Helen has just retired with $500 000 in her account and wishes to start an annuity. She intends to withdraw $M at the end of each year for 30 years. Reducible interest is calculated at 6% p.a., compounded annually just before each withdrawal.
a) Calculate $M, the amount of each withdrawal.
b) After her 15th withdrawal, the interest rate on the annuity reduces to 4% p.a. Helen is informed that to continue to withdraw $M, the life of the annuity must also reduce. For how many more years will Helen be able to withdraw money from this annuity.
Can someone please confirm my answer for b). I got 12.5 years. If it is wrong can you please provide the solution for b) thanks.
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To solve this problem, we can use the formula for the present value of an ordinary annuity:
PV = PMT * [(1 - (1 + r)^(-n)) / r]
where PV is the present value (initial amount), PMT is the withdrawal amount, r is the interest rate, and n is the number of withdrawals.
a) To calculate the amount of each withdrawal, we need to solve for PMT in the formula above. Given that Helen has $500,000 in her account and intends to withdraw $M for 30 years, the equation becomes:
$500,000 = $M * [(1 - (1 + 0.06)^(-30)) / 0.06]
Solving for $M:
$M = $500,000 / [(1 - 1.853) / 0.06]
$M = $500,000 / (0.853 / 0.06)
$M ≈ $8,354.76
So, Helen needs to withdraw approximately $8,354.76 at the end of each year.
b) After the 15th withdrawal, the interest rate on the annuity reduces to 4% p.a. We need to find out for how many more years Helen will be able to withdraw money from this annuity.
To calculate the remaining number of withdrawals, we can use the present value formula again, but now with a reduced interest rate of 4%:
$500,000 = $M * [(1 - (1 + 0.04)^(-n)) / 0.04]
Using the value of $M calculated in part a), we have:
$8,354.76 * [(1 - 1.04^(-n)) / 0.04] = $500,000
[(1 - 1.04^(-n)) / 0.04] ≈ $500,000 / $8,354.76
[(1 - 1.04^(-n)) / 0.04] ≈ 59.92
1 - 1.04^(-n) ≈ 0.04 * 59.92
1 - 1.04^(-n) ≈ 2.3968
1.04^(-n) ≈ 1 - 2.3968
1.04^(-n) ≈ -1.3968
Taking the logarithm of both sides:
-n * log(1.04) ≈ log(-1.3968)
n ≈ log(-1.3968) / log(1.04)
This equation doesn't have a valid solution because the logarithm of a negative number is undefined.
Therefore, it seems there is an error in the calculations. To find the correct solution for part b), we need to reevaluate the problem.
To calculate the amount of each withdrawal, we can use the formula for the present value of an annuity:
P = M (1 - (1 + r)^(-n)) / r
Where:
P = Present value of the annuity
M = Amount of each withdrawal
r = Interest rate
n = Number of years
a) We want to find M, so let's rearrange the formula:
M = P * r / (1 - (1 + r)^(-n))
Given that Helen has $500,000 in her account, and she wants to withdraw this amount every year for 30 years, we can plug in the values:
M = $500,000 * 0.06 / (1 - (1 + 0.06)^(-30))
M = $500,000 * 0.06 / (1 - 0.98806591)
M = $500,000 * 0.06 / 0.01193409
M ≈ $31,543.41
Therefore, the amount of each withdrawal should be approximately $31,543.41.
b) After the 15th withdrawal, the interest rate reduces to 4% p.a. To determine for how many more years Helen will be able to withdraw money, we need to calculate the present value of the remaining annuity based on the new interest rate.
First, let's calculate the present value of the annuity after the 15th withdrawal:
PV = M (1 - (1 + r)^(-n)) / r
Using the new interest rate of 4%, we can plug in the values:
PV = $31,543.41 (1 - (1 + 0.04)^(-15)) / 0.04
PV ≈ $31,543.41 (1 - 0.518167105) / 0.04
PV ≈ $31,543.41 (0.481832895) / 0.04
PV ≈ $378,146.97 / 0.04
PV ≈ $9,453,674.25
Now, let's find out for how many more years Helen can withdraw money if she continues to withdraw $31,543.41 annually.
PV = M (1 - (1 + r)^(-n)) / r
$9,453,674.25 = $31,543.41 (1 - (1 + 0.04)^(-n)) / 0.04
$9,453,674.25 * 0.04 = $31,543.41 (1 - 1.04^(-n))
$378,146.97 = $31,543.41 (1 - 1.04^(-n))
Now, we can solve for n by rearranging the equation:
1.04^(-n) = 1 - ($378,146.97 / $31,543.41)
1.04^(-n) = 1 - 11.9999587
1.04^(-n) = -10.9999587
In this case, the resulting value is negative, which is incorrect. It seems there might be an error in calculating the number of remaining years (b). Perhaps a recalibration of the solution is required. It is possible that the formula for the present value is not being applied correctly or there might be a mistake with the interest rate. To ensure accuracy, please double-check your calculations and reassess the problem.