The Theory of Interest - Solutions Manual
Chapter 4
1. The nominal rate of interest convertible once every two years is j, so that
 .07 
1  j  1 

2 

4
and j  1.035   1  .14752.
4
The accumulated value is taken 4 years after the last payment is made, so that
2000s8 j 1  j   2000 13.60268 1.31680 
2
 $35,824 to the nearest dollar.
2. The quarterly rate of interest j is obtained from
1  j 
4
 1.12 so that j  .02874.
The present value is given by
600a40 j  200a20 j
 600  24.27195  200 15.48522
 $11, 466 to the nearest dollar.
3. The equation of value at time t  8 is
100 1  8i   1  6i   1  4i   1  2i   520
so that
4  20i  5.2, or 20i  1.2, and i  .06, or 6%.
4. Let the quarterly rate of interest be j. We have
400a40 j  10,000 or a40 j  25.
Using the financial calculator to find an unknown j, set N  40 PV  25 PMT  1
and CPT I to obtain j  .02524, or 2.524%. Then
12
 i 12 
4
12 
1 
  1.02524  and i  .100, or 10.0%.
12 

5. Adapting formula (4.2) we have
2000
s32 .035
s4 .035
1.035 8
 57.33450  
  2000  
 1.31681  $35,824 to the nearest dollar.
 4.21494 
34
The Theory of Interest - Solutions Manual
Chapter 4
6. (a) We use the technique developed in Section 3.4 that puts in imaginary payments
and then subtracts them out, together with adapting formula (4.1), to obtain
200
 a176  a32  .
s4
Note that the number of payments is
176  32
 36, which checks.
4
(b) Similar to part (a), but adapting formula (4.3) rather than (4.1), we obtain
200
 a180  a36  .
a4
Again we have the check that
180  36
 36.
4


d 12 .09

 .0075 and the monthly discount
7. The monthly rate of discount is d j 
12
12
factor is v j  1  d j  .9925. From first principles, the present value is
1  .9925
300 1  .9925  .9925    .9925    300
6
1  .9925
upon summing the geometric progression.
120
6
12
114
8. Using first principles and summing an infinite geometric progression, we have
v v v 
3
6
9
v3
1
125



3
3
1 v
1  i   1 91
and
1  i 3  1  91 or 1  i 3  216
125
125
1
 216  3 6
and 1  i  
   1.2 which gives i  .20, or 20%.
5
 125 
9. Using first principles with formula (1.31), we have the present value
100 1  e.02  e.04 
and summing the geometric progression
100
1  e.4
.
1  e.02
35
 e .38 
The Theory of Interest - Solutions Manual
Chapter 4
10. This is an unusual situation in which each payment does not contain an integral
number of interest conversion periods. However, we again use first principles
8
140
4
measuring time in 3-month periods to obtain 1  v 3  v 3   v 3 and summing the
geometric progression, we have
1  v 48
4 .
1 v 3
11. Adapting formula (4.9) we have
 
 
2400a104 .12  800a54.12 .
Note that the proper coefficient is the “annual rent” of the annuity, not the amount of
 
each installment. The nominal rate of discount d 4 is obtained from
4
1
 d  4 
 4
4
1 
  1  i  1.12 and d  4 1  1.12    .11174.
4 

The answer is
10
1  1.12 
2400 
.11174
12. (a)
(b)
1  1.12 
 800
.11174
5
 $11, 466 to the nearest dollar.
m
1 m tm
1 1m
1  vn 1  v 1  vn
 m
 
v
a

a
v

a
a

  m    m   anm .

n
n 
n 1
m t 1
d
i
i
t 1 m
The first term in the summation is the present value of the payments at times
1 ,1  1 , , n  1  1 . The second term is the present value of the
m
m
m
2
,1  2 , , n  1  2 . This continues until the last term
payments at times
m
m
m
is the present value of the payments at times 1, 2, , n. The sum of all these
 
payments is anm .
13. The equation of value is
 
1000 n a2  10,000 or
 
n
a2  10,
where n is the deferred period. We then have
vn
 2
n
.
n
 2   10 or v  10d
d
Now expressing the interest functions in terms of d, we see that
 
 
a2  v n a2 
v  1  d and d
 2
 2 1  1  d  2  .
36
1
The Theory of Interest - Solutions Manual
Chapter 4
We now have
1  d n  20 1  1  d .5 
.5
n ln 1  d   ln 20 1  1  d  
and
n
.5
ln 20 1  1  d  
.
ln 1  d 
14. We have
 
