Mathematics Advanced • Year 12 • Module 7 • Lesson 12

Superannuation Modelling

Build fluency computing the net return, annual contribution and projected super balance with the recurrence and closed-form.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the net return formula:

r_net = ____________________

Q1.2 A worker earns $70,000 with an 11.5% employer contribution. State the annual super contribution C.

C = ____________________

Q1.3 Write the annual super recurrence relation in terms of A_n, r_net and C.

Stuck? Revisit lesson § Formula Reference and § Super Model.

2. Worked example — Project a super balance to age 65

Follow every line. Each step has a short reason.

Problem. A 30-year-old has $25,000 in super. Salary $75,000. Employer contributes 11.5%. Fund return 6.8% p.a. Fees 1.1% p.a. Find the projected balance at age 65 (35 years).

Step 1 — Compute the annual contribution.

C = 75,000 × 0.115 = $8,625 per year

Reason: employer contribution rate × salary.

Step 2 — Compute the net annual return.

r_net = 0.068 − 0.011 = 0.057 (5.7%)

Reason: fees are paid as a percentage of the balance each year, so they directly reduce growth.

Step 3 — Substitute into the closed form with n = 35.

A₃₅ = 25,000(1.057)³⁵ + 8,625 × [(1.057)³⁵ − 1] ÷ 0.057

(1.057)³⁵ ≈ 6.840

= 25,000(6.840) + 8,625 × (5.840 ÷ 0.057)

= 171,000 + 8,625 × 102.456 = 171,000 + 883,683 ≈ $1,054,683

Conclusion. The projected balance at age 65 is approximately $1.05 million. The contributions paid in total only $8,625 × 35 = $301,875 — about 29% of the final balance. The remaining 71% is compound growth.

3. Faded example — fill in the missing steps

Starting balance $15,000. Salary $60,000. Contribution rate 11.5%. Return 7%. Fees 0.8%. Project the balance after 30 years. 4 marks

Step 1 — Annual contribution:

C = 60,000 × ______________ = $______________

Step 2 — Net return:

r_net = ______________ − ______________ = ______________

Step 3 — Closed form, n = 30:

A₃₀ = 15,000(1.062)³⁰ + ______________ × [(1.062)³⁰ − 1] ÷ ______________

= 15,000(______________) + ______________ × (______________)

= ______________ + ______________ = $______________

Conclusion. Projected balance ≈ $______________.

Stuck? Revisit lesson § Worked Example — Try It Now.

4. Graduated practice

Use the appropriate formula. Round monetary answers to the nearest dollar unless cents are required.

Foundation — direct substitution (4 questions)

Q	Scenario	Working & answer
4.1 1	Salary $80,000, contribution rate 11.5%. State C.
4.2 1	Return 7%, fees 1%. State r_net.
4.3 1	Return 6.5%, fees 0.5%. State r_net.
4.4 1	Write the super recurrence with r_net = 0.06 and C = $9,000.

Standard — typical HSC difficulty (6 questions)

Show at least one substitution line and one evaluation line.

4.5 A₀ = $20,000, salary $80,000, contribution 11.5%, return 7%, fees 1%, n = 35. Use the closed form to find A₃₅. 2 marks

4.6 A₀ = $30,000, salary $85,000, contribution 11.5%, return 7%, fees 1.2%, n = 30. Find A₃₀. 2 marks

4.7 Same starting figures as 4.5 but fees rise to 1.5%. Find A₃₅. 2 marks

4.8 A₀ = $10,000, salary $50,000, contribution 11.5%, return 6.5%, fees 0.5%, n = 40. Find A₄₀. 2 marks

4.9 Use the super recurrence with A₀ = $20,000, r_net = 0.058 and C = $9,200 to find A₁ and A₂ by iteration. 2 marks

4.10 A worker's salary $90,000, contribution 11.5%, return 7%, fees 0.8%, A₀ = $40,000, n = 25. Find A₂₅. 2 marks

Extension — combine concepts (2 questions)

4.11 Two funds offer the same 7% gross return for 40 years on A₀ = $20,000 with constant C = $9,200. Fund X: fees 0.5%. Fund Y: fees 1.5%. Find both final balances and the dollar gap between them. 3 marks

4.12 A balance is to grow from A₀ = $0 by contributions only (C = $10,000/year) at r_net = 6% for 30 years. Find A₃₀. Then identify what fraction of A₃₀ is contributions and what fraction is growth. 3 marks

Stuck on 4.11? Compute both balances using the closed form A₀(1+r_net)ⁿ + C × [(1+r_net)ⁿ − 1] ÷ r_net with the two different r_net values.

5. Self-check the easy 3

Tick the first three once you have checked the method works.

For 4.1 I got C = 80,000 × 0.115 = $9,200.

For 4.2 I got r_net = 0.07 − 0.01 = 0.06.

