# MATH2805 Solution1 ```MATH2805 (Spring 2021)
Solutions to Homework #1
1. (a) For the given A(t), we have
a(t) =
1
A(t)
= (t2 + 2t + 3).
A(0)
3
(b) First, a(0) = 1. Second, noting that a0 (t) = (2t + 2)/3 &gt; 0 for any t &gt; 0, a(t)
is an increasing function. Third, a(t) is a continuous function. Therefore, a(t)
satisfies the three properties of an accumulation function.
(c) In = A(n) − A(n − 1) = 2n + 1.
2. For the form a(t) = at2 + b, by letting t = 0 we have a(0) = b. This shows that
b = 1. Also by A(3) = 100a(3) = 172, we a(3) = 9a + 1 = 1.72. This leads to
a = 0.08 so that the accumulated function is
a(t) = 0.08t2 + 1.
Thus, the accumulated value at time 10 of \$100 invested at time 5 is
A(5) = 100a(5) = \$300.
3. (a) For A(t) = 100 + 5t, we have
A(5) − A(4)
1
= ,
A(4)
24
A(10) − A(9)
1
= .
=
A(9)
29
i5 =
i10
(b) For A(t) = 100(1.1)t , we have
A(5) − A(4)
= 0.1,
A(4)
A(10) − A(9)
=
= 0.1.
A(9)
i5 =
i10
4. By definition, we have
A(7) = A(4)(1 + i5 )(1 + i6 )(1 + i7 )
= 1000(1 + 0.01 ∗ 5)(1 + 0.01 ∗ 6)(1 + 0.01 ∗ 7)
= 1190.91.
1
5. (a) Let the interest rate be i. We have
1
500(1 + 2 ∗ i) = 615.
2
This leads to i = 9.2%.
(b) Let the number of years be n. We have
500(1 + 7.8% ∗ n) = 630.
This leads to n = 3 13 .
6. For simple interest, we have a(t) = 1 + it. The effective rate at time n is
in =
i
4%
a(n) − a(n − 1)
=
=
= 2.5%.
a(n − 1)
1 + i(n − 1)
1 + 4%(n − 1)
This results in n = 16.
7. For convenience, we define f (i) = (1 + i)t − (1 + it). Then, f (0) = 0 and f 0 (i) =
t(1 + i)t−1 − t = t[(1 + i)t−1 − 1].
(a) If 0 &lt; t &lt; 1, then f 0 (i) &lt; 0, i.e. f (i) is decreasing in (0, 1), so f (i) &lt; f (0) = 0,
i.e. (1 + i)t &lt; 1 + it.
(b) If t &gt; 1, then f 0 (i) &gt; 0, so f (i) is increasing in (0, 1) and f (i) &gt; f (0) = 0, i.e.
(1 + i)t &gt; 1 + it.
8. For compound interest, we have
600(1 + i)2 = 600 + 264.
This leads to i = 0.2. Consequently, the accumulated value of \$2000 invested for
three years is
2000((1 + 0.2)3 = \$3456.
9. We have 1 = v n + v 2n . This leads to
√
n
v =
Further,
(1 + i)2n = (v n )−2
5−1
.
2
√
2
3
+
5
= (√
)2 =
.
2
5−1
10. We have 500(1 + i)30 = 4000. This leads to (1 + i)10 = 2, or equivalently, v 10 = 0.5.
Further, the sum of present value is
10, 000(v 20 + v 40 + v 60 ) = 10, 000(0.52 + 0.54 + 0.56 ) = 3281.25.
2
11. From the question, we have Ai = 336 and Ad = 300. Note also that
i=
d
.
1−d
Solving the above equations, we have d = 3/28 and
A=
300
28
= 300 &times;
= 2800.
d
3
12. Define f (d) = (1 − d)t − (1 − dt), then f (0) = 0 and f 0 (d) = −t(1 − d)t−1 + t =
t[1 − (1 − d)t−1 ].
(a) If 0 &lt; t &lt; 1, then f 0 (d) &lt; 0, i.e. f (d) is decreasing in (0, 1), so f (d) &lt; f (0) = 0,
i.e. (1 − d)t &lt; 1 − dt.
(b) If t &gt; 1, then f 0 (d) &gt; 0, so f (d) is increasing in (0, 1) and f (d) &gt; f (0) = 0, i.e.
(1 − d)t &gt; 1 − dt.
h
13. (a) 1 +
i(3)
3
h
(b) 1 +
i(6)
6
i3
i6
h
= 1−
d(4)
4
h
= 1−
d(2)
2
i−4
i−2
− 34 (3)
i
d(4) = 4 1 − 1 + 3
.
− 13
d(2)
(6)
i =6 1− 2
−1 .
⇒
⇒
14. The accumulated value of \$100 after two years is
8
6%
= 112.65.
100 &times; 1 +
4
15. By the formula 1 + i(m) /m = (1 − d(m) /m)−1 , we have
i(m) d(m)
i(m) d(m)
−
=
&middot;
m
m
m
m
Plugging in the values,
0.1844144 0.1802608
0.1844144 0.1802608
−
=
&middot;
m
m
m
m
This leads to m = 8.
0
(t)
16. (a) By definition, we have δt = aa(t)
= dtd ln a(t). Then,
Z n
Z n
d
δt dt =
ln a(t)dt
0
0 dt
= ln a(n) − ln a(0)
= ln(1 + i)n
= − ln v n .
3
(b) As δt =
A0 (t)
,
A(t)
n
Z
0
we have
Z n
A0 (t)
A(t)
A0 (t)dt
dt =
A(t)
0
= A(n) − A(0)
= [A(n) − A(n − 1)] + &middot; &middot; &middot; + [A(1) − A(0)]
= I1 + I2 + &middot; &middot; &middot; + In .
17. For the given question, we have
(1 + i)3 = (1 − 0.08)−1 (1 − 0.07)−1 (1 − 0.06)−1 .
This leads to i = 7.53%.
18. For ease of notation, let wi = si /(s1 + s2 + &middot; &middot; &middot; + sn ) for i = 1, . . . , n. Following the
lecture note, the value of t̄ is
t̄ =
s1 t1 + s2 t2 + . . . + sn tn
= w1 t1 + w2 t2 + &middot; &middot; &middot; + wn tn .
s1 + s2 + &middot; &middot; &middot; + sn
Further, we have
v t̄ = v w1 t1 +w2 t2 +&middot;&middot;&middot;+wn tn = (v t1 )w1 (v t2 )w2 &middot; &middot; &middot; (v tn )wn .
On the other side, for the true value of t,
vt =
s1 v t1 + s2 v t2 + &middot; &middot; &middot; + sn v tn
= w1 v t1 + w2 v t2 + &middot; &middot; &middot; + wn v tn .
s1 + s2 + &middot; &middot; &middot; + sn
Treating v t1 , v t2 , . . . , v tn as the variables, together with the well-known inequality
that “the arithmetic mean is always greater than or equals to the geometric mean”
(also referred to as the Weighted AM-GM inequality), we have
(v t1 )w1 (v t2 )w2 &middot; &middot; &middot; (v tn )wn ≤ w1 v t1 + w2 v t2 + &middot; &middot; &middot; + wn v tn .
or equivalently, v t̄ ≤ v t . Finally, noting that 0 &lt; v &lt; 1, we have t̄ ≥ t.
4
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