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How to Calculate Future Values Using Timelines

In economy it’s important to know about future values, as the value of money changes over time. $ $1$ today is not worth the same as $ $1$ will be worth in a year. Inflation and the general economic development in the world is affecting the value of money. This is important to take into consideration when looking at money from different time periods.

Theory

Future Value

When we talk about future values, we are describing how much money will be worth some time in the future. Future values are often related to saving and repayment of loans, and has a quotient

k = growth factor = {(1 + \frac{p}{100})}^{n} .

The future value $K_{n}$ of a sum of money $K_{0}$ you want to use in $n$ periods of time is given as

K_{n} = K_{0} \cdot {(1 + \frac{p}{100})}^{n},

where $p$ is the interest in percentages.

Example 1

Given that you deposit $$ 5000$ into a fund for saving that guarantees an annual return of $7.5 %$ for the next 20 years, how much money will you have in 20 years?

\begin{array}{l} K_{n} & = 5000 \cdot {(1 + \frac{7.5}{100})}^{20} \\ = 21 239.26 $ \end{array}

K_{n} = 5000 \cdot {(1 + \frac{7.5}{100})}^{20} = 21 239.26 $

This means that if you deposit $

5000

into the fund today, that money will be worth $

21 239

in 20 years.

Installment and Savings

When we’re looking at installments and savings, it’s often smart to draw timelines. A timeline helps you keep control of how many periods of time give you interest on your money. Here is an example of future values connected to saving.

Example 2

How much money will you have in the bank if you save $$ 100$ each month for 15 years? You deposit the money once a year, and the interest is $4.7 %$ .

For this type of exercises it is a good idea to use timelines. But first we need to find out how much money you will save each year:

100 \cdot 12 = $ 1200

Then the timeline looks like this:

Timeline showing the future value of the yearly deposits

This gives you the geometric series

\begin{array}{l} 1200 & \cdot 1.047 + 1200 \cdot 1.04 7^{2} \\ + \dots + 1200 \cdot 1.04 7^{15}, \end{array}

1200 \cdot 1.047 + 1200 \cdot 1.04 7^{2} + \dots + 1200 \cdot 1.04 7^{15},

with

a_{1} = 1200 \cdot 1.047, k = 1.047, n = 15 .

a_{1} = 1200 \cdot 1.047, k = 1.047, n = 15 .

You can then insert these numbers into the formula for the sum of a geometric series:

\begin{array}{l} S_{15} & = (1200 \cdot 1.047) \cdot \frac{1.04 7^{15} - 1}{1.047 - 1} \\ = $ 26 507.13 \end{array}

S_{15} = (1200 \cdot 1.047) \cdot \frac{1.04 7^{15} - 1}{1.047 - 1} = $ 26 507.13

This means that you save $

26 507.13

15

years. If there was no interest on the money, you would have saved

1200 \cdot 15 = $ 18 000

. That makes the total amount of interest you’ve gotten

$ 26 507.13 - $ 18 000 = $ 8507.13,

which is quite amazing!

Example 3

You created a bank account on 01.01.2015 with an interest rate of $4.5 %$ and deposited $$ 1000$ into the account. You will continue to deposit $$ 1000$ at the start of every new year.

Part 1 – You want to save a total of $$ 25 000$ . How many years will you have to save to get $$ 25 000$ in the account, given that the interest rate doesn’t change from $4.5 %$ ?

This gives you the geometric series

\begin{array}{l} 1000 & \cdot 1.045 + 1000 \cdot 1.04 5^{2} \\ + \dots + 1000 \cdot 1.04 5^{n} . \end{array}

1000 \cdot 1.045 + 1000 \cdot 1.04 5^{2} + \dots + 1000 \cdot 1.04 5^{n} .

The geometric series can also be shown as a timeline like this:

Timeline of saving 1000 dollars per year over n years

Where $a_{1} = 1000 \cdot 1.045$ and $k = 1.045$ . The final sum is set to be $S_{n} = 25 000$ , which makes $n$ the unknown you need to find. That means you need to use the formula for the sum of a geometric series and solve the equation for $n$ . You will then get the number of years you need to save in order to get the final sum you wanted.

\begin{array}{l} S_{n} & = a_{1} \frac{k^{n} - 1}{k - 1} \\ 25 000 & = 1000 \cdot 1.045 \cdot \frac{1.04 5^{n} - 1}{1.045 - 1} \end{array}

This equation can be input into a digital tool like CAS in GeoGebra. That will give you the solution $n = 16.6$ . This means that you have to save for 17 years to reach the total of $ $25 000$ you wanted in your account.

Part 2 – You would like to achieve the total sum of $$ 25 000$ in just 15 years. If the interest rate stays the same at $4.5 %$ , how much would you have to deposit every year to achieve this?

This can be expressed as the geometric series

\begin{array}{l} x \cdot 1.045 + x \cdot 1.04 5^{2} + \dots \\ + x \cdot 1.04 5^{14} + x \cdot 1.04 5^{15}, \end{array}

x \cdot 1.045 + x \cdot 1.04 5^{2} + \dots + x \cdot 1.04 5^{14} + x \cdot 1.04 5^{15},

which can also be represented as the following timeline:

Timeline of saving x dollars per year over 15 years

This time the unknown is the annual deposit $x$ . This gives you

\begin{array}{l} a_{1} & = x \cdot 1.045, k = 1.045, \\ n = 15, S_{n} = 25 000 . \end{array}

a_{1} = x \cdot 1.045, k = 1.045, n = 15, S_{n} = 25 000 .

You can use the same formula as in part 1, but now you solve it for

x

\begin{array}{l} S_{n} & = a_{1} \frac{k^{n} - 1}{k - 1} \\ 25 000 & = x \cdot 1.045 \cdot \frac{1.04 5^{15} - 1}{1.045 - 1} \end{array}

You can solve this equation with a digital tool, which will give you that the annual deposit has to be $ $1151.05$ to get $ $25 000$ in the account after 15 years.

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