Future Value of deposits functions

The future value of deposits are pure interest calculations. It means that you deposit some money on the bank, and do nothing.

If you are looking for functions around savings, where some money are periodical put into your account or withdrawn, go to the "future value of savings" functions.

The first function, is calculating the future value:

FutureValueOfDeposit(x; y; z )

Where

• x = Present value
• y = Compounding periods (years)
• z = interest rate

The following formula is used:

$$\begin{aligned} x &\colon\quad \text{present value} \\ y &\colon\quad \text{compounding periods (years)} \\ z &\colon\quad \text{interest rate} \\[0.6em] \mathrm{result} &= x\cdot \left(1 + \frac{z}{100}\right)^{y} \end{aligned}$$

Example:

You have 10.000 on a bank account. The interest rate is 2%. What is the future value after 10 years.

FutureValueOfDeposit(10000;10;2)=12189.9442

*The next function is the interest needed for a given future value. *

InterestPVFV(x; y; z )

• x = present value
• y = future value
• z = compounding periods (years)

$$\begin{aligned} x &\colon\quad \text{present value} \\ y &\colon\quad \text{future value} \\ z &\colon\quad \text{compounding periods (years)} \\[0.6em] \mathrm{result} &= \left(\left(\frac{y}{x}\right)^{\frac{1}{z}} - 1\right)\cdot 100 \end{aligned}$$

Example:

What is the interest needed for getting the 10.000 to be 15.000 after 10 years?

InterestPVFV(10000;15000;10)=4.137974399

Let us check this with the previous function:

FutureValueOfDeposit(10000;10;4.137974399)=15000.000000

This seams to be pretty accurate.

The next function is for calculating the number of years for a given future value:

YearsToFutureValue(x; y; z )

Where

• x = Present value
• y = Future value
• z = annual interest rate

The following formula is used:

$$\begin{aligned} x &\colon\quad \text{present value} \\ y &\colon\quad \text{future value} \\ z &\colon\quad \text{annual interest rate} \\[0.6em] \mathrm{result} &= \frac{\ln\left(y\right) - \ln\left(x\right)}{\ln\left(1 + \frac{z}{100}\right)} \end{aligned}$$

Example:

Let us say, that we test this with the values from the previous example, and see if we get 10 years:

YearsToFutureValue(10000;15000;4.137974399)=10.000000

*With a little rounding, this seems to be pretty accurate. *

Present value

PresentValue(x; y; z ) calculates the present value corresponding to a future value.

$$\begin{aligned} x &\colon\quad \text{future value} \\ y &\colon\quad \text{number of years} \\ z &\colon\quad \text{annual interest rate} \\[0.6em] \mathrm{result} &= \frac{x}{\left(\frac{z}{100} + 1\right)^{y}} \end{aligned}$$

Compound interest

CompoundInterest(x; y; z ) calculates the accumulated interest, excluding the original balance.

$$\begin{aligned} x &\colon\quad \text{original balance} \\ y &\colon\quad \text{number of compounding periods} \\ z &\colon\quad \text{period interest rate} \\[0.6em] \mathrm{result} &= x\cdot \left(\left(1 + \frac{z}{100}\right)^{y} - 1\right) \end{aligned}$$

This completes the deposit functions. The next series is Future value of savings.

RELATED TOPICS

• Financial functions
• Annuity loan calculations
• Future value of savings functions