Mathematics Advanced • Year 12 • Module 7 • Lesson 3
Effective Annual Rate of Interest
Apply EAR to real product comparisons, marketing critique, and continuous-compounding limits.
Problem 1 — Two savings accounts (the "Think First" scenario)
Two accounts both advertise "6% p.a." Account A compounds annually; Account B compounds monthly.
Set up: What are we solving for?
(i) Find the balance after one year for a $10,000 deposit in each account. 2 marks
(ii) Compute the EAR of each account and state the dollar difference after one year. 2 marks
(iii) Project the dollar difference after 20 years (still $10,000 deposit, no further contributions). Use the EAR of each account to compute final balances. 3 marks
Stuck? Revisit lesson § Nominal vs Effective.Problem 2 — Rank four products (4-way EAR comparison)
A: 5.80% p.a. compounded semi-annually.
B: 5.70% p.a. compounded monthly.
C: 5.65% p.a. compounded daily.
D: 5.90% p.a. compounded annually.
Set up: What are we solving for?
(i) Compute the EAR of each product to 2 dp. 3 marks
(ii) Rank from highest to lowest EAR and identify the best deposit product. 2 marks
(iii) Comment in one sentence on the headline rate vs the winner — does the highest nominal rate always win? 1 mark
Problem 3 — Credit-card EAR (financial trap alert)
A typical Australian credit card advertises 19.99% p.a. compounded daily on unpaid balances.
Set up: What are we solving for?
(i) Compute the EAR to 2 dp. 2 marks
(ii) On a $5,000 balance carried for one year, compute the "hidden" interest — the difference between the EAR-implied interest and the nominal-implied interest. 2 marks
(iii) Write a 1-2 sentence paragraph you could send to a family member explaining why "19.99% p.a." is misleading. 2 marks
Stuck? Revisit lesson § Financial Trap Alert.Problem 4 — Designing an equivalent product (reverse EAR)
Product A is offered at 7.2% p.a. compounded quarterly. A bank wants to launch Product B, with monthly compounding, that has exactly the same EAR as Product A.
Set up: What are we solving for?
(i) Find the EAR of Product A to 2 dp. 1 mark
(ii) Write an equation in rnominal (the unknown nominal rate of Product B) such that B has the same EAR. Solve for the required nominal rate, to 2 dp. 3 marks
(iii) Explain in one line why the nominal rate of B is lower than that of A even though they are equivalent. 1 mark
Problem 5 — Continuous compounding limit (n → ∞)
As compounding becomes infinitely frequent, the EAR approaches reff = er − 1.
Set up: What are we solving for?
(i) For a nominal rate of 10% p.a., compute the EAR under (a) annual, (b) monthly, (c) daily, and (d) continuous compounding. 3 marks
(ii) By how many percentage points does daily compounding fall short of continuous? 1 mark
(iii) Explain in one sentence why no real-world bank can offer "more" than the continuous-compounding rate at a given nominal — and what this implies about the practical ceiling on compounding-frequency competition. 2 marks
Stuck? The continuous EAR is er − 1 — a hard ceiling that no finite-n formula can exceed.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Two 6% accounts
Set up. We are comparing one-year and long-term returns of nominal-equivalent products with different compounding frequencies.
(i) Account A: A = 10,000(1.06) = $10,600.00. Account B: A = 10,000(1.005)¹² = 10,000 × 1.06168 = $10,616.78.
(ii) EARA = 6.00%. EARB = (1.005)¹² − 1 = 6.17%. Difference after 1 yr = 10,616.78 − 10,600 = $16.78.
(iii) After 20 years: A = 10,000(1.06)²⁰ = 10,000 × 3.20714 = $32,071.35. B = 10,000(1.005)²⁴⁰ = 10,000 × 3.31020 = $33,102.04. Difference = $1,030.69 — small at one year, ~$1,000 over 20 years.
Problem 2 — Rank four products
Set up. We are putting four products on a common annual basis (EAR) and ranking them.
(i) A: (1 + 0.058/2)² − 1 = (1.029)² − 1 = 5.88%. B: (1 + 0.057/12)¹² − 1 = (1.00475)¹² − 1 = 5.85%. C: (1 + 0.0565/365)³⁶⁵ − 1 = 5.81%. D: 5.90% (already annual).
(ii) Ranking: D (5.90%) > A (5.88%) > B (5.85%) > C (5.81%). Best deposit product is D.
(iii) Yes — at these closely-spaced rates, the highest nominal rate wins because the EAR gap from compounding frequency (a few basis points) is smaller than the nominal-rate gap (10-25 basis points).
Problem 3 — Credit card EAR
Set up. We are quantifying the hidden cost of daily compounding on a credit-card balance and explaining it in plain language.
(i) reff = (1 + 0.1999/365)³⁶⁵ − 1 = 1.22125 − 1 = 22.13%.
(ii) Hidden interest on $5,000 = 5,000 × (0.2213 − 0.1999) = 5,000 × 0.0214 = $107.00 per year.
(iii) Sample: "When a card advertises 19.99% p.a., the actual annual cost on an unpaid balance is closer to 22.13% because the bank adds interest each day and then charges interest on that interest. On a $5,000 balance you pay about $107 more per year than the headline rate suggests."
Problem 4 — Equivalent product
Set up. We are finding the nominal rate of a monthly-compounded product whose EAR matches a quarterly-compounded product.
(i) EARA = (1 + 0.072/4)⁴ − 1 = (1.018)⁴ − 1 = 1.07396 − 1 = 7.40%.
(ii) Solve (1 + r/12)¹² − 1 = 0.0740 ⇒ (1 + r/12)¹² = 1.0740 ⇒ 1 + r/12 = 1.07401/12 = 1.005967 ⇒ r/12 = 0.005967 ⇒ r = 12 × 0.005967 = 0.07161 = 7.16% p.a.
(iii) Monthly compounding adds interest 12 times rather than 4 times per year, so the same EAR requires a lower nominal rate — more frequent compounding squeezes more value from each percentage point.
Problem 5 — Continuous compounding
Set up. We are placing the discrete EAR formula against its continuous-compounding limit.
(i) (a) Annual: 10.00%. (b) Monthly: (1 + 0.1/12)¹² − 1 = 10.47%. (c) Daily: (1 + 0.1/365)³⁶⁵ − 1 = 10.516%. (d) Continuous: e0.10 − 1 = 0.10517 = 10.517%.
(ii) Daily falls short of continuous by about 0.001 percentage points (10.517% − 10.516%) — essentially nil.
(iii) No real-world product can exceed the continuous-compounding EAR at a given nominal because er − 1 is the exact mathematical limit; once compounding is daily, increasing the frequency further (hourly, by the second) adds essentially zero — so banks compete on nominal rates rather than on frequency.