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Annuity Investments and Superannuation - TCE Mathematics Applications (Tasmania)

Syllabus dot point

“Model annuity investments and superannuation where regular contributions are added to a fund that earns compound interest.”

TCEMathematics ApplicationsUnit 48 min read

Quick answer

You must model an annuity investment, a fund that earns compound interest while a fixed regular contribution is added each period, using Vn+1=Vn(1+r)+dV_{n+1} = V_n(1 + r) + d, and use it to find the future value of savings such as superannuation.

What this dot point is asking

An annuity investment is the savings mirror of a loan. Rather than paying a debt down, you build a balance up: the fund earns interest and you add a regular deposit on top. Superannuation is the most common example, where contributions accumulate over a working life.

Annuity investment

A fund that earns compound interest each period and also receives a fixed regular contribution. The balance grows from both the interest on what is already there and the new money added.

The recurrence relation

The order of operations is interest first, then the contribution. The plus sign on dd is what separates a growing investment from a draining annuity.

Annuity investment recurrence

Vn+1=Vn(1+r)+d,V0=initial balanceV_{n+1} = V_n(1 + r) + d, \quad V_0 = \text{initial balance}

Each period the balance grows by interest at rate rr, then the contribution dd is added. Both effects push the balance upward, so the fund grows faster than interest alone.

Building a superannuation balance

A fund starts at $20 000\$20\,000, earns 0.4 percent per month, and receives $500\$500 at the end of each month.

Step 1: Identify the parameters and set up the recurrence

The monthly interest rate is r=0.004r = 0.004, the regular contribution is d=500d = 500, and the starting balance is V0=20000V_0 = 20000. Each month we apply interest to the existing balance first and then add the contribution, which matches the recurrence Vn+1=Vn(1.004)+500V_{n+1} = V_n(1.004) + 500. Setting up these values before calculating prevents errors later.

Step 2: Iterate the recurrence across three months

Apply the rule three times in sequence, carrying forward the exact result each time.
V1=20000(1.004)+500=20080+500=20580V_1 = 20000(1.004) + 500 = 20080 + 500 = 20580.
V2=20580(1.004)+500=20662.32+500=21162.32V_2 = 20580(1.004) + 500 = 20662.32 + 500 = 21162.32.
V3=21162.32(1.004)+500=21246.97+500=21746.97V_3 = 21162.32(1.004) + 500 = 21246.97 + 500 = 21746.97.

Step 3: Interpret the result

After three months the fund holds about $21 746.97\$21\,746.97, growing from both interest and the steady contributions. Notice that each month's interest is calculated on a slightly larger balance than the month before, so the interest portion itself increases over time even though the contribution stays constant.

Interest versus contributions

Early on, most of the growth comes from the contributions, because the balance is small and earns little interest. Over many years the balance becomes large enough that the interest it earns can exceed the contributions, which is the power of compounding over a long working life.

Comparing the two sources of growth

In month 3 above, the fund earned interest of 21162.32×0.004=$84.6521162.32 \times 0.004 = \$84.65 and received a $500\$500 contribution.

Interpret

The $500\$500 contribution dwarfs the $84.65\$84.65 of interest at this early stage. After decades, when the balance might be hundreds of thousands of dollars, the monthly interest would far exceed the contribution, which is why starting super early matters so much.

Add the contribution after the interest

Apply interest to the existing balance first, then add the new contribution. Adding the contribution before applying interest would credit a full period of interest on money that was only just deposited, overstating the balance.

Future value over many periods

To find the balance after a long time, iterate the recurrence with technology or a financial solver. The future value depends on the starting balance, the rate, the contribution, and the number of periods. Increasing any of these increases the final balance, but the rate and the number of periods have the most dramatic effect because they drive compounding.

What grows a fund fastest

Higher contributions help, but a longer time in the fund and a higher interest rate matter more, because compounding multiplies the balance period after period. Time is the single most powerful factor in long-term saving.

Confusing the sign with an annuity that pays out

An annuity investment adds the contribution (+d+d), so the balance grows. A drawdown annuity subtracts a withdrawal (−d-d), so the balance falls. Using a minus sign here would model spending the fund rather than building it. Keep dd positive for contributions.

The future-value formula

For many periods, iterating by hand is slow, so an explicit future-value formula is used. For an ordinary annuity (contributions at the end of each period),

FV=d (1+i)n−1i,FV = d\,\frac{(1 + i)^n - 1}{i},

where dd is the regular contribution, ii is the per-period rate, and nn is the number of contributions. If the contributions are made at the start of each period (an annuity in advance), multiply the whole expression by an extra (1+i)(1 + i), because every deposit then earns one more period of interest. Any starting balance V0V_0 grows separately as V0(1+i)nV_0(1 + i)^n and is added on.

