Annuities and superannuation in WACE Mathematics Applications Unit 4

How long a pension lasts

A fund of $\$200\,000$ earns $6\%$ per annum compounding monthly. The retiree withdraws $\$1800$ at the end of each month. Trace the first three months and explain how to find when it runs out.

Step 1: Convert the annual rate to a monthly rate and write the recurrence.

The annual rate of $6\%$ must be divided by $12$ to get the per-period rate: $i = \dfrac{0.06}{12} = 0.005$, so the growth factor is $R = 1.005$. Because money is being withdrawn, the payment is subtracted each month, giving $A_{n+1} = 1.005 A_n - 1800$ with $A_0 = 200000$.

Step 2: Iterate the recurrence for the first three months.

Each month we multiply the current balance by $1.005$ to add interest, then subtract the $\$1800$ withdrawal. Applying this three times gives:

$A_1 = 1.005(200000) - 1800 = 201000 - 1800 = 199200$.

$A_2 = 1.005(199200) - 1800 = 200196 - 1800 = 198396$.

$A_3 = 1.005(198396) - 1800 = 199387.98 - 1800 = 197587.98$.

Step 3: Check whether the fund depletes and find how long it lasts.

The first month's interest is $0.005 \times 200000 = \$1000$, which is less than the $\$1800$ withdrawal, so the balance falls each month and the fund will eventually run out. Using the finance solver with $PV = 200000$, $PMT = -1800$ and $I\% = 6$ monthly, solving for $N$ shows the fund lasts about $158$ months (just over $13$ years), with a smaller final payment in the last month.

Reaching a savings target

How much must be deposited monthly to reach $\$50\,000$ in $5$ years at $4.8\%$ per annum compounding monthly, from a zero start?

Step 1: Identify all known quantities and convert to monthly terms.

Because the account compounds monthly, the per-period rate is $i = \dfrac{0.048}{12} = 0.004$ and the number of periods is $n = 60$ months. The starting balance is $PV = 0$ and the desired future value is $FV = 50000$; the unknown is the regular deposit $PMT$.

Step 2: Use the finance solver to find the required monthly deposit.

Enter $PV = 0$, $FV = 50000$, $I\% = 4.8$ (monthly compounding), $N = 60$ into the finance solver and solve for $PMT$. The solver returns a monthly deposit of about $\$737$.

Step 3: Interpret the result by splitting the final balance into contributions and interest.

Depositing about $\$737$ each month for $5$ years, with interest compounding at $4.8\%$ per annum, builds the fund to $\$50\,000$. The total amount actually deposited is $737 \times 60 = \$44\,220$, so the remaining $\$5780$ is interest earned by the fund over the five years.