Mathematics Advanced • Year 12 • Module 7 • Lesson 4
Depreciation
Apply flat-rate and reducing-balance depreciation to vehicles, technology, machinery and tax planning.
Problem 1, New car drives off the lot (the "Think First" scenario)
A new car is purchased for $40,000. It depreciates at 20% per year.
Set up: What are we solving for?
(i) Using flat-rate depreciation, find the book value after 5 years. Is the figure realistic? 2 marks
(ii) Using reducing-balance depreciation, find the book value after 5 years. 2 marks
(iii) A used-car dealer offers $14,500 for the car after 5 years. Comment in one sentence on whether that price is fair for each depreciation model. 2 marks
Stuck? Revisit lesson § Think First and § Reducing Balance.Problem 2, Delivery van for a small business
A small business purchases a delivery van for $45,000. The accountant models depreciation at 15% per year.
Set up: What are we solving for?
(i) Find the book value at the end of year 4 under flat rate and reducing balance. 2 marks
(ii) Find the total dollar depreciation over the 4-year period under each method. 2 marks
(iii) Explain in 1-2 sentences why a delivery business might prefer the reducing-balance method for tax purposes. 2 marks
Problem 3, Industrial machinery (find the rate)
A factory purchased industrial machinery for $120,000. After 6 years its book value is $28,000 under reducing balance.
Set up: What are we solving for?
(i) Find the annual depreciation rate r to two decimal places. 2 marks
(ii) Project the book value at year 10 using your rate from (i). 2 marks
(iii) If the factory wants to dispose of the machine when its book value first falls below $10,000, use logarithms to find the year (to nearest whole). 3 marks
Stuck? Revisit lesson § Worked Example.Problem 4, Office IT fleet (technology obsolescence)
A company buys $80,000 of laptops. Technology depreciates at 30% p.a. reducing balance (rapid obsolescence). The same fleet under flat rate would depreciate at 20% p.a.
Set up: What are we solving for?
(i) Find the book value at the end of years 1, 3 and 5 under reducing balance. 3 marks
(ii) Find the book value at the same years under flat rate (20% p.a.). At which year do the two methods cross over (i.e. give the same book value)? 3 marks
(iii) Recommend which method best models technology obsolescence and justify in one sentence. 1 mark
Problem 5, Salvage value and end of life
An accountant uses reducing-balance depreciation for a $50,000 piece of equipment at 15% p.a., with a salvage value of $5,000 (the lowest value the asset can have on the books).
Set up: What are we solving for?
(i) Compute the book value at year 5, 10 and 15. 3 marks
(ii) Use logarithms to find the first year in which the unadjusted book value drops below the salvage value $5,000. 2 marks
(iii) Explain in one sentence why reducing balance approaches but never reaches zero, and why salvage values are imposed in practice. 2 marks
Stuck? Solve V₀(1 − r)ⁿ < salvage and take logs.How did this worksheet feel?
What I'll revisit before next class:
Problem 1, Car at 20% per year
Set up. We are valuing the car under each model and judging which matches a real dealer offer.
(i) Flat: S = 40,000 − 0.20 × 40,000 × 5 = 40,000 − 40,000 = $0. Not realistic, a 5-year-old car still has trade-in value.
(ii) RB: S = 40,000(0.80)⁵ = 40,000 × 0.32768 = $13,107.20.
(iii) The dealer's $14,500 offer is generous against reducing balance ($13,107) and absurdly generous against flat rate ($0), only the RB model gives a price comparison that makes sense for vehicles.
Problem 2, Delivery van at 15% per year
Set up. We are comparing the dollar depreciation under each method and explaining the tax incentive.
(i) Flat: D = 0.15 × 45,000 = $6,750/yr; S4 = 45,000 − 27,000 = $18,000. RB: S4 = 45,000(0.85)⁴ = 45,000 × 0.52201 = $23,490.28.
(ii) Total depreciation: flat = $27,000; RB = 45,000 − 23,490.28 = $21,509.72. Flat rate "expenses" more in total over 4 years.
(iii) Reducing balance front-loads the depreciation expense (largest dollar amount in year 1), which reduces taxable income sooner, and a tax saving today is worth more than the same saving in the future (time value of money).
Problem 3, Machinery (find rate then project)
Set up. We are inferring r from two book values, projecting forward, and inverting with logs.
(i) (1 − r)⁶ = 28,000/120,000 = 0.23333; 1 − r = 0.233331/6 = 0.79263; r = 0.20737 ≈ 20.74% p.a.
(ii) S10 = 120,000(0.79263)¹⁰ = 120,000 × 0.09995 ≈ $11,994.
(iii) Solve 120,000(0.79263)ⁿ < 10,000 ⇒ (0.79263)ⁿ < 0.0833 ⇒ n > ln 0.0833 / ln 0.79263 = −2.485 / −0.23253 = 10.69 years. First year below $10,000 is year 11.
Problem 4, IT fleet (technology obsolescence)
Set up. We are tracing two depreciation paths to find the crossover and recommend a method.
(i) RB at 30%: yr 1 = 80,000 × 0.7 = $56,000; yr 3 = 80,000 × (0.7)³ = 80,000 × 0.343 = $27,440; yr 5 = 80,000 × (0.7)⁵ = 80,000 × 0.16807 = $13,445.60.
(ii) Flat at 20%: yr 1 = 80,000 − 16,000 = $64,000; yr 3 = 80,000 − 48,000 = $32,000; yr 5 = 80,000 − 80,000 = $0. Crossover: Year 5 flat hits