# Compound Interest: Final Value

How to find the final value of the compound interest investment (yearly, monthly, continuously): 3 formulas, 3 examples, and their solutions.

## Formula: Yearly Compounded Interest

### Formula

A compounded interest means

add the principle and the interest,

calculate the next interest,

and repeating this process.

The amount of money shows exponential growth.

To find the final value

of a compound interest investment,

use the exponential growth formula:

A = A_{0}(1 + r)^{t}.

Use this formula when

the unit of the rate r [per year]

and the unit of the compound period [yearly]

are the same.

## Example 1

### Example

### Solution

The initial value of the investment is $1,000.

So A_{0} = $1000.

The investment is at a rate of 6% per year.

So r = 0.06/year.

Write the unit [per year].

The final value is the value 5 [years] later.

And the unit of the rate is [per year].

So write 5 years in [years]:

t = 5 years.

A_{0} = 1000

r = 0.06/year

t = 5 years

The investment is compounded yearly.

The unit of the rate r [per year]

and the unit of the compound period [yearly]

are the same.

Then the final value A is

A = 1000(1 + 0.06)^{5}.

(1 + 0.06) = 1.06

It says

assume 1.06^{5} = 1.338.

So 1000⋅1.06^{5} = 1000⋅1.338.

1000⋅1.338 = 1338

The initial value A_{0} is in $.

So the final value A is $1,338.

So $1,338 is the answer.

## Formula: Monthly, Daily Compounded Interest

### Formula

If the units of the rate r and t (year)

and the unit of the compound period (monthly or daily)

are different,

use this formula:

A = A_{0}(1 + r/n)^{t⋅n}.

Change the units of r and t

to the unit of the compound period:

r (per year) → r/n (per month, day)

t (years) → t⋅n (months, days).

n: [12 months/year], [365 days/year], ...

## Example 2

### Example

### Solution

The initial value of the investment is $1,000.

So A_{0} = $1000.

The investment is at a rate of 6% per year.

So r = 0.06/year.

The final value is the value 5 [years] later.

And the unit of the rate is [per year].

So write 5 years in [years]:

t = 5 years.

A_{0} = 1000

r = 0.06/year

t = 5 years

The investment is compounded monthly.

The units of r and t [year]

and the unit of the compound period [monthly]

are different.

Then the final value A is

A = 1000(1 + 0.06/12)^{5⋅12}.

r: 0.06 per year = 0.06/12 per month

t: 5 years = 5⋅12 months

0.06/12 = 0.01/2

5⋅12 = 60

0.01/2 = 0.005

(1 + 0.005) = 1.005

It says

assume 1.005^{60} = 1.349.

So 1000⋅1.005^{60} = 1000⋅1.349.

1000⋅1.349 = 1349

The initial value A_{0} is in $.

So the final value A is $1,349.

So $1,349 is the answer.

Compare this answer

to the answer of the previous example

(compounded yearly: $1,338).

As the compound period gets shorter

(yearly → monthly),

the total investment gets bigger

($1,388 → $1,349)

## Formula: Continuously Compounded Interest

### Formula

As the compound period gets shorter,

the total investment gets bigger.

So, to maximize the investment,

the compound period should be minimized (r/n → 0)

and the number of period should be maximized (t⋅n → ∞).

This is the case of

continuously compounded interest.

To find the final value

of a continuous compound interest investment,

use the continuous exponential growth formula:

A = A_{0}e^{rt}.

## Example 3

### Example

### Solution

The initial value of the investment is $1,000.

So A_{0} = $1000.

The investment is at a rate of 6% per year.

So r = 0.06/year.

The final value is the value 5 [years] later.

And the unit of the rate is [per year].

So write 5 years in [years]:

t = 5 years.

A_{0} = 1000

r = 0.06/year

t = 5 years

The investment is compounded continuously.

Then the final value A is

A = 1000⋅e^{0.06⋅5}.

0.06⋅5 = 0.3

It says

assume e^{0.3} = 1.350.

So 1000⋅e^{0.3} = 1000⋅1.350.

1000⋅1.350 = 1350

The initial value A_{0} is in $.

So the final value A is $1,350.

So $1,350 is the answer.

Compare this answer

to the answers of the previous examples

(compounded yearly: $1,338, compounded monthly: $1,349).

As you can see,

the total investment is maximized

by continous compounding.