
1046.72 100 30 1000
4.2036%
A
A
j
j
i
i
Pa v
P
av
j
+
=
= +
=
−= +
=
74
272. Solution: E
Value at time 5 years:
20
20 0.02
0.02
100,000(1.02) 2500
87,851.32 5000
m
s
a
−
=
m = 21.86, using the BA II Plus. Since we want a balloon payment, use m = 21.
21
21 0.02
87,851.32 5000a Bv= +
B = 4236.70, so balloon equals 5000 + 4236.70 = 9236.70
273. Solution: B
40
40 0.05
20
20 0.04
20
20 0.04
900 1000 900
Using BA II Plus:
1000 45
45 900
1022.31
1022.31 1100
38.28
ra v
r
Pa v
P
Fra v
Fr
= +
=
= +
=
= +
=
274. Solution: C
Let Y indicate the nominal value of the two-year bond, then:
2
0.05 1.05
9465= + , so 10,000.
1.08
1.08
YY
Y =
Thus, the amount of liability at the end of the second year is 10,500.
Hence, the liability at the end of the first year is:
10,500
= 5250.
2
So, the amount invested in the one-year bond is:
5250 10,000(0.05)
4481.
1.06
−
=
75
275. Solution: C
Macaulay duration of the liability is 3. Asset duration must equal 3.
Let P
1
and P
4
be the present values of the two assets.
14
41
14
3
11
14
4
41
4
3
14
3, then 2
20,000
20,000 50,000 1
1
, , :
1 0.8(1 ) :
50,000
1 22
(1 )
(1 )
(1 ) 1.25; (1 ) 1.077217
PP
PP
PP
PP
i
PP i
i PP
i
i
ii
+
= =
+
+
= = =
=
= = +
+
+
+
+ = +=
PV of assets must equal PV of liabilities. So, PV of assets equals:
111
20,000
2 3 3 55,699.07.
1.077217
PPP+== =
Amount of liability equals: 55,699.07(1.077217)
3
= 69,623.83.
276. Solution: D
( )
10 29
0.06
10
19 0.06
10 29
1
20
0.06
147.20 130.70 16.67
147.20 72.98 3.08 223.26
a v Da v
vv
++
=++
= + +=
277. Solution: C
14 13 2 12 14
14 15
2000 (1.08) (1.03)(1.08) (1.03) (1.08) (1.03)
1.08 1.03 /1.08
1 1.03 /1.08
32.284
61.95
XX X X
X
X
X
= + + ++
−
=
−
=
=
278. Solution: E
The present value of annuity X is
10
10
1
1.0331 1.0331 .
v
a
i
−
=
The present value of annuity Y is
2 12 10
2 10
22
1
() .
1 (1 ) 1
vv v
Pv v P P
vi
−−
++ = =
− +−
Equating the present values and solving,
2
2
(1 ) 1
1.0331 1.0331 1.0331(2.075) 2.14.
i
Ps
i
+−
= = = =
76
279. Solution: D
12
(12)
(12)
60 3 1
60 33 1
33
1.12 1
12
0.00948879
12
1
900
1.00948879
1556.43
1
1556.43 1194.78
1.00948879
i
i
P
P
P
−+
−+
= +
=
=
=
= =
280. Solution: C
( )
10 0.05
10 0.05
5000 10 0.08
10
5000 10 0.08
0.05
75,654.30
X Is
s
X
X
=
+
−
= +
=
281. Solution: B
8
12
8 0.03
38 1000
1056.16
BV a v= +
=
282. Solution: D
2
34 1
2
2 24
3
2 24
3
2 2 43
2 22 22
2 22
2
2
(1 ) ( )
(1 ) (1 ) (1 ) ( )
(1 1 ) (1 ) ( )
(1 1 ) ( )
2 ()()
2 2( )
1
10
0.61803
1 1.61803
0.61803
I I i PP
v v iv v
v v iv v
v v v vv
v vv v vv v
v vv v
vv
vv
v
i
i
+=+ +
− +− =+ +
− +− = + +
− +− = +
− += +
= +
= +
+−=
=
+=
=
77
283. Solution: C
2
2
2
2
2
500(1 ) 600
0.095445
100(1 ) 100(1 )
600(1.10) 100(1.10) 600
236
100(1 ) 100(1 ) 236
(1 ) (1 ) 2.36 0
1 1 9.44
(1 )
2
0.115549
0.0201
X
X
Y YZ
Z
Z
YY
YY
Y
Y
YX
+=
=
+ + +=
+=+
=
+ + +=
+ ++ − =
−± +
+=
=
−=
284. Solution: B
Assuming the loss of interest is the loss of the last 9 months interest:
( ) ( ) ( )
3 0.75 2.25 3
1.08 1.08 1.189 1
5.94%
j
j
−
= = = +
=
285. Solution: C
20 0.0925
10
360 %
12
1000 57, 485.26
500 57, 450.21
57, 485.26 57, 450.21 35.05
s
a
=
=
−=
78
286. Solution: B
Let j equal the quarterly rate and let k equal the four-year rate.
