Future value of savings functions

The savings functions differ from the previous functions, that in addition to the interest, you also have periodic deposits or withdrawals from the account.

Those functions can be used in different ways, depending on the situations. We go through them all on this page.

The first function is for calculating the future value of savings.

FutureValueOfSavings(x; y; z; a; b )

Where

x = Present value
y = Periodical payment amount
z = No of annual payments(12)
a = No of years
b = annual interest rate

The following formulas are used:

$\begin{aligned} x &\colon\quad \text{present value} \\ y &\colon\quad \text{periodical payment amount} \\ z &\colon\quad \text{number of annual payments (12)} \\ a &\colon\quad \text{number of years} \\ b &\colon\quad \text{annual interest rate} \\[0.6em] r &= \frac{\frac{b}{z}}{100} \\ c &= z\cdot a \\ \mathrm{result} &= x\cdot \left(1 + r\right)^{c} + \frac{y}{r}\cdot \left(\left(1 + r\right)^{c} - 1\right) \end{aligned}$

Example:

Lets say, that we have 10.000 on a bank account now. Then we deposit 500 on this account every month. The interest rate on the account is 2%. What is the future value after 10 years.

FutureValueOfSavings(10000;500;12;10;2)=78571.824497

The next function calculates the payment needed to reach a specific future value.

PaymentSavings(x; y; z; a; b )

Where

x = Savings target
y = Present value
z = No of annual payments(12)
a = No of years
b = annual interest rate

This function uses the following formulas:

$\begin{aligned} x &\colon\quad \text{savings target} \\ y &\colon\quad \text{present value} \\ z &\colon\quad \text{number of annual payments (12)} \\ a &\colon\quad \text{number of years} \\ b &\colon\quad \text{annual interest rate} \\[0.6em] r &= \frac{\frac{b}{z}}{100} \\ c &= z\cdot a \\ \mathrm{result} &= \frac{x\cdot r - y\cdot r\cdot \left(1 + r\right)^{c}}{\left(1 + r\right)^{c} - 1} \end{aligned}$

Example:

Lets say, that we have 10.000,- on a savings account. We want this to be 100.000,- in the next 10 years. The interest rate is 2%. How much do we need to deposit every month for this to happen.

PaymentSavings(100000;10000;12;10;2)=661.454418

Another example:

Suppose a savings account contains 1,000,000. We want to withdraw a monthly amount while making the balance last for ten years. What is the maximum monthly withdrawal?

PaymentSavings(0;1000000;12;10;2)=-9201.345384

After that, the account will be empty. The result is negative because money is withdrawn rather than deposited.

Next function in this series, is "Years to reach a savings target".

SavingsYearsToTarget(x; y; z; a; b )

Where

x = Savings target
y = Present value
z = No of annual payments(12)
a = Periodic payment sum
b = annual interest rate

The formulas used for this function are:

$\begin{aligned} x &\colon\quad \text{savings target} \\ y &\colon\quad \text{present value} \\ z &\colon\quad \text{number of annual payments (12)} \\ a &\colon\quad \text{periodic payment sum} \\ b &\colon\quad \text{annual interest rate} \\[0.6em] r &= \frac{\frac{b}{z}}{100} \\ \mathrm{result} &= \frac{\frac{\log\left(\frac{a + x\cdot r}{a + y\cdot r}\right)}{\log\left(1 + r\right)}}{z} \end{aligned}$

Example:

We have 10.000 on a bank account. We deposit 2.000 each month on this. The bank interest rate is 2%. How long time will it take to get 1.000.000 on that account.

SavingsYearsToTarget(1000000;10000;12;2000;2)=29.916753

This completes the savings functions.

*There is much more to see in the Geometric section if that would be of interest.

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