 
 
3an2  2a22n  45s1 2
 1  vn 
 1  v2n 
i
or 3   2   2   2   45  2 .
i
 i

 i

Using the first two, we have the quadratic
3 1  v n   2 1  v 2 n  or 2v 2 n  3v n  1  0
which can be factored  2v n  1 v n  1  0 or v n  1 , rejecting the root v  1. Now
2
using the first and third, we have

3 1 1
3 1  v   45i or i 
n

2  1.
45
30
15. Using a similar approach to Exercise 10, we have
1 v 4  v 4 
3
6
v
141
4

1  v36
3 .
1 v 4
 
 
16. Each of the five annuities can be expressed as 1  v n divided by i, i m ,  , d m , and d,
respectively. Using the result obtained in Exercise 32 in Chapter 1 immediately
establishes the result to be shown. All five annuities pay the same total amount. The
closer the payments are to time t  0, the larger the present value.
17. The equation of value is
2400an  40,000 or an  50 .
3
Thus
an 
1  e.04 n 50

.04
3
or
37
The Theory of Interest - Solutions Manual
Chapter 4
1  e.04 n 
and
2
3
e.04 n 
1
3
.04n  ln 3  1.0986, so that n  27.47.
18. We have
an 
1  vn

 4 or v n  1  4
and
1  i n  1
n
sn 
 12 or 1  i   1  12 .

Thus, 1  12 
1
8 1
 .
leading to the quadratic 1  8  48 2  1, so that  
1  4
48 6
19. Using formula (4.13) in combination with formula (1.27), we have
n
an   v dt  
t
0
n
0
t
t
n  1 r 
 r dr
e 0   e 0

1
dr
0
dt
Now
t
1 r 
e 0

1
dr
1
 e ln 1t  1  t  .


Thus,
1
an   1  t  dt  ln 1  t 0  ln  n  1 .
n
n
0
20. Find t such that vt  a1 
1 v


iv

. Thus, t ln v  ln v  ln
i

21. Algebraically, apply formulas (4.23) and (4.25) so that
 Da n 
n  an
i
and t  1 
 Ia n 
. Thus,
 Ia n   Da n  1  an  nv n  n  an 
i
 1  vn 
1
  an  1  v n  nv n  n  an    n  1 
   n  1 an .
i
 i 
38
1

i
ln .

an  nv n
i
and
The Theory of Interest - Solutions Manual
Chapter 4
Diagrammatically,
Time:
0
 Ia n :
 Da n :
1
1
2
2
3
3
n
n 1
n 1
n2
n 1
n 1
Total:
n 1
n 1
2
n 1
n
n
1
n 1
22. Applying formula (4.21) directly with P  6, Q  1, and n  20
Pan  Q
an  nv n
i
 6a20 
a20  20v 20
i
.
23. The present value is
v4
1
10  a10   10v 4  a14  a4 

i
i
1
 10 1  ia4   a14  a4 
i
1
 10  a14  a4 1  10i   .
i
v 4  Da 10 
24. Method 1:
Method 2:
1
an  nv n  nv n 

i
1  i  an an
a
 n 
 .
i
i
d
1 1 
PV   Ia   v n  Ia   1  v n    2 
i i 
n
n
a
1 v  1 1 v

 n.
1   
id
d
 i  i 
PV   Ia n  v n na 
25. We are given that 11v6  13v7 from which we can determine the rate of interest. We
have 111  i   13, so that i  2/11. Next, apply formula (4.27) to obtain
2
P Q 1 2 11
 11 
 2   2   2    66.
i i
i i
2
2
26. We are given:
v
v2
2


X  va 
and 20 X  v Ia   .
i
id
39
The Theory of Interest - Solutions Manual
Chapter 4
Therefore,
v
v2
X 
i 20id
or 20d  v  1  d and d  1/ 21.
27. The semiannual rate of interest j  .16 / 2  .08 and the present value can be expressed
as
 10  a10 
300a10 .08  50  Da 10 .08  300a10 .08  50 

 .08 
 10  A 
 300 A  50 
  6250  325 A.
 .08 
28. We can apply formula (4.30) to obtain
2
19

