Mathematics Advanced • Year 12 • Module 7 • Lesson 15
Extra Repayments, Offset Accounts and Redraw
Apply extra-repayment, offset and redraw mathematics to debt-reduction strategy problems.
Problem 1 — An extra $353/month saves how much? (Think First scenario)
A $400,000 loan at 5% p.a. compounded monthly over 30 years has a minimum monthly repayment of Mold ≈ $2,147. The borrower pays $2,500 instead — an extra $353/month.
Set up: What are we solving for?
(i) Use nnew = −ln(1 − Pr ÷ Mnew) ÷ ln(1 + r) to find the new term. 3 marks
(ii) Compute interest saved = Mold × nold − Mnew × nnew. State the years saved as well. 3 marks
(iii) Compute the return on investment: interest saved ÷ total extra paid ($353 × nnew). Express as a percentage. 2 marks
Stuck? Revisit lesson § Think First and § Extra Repayments table.Problem 2 — Offset account vs taxable savings
A borrower has $50,000 spare cash and a $400,000 home loan at 5.2% p.a. compounded monthly. They can either (i) keep the $50,000 in an offset account against the loan, or (ii) put it in a high-interest savings account paying 4.5% p.a. with interest taxed at the marginal rate of 32.5%.
Set up: What are we solving for?
(i) Compute the annual interest saving from keeping $50,000 in offset. 2 marks
(ii) Compute the after-tax annual interest earned from the savings account. 2 marks
(iii) State which option wins and by how many dollars. Explain in 1 sentence why offset is sometimes called a "tax-free guaranteed return at the loan rate". 2 marks
Problem 3 — Why early extra payments save more than late ones
A $400,000 loan at 5% p.a. compounded monthly, minimum M = $2,147 over 30 years. Compare two strategies that both contribute the same extra amount in total: (A) extra $500/month for the first 60 months only, then minimum; (B) minimum for 300 months, then extra $500/month for the final 60 months. Both add 60 × $500 = $30,000 of extra payments.
Set up: What are we solving for?
(i) For Strategy A, the first 60 months use Mnew = $2,647. Approximate the balance after 60 months using A60 = P(1+r)60 − Mnew × [(1+r)60 − 1] ÷ r. 3 marks
(ii) Qualitatively, which strategy clears the loan sooner — A or B? Explain in 1-2 sentences referencing how reducing principal early prevents years of compound interest accruing on that principal. 2 marks
(iii) Estimate the difference in years saved: the lesson states extra-then-minimum vs minimum-then-extra at the same $500/month for 60 months will differ by roughly several years. Make a 1-sentence quantitative claim. 2 marks
Stuck? Revisit lesson § Misconceptions — early repayments save more.Problem 4 — Redraw or stay put?
A borrower has paid an extra $40,000 into their $300,000 / 5% / 25-year home loan over the past 3 years. Their effective balance is $260,000. An emergency car repair needs $8,000. They can either redraw $8,000 from the loan or use a credit card at 18% p.a.
Set up: What are we solving for?
(i) If they redraw $8,000, the effective loan balance rises from $260,000 to $268,000. State the annual extra interest cost. 1 mark
(ii) If they use the credit card and pay $8,000 over 12 months, approximate the interest cost using the simple I = Pr formula on the average outstanding balance (≈$4,000). 2 marks
(iii) Recommend the redraw or the credit card in 1-2 sentences, citing the dollar gap and the practical drawback of each. 2 marks
Problem 5 — Design a debt-reduction strategy
A $350,000 loan at 4.5% p.a. compounded monthly is taken over 30 years. The borrower has $25,000 spare cash and can comfortably afford an extra $250/month above the minimum repayment.
Set up: What are we solving for?
(i) Compute the minimum monthly repayment Mmin using M = Pr ÷ [1 − (1+r)−n]. 2 marks
(ii) With $25,000 in an offset account from day one, the interest charge is computed on $325,000. Using Mmin + $250 as the actual monthly payment, find nnew (months and years) — set Peffective = $325,000 and Mnew = Mmin + 250 in the formula. 3 marks
(iii) State the years saved against the 30-year baseline and the approximate interest saved. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Extra $353/month on the $400,000 loan
Set up. Solve for the new term, compute interest saved, then convert to a return on extra dollars contributed.