( )
4
1
4
1
300 (1 )
1
299
1
11
299
0.054875
(1 )
300
15.61
j
j
j
k
k
Xk
k
X
= +
=
+=+
=
+
=
=
287. Solution: A
The outstanding loan balance at any point in time is equal to the present value (at that same point
in time) of the remaining installment payments.
20
40
40
0.98
1
1.0075
1000(0.98) 6889.11
0.0075 0.02
B
−
= =
+
288. Solution: C
5
10 0.06
5
400 1000 1944.03
1944.03 0.06
461.51
OB a
Xa
X
= −=
=
=
79
289. Solution: C
Option 1
20 0.05
1500
120.3639
Total Payment 120.3639(20) 2407.28
Pa
P
=
=
= =
Option 2
( )
( )
( )
19
0
Total Payment 75* 20 1500 1500 75
1500 2*75 (1500 19*75)
1500 1500 (20) 75
19 20
1500 30,000 75
2
1500 15, 750
k
ii
ii
i ik
ii
i
=
= + + −+
− ⋅+ + −
= + −⋅
= + −⋅
= +
∑
2407.28 1500 15,750
0.0576
i
i
= +
=
290. Solution: C
(2)
15
5
15 / 2
(2)
(2)
1,100,000 1,000,000 40,000
.03153
2
0.06306
i
MV v a
i
i
= = +
=
=
291. Solution: B
2
2
2
4
(4)
(4)
150 100 80
80 100 150 0
100 100 4(80)( 150)
160
0.880199
0.136106
1.136106 1
4
0.129664
vv
vv
v
v
i
i
i
= +
+ −=
−± − −
=
=
=
= +
=
80
292. Solution: B
Since the bond is bought at a premium and redemption will occur to the investor’s greatest
disadvantage, assume the bond is called at the earliest possible redemption date, or simultaneous
with the coupon payment occurring at the end of the 15
th
year. The cash flows then become a
bond paying semiannual coupons of 40 for 15 years and returning 1000 at the end of the 15
th
year. At a yield of 7% effective annually:
2
(2)
(2)
30
30
1.07 1
2
0.034408
2
40 1000
1103.61
i
i
Pa v
P
= +
=
= +
=
293. Solution: A
( )
( ) ( )
( ) ( )
45
34
1500 1 4000 1
()
100 500 1 1000 1
Pi i
i
Pi
ii
−−
−−
′
+− +
−=
+ +− +
294. Solution: E
422
1.05 1.04 (1 )
0.0601
f
f
= +
=
295. Solution: B
Besides present values of assets and liabilities matching, 1) their modified durations must also
match, and 2) the convexity of the assets must exceed the convexity of the liabilities, in order for
the company’s position to be immunized against small changes in interest rate. Only company V
satisfies all these conditions.
296. Solution: B
Let j be the monthly interest rate.
0.012
50,000(1 ) 800 49,800
0.012
50,000 800
116.22
n
j
j
a
n
+− =
=
=
=
Drop payment at payment number 117.
81
297. Solution: C
27
27
27 27
27
21
0.75
37
21
0.75 (1 )
37
16
0.18243
37
0.03248
0.75
4
(1 )
3
9
i
n
n
F iFa Fv
vv
v
i
F Fv
i
n
= +
= −+
=
=
=
+=
=
298. Solution: A
5
5
1.04
1
1.04 1
1.12
3.87 17.26 21.13
0.12 0.04 1.12 0.12 0.08
−
+ =+=
−−
299. Solution: B
Let x be the amount invested in the one-year bond and y the amount invested in the four-year
bond. First match the present value of assets and liabilities:
2000
AL
PV PV
xy
=
+=
Second, the durations of assets and liabilities should also match:
14
1 4(2000 )
3
2000
666.67
A
AL
xy
D
xy
xx
DD
x
+
=
+
+−
= =
=
=
Convexity of the assets is:
22
666.67(1 ) 1333.33(4 )
11
2000
+
=
Convexity of the liability is:
2
39=
.
Convexity of assets is greater than convexity of liabilities so Reddington immunization is met.
82
300. Solution: C
40 0.02
20
20
20 0.02
20
0.02
5,000,000
182,778.74
5,000,000(1.02) 183, 000
2,983,318.31
2,983,318.31 200,000
17.89
n
Xa
X
OB s
OB
a
n
=
=
= −
=
=
=
20 original payments plus 18 with the drop payment equals 38 total payments.
