Let’s calculate how much $1000 invested for 30 years would grow to if invested with a simple interest rate of 8% vs an annual compound interest rate of 8%.

PV <- 1000
i <- .08
m <- 1
t<- 30

simpleFV <- PV*(1+(i*t))
compoundFV <- PV*((1+(i/m))^(m*t))
dollarSimpleFV<- sprintf("$%.2f", simpleFV)
dollarCompoundFV<- sprintf("$%.2f", compoundFV)
cat("Future value invested with simple interest =", dollarSimpleFV, "\n",
    "Future value invested with compound interest =", dollarCompoundFV, "\n")

## Future value invested with simple interest = $3400.00 
##  Future value invested with compound interest = $10062.66

Here we see that even though you’re investing the same amounts, compounded interest grows to nearly 3 times as much as the simple interest value.

Here we will view a graph of the previous example to visually see how money grows with simple interest vs compound interest.

As time progresses, these two values grow further and further apart. This is because compounded investments grow exponentially.

We will look at how the time at which someone begins investing contributes greatly to how much their investment grows to. Let’s consider the following

• Scenario 1: I begin investing $5000 per year until 2055.
• Scenario 2: I wait 10 years then begin investing $5000 per year until 2055.

We can see from this plot that in the scenario that I begin investing in 2025, I invest $50,000 more than the other scenario. However in the end, I end up with $364,614 more than if I delayed investing by 10 years. The sooner I begin investing, the greater the effect of compound interest.

I would like to retire at age 60. To retire, I would like to be able to live off my investments. I would like my to have an annual investment income of $100,000 to live off.

• We will assume my investments will earn 8% interest each year. First I need to calculate how much I need in investments to receive $100,000 as an 8% return.

• The formula \(I=Pit\) Will give me this value where \(I\) = $100,000 of interest earned in one year. \(t\) = 1 as we are calculating for 1 year, \(i\) =.08, and \(P\) will represent how much money I need to have at age 60 to retire. Rearranging this equation we get \[ \frac{1000000}{.08} = P = 125000\]

• Next to calculate how much I need to invest, I will use the equation for the future value of an annuity. \[ FV = X* \frac{{{(1+i)}^n}-1}{i} \]

• Where \(X\) equals the amount I need to invest each year and \(FV\) is the 125000 we just solved for.

• Rearranging this equation and plugging in our values we get \[ X = \frac{(1250000)(.08)}{(1+.08)^{30}-1} = 11034.30 \]

• To reach this retirement goal, I will need to invest $11,034.30 annually.