What is annual rate in a monthly model?

Divide the nominal rate by the periods per year first.

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WAMathematics ApplicationsSyllabus dot point

How do regular payments build up or draw down a fund over time?

Model annuities and annuity-investments with recurrence relations and find balances, payments and the time to exhaust or reach a target.

How to model an annuity that pays out and an annuity-investment that builds up, using recurrence relations, and find balances, payment sizes and how long a fund lasts or takes to reach a target.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Two related models
Draw-down annuities
Annuity-investments (superannuation)
The effect of contributions and time

What this dot point is asking

You must model both kinds with recurrences, find balances, payments and the time to exhaust a fund or reach a target.

Both are the same recurrence with opposite signs on the regular payment.

As always, convert the nominal annual rate to the per-period rate (divide by periods per year) and count $n$ in those periods.

Draw-down annuities

A retiree puts a lump sum into an account and withdraws a regular amount. Each period interest is added, then the withdrawal is taken out. The fund lasts until the balance reaches zero. If the withdrawal is smaller than the interest earned, the balance grows instead (it behaves like a perpetuity or better); if larger, it depletes.

Worked example

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 Because money is being withdrawn, the payment is subtracted each month, giving $A_{n+1} = 1.005 A_n - 1800$ with $A_0 = 200000$ .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.

Annuity-investments (superannuation)

Here regular deposits are added to an interest-earning fund, so it grows by both interest and contributions. This models superannuation: each period interest is added, then the deposit. The finance solver finds the future value, the required deposit, or the time to reach a target. Note the solver sign convention: a deposit is entered as a negative $PMT$ (money leaving your pocket into the fund).

Worked example

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.

The effect of contributions and time

In a superannuation fund, the longer the money stays invested the larger the share of the final balance that comes from interest rather than from contributions, because early deposits compound for the most periods. This is why starting contributions early matters so much: a deposit made in year one earns interest every remaining year, while a deposit made near the end earns almost none. SCSA questions sometimes ask you to split a final balance into total contributions and total interest, which is total deposited (payment times number of payments) subtracted from the final value.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksA retiree invests

\

300\,000

in an annuity paying

4.8\%

per annum compounding monthly, and withdraws

2500

at the end of each month. (a) Write a recurrence relation for the balance

A_n

. (b) Find the balance after

3

months. (c) Explain how to find how long the annuity lasts.

Show worked answer →

An annuity adds interest then subtracts the withdrawal each month.

(a) Monthly rate $= \dfrac{0.048}{12} = 0.004$ , so $A_{n+1} = 1.004 A_n - 2500$ , with $A_0 = 300000$ . (2 marks)

(b) $A_1 = 1.004(300000) - 2500 = 301200 - 2500 = 298700$ . $A_2 = 1.004(298700) - 2500 = 299894.80 - 2500 = 297394.80$ . $A_3 = 1.004(297394.80) - 2500 = 298584.38 - 2500 = 296084.38$ , so $\$296\,084.38$ . (3 marks)

(c) Iterate (or use the finance solver) until the balance first reaches zero; the number of months at that point, with a smaller final withdrawal, gives the duration of the annuity. (2 marks)

Markers reward the annuity recurrence with a subtracted withdrawal, accurate iteration, and the depletion method.

WACE 20245 marksA worker deposits

\

500

at the end of each month into a superannuation fund earning

6\%

per annum compounding monthly, starting from

0

. Find the balance after

2

years and state how this annuity-investment differs from a draw-down annuity.

Show worked answer →

An annuity-investment adds interest then adds a deposit each month.

Monthly rate $= \dfrac{0.06}{12} = 0.005$ , so $A_{n+1} = 1.005 A_n + 500$ , with $A_0 = 0$ , over $n = 24$ months. Iterating (or the finance solver with $PV=0$ , $PMT=-500$ , $n=24$ , $i=0.5\%$ ) gives about $\$12\,718$ . (3 marks)

A draw-down annuity subtracts regular withdrawals and runs down towards zero; an annuity-investment adds regular deposits and builds up. The sign of the regular payment is positive (deposit) rather than negative (withdrawal). (2 marks)

Markers reward the annuity-investment recurrence with a positive deposit and the build-up versus draw-down distinction.