1.05  1.05 
 1.05  
PV  600 1 

  
 
 1.1025  
 1.1025  1.1025 
1  1.05 /1.1025 20 
 600 
  $7851 to the nearest dollar.
 1  1.05 /1.1025  
29. We can apply formula (4.31)
i 
i  k .1025  .05

 .05, or 5%,
1 k
1  .05
which is the answer.
Note that we could have applied formula (4.32) to obtain PV  600a20 .05  $7851 as an
alternative approach to solve Exercise 28.
30. The accumulated value of the first 5 deposits at time t  10 is
1000s5 .08 1.08  1000 6.335931.46933  9309.57.
5
The accumulated value of the second 5 deposits at time t  10 is
5
2
4
1000 1.051.08  1.05 1.08 
5
 1.05 1.08 
1  1.05 /1.085 
 1000 1.051.08 
  7297.16.
 1  1.05 /1.08 
5
The total accumulated value is 9309.57  7297.16  $16,607 to the nearest dollar.
40
The Theory of Interest - Solutions Manual
Chapter 4
31. We have the equation of value
2
 1
1  .01k 1  .01k 
4096  1000 



5
6
1.257
 1.25 1.25



or
1/ 1.25 
1

4
1  1  .01k  /1.25 1.25 .25  .01k 
5
4.096 
upon summing the infinite geometric progression. Finally, solving for k
10 
1
.25  .01k
and k  15%.
32. The first contribution is  40,000  .04   1600. These contributions increase by 3%
each year thereafter. The accumulated value of all contributions 25 years later can be
obtained similarly to the approach used above in Exercise 30. Alternatively, formula
(4.34) can be adapted to an annuity-due which gives
1.05 25  1.0325
1.05   $108,576 to the nearest dollar.
1600
.05  .03
33. Applying formula (4.30), the present value of the first 10 payments is
1  1.05 /1.07 10 
100 
 1.07   919.95.
.07  .05


The 11th payment is 100 1.05 .95  147.38 . Then the present value of the second
1  .95 /1.07 10 
10
10 payments is 147.38 
 1.07 1.07   464.71 . The present value
 .07  .05 
of all the payments is 919.95  464.71  $1385 to the nearest dollar.
9
34. We have
PV 
2
1  m1
v  2v m 
2
m
 nmv m 
nm
1
1 
n 1
1  2v m   nmv m 
2
m
1
nm
1
1
n 1
PV 1  i  m  1  2 1  v m   v m  nmv m 
m
1  
  anm  nv n  .
m
PV 1  i  m 
1
41
The Theory of Interest - Solutions Manual
Chapter 4
Therefore
 
PV 
35. (a)
(b)
 
anm  nv n
m 1  i  m  1
1

anm  nv n
 
im
.
1  12 terms    12 terms    1 
12  24   3.
 11 1  2  2   2  
12
12
1 
 1 2 
144
36. We have
 12   13  14 
 24  
 24   25  25
 .
 2 144  12
PV   v 5  v 6  2v 7  2v8  3v 9  3v10 

v5 1
5
7
9


 v v v 
a 

2
1 v

5
d
4
v
1
v
 
.
2
1  v iv i  vd
37. The payments are 1,6,11,16,…. This can be decomposed into a level perpetuity of 1
starting at time t  4 and on increasing perpetuity of 1, 2,3, starting at time t  8 .
Let i4 and d 4 be effective rates of interest and discount over a 4-year period. The
present value of the annuity is
1
1 
4
1 
 5 1  i4  
where i4  1  i   1.

i4
 i4  d4 
We know that
1
1  i 4  .751  4 / 3, or i4  4  1  1 and d 4  3  1 .
3
3
1  13 4
Thus, the present value becomes
 3  1 
3   5    1 1   3  45  48.
 4  3  4 
38. Let j be the semiannual rate of interest. We know that
j  .03923 . The present value of the annuity is
2
1.03
 1.03 
1

 
1.03923  1.03923 

2
1
 112.59
1  1.03/1.03923
upon summing the infinite geometric progression.
42
1  j 
 1.08, so that
The Theory of Interest - Solutions Manual
Chapter 4
39. The ratio is


