# Compound Savings Calculator

Consistent investments over a number of years can be an effective strategy to accumulate wealth. Even small additions to your savings add up over time. This calculator demonstrates how to put this savings strategy to work for you.

## Compound Interest Formula

A = P (1 + rn ) nt

Where:

• P = initial principal
• r = interest rate as a decimal
• t = number of years invested
• n = number of times the money is compounded per year
• A = final amount, including the initial principal and all interest earned over n years

### Example

Jane deposits \$3,700 at a 6.5% rate of interest. The interest is compounded daily. How much money does she have saved after 5 years?

### Solution

A = \$3,700 (1 + 0.065365 ) 365*5

A = \$5,121

## More Complex Problems

### Working Backward

If you start with your end savings goal in mind you can change the above calculation to calculate the amount you would need to save to reach that goal.

P = A ⁄ (1 + rn ) nt

Let's say you want to have \$5,000 saved in 3 years and want to know how much you would need to deposit to end up with \$5,000 in savings at 6.5% interest compounded weekly over 3 years.

The calculation would be 5000/(1+0.065⁄52) 52*3 = \$4,114.67 ### Multiple Regular Deposits

If you start with an initial amount & deposit money at the beginning of each period, the formula for compounded savings is:

A = P (1 + r) t + Df ((1 + r)t+1 -(1 + r))/ r

Where:

• P = initial principal
• D = deposit
• f = frequency of deposit, per year
• r = interest rate as a decimal
• t = number of years invested
• A = final amount, including the initial principal and all interest earned over t years

### Example

John deposits \$3,000 at a 5% rate of interest & \$1,200 at the beginning of each year. How much money does he have saved after 4 years?

### Solution

A = \$3,000 (1 + 0.05)4 + \$1,200 ((1 + 0.05)4+1 - (1 + 0.05)) / 0.05
A = \$3646.52 + \$1,200*4.52563125
A = \$3646.52 + \$5,430.76
A = \$9,077

### Irregular Deposits & Interest Rate Changes

If a person makes multiple irregular deposits over a period of time, then one would do the same sort of calculation numerous times, or one could use the above calculator to automatically do the calculations. If interest rates change over a period of time or deposits are irregular one could break the savings calculation into multiple steps.

• Calculate the problem for the first period where no variables change.
• Use the solution to that calculation as an input to the next calculation.

For example, if a person planned to save for 9 years but expects the rate of interest to change after 3 years then they can calculate their savings after 3 years & use that savings amount as an input for the initial savings amount on a subsequent 6 year interest calculation.

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