# Recurring investments - Present value and final value

- tsp
Last update 26 Jul 2024
10 mins

Understanding how recurring investments and pensions work is crucial for making informed financial decisions. This article will guide you through the basics of percentages, compound interest, and how to calculate the present and final values of recurring investments and pensions. By the end, you’ll be equipped with the knowledge to plan your savings and investments effectively.

This is one of the more basic topics usually taught at school.

## Recalling the basics - percentages and compound interest

First let’s recall how percentages are calculated. Assuming a basis value $G$ represents $100%$ then we can calculate $p$ percent simply by utilizing linearity:

• We know $G$ equals $100 %$
• Thus we know that $\frac{G}{100}$ equals $1%$
• This means $\frac{G}{100} * p$ equals $p%$

Keep in mind the basis value $G$ is always the value before the process happened.

Now we are mostly interested in growth by $p%$ or shrinkage by $p%$. This is easy - we just calculate $p%$ and add or subtract it from our basis value:

For growth we add it to the basis value:

[ \begin{aligned} K_1 &= K_0 + \frac{K_0}{100} * p \\ K_1 &= K_0 + K_0 * \frac{p}{100} \\ K_1 &= K_0 * \underbrace{(1 + \frac{p}{100})}_{q} \end{aligned} ]

From this we can see that we multiply by a compound value $q$ that represents the percentage remaining after the process including the basis value divided by 100. If we for example grow by $25%$ we get the value

[ \begin{aligned} q &= 1 + \frac{25}{100}\\ &= 1.25 \\ &= \frac{125}{100} \end{aligned} ]

Since we modeled growth by $25%$ and add it to $100%$ we now calculate the remaining $125%$ after the growth process.

The same works for shrinkage. If we want to model shrinkage by $25%$ we are going to calculate $100% - 25% = 75%$ that are remaining:

[ \begin{aligned} q &= 1 - \frac{25}{100} \\ &= 0.75 \\ &= \frac{75}{100} \end{aligned} ]

Please keep in mind here it looks like we are adding and subtracting percentages - this only works because it’s exactly the same process that we are describing. You can never ever add or subtract percentages of processes happening after each other.

## Recurring interest

So now lets take a look at typical recurring interest. This is the same process that also happens during decay of nuclear substances, growth of bacteria or when leaving money on your bank account. The assumption is that for each time step that we are looking at a given percentage of our quantity is growing or shrinking. As we will see this will yield a typical exponential function.

So let’s recall from before that growth (or for decay just flip the sign) by $p%$ we can model a single step by:

[ \begin{aligned} K_1 &= K_0 + \frac{p}{100} * K_0 \\ &= K_0 \underbrace{(1 + \frac{p}{100})}_{q} \\ &= K_0 * q \end{aligned} ]

Now lets assume we model a second time step. The only difference is that our base value is not $K_1$ instead of $K_0$:

[ \begin{aligned} K_2 &= K_1 + \frac{p}{100} * K_1 \\ &= K_1 * (1 + \frac{p}{100}) \\ &= K_1 * q \end{aligned} ]

If we now insert the expression for $K_1$ from above:

[ \begin{aligned} K_2 &= K_1 * q \\ &= \underbrace{(K_0 * q)}_{K_1} * q \\ &= K_0 * q^2 \end{aligned} ]

Now we can repeat this process for more time steps:

[ \begin{aligned} K_3 &= K_2 * q \\ &= K_0 * q^2 * q \\ &= K_0 * q^3 \end{aligned} ]

Since we know that

[ \begin{aligned} K_{n+1} &= K_n * q \end{aligned} ]

We can see that

[ K_n = K_0 * q^n ]

This is a typical exponential function.

As a quick example let’s assume we put $100 Eur$ into our bank account and we have an interest rate of $0.4%$ per year. Then we calculate the amount of money after $t$ years simply by

[ K(t) = 100 * 1.004^t ]

This doesn’t look that impressive - let’s take a look what would happen with 4% growth

Here we see the typical recurring interest effect. Please note that for proper financial calculations you would need to honor also takes - in many countries there is a flat rate of 25% on all profits. This can simply be added to the formula:

[ K(t) = K_0 * (1 + \frac{p}{100} * 0.75) ]

Here we used the term $0.75$ since we are allowed to keep $75%$ of the profit when we have to pay $25%$ taxes (this is usually called capital gains tax)

Please note that this also did not take into account for inflation rates, etc. - it’s just a very simple model.

