Jaime Frade
Computational Finance: Dr. Kopriva
Homework #3
0.
Executive Summary
In this assignment there is a need to determine the minimal interest rate,
i
, at which a person can invest $1000
per month for 15 years to ensure retirement nest egg of $1 million. In order to calculate
i
three rootfinding
algorithms, bisection method, NewtonRapshon, and a combination, were programmed in C++.
It will be shown that even though the bisection method does provide solutions, it suffers from a slow error linear
convergence. On the other hand, using Newton’s method will show fast quadratic convergence, however, highly
depends on the choosing initial starting point appropriately. A combination of these methods will combine the
speed and convergence stability needed to solve the problem. In order to reassure the person of my calculations
and show a comparison of each of the three algorithms, I will test several functions with known roots. Results
will be shown. This was done by advice of my collegue.
From the calculations, it was found that an monthly interest rate,
i
, of approximately 1.5796708% under the
IEEE standard of single precision. Therefore, to create this “nest egg,” the investor will need to find a fund
offering an annual interest rate of 18.95% compounded monthly. Unfortunately, as of today, retirement will be
impossible due to investments, such as CDs/savings, only offering rates 34%.
1.
Statement of Problem
Determine the interest rate which will satisfy the following equation:
A
=
P
i
[(1 +
i
)
n

1]
,
(1)
where
A
⇒
amount in the account,
P
⇒
deposit amount at each month, and
i
⇒
interest rate per period.
For the “nest egg,”
n
= 15 years (180 month since compounded monthly),
P
= 1
,
000 and
A
= 1
,
000
,
000.
To determine
i
, will find a root using the bisection method, NewtonRapshon, and a combination of the two.
2.
Description of the Mathematics
From lecture, the problem is to find a root for a function,
f
(
i
) :
f
(
i
) =
A

P
i
[(1 +
i
)
n

1]
,
i
= 0
(2)
such that it solves
f
(
i
) = 0. The root exists by the Intermediate value theorem, if
f
(
i
) is a continuous function
on an closed interval, [a,b], and
f
(
a
)
f
(
b
)
<
0
.
Then there
∃
an
i
such that it satisfies, (2),
f
(
i
) = 0 for some
i
∈
[
a, b
].
Bisection method will repeatedly check for an
i
, by making an initial guess such that
i
=
a
+
b
2
.
Newton’s
method will use functional values to determine the
i
which satisfies
f
(
i
) = 0
.
If
i
is an approximation to
ˆ
i
satisfying
f
(
ˆ
i
) = 0, then by using simplification on the Taylor series expansion on the function (2) there exists
a better approximation for
ˆ
i
such that
ˆ
i
=
i
n

f
(
i
n
)
f
(
i
n
)
,
where
(3)
f
(
i
) =
Pni
(1 +
n
)
n

1

P
(1 +
i
)
n
+ 1
i
2
(4)
1