9514 1404 393
Answer:
- 346.1 monthly payments; 28.8 years
- 18.5 years; infinite years
Step-by-step explanation:
Two formulas come into play here. One is the future value of a series of payments. The other is the payment amount available from an ordinary annuity.
FV = An/r((1 +r/n)^(nt) -1) . . . . future value of payments of A made n times per year for t years
A = P(r/n)/(1 -(1 +r/n)^(-nt)) . . . . payment from principal P at invested at rate r for t years, compounded n times per year.
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Problem 1
You want FV = $350,000, A = $400, r = 0.057, n = 12, and we want to find t. We can solve the FV equation for t to get ...
log(FV·r/(An) +1)/log(1 +r/n) = nt . . . . . number of months
log(350000·0.057/(400·12) +1)/log(1 +0.057/12) = nt ≈ 346.1
The retirement account can be funded by 346.1 monthly payments. That will take 28.8 years.
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Problem 2
Again, solving for t, we get ...
log(1 -Pr/(An))/(-n·log(1 +r/n)) = t
The parameters for this are P = 700,000, A = 5000, n = 12, r = 0.054, so the account can be expected to last for ...
log(1 -700000·0.054/(5000·12))/(-12·log(1 +0.054/12)) ≈ 18.5 . . . years
Changing the withdrawal to $3000 per month makes the account last forever. It earns $3150 in interest each month, so the account balance continues to increase.
The account will support withdrawals of $5000 per month for 18.5 years.
The account will support withdrawals of $3000 per month forever.