Future value of a savings plan

Deposit $300\$300 at the end of each month into an account earning 4.8%4.8\% p.a. compounding monthly for 5 years.

Step 1: Identify and convert the inputs

d=300d = 300, i=0.04812=0.004i = \frac{0.048}{12} = 0.004, n=5×12=60n = 5 \times 12 = 60. The annual rate must be divided by 12 to match the monthly compounding period, and the time in years must be converted to the same number of monthly periods. Mismatching the rate and period units is the most common error in annuity calculations.

Step 2: Apply the future-value formula and evaluate

Because deposits occur at the end of each month, this is an ordinary annuity and the formula applies directly without adjustment. Substitute the identified values and simplify step by step.

FV=300×(1.004)60−10.004=300×1.27024−10.004=300×67.56=$20 268.FV = 300 \times \frac{(1.004)^{60} - 1}{0.004} = 300 \times \frac{1.27024 - 1}{0.004} = 300 \times 67.56 = \$20\,268.

Step 3: Interpret the result

After 5 years the plan is worth about $20 268\$20\,268, of which $18 000\$18\,000 is your own deposits and the rest is interest. This demonstrates how compounding amplifies even modest regular deposits over a multi-year period, turning a straightforward saving habit into meaningfully more wealth than the sum of contributions alone.

Making the deposit the subject

A common task gives a savings target and asks for the required regular deposit. Rearrange the future-value formula to make dd the subject by dividing the target FVFV by the annuity factor (1+i)n−1i\frac{(1 + i)^n - 1}{i} (times (1+i)(1 + i) for an annuity in advance). This is exactly the algebra in the second exam answer above, and the marker rewards showing the rearrangement, not just the final number.

A complete answer models the fund with Vn+1=Vn(1+r)+dV_{n+1} = V_n(1 + r) + d using a positive contribution, applies interest before adding the contribution, and finds the future value by iterating or by the future-value formula over the correct number of periods.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice9 marks
A business pays $2500\$2500 every quarter into a fund paying 5.8%5.8\% p.a. compounding quarterly. a) Model the fund with a difference equation (use T0T_0). b) How much is in the fund after 10 years? The company instead makes an initial deposit of $10 000\$10\,000 then the quarterly instalments. c) Remodel. d) How much after 10 years? e) What is the present value of the total?
Show worked answer →

a) (2 marks) Quarterly rate i=0.0584=0.0145i = \frac{0.058}{4} = 0.0145. Each quarter the balance earns interest then $2500\$2500 is added: Tn+1=1.0145 Tn+2500T_{n+1} = 1.0145\,T_n + 2500, T0=0T_0 = 0.

b) (2 marks) Iterate for 10 years =40= 40 quarters (or use the future-value annuity formula). The fund reaches about $134 244\$134\,244.

c) (2 marks) Only the starting value changes: Tn+1=1.0145 Tn+2500T_{n+1} = 1.0145\,T_n + 2500, T0=10000T_0 = 10000.

d) (2 marks) The $10 000\$10\,000 also grows for 40 quarters, adding 10000×1.01454010000 \times 1.0145^{40} to the previous total, giving about $152 030\$152\,030.

e) (1 mark) Present value discounts the final amount back 40 quarters: PV=1520301.014540=$85 477PV = \frac{152030}{1.0145^{40}} = \$85\,477 (approximately). It is the single deposit today that would grow to the same total.

Practice3 marks
You want to save $6000\$6000 over 2 years. Your account pays 3%3\% p.a. compounding monthly. Use a formula and algebra to find how much you must deposit each month.
Show worked answer →

(3 marks) This is a future-value annuity (in advance). Monthly rate i=0.0312=0.0025i = \frac{0.03}{12} = 0.0025 and the number of deposits n=24n = 24.

Using the annuities-in-advance future-value formula FV=d(1+i)(1+i)n−1iFV = d(1 + i)\frac{(1 + i)^n - 1}{i}, substitute FV=6000FV = 6000:

6000=d(1.0025)(1.0025)24−10.00256000 = d(1.0025)\frac{(1.0025)^{24} - 1}{0.0025}.

Compute the bracket: (1.0025)24−1=0.06176(1.0025)^{24} - 1 = 0.06176, so the factor is 1.0025×0.061760.0025=24.76461.0025 \times \frac{0.06176}{0.0025} = 24.7646. Then d=600024.7646=$242.28d = \frac{6000}{24.7646} = \$242.28 per month. Markers want the substitution, the algebra to make dd the subject, and the rounded monthly deposit.

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