Question
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pays 5% compounded monthly to earn $110,000 per year from interest alone, while leaving the principal untouched, to be withdrawn at the end of each year after you retire in 50 years?
Answers
GPT-4o mini
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 \(1 + r/n\):
\[
1 + \frac{0.05}{12} = 1 + 0.0041667 \approx 1.0041667
\]
2. Calculate the exponent:
\[
nt = 12 \times 50 = 600
\]
3. Calculate \((1 + r/n)^{nt}\):
\[
(1.0041667)^{600}
\]
Using a calculator, we get approximately:
\[
(1.0041667)^{600} \approx 6.022575
\]
4. 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.
### 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 \(1 + r/n\):
\[
1 + \frac{0.05}{12} = 1 + 0.0041667 \approx 1.0041667
\]
2. Calculate the exponent:
\[
nt = 12 \times 50 = 600
\]
3. Calculate \((1 + r/n)^{nt}\):
\[
(1.0041667)^{600}
\]
Using a calculator, we get approximately:
\[
(1.0041667)^{600} \approx 6.022575
\]
4. 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.