How much should you deposit at the end of each month in an IRA that pays 5% compounded monthly to earn $110,000 per year from interest alone, while leaving the principal untouched, to be withdrawn at the end of each year after you retire in 50 years?

1 answer

To determine how much you should deposit at the end of each month into an IRA that pays 5% compounded monthly, such that you can withdraw $110,000 annually from the interest after retirement in 50 years, we can follow these steps:

Step 1: Calculate the necessary principal at retirement

Since you want to withdraw $110,000 annually from the interest, first, we need to find out how much principal is required to yield this amount at a 5% annual interest rate.

Using the formula for interest:

\[ I = P \times r \]

Where:

  • \(I\) is the annual interest desired ($110,000).
  • \(P\) is the principal amount required.
  • \(r\) is the annual interest rate (0.05).

Rearranging the formula to find \(P\):

\[ P = \frac{I}{r} = \frac{110,000}{0.05} = 2,200,000 \]

So, you will need $2,200,000 as the principal at retirement to withdraw $110,000 annually.

Step 2: Determine the monthly deposits

Now we need to determine how much to deposit monthly to reach a total of $2,200,000 in 50 years using the future value of a series formula (the future value of an annuity):

The future value of a series (annuity) is given by:

\[ FV = PMT \times \frac{(1 + r/n)^{nt} - 1}{(r/n)} \]

Where:

  • \(FV\) is the future value of the annuity ($2,200,000).
  • \(PMT\) is the monthly payment (what we are solving for).
  • \(r\) is the annual interest rate (0.05).
  • \(n\) is the number of compounding periods per year (12).
  • \(t\) is the number of years (50).

Substituting into the formula, we have:

\[ 2,200,000 = PMT \times \frac{(1 + 0.05/12)^{12 \times 50} - 1}{(0.05/12)} \]

Calculating the components:

  1. Calculate \(1 + r/n\):

\[ 1 + \frac{0.05}{12} = 1 + 0.0041667 \approx 1.0041667 \]

  1. Calculate the exponent:

\[ nt = 12 \times 50 = 600 \]

  1. Calculate \((1 + r/n)^{nt}\):

\[ (1.0041667)^{600} \]

Using a calculator, we get approximately:

\[ (1.0041667)^{600} \approx 6.022575 \]

  1. Now substitute back in to find \(FV\):

The factor now becomes:

\[ \frac{(1.0041667)^{600} - 1}{(0.0041667)} \approx \frac{6.022575 - 1}{0.0041667} \approx \frac{5.022575}{0.0041667} \approx 1207.027 \]

Step 3: Solve for \(PMT\)

Now substitute into the future value formula:

\[ 2,200,000 = PMT \times 1207.027 \]

To isolate \(PMT\):

\[ PMT = \frac{2,200,000}{1207.027} \approx 1824.80 \]

Conclusion

You need to deposit approximately $1,824.80 at the end of each month for 50 years in an IRA that pays 5% compounded monthly to accumulate a principal amount of $2,200,000, which will allow you to withdraw $110,000 annually from the interest alone after retirement.