Which is the final capital after 10 years?

Question

We want to calculate the final capital after 10 years for a savings plan with the following data:

Monthly deposit at the beginning of the month 200 Euro; Interest rate coupon annually; Interest rate 4.5% p.a.

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I have done the following:

The intereset rate is $q=1+\frac{p}{100}=1+\frac{4,5}{100}=1+0,045=1,045$.

Since we are paying every month for $10$ years, we are paying for $12\cdot 10=120$ months.

Using the formula \begin{equation*}\overline{R_m}=\frac{r\cdot q_m\cdot \left (q_m^m-1\right )}{q_m-1}\end{equation*}

with $m=120$, $r=200$ und $q_m=\sqrt[12]{q}=\sqrt[12]{1,045}=1.003675$,

we get the following

\begin{align*}\overline{R_m}&=\frac{200\cdot 1.003675\cdot \left (1.003675^{120}-1\right )}{1.003675-1}\approx 30206.10\end{align*}

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My result is not the same as the answer of the book. What have I done wrong? Do we not use that formula?

Answer 1

To get the required result in the book you first have to calculate the amount of the equivalent annual payment. The formula for payments at the beginning of every month is

$C_1=12\cdot r+\frac{\color{blue}{13}\cdot r\cdot i}{2}$

In case of payments at the end of every month $\color{blue}{13} $ has to be replaced by $11$.

$C_1=12\cdot 200+\frac{\color{blue}{13}\cdot 200\cdot 0.045}{2}=2458.5$

To get the Future value after 10 years we use the formula for annual payments.

$C_{10}=2458.5\cdot \frac{1-1.045^{10}}{1-1.045}=30210.56$

But in general I wouldn´t say that your method is worse then the method above. Your result differs from my result about $0.015\%$ only.

Answer 2

I have some doubts about your question, but I am assuming that there is a new deposit made every month for 10 years, so 120 deposits of 200 Eur each. This means that the 1st deposit will earn interest for 120 periods (months) while the last, just for one period.

If the rate is 4.5% p.a nominal, than the real is just 4.5% / 12 = 0.375% Interest Rate Calculator

This being said, the total amount should be 30353.01329 Eur

Answer 3

I'm not sure about the formula you are using, but this is how I'd do it. Say you start of by putting in $c$ the first month and then you have a monthly interest on that at $4.5$%. In the start your capital is $y_0=c$, one, two and $n$ months later, respectively, right after the deposit, you have $$\begin{array}{lcl} y_1 &=& 1.003675c+c = (1.003675+1)c \\ y_2&=& 1.003675(1.003675c+c) = (1.003675^2+1.003675+1)c \\ &\vdots \\ y_n&=& (1.003675^n+1.003675^{n-1}+...+1)c= \frac{(1.003675^{n+1}-1)c}{1.003675-1} & \end{array} $$

By the geometric sum. Now plug in $n=120$ and $c=200$.