Some Homework 2 Solutions
#2.1
Without loss of generality, we may assume that both funds begin with $1. After 7 years,
j
fund x will have a value of e0.49 . Fund y will have a value of (1 + )28 . This give us the
4
following:
j
e0.49 = 1.06(1 + )28
4
j 28
1.53992 = (1 + )
4
j
1.01554 = 1 +
4
0.06215 = j
#2.2
We know that
a(t) = e
In this case,
Z
3
s0.03 ds =
0
Rt
0
δ(s)ds
31.03
, so we have
1.03
31.03
1
= e− 1.03
a(3)
1
= 0.0493
a(3)
#2.4
We must set our payments equal to our liabilities at a specific point in time. In this case,
I will use year 4.
15, 000(1.04)4 + x = 80, 000e−
R8
1
5 t+1 dt
1
1.04
17, 547.88 + x = 80, 000e−(ln(9)−ln(6))
17, 547.88 + x = 80, 000
x = 33, 734
2 1
3 1.04
1
1.04
Some Homework 2 Solutions
#2.5
2
0.1
Over the first year the $400 accrues to 400 1 +
= 441. Adding the deposit at
2
time 1 gives us the accumulated value of $483 at time 1. Then we must use the force of
interest.
483e
This gives us that
R2
1
1 k+t dt
k+2
= 483eln k+1
k+2
= 483
k+1
= 552 by the fourth assumption
552
k+2
=
. Solving for k, we arrive at the solution k = 6
k+1
483
#2.7
The investment is made at time 1, so the integral’s lower bound is 1. The upper bound
is 9.
14e
R9
1
√
0.3 tdt
t1.5 9
= 14e0.3 1.5 |1
= 14e0.2(27−1)
= 14e5.2
= 2537.81
#2.8
The accumulated value of the fund is
1000e
R4
0
δ(t)dt
= 1000e
R3
0
0.02tdt+
R4
2 t=3
3
0.045dt
= 1000e0.01t |t=0 +0.045
= 1000e0.135
= 1144.54
This gives a nominal rate compounded quarterly of
16
i(4)
1+
= 1.14454
4
i(4)
= 1.00847
1+
4
i(4)
= 0.00847
4
i(4) = 0.03389
Some Homework 2 Solutions
#2.9
By our assumptions, we have the following equation:
5000 + 3000v 12∗1 + 2000v 12∗5 = 10000v 12∗t
where v =
gives us
1
1
. Evaluating the three terms on the left hand side of our equation
.12 =
1.01
1 + 12
8763.26
= v 12t
10000
8763.26
ln
= 12t ln v
10000
8763.26
1
=t
ln
12 ln v
10000
Evaluating this term gives us that t = 1.056