(i) r = 0.004167. Pr ÷ Mnew = 400,000 × 0.004167 ÷ 2,500 = 0.6667. nnew = −ln(0.3333) ÷ ln(1.004167) = 1.0986 ÷ 0.004158 ≈ 264 months (22.0 years).
(ii) Old total = 2,147 × 360 ≈ $773,000 → old interest ≈ $373,000. New total = 2,500 × 264 = $660,000 → new interest ≈ $260,000. Interest saved ≈ $113,000, years saved ≈ 8 years.
(iii) Total extra paid above the minimum = 353 × 264 ≈ $93,192. ROI = 113,000 ÷ 93,192 ≈ 121%. Each extra dollar paid avoids about $1.21 of compound interest over the loan life.
Problem 2 — Offset vs taxable savings on $50,000
Set up. Compute each option's annual benefit and compare.
(i) Offset on a 5.2% loan: tax-free annual saving = 50,000 × 0.052 = $2,600/yr.
(ii) Savings 4.5% taxable at 32.5%: after-tax interest = 50,000 × 0.045 × (1 − 0.325) = 50,000 × 0.045 × 0.675 = 50,000 × 0.030375 = $1,518.75/yr.
(iii) Offset wins by 2,600 − 1,518.75 = $1,081.25/yr. Offset is called a "tax-free guaranteed return at the loan rate" because every dollar in the offset reduces the interest you would otherwise pay, and that saving is not income (no tax) and not subject to market risk (guaranteed).
Problem 3 — Early vs late extra payments
Set up. Project the early-extra strategy's balance after 5 years and reason about the late strategy.
(i) r = 0.004167, n = 60, Mnew = $2,647. (1.004167)⁶⁰ ≈ 1.28336. A60 = 400,000(1.28336) − 2,647 × (1.28336 − 1) ÷ 0.004167 = 513,344 − 2,647 × 68.005 = 513,344 − 180,009 ≈ $333,335. So after 5 years of extra $500/month, the balance is roughly $333,000 (vs ~$367,000 at the minimum schedule).
(ii) Strategy A clears the loan sooner. Reducing principal early means a smaller balance accrues interest for the remaining 25 years; the $30,000 of extra payments in years 1–5 displaces decades of compound interest, whereas the same $30,000 in years 26–30 only displaces 5 years of interest charges.
(iii) Sample claim: Strategy A clears the loan roughly 5–7 years sooner than Strategy B (despite both paying the same extra $30,000 in total), because the early extra payments shrink the balance that compounds for the longest time.
Problem 4 — Redraw vs credit card
Set up. Compare the two interest costs over a 1-year horizon.
(i) Extra $8,000 in the loan at 5% p.a. costs 8,000 × 0.05 = $400/yr in extra interest.
(ii) Credit card 18% on average $4,000 outstanding ≈ 4,000 × 0.18 = $720 for the year. (Realistic estimates including fees make it higher.)
(iii) Recommend the redraw: $400/yr vs $720/yr saves $320/yr. The drawback is that the redrawn $8,000 will sit on the loan balance and accrue interest indefinitely if not repaid quickly, whereas the credit-card debt would have been forcibly cleared in 12 months — so the borrower should treat the redraw as an obligation to top up future repayments to clear the $8,000 within 12 months too.
Problem 5 — $350,000 / 4.5% / 30-year with offset and extra
Set up. Compute baseline M, then solve for the new term given the offset and extra repayment.
(i) r = 0.045 ÷ 12 = 0.00375; n = 360. Pr = 350,000 × 0.00375 = $1,312.50. (1.00375)−360 ≈ 0.25946. Denom = 0.74054. Mmin = 1,312.50 ÷ 0.74054 ≈ $1,772.66/month.
(ii) Effective balance = $325,000. Mnew = 1,772.66 + 250 = $2,022.66. Pr ÷ Mnew = 325,000 × 0.00375 ÷ 2,022.66 = 1,218.75 ÷ 2,022.66 ≈ 0.6026. nnew = −ln(0.3974) ÷ ln(1.00375) = 0.9226 ÷ 0.003743 ≈ 246.5 months (≈ 20.5 years).
(iii) Years saved ≈ 30 − 20.5 = 9.5 years. Total repaid: 2,022.66 × 246.5 ≈ $498,585; baseline 1,772.66 × 360 ≈ $638,158. Approximate interest saved ≈ $140,000.