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April 3rd, 2014, 05:30 PM   #1
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Finance & Compounding Interest

It's been years since I studied math. I've forgotten the basics.

I've come across this question, and I can't solve it (at least I'm not getting the same solution as a book). Here's the problem:

How much money do you have to deposit into an account monthly, for 30 years, at an interest rate of 12%, compounded monthly, in order to have $1.5 million at the end of the 30 years?

Here's what I have:

y = ca^x, where c is a constant, a is the growth factor, and x is the number of months

c E (1.12)^n (i = 1, n = 360)

Therefore,

1.5x10^6 = c[1.12^360 + 1.12^359 + 1.12^358 + ... + 1.12]


Solving for c (after computing the summation with a calculator) yields a very small number. I'm not sure my Sigma formula is correct, and I don't know how to calculate it by hand. I'd like to know how to manually do it. I totally forgot how to compute this like a human.
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April 3rd, 2014, 07:45 PM   #2
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I would model this using a linear difference equation. Let $A_t$ be the amount in the account at time $t$ measured in months and $D$ be the monthly deposit. We will let the clock begin at the moment the first deposit is made. Hence, our difference equation is:

$\displaystyle A_{t+1}-1.01A_{t}=D$

The homogeneous solution is:

$\displaystyle h_t=c_1(1.01)^t=c_1\left(\frac{101}{100} \right)^t$

and the particular solution is:

$\displaystyle p_t=B$

Using the method of undetermined coefficients, we find:

$\displaystyle B=-100D$

Thus, by the principle of superposition, the general solution is:

$\displaystyle A_t=c_1\left(\frac{101}{100} \right)^t-100D$

Using the initial value $A_0=D$, we may determine the value of the parameter:

$\displaystyle A_0=c_1-100D=D\implies c_1=101D$

And so the solution satisfying the model is:

$\displaystyle A_t=D\left(101\left(\frac{101}{100} \right)^t-100 \right)$

Now, to find find the value of $D$ such that:

$\displaystyle A_{360}=1500000$

we may state:

$\displaystyle D\left(101\left(\frac{101}{100} \right)^{360}-100 \right)=1500000$

Solving for $D$, we find:

$\displaystyle D=\frac{1500000}{101\left(\frac{101}{100} \right)^{360}-100}\approx424.82$
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April 4th, 2014, 12:11 AM   #3
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Quote:
Originally Posted by MarkFL View Post
I would model this using a linear difference equation. Let $A_t$ be the amount in the account at time $t$ measured in months and $D$ be the monthly deposit. We will let the clock begin at the moment the first deposit is made. Hence, our difference equation is:

$\displaystyle A_{t+1}-1.01A_{t}=D$

The homogeneous solution is:

$\displaystyle h_t=c_1(1.01)^t=c_1\left(\frac{101}{100} \right)^t$

and the particular solution is:

$\displaystyle p_t=B$

Using the method of undetermined coefficients, we find:

$\displaystyle B=-100D$

Thus, by the principle of superposition, the general solution is:

$\displaystyle A_t=c_1\left(\frac{101}{100} \right)^t-100D$

Using the initial value $A_0=D$, we may determine the value of the parameter:

$\displaystyle A_0=c_1-100D=D\implies c_1=101D$

And so the solution satisfying the model is:

$\displaystyle A_t=D\left(101\left(\frac{101}{100} \right)^t-100 \right)$

Now, to find find the value of $D$ such that:

$\displaystyle A_{360}=1500000$

we may state:

$\displaystyle D\left(101\left(\frac{101}{100} \right)^{360}-100 \right)=1500000$

Solving for $D$, we find:

$\displaystyle D=\frac{1500000}{101\left(\frac{101}{100} \right)^{360}-100}\approx424.82$
That's pretty neat. The book's answer was $429, but that could be an error (they don't show their work).

I am not familiar with your method (despite what I read on Google a few minutes ago).

As for my Summation Notation Equation, was my equation incorrect? That is:

cE1.12^n (i=1, n=360)
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April 4th, 2014, 12:31 AM   #4
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It could be that they intended an effective rate of 12% which means we would get the following monthly interest rate instead:

$\displaystyle r=0.94888\overline{3}\%$

And this would change the deposit amount slightly...but I don't want to go through the derivation again.

You need to use a monthly rate instead of the annual rate in your summation, that's why you are getting a small value.
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