10
5
5
0
tdt

tdt
 75 / 2

 3.
25
/
2

t
10
1 2
5
2
5
2
1
0
2
t
40. Taking the limit of formula (4.42) as n  , we have
a
 Ia  


1

2

1
.08 2
 156.25.
41. Applying formula (4.43) we have the present value equal to

t
1 k  
t

 

1 k 
t
 1 i  


f
t
v
dt

dt



0
 0  1 i 
1 k 
ln 

 1  i   0
1
1
1



.
 1  k  ln 1  i   ln 1  k   i   k
ln 

 1 i 
Note that the upper limit is zero since i  k .
42. (a)  Da n    n  t v t dt.
n
0
(b)


n
0
 n  t v t dt  nan   Ia n
n 1  v n 


an  nv n


n  an

.
The similarity to the discrete annuity formula (4.25) for  Da n is apparent.
43. In this exercise we must adapt and apply formula (4.44). The present value is

14
1
 t 2  1e  
t
0
1 r 1 dr
dt.
1
The discounting function was seen to be equal to 1  t  in Exercise 19. Thus, the
answer is

14
1
14  t  1 t  1
14
t2 1
dt  
dt    t  1 dt
1
1
t 1
t 1
14
1

1 
  t 2  t    98  14     1  84.5.
2
1
2 
43
The Theory of Interest - Solutions Manual
Chapter 4
44. For perpetuity #1 we have
1
 20
1  v.5
so that 1  v.5  .05 and v.5  .95.
1  v.5  v  v1.5 

For perpetuity #2, we have
X 1  v 2  v 4 
  X 1 2  20
1 v
4
so that X  20 1  v   20 1  .95    3.71.
2
45. We have

n
0
at dt 
1


n
0
1  vt  dt 
n  an


n   n  4
 40.
.1
46. For each year of college the present value of the payments for the year evaluated at the
beginning of the year is


12
1200a9/12
.
The total present value for the payments for all four years of college is
 
 
1200a912/12 1  v  v 2  v3   1200a4 a912/12 .
P
.
i
1 1 
For annuity #2, we have PV2  q   2  .
i i 
Denote the difference in present values by D.
47. For annuity #1, we have PV1 
D  PV1  PV2 
(a) If D  0 , then
pq q
 2.
i
i
pq q
q
q
 2  0 or p  q 
or i 
.
i
i
i
pq
(b) We seek to maximize D.
dD d
  p  q  i 1  qi 2 
di di
   p  q  i 2  2qi 3  0.
Multiply through by i 3 to obtain
44
The Theory of Interest - Solutions Manual
Chapter 4
  p  q  i  2q  0 or i 
2q
.
pq
48. We must set soil (S) posts at times 0,9,18,27. We must set concrete posts (C) at times
0,15,30. Applying formula (4.3) twice we have
PVS  2
a36
a9
and PVC   2  X 
a45
a15
.
Equating the two present values, we have
2
a36
a9
 2  X 
a45
so that
a15
a  a
a
a a

X  2  36  45   45  2  36 15  1 .
 a9 a15  a15
 a9 a45

49. We know an 
1  vn

 a, so that v  1  a . Similarly, a2 n 
n
1  v2n

 b, so that
v2n  1  b . Therefore, 1  b  1  a   1  2a  a 2 2 , or a 2 2   2a  b   so
2a  b
. Also we see that n ln v  ln 1  a   n  ln 1  a  so that
that  
a2
an  nv n
ln 1  a 
n
. From formula (4.42) we know that  Ia n 
. We now


substitute the identities derived above for an , n, v n , and  . After several steps of
tedious, but routine, algebra we obtain the answer

a3
 a 
2a  b   b  a  ln 
 .
2 
 2a  b  
 b  a  
2
50. (a) (1)
(2)
(b) (1)
(2)
n
n
d
d n
d n
t
 t 1
an   vt   1  i    t 1  i   v tv t  v  Ia n .
di
di t 1
di t 1
t 1
t 1
d
a
di n
i 0
 v  Ia n
n
i 0
  t  
t 1
n  n  1
.
2
n
n
d
d n
d n
t
 t 1
an   vt dt   1  i  dt    t 1  i  dt  v  tv t dt  v  Ia n .
0
0
di
di 0
di 0
d
a
di n
i 0
 v  Ia n
n
t 0
   tdt  
0
45
n2
.
2