For 4.4 my recurrence is A_n+1 = 1.06A_n + 9,000.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Net return

r_net = r_return − r_fees.

Q1.2 — Annual contribution at $70,000 × 11.5%

C = 70,000 × 0.115 = $8,050.

Q1.3 — Super recurrence

A_n+1 = (1 + r_net)A_n + C.

Q3 — Faded example: $15,000, salary $60,000, 11.5%, return 7%, fees 0.8%

C = 60,000 × 0.115 = $6,900. r_net = 0.07 − 0.008 = 0.062. (1.062)³⁰ ≈ 6.099.
A₃₀ = 15,000(6.099) + 6,900 × (5.099 ÷ 0.062) = 91,485 + 6,900 × 82.241 = 91,485 + 567,463 ≈ $658,948.

Q4.1 — Annual contribution at $80,000 × 11.5%

C = 80,000 × 0.115 = $9,200.

Q4.2 — r_net for 7% return, 1% fees

r_net = 0.07 − 0.01 = 0.06 (6%).

Q4.3 — r_net for 6.5% return, 0.5% fees

r_net = 0.065 − 0.005 = 0.06 (6%).

Q4.4 — Super recurrence with r_net = 0.06, C = 9,000

A_n+1 = 1.06A_n + 9,000.

Q4.5 — $20,000, salary $80,000, return 7%, fees 1%, n = 35

C = $9,200. r_net = 0.06. (1.06)³⁵ ≈ 7.6861.
A₃₅ = 20,000(7.6861) + 9,200 × (6.6861 ÷ 0.06) = 153,722 + 9,200 × 111.435 = 153,722 + 1,025,202 ≈ $1,178,924.

Q4.6 — $30,000, salary $85,000, return 7%, fees 1.2%, n = 30

C = 85,000 × 0.115 = $9,775. r_net = 0.058. (1.058)³⁰ ≈ 5.404.
A₃₀ = 30,000(5.404) + 9,775 × (4.404 ÷ 0.058) = 162,120 + 9,775 × 75.931 = 162,120 + 742,222 ≈ $904,342.

Q4.7 — As 4.5 but fees 1.5% (so r_net = 0.055)

(1.055)³⁵ ≈ 6.5138.
A₃₅ = 20,000(6.5138) + 9,200 × (5.5138 ÷ 0.055) = 130,276 + 9,200 × 100.250 = 130,276 + 922,300 ≈ $1,052,576. The extra 0.5% fee cost about $126,000 over the term.

Q4.8 — $10,000, salary $50,000, return 6.5%, fees 0.5%, n = 40

C = 50,000 × 0.115 = $5,750. r_net = 0.06. (1.06)⁴⁰ ≈ 10.2857.
A₄₀ = 10,000(10.2857) + 5,750 × (9.2857 ÷ 0.06) = 102,857 + 5,750 × 154.762 = 102,857 + 889,881 ≈ $992,738.

Q4.9 — Iterate the recurrence A_n+1 = 1.058A_n + 9,200, A₀ = 20,000

A₁ = 1.058(20,000) + 9,200 = 21,160 + 9,200 = $30,360.
A₂ = 1.058(30,360) + 9,200 = 32,120.88 + 9,200 = $41,320.88.

Q4.10 — $40,000, salary $90,000, return 7%, fees 0.8%, n = 25

C = 90,000 × 0.115 = $10,350. r_net = 0.062. (1.062)²⁵ ≈ 4.503.
A₂₅ = 40,000(4.503) + 10,350 × (3.503 ÷ 0.062) = 180,120 + 10,350 × 56.500 = 180,120 + 584,775 ≈ $764,895.

Q4.11 — Two funds, 0.5% vs 1.5% fees, n = 40

C = $9,200. Fund X (r_net = 0.065): (1.065)⁴⁰ ≈ 12.4161. A₄₀ = 20,000(12.4161) + 9,200 × (11.4161 ÷ 0.065) = 248,322 + 9,200 × 175.633 = 248,322 + 1,615,823 ≈ $1,864,145.
Fund Y (r_net = 0.055): (1.055)⁴⁰ ≈ 8.5133. A₄₀ = 20,000(8.5133) + 9,200 × (7.5133 ÷ 0.055) = 170,266 + 9,200 × 136.605 = 170,266 + 1,256,766 ≈ $1,427,032.
Gap = 1,864,145 − 1,427,032 ≈ $437,000. A 1% fee gap costs roughly $437k over 40 years.

Q4.12 — A₀ = 0, C = $10,000/yr, r_net = 6%, n = 30

(1.06)³⁰ ≈ 5.7435. A₃₀ = 0 + 10,000 × (4.7435 ÷ 0.06) = 10,000 × 79.058 = $790,582.
Total contributions = 10,000 × 30 = $300,000, about 38% of A₃₀. Compound growth supplies the remaining 62% (~$490,000).