## Recurring investments and pensions

Often we are interested in what would happen if we would invest a fixed value each month into some kind of investment instrument or a bank account. This is a little bit harder. Let’s see why by looking into how we would naively calculate the capital after a few years when we invest an amount of $R$ each time step:

• At the beginning we start with $K_0 = R$
• After one year we also invest $R$ so $K_1 = K_0 * q + R$
• One year later we do the same:
[ \begin{aligned} K_2 &= K_1 * q + R \\ &= (R * q + R) * q + R \\ &= R * q^2 + R * q + R \\ &= R * (q^2 + q^1 + \underbrace{q^0}_{1}) \end{aligned} ]
• Investing one year later looks exactly the same:
[ \begin{aligned} K_3 &= K_2 * q + R \\ &= (R * q^2 + R * q + R) * q + R \\ &= R * (q^3 + q^2 + q^1 + q^0) \end{aligned} ]

As we can see the term in the parenthesis forms the sum over a geometric series. Let’s make a short excursion into calculating the sum of a geometric series

### Sum over geometric series

A geometric series is a series of the form

[ 1 + q + q^2 + q^3 + q^4 + \ldots ]

There is a simple way to calculate the partial sums of such a series. The partial sum is the sum of the first $n$ elements:

[ \begin{aligned} s_n &= \underbrace{q^0}_{1} + q^1 + q^2 + q^3 + \ldots + q^n \\ &= \sum_{i=0}^{n} q^i \end{aligned} ]

When we take a look at the recursive structure of this sum we can calculate a very simple formula:

[ \begin{aligned} s_n &= 1 + q^1 + q^2 + \ldots + q^n \\ &= 1 + q \underbrace{(1 + q^1 + q^2 + q^3 + \ldots + q^{n-1})}_{s_{n-1}} \end{aligned} ]

Using that $s_{n-1} = s_n - q^N$ we can further simplify:

[ \begin{aligned} s_n &= 1 + q^1 + q^2 + \ldots + q^n \\ &= 1 + q \underbrace{(s_n - q^n )}_{s_{n-1}} \\ \to s_n &= 1 + q * s_n - q^{n+1} \\ s_n - q * s_n &= 1 - q^{n+1} \\ s_n (1 - q) &= 1 - q^{n+1} \\ \to s_n &= \frac{1 - q^{n+1}}{1 - q} \end{aligned} ]

This is the summation formula for a geometric series.

### Applying the geometric series to recurring investments

#### Final value

As we have seen before a recurring investment yields

[ K(t) = R * (1 + q^1 + q^2 + q^3 + \ldots + q^t) ]

Now we just insert the formula for recurring interest:

[ K(t) = R * \frac{1 - q^{t+1}}{1 - q} ]

This is what usually is called the final value of an investment. Usually finance folks now also differentiate between advance and retroactive interest. Retroactive interest happens always after we have paid our next rate. For advance interest one need to account for another time step:

$R * \frac{1 - q^{n}}{1 - q}$ $R * \frac{1 - q^{n}}{1 - q} * q$

Note that in some textbooks the signs look different. This simply is a multiplication with $-1$ in the nominator and denominator - so it’s the same equation

$R * \frac{q^{n} - 1}{q - 1}$ $R * \frac{q^{n} - 1}{q - 1} * q$

As an simple example let’s assume we have a yearly interest rate of $4%$. Lets assume that in the beginning of each year we invest $1200 Eur$ and take a look at the development over time:

[ K(t) = 1200 * \frac{1.04^t - 1}{1.04 - 1} ]

If we want to take a look at a monthly investment but know the yearly interest rate we have to adjust for the monthly rate. We know that recurring application (12 times) or the monthly rate has to yield the yearly rate:

[ \begin{aligned} q_{12}^12 &= q \\ q_{12} &= q^\frac{1}{12} \end{aligned} ]

Let’s take a look what happens if we invest $100 Eur$ per month with the same parameters as above:

[ K(t) = 100 * (q^\frac{1}{12})^{12 * t} ]

Note we also had to change the time unit to months while $t$ still is supplied in years.

#### Present value

Now lets assume we want to know how much we need when we want to withdraw a fixed rate $R$ from some investment method with a fixed interest rate every time step and want to know how much money we need for this. This is a typical example of a pension calculation. This works exactly the same - we just don’t need to know the value of a recurring investment at the end but in it’s beginning. For that we can simply calculate the present value by discounting over the whole timespan. Assuming the final value is

[ E(t) = K_0 * \frac{1 - q^t}{1 - q} ]

Then we can calculate the value at the beginning of the period assuming constant recurring interest:

[ \begin{aligned} B(t) &= E(t) * \frac{1}{q^t} \\ &= K_0 * \frac{1 - q^t}{1 - q} \frac{1}{q^t} \end{aligned} ]

Again finance folks distinguish between advance and retrospective interest rates - the only difference again being one time interval more or less.

Final value $R * \frac{q^{n} - 1}{q - 1}$ $R * \frac{q^{n} - 1}{q - 1} * q$
Present value $R * \frac{q^{n} - 1}{q - 1} \frac{1}{q^n}$ $R * \frac{q^{n} - 1}{q - 1} \frac{1}{q^{n-1}}$