301. Solution: A
Since the coupon rate per coupon payment period 4% is greater than the effective rate of interest
per coupon payment period 2.9563%, it is to the disadvantage of the bond holder to have the
bond redeemed at an early date. Hence, we only need to calculate the present value of such a
bond at the worst-case scenario, which is that the bond is called at the end of the 5
th
year.
10
10 0.029563
4 100
108.92
Pa v
P
= +
=
302. Solution: D
20
20
0.9
1
1.03
100 (
1.03)
0.03
0.10
1295.80
FV
FV
−
=
+
=
Alternatively,
19 20 1
19 18 19 0
1
(1.03) (0.9) (1.03)
100[(1.03) 0.9(1.03) (0.9) (1.03) ] 100 1205.80.
1 0.9(1.03)
FV
−
−
−
= + ++ = =
−
303. Solution: D
First, find
()at
2
00
( ) exp exp exp /100
50
tt
r
r
a t dr dr t
δ
= = =
∫∫
The balance in the account at time 10 is:
X
a
a
Xa 31637.248.815
)4(
)10(
)10(300 +=+
The total interest earned from
0 10t to t= =
is:
815.48 2.31637 (300 ) 4 192.08.X X XX+ − + = =>=
83
304. Solution: A
Since Asset Y provides a cash flow at the same time that the liability is due (t = 4), we can apply
its 250,000 value to reducing the liability amount from 750,000 to 500,000. Then, we can
establish the following two equations, both using t = 4 as the reference point for all cash flows.
22
500,000 (0.95) (0.95)
X ZX Z
Av Av A A
−−
= += +
Second, taking the derivative (with respect to v) of both sides of the first equation, we have:
33
2
0 2 2 (0.95)
0 2 (0.95) (0.95)
XZX Z
XZ
Av A A A
AA
−−
−
=− +=− +
=−+
Then, subtracting the second equation from the first equation yields:
2
500,000 3 (0.95)
150,416.67
X
X
A
A
−
=
=
305. Solution: D
4
(4)
(4)
5
1.06 1
4
0.014674
4
50,000(1.014674) 2,125
31.86
n
i
i
a
n
= +
=
=
=
There will be 31 payments of 2125.
306. Solution: A
The PV and duration of the liability payments using 7% rate are
12
1,750,000 777,021PV v= =
and duration 12.
The amount invested in the 5-year bond is
5
242,180
172,671
1.07
=
,
Thus, the amount invested in the 14 year bond is 777,021-172,671=604,350.
The maturity value of the 14-year bond is
14
604,350(1.07) 1,558,337.=
The surplus if the interest rate moves to 4% is:
5 14 12
242,180 1,558,337 1,750,000
5,910
1.04 1.04 1.04
AL
PV PV
−= + − =
.
84
307. Solution: A
Let
() () ()
AL
h i PV i PV i= −
.
Full immunization of a single liability requires both equations:
() 0, '() 0hi h i= =
32
13
24 3
13
20,000 0
3 40,000 0
Av Av v
Av Av v
+− =
+− =
1
.
1.055
v =
Solve these two equations in two unknowns to get
1
A
= 9478.67
308. Solution: D
Let
123
,,FFF
be the redemption amounts of each bond for purchase.
To exactly match the liabilities with cash income:
3
3
2
2
1
1
1000
1000
1000 (1.02)
980.39
1000 980.39(0.02) (1.01)
970.69
F
F
F
F
F
F
=
=
=
=
−=
=
The total purchase price is
23
970.69(1.01) 980.39(0.02) 980.39(1.02) 1000
2241.82.
1.14 1.15 1.15 1.18
+ + +=
309. Solution: B
The PV of the annuity following the 11
th
payment is:
9 0.06
10 68.0169.a =
The effective semi-annual rate is
(2)
1/2
1.06 1 0.02956301.
2
i
j = = −=
Next,
1
68.0169
0.02956301 0.005
1.67
PV K
K
= =
−
=
85
310. Solution: B
Split this into three perpetuities with payments 3 years apart. Find the three-year interest rate:
( )
1
3
1.125 1 , 0.423828jj=+=
.
The present value of the three perpetuities, starting at times 0, 1, and 2 is:
2
100(1.423828) (1.423828) 100(1.423828)
9450
0.423828 0.423828(1.125) 0.423828(1.125)
2963.19.
X
X
++ =
=
311. Solution: E
Note that had the borrower 1) charged X at the end of month 0, and 2) paid off the remaining
3000 at the beginning of month 16, then the initial and final balances would have become 0. In
that situation, the present value of the amounts charged to the credit card, minus the present
values of the payments, would have been 0 in the 15-month period.
The amounts charged to the card were 79.99 at each of times
0.5
12
n
t
−
=
, for each whole number
of n from 1 to 15.
The monthly payments were 250 at each of times
12
n
t =
, for each whole number n from 1 to 15
inclusive.
Then to make the final balance 0, the final (additional) payment would have been 3000 at time
15 5
12 4
t = =
.
Therefore, we have
( )
( )
( )
( ) ( )
( )
15 15
0.5 5
11
12 12 4
15 15
0.5 5
11
12 4 12
79.99 250 3000
0
1.168 1.168 1.168
79.99 3000 250
1.168 1.168 1.168
nn
nn
nn
nn
X
X
−
= =
−
= =
+ − −=
+=+
∑∑
∑∑
.
312. KEY: A
The accumulated value is
2
0
( ) exp 0.03 0.005 exp[0.03 0.0025 ]
t
a t r dr t t
=+=+
∫
, The
account balance is
(2) 1.0725
100 (2) 100 100(1.0725) 100 211.07.
(1) 1.0330
a
a
a
+= + =
86
313. Solution: E
Break this into two parts – the first 30 increasing payments and the remaining level perpetuity.
( ) ( )
30
29
29
1.03
1
1
1.07
350 1.07 350 1.03 8033.38
0.07 0.03 0.07
v
−
+=
−
314. Solution: E
Let n index the payment times in months. Then for a payment at time n, the discount factor is
( )
/6
1
1.045
n
. The end-of-month payment at month n is be
500 ( 1)+−nX
.
Therefore, we have
( )
60
/6
1
500 ( 1)
30,000
1.045
n
n
nX
=
+−
=
∑
.
315. Solution: A
Let i be the yield rate, v = 1/(1 + i), and let n be the term. For Bond A,
20,000 10,000
n
v =
and
so
0.5
n
v =
. For Bond B,
10,835.58( 0.04 ) 10,000
n
ni
va+=
and so
10,000
0.5 / 0.04 10.5721
10,835.58
ni
a
= −=
.
For Bond C,
10,000 ( 0.03 ) 0.81716 12, 237.51.
n
ni
Xv a X X= + = ⇒=
316. Solution: C
[ ]
10
2
10 0.052
11
50,000 ( 200) 200
0.052 0.052
50
,000 (7.647284) (0.602341) 19.230769 3846.153846 73964.50
50,000 19.230765 46,868.54
162.83
ka v k
kk
k
k
= ++ +
=+ ++
= +
=
87
317. Solution: B
Let
A
r
represent the coupon rate of bond A. The coupon rate of bond B is then
0.005+
A
r
.
From the given information,
( )
30 30
30 0.07 30 0.07
30
30
0.07 30 0.07
11
3000 1000 0.005
(1.07) (1.07)
2
3 2
0.005
(1.07
)
3 0.26273 24.81808 0.06205
3 0.26273 0.06205
0.1078 10.78%
24.81808
AA
A
A
A
ra r a
ra a
r
r
= + + ++
=++
=++
−−
= =
=
.
318. Solution: C
[ ]
20 12 1
12
9
7 ()
1
7
0.0375
9.7497
9.7497 (0.0375) 35
1193.33
P Ci gv
Ci Cg
Ci Fr
C
C
−+
= = −
= −
= −
= −
=
319. Solution: C
40
40 0.025
0.025
350 1.1
8785.97 (0.409674)
14,883.24
1.1 16,371.57
14,724.91 350 16,371.57
48
24
AA
AA
A
A
m
m
P a Pv
PP
P
RP
av
m
n
= +
= +
=
= =
= +
=
=
88
320. Solution: A
Let i represent the common yield rate of the two bonds.
Since the modified duration is the Macaulay duration divided by (1 + i) and i > 0, the Macaulay
duration of each bond is greater than its modified duration. Since a < d < b, the Macaulay
duration of d years must be associated with the bond with modified duration a years.
Since the bonds have the same yield rate, the ratio of the two types of duration is the same for
each bond. So if x represents the Macaulay duration of the other bond in years, we have
d / a = x / b implies ax = bd implies x = bd / a.
The Macaulay duration of the other bond is bd / a years.
321. Solution: C
( )
( )
3x4 2 4
0.05 3
0.04 0.06
1000 1 1
42
1670.42
x
e
−
+−
=
322. Solution: C
( )
( )
( )
45
45
45
45
45 0.0525
75.6
7
1.2
71 1
75.6
1.2
7 1 7 63
1 10
0.0525
7 1200
i
s
i
i
ii
i
i
i
Xs
=
+−
=
+ −=
+=
=
= =
89
323. Solution: C
10
89
8
2
1.06
1
1.10
20,000
0.10 0.06
2584.39
1.06 1.06
7353
1.10 1.10
X
X
LB X
−
=
−
=
= +=
324. Solution: D
[ ]
0
0
5
1
( ) exp
8
exp ln( 8)
exp ln( 8) ln(8)
88
exp ln
88
(5) 13 / 8
1.08333
(4) 12 / 8
0.08333
t
t
a t dr
r
t
t
tt
a
a
i
=
+
= +
= +−
++
= =
= =
=
∫
325. Solution: D
2
2 /2
98.5 3 100
0.037929
2
0.07586
j
av
j
j
= +
=
=
326. Solution: D
( )
( )
( )
1
13
0.04
20
4
20 0.04
20
4
20 0.04
Adjustment in book value
0.08 0.04 43.24
1800
1800 0.08 1800
144 1800 2778.50
nt
i
g iv F
vF
F
P av
av
−+
= −
=−=
=
= +
= +=
90
327. Solution: A
1
1
1.0614
5.5554
0.0614
7
n
n
−
=
=
328. Solution: C
36
36 0.01
36 0.01
8000
15,000 1000
1.01
8408.60
8408.60 (30.10751)
279.29
Xa
Xa
X
X
−=+
=
=
=
329. Solution: B
1
10
20000 20000
9.562
1.1
1
1.09
0.09 0.1
2091.61
R = =
−
−
=
The fourth payment is
( )
2091.61 1.1 2300.71=
The principal outstanding before the fourth payment is
9
1.1
1
1.09
2300.71 19708.94
0.09 0.1
−
=
−
Interest in the fourth payment is
( )
19708.94 0.09 1773.80= =
Principal repaid 2300.71 1773.80 526.91X = =−=
330. Solution: E
Interest earned on the first account is
( ) ( )
32
00
11
3 3 1 3 1 2 3.
dt dt
tt
ee
++
∫∫
− = + −+ =
Interest on the second account is
( )
32
1.05 1.05 0.055125XX−=
.
3 0.055125 54.42XX= ⇒=
.
91
331. Solution: D
5 32
1.0725 1.0575 (1 )
0.0954
f
f
= +
=
332. Solution: C
( )
5
5 0.08 2 0.08
150,000 14,000 1.08
150,000 (88,703.01 2.2464 )1.46933
5,958
s Xs
X
X
= +
= +
=
333. Solution: A
I is false as it is true for small parallel changes in interest rates only.
II is false, as the durations will be equal.
III is false as the convexity of assets should be greater than the convexity of liabilities.
334. Solution: B
5
20 0.04
5
5
9550(1 ) 756.97
9550(1 ) 10,698.97
(1 ) 1.12031
0.02298
ia
i
i
i
+=
+=
+=
=
335. Solution: B
( )
600(1 ) 1000
10
(1 )
6
0.6
600
2
1
600
2
1
600 1 0.6 0.6
2
750
n
n
n
n
ni
n
n
i
i
v
i
F a Fv
iv
Fv
i
F
F
+=
+=
=
= +
−
= +
= −+
=
92
336. Solution: B
29 0.03
29 0.03
29
40( ) 30(1000) 40 30(1000) 53,
433.89
0.03
s
Is
−
+= +=
337. Solution: C
0.0625
33
33 0.0625
33
33
5020 360
33.85
Use 33
5020 360
5020 4980.95
39.05
288.75
360 288.75 648.75
n
a
n
n
a Xv
Xv
Xv
X
B
=
=
=
= +
= +
=
=
=+=
338. Solution: A
For a 7.5% yield rate, the present value and Macaulay duration of the assets are, respectively,
30,000 + 20,000 = 50,000 and
30,000(28) 20,000(35)
30.8
30,000 20,000
+
=
+
.
The present value and Macaulay duration, of the liabilities are, respectively,
50,000(1.075)
50,000
(1.075)
y
y
=
and y.
Note that the present values of assets and liabilities already match. Since Macaulay durations
must match, y = 30.8.
339. Solution: B
Use the full immunization equations and let N be the maturity value of the asset maturing in n
years.
7 ( 12)
7 ( 12)
242,180(1.07) (1.07) 1,750,000 0
242,180(7)(1.07) ( 12)(1.07) 0
n
n
N
Nn
−−
−−
+ −=
−− =
From the first equation:
( 12) 7
(1.07) 1,750,000 242,180(1.07) 1,361,112.
n
N
−−
=−=
Substituting this in the second equation:
7
12 242,180(7)(1.07) /1,361,112 2n −= =
and so n = 14.
93
340. Solution: B
2 2 3 29 30 30
30
30
2895.28 (1.01) (1.01) ... (1.01) 1000
1.01
1
1.031
1000 2895.28
0.031 0.01
113.75
Rv R v R v R v v
Rv
R
= + + ++ +
−
=> +=
−
=>=
341. Solution: B
( )
0.12
1
0.12749685
1
3000 1.12749685
0.12749685 0.07
58,829.14
ie
i
= −
=
−
=
342. Solution: E
If accumulating the payments, they accumulate for 0, 1, …, 39 periods. This accumulation is
reflected in the left-hand sides of (A) and (D). Since the accumulated value is X, neither right-
hand side is correct. If discounting the payments to time zero, the payments are discounted for 1,
2, …, 40 periods as reflected in the left-hand side of (E). The right-hand side of (E) discounts the
accumulated value of X for 40 periods and hence is the correct value. Answers (B) and (C) do
not reflect the time-zero present value.
343. Solution: D
The cost to purchase the bond at the end of year 3 is
19.1885)03.1(2000
2
=×
−
. Subtracting this
cost from 2260.19 we get 375, the amount of interest paid which is 3.75% of 10000. Thus the
interest rate is 3.75%.
94
344. Solution: A
Let a, b, and c represent the face values of the three bonds. One, two, three, and four years from
now, respectively:
the 1-year bond provides payments of 1.01a, 0, 0, 0;
the 3-year bond provides payments of 0.05b, 0.05b, 1.05b, 0; and
the 4-year bond provides payments of 0.07c, 0.07c, 0.07c, 1.07c.
The total payments one, two, three, and four years from now must match the liabilities.
Therefore, we have
1.01 0.05 0.07 5766
0.05 0.07
1.05 0.07 15421
1.07 7811
++=
+=
+=
=
abc
b cX
bc
c
Note that to find X, we do not need the first equation.
Solving the fourth equation for c yields
7811
7300
1.07
= =c
.
Substituting this value of c into the third equation and solving for b yields
15421 0.07(7300)
14200
1.05
−
= =b
.
Finally, substituting these values of b and c into the second equation yields
0.05(14200) 0.07(7300) 1221
= +=X
.
345. Solution: E
(2)
(2)
2
2 0.059
49
(2) 0.10
980
0.10 0.018 0.118
0.059
2
915.70 49 1000
2 12
6
n
n
r
i
i
av
n
n
= =
=+=
=
= +
=
=
95
346. Solution: B
The price of a one-year bond for 1000 is
2
30 1030
1017.45
1.021
1.021
+=
.
Therefore, to match the payment at time 2 we need to invest
1500
1017.45 1481.72
1030
=
in the one-
year bond.
The one-year bond gives a payment of
1500
30 43.69
1030
=
at time 0.5. Therefore, the amount that
needs to be invested in the six month zero coupon bond is =
66.1922
0175.1
69.432000
=
−
.
The total cost of the dedicated portfolio is: 1481.72 + 1922.66 = 3404.38.
347. Solution: A
The current value for couch #2 is
0.1(4/12)
260 1500 1450.82 1190.82e X XX
−
= −= −⇒=
.
The current value for couch #1 is
0.1(6/12) 0.1(2/12)
1500 1190.82 1426.84 1210.83 216.ee
−
− =−=
348. Solution: A
20
1.05
1
1.04
2500 52
,732.61
0.04 0.05
X
PV
−
= =
−
Annuity Y has the same increasing percentage as interest rate, so:
30
Y
PV k=
1757.75
XY
PV PV k= =>=
349. Solution: A
Solution:
0.5(0.4)
(1/ 2)
2.4
2.0
2
1 exp
0.5 10
d
dt
t
−
−=
−
∫
( )
( )
0.2
(1/ 2) 2.4
2.0
2
0.2
(1/
2)
(1/ 2)
1 2 exp 2ln(10 ) |
8
12
7.6
0.20063
dt
d
d
−
−
− =−−
−=
=
96
350. Solution: D
The price of Bond A is
123 3
60(1.04 1.04 1.04 ) 1000(1.04 ) 1055.50,
−−− −
++ + =
while the
Macaulay duration of Bond A is
12 3 3
60[1.04 2(1.04 ) 3(1.04 )] 3(1000)(1.04 )
2.838.
1055.50
−− − −
++ +
=
Note that the one-year zero-coupon bond has duration 1.
Let w denote the proportion of wealth to invest in Bond A; then,
1 w−
is the proportion of wealth
invested in Bond B. Then
2 2.838 1(1 ),ww= +−
or w = 0.5440.
351. Solution: A
Let K be the amount that Bank B paid. Equating the amount borrowed (the 16 payments
discounted at 7%) to the actual payments received (using the 6% yield rate) gives the equation
8
16 7% 8 6%
1000 1000
1.06
K
aa= +
.
Then,
8
16 7% 8 6%
8
1000( )1.06
1000(9.44665 6.20979)1.06
3236.86(1.59385) 5159.06.
K aa= −
= −
= =
352. Solution: A
( )
640(1 ) 1000
(1 ) 1.5625
0.64
640
2
1
640
2
1
640 1 0.64 0.64
2
780.49
n
n
n
n
ni
n
n
i
i
v
i
F a Fv
iv
Fv
i
F
F
+=
+=
=
= +
−
= +
= −+
=
353. Solution: E
The price is P = 950. The modified duration is
' 4750
5
950
P
P
−
−=− =
.
Macaulay duration is
(1.09)(5) 5.45=
97
354. Solution: B
Because Bond A sold for its fact amount, the yield rate is the coupon rate of 6% per year.
The present values of the three bonds are:
Bond A: 1000 (given)
Bond B:
5
1000(1.06) 747.26
−
=
Bond C:
10
1000(1.06) 558.39
−
=
The durations are:
Bond A: 7.8017
Bond B: 5
Bond C: 10
The portfolio duration is the average of these three durations, weighted by the bond prices.
Duration =
426.7
39.55826.7471000
)10(39.558)5(26.747)8017.7(1000
=
++
++
355. Solution: A
The present value of the two payments is:
5
1,000,000
1,000,000 1
,821,927.07
1.04
+=
The present value of the perpetuity is:
2
2
1000
(1.04) 1
,821,927.07
0.04 0.04
1000
1,751,852.99
0.04 0.04
1,126,852.99
0.04
45,074.12
X
X
X
X
+=
+=
=
=
356. Solution: C
( )
6
2
6
2
4(2)
0.5
1000exp
5 0.5
1000exp ln(5 0.5 ) |
8
1000exp ln 8 ln 6 1000 1333.33
6
0.08
1333.33 1
4
1134.35
X dt
t
t
Y
Y
−
=
+
= +
= −= =
= −
=
∫
98
357. Solution: D
The quarterly interest rate is
1/4
1.08 1 0.01943−=
( )
135
24 0.01943
1000 1.08 1.08 1.08
19,406.51 2.40034
8085
aX
X
X
−−−
= ++
=
=
358. Solution: E
365 20
1
1 44.6%
0.98
−=
359. Solution: C
First calculate the three-year interest rate.
3
1.06 1 0.191016−=
This is an arithmetic increasing perpetuity.
2
11
655.56
0.191016 0.191016
20.08
X
X
+=
=
360. Solution: D
The present value of the first year’s payments is:
1
12
12 0.006434
1.08 1 0.00643403
500 5756.43a
−=
=
This perpetuity can be thought of as a geometrically increasing perpetuity-due with first payment
5756.43. The present value is:
1
5756.43 1.08
0.08 0.05
207, 231.44
−
=
99
361. Solution: D
( )
( )
( )
12
12
60
5000 1.01
5000 5000
1
1.01 1
44,423.95 81.67
543.94
i
Xs i
di
X
X
= = +=
−
=
=
362. Solution: B
10 0.04
10 0.04
10
10
10,000(10) 600 600 137, 295.27
0.04
55,400(1 ) 137, 295.27
0.095.
s
s
i
i
−
++ =
+=
=
363. Solution: E
5 0.04
5 0.04
5
1000(5) 50 50 5791.22
0.04
s
s
−
++ =
364. Solution: A
The total amount paid is n. The initial loan amount is
ni
a
The total interest paid equals
ni
na−
.
365. Solution: C
( )
( )
( )
( )
5
5
2
2
2
5
5
1
100 1 0.05 exp 100 1.1025 exp ln(1 )
1
6
110.25 220.50
3
100
220.50
1
100
1 0.453515
220.50
1 0.853726
0.1463
t
t
d
d
d
d
+= +
+
= =
=
−
−= =
−=
=
∫
100
366. Solution: A
( )
5
5 0.13 3 0.13
12
100 50 1.13 1703.81
1703.81
1.12 1
99.33
ss
x
X
+=
=
−
=
367. Solution: B
12
240 0.006434
30 0.07
(1 ) 1.08
0.006434
225 128,848.51
128,848.51
128,848.51 (13.27767)
9704.15
j
j
s
Xa
X
X
+=
=
=
=
=
=
368. Solution: C
20 0.015
4
16 0.015
4
10,000
582.46
582.46
8230.86
Pa
P
OB a
OB
=
=
=
=
Remaining payments total:
16(582.46) 9319.32
9319.32 8230.86 1088.46.
=
−=
369. Solution: A
12
180 0.0073
59
121 0.0073
59
1.0912 (1 )
0.0073
100,000
1000.02
1000.02
80,174.59
j
j
Pa
P
OB a
OB
= +
=
=
=
=
=
12
1.0576 (1 )
0.004678
j
j
= +
=
Interest portion of first revised payment:
80,174.59(0.004678) 375.04.=
101
370. Solution: A
Define
i
′
as the quarterly effective interest rate for the loan
Solve for
i
′
20
291.23 5000
1.5%
i
a
i
′
=
′
=
Then, solve for j using the formula
( )
12
4
1 1 0.015
12
0.0597.
j
j
+=+
=
371. Solution: B
All prices imply an interest rate of 6% indicating that the yield rate does not change with
duration, that is. the yield curve is flat.
372. Solution: E
Macaulay duration at an interest rate of 5% is
( )
5
5
5
5
100 5000
100 1000
100(12.56639) 3917.63
4.2535
100(4.32947) 783.53
Ia v
av
+
+
+
= =
+
373. Solution: C
12
24 0.0075
0.0075
1000 21,889.15
21,889.15 2000
11.46
n
OB a
a
n
= =
=
=
Drop payment will be made at the 12
th
month.
102
374. Solution: E
120 0.0075
0.0075
30,000
380.03
33, 000 380.03
141
n
Pa
P
a
n
=
=
=
=
375. Solution: B
1
6
5.76
1
0.041667
5
4.8
1.041666
d
v
r
r
r
v
=
+
=
+
=
= =
376. Solution: D
( )
10
10
10
9.26
1
0.079914
32.70395
4.8720.
6.71270
i
i
Ia
d
a
=
+
=
= = =
377. Solution: E
0.01 200 0.0201
0.01 200
0.0201
0.01
2.01
2.01
98.13168
400
n
n
n
Xa Xa
aa
a
n
=
=
=
=
378. Solution: C
Let r be the coupon rate.
20
20 0.07
2300 2000 2000
2000 168.32
ra v
r
= +
=
Bond is bought at a premium, so assume called as early as possible at year 18.
18
18 0.07
168.32 2000
2284.85
Pa v
P
= +
=
103
379. Solution: C
Since the annual effective discount rate is 3.2%, the present value of an amount is calculated by
multiplying it by a discounting factor of
(1 0.032) (0.968)
tt
−=
, where t is the number of years
since the deposit.
At time t = 0, an initial deposit of 50,000 is made just after the balance is 0 (the account is new
just before the deposit). The withdrawals are then X at each of times t = 2, 4, 6, 8, 10, 12 or
equivalently at time
2tk=
for each whole number k from 1 to 6 inclusive.
Then to make the final balance 0, an additional withdrawal of 45,000 at time t = 12 would be
needed.
Since the net present value of the cash flows (withdrawals minus deposits) must be zero, in a
time period from a zero balance to another zero balance, we have
6
12 2
1
45,000(0.968) (0.968) 50,000 0
k
k
X
=
+ −=
∑
6
12 2
1
45,000(0.968) (0.968) 50,000
k
k
X
=
+=
∑
.
380. Solution: E
0.045
3400 240
23.05
n
a
n
=
=
Use 23 for the number of full payments of 240. X will be paid at time 24.
24
23 0.045
24
24
3400 240
3400 3395.47
4.53
13.04
a Xv
Xv
Xv
X
= +
= +
=
=
381. Solution: C
The minimum yield will occur at the earliest redemption date since the bond is brought at a
premium. Therefore n = 18.
18
18
1321 111.10 1111
0.07985
i
av
i
= +
=
104
382. Solution: A
Conditions for full immunization of a single liability cash flow are:
1. PV (Assets) = PV (Liability)
2. Duration (Assets) = Duration (Liability)
3. The asset cash flows occur before and after the liability cash flow.
Checking Condition 1, only I and II are present value-matched, so can rule out III at this point.
Checking Condition 2, only I is duration-matched, so can rule out II at this point.
Checking Condition 3, the asset cash flows do occur before and after the liability for I.
So only I fully immunizes the liability.
383. Solution: B
Let i represent the effective market annual yield rate and
1
1
v
i
=
+
. The Macaulay duration is
3.70 years, which is equal to the present-value-weighted times of the liabilities. Therefore, we
have
55
55
55
5
20,000(0) 100,000 (5) 25
3.70
20,000 100,000 1 5
3.70 18.5 25
3.70 6.5
0.89342
1 1.11929
vv
vv
vv
v
v
i
+
= =
++
+=
=
=
+=
.
Modified duration equals Macaulay duration divided by (1 + i), so the modified duration is
3.70
3.30567
1.11929
=
years.
384. Solution: A
Using the basic formula:
( )
( )
23
58 57
57|
58|
6.88
87.5 87.5 6.88
39.0667 0.00625 6.88
5150
BV BV
a Cv a Cv
C
C
−=
+− +=
−=
=
105
385. Solution: E
12
12 0.02
12
12 0.02
1000
500 2 1000
P Xa v
P Xa v
= +
+= +
Subtract the first equation from the second.
12 0.02
500
47.28
Xa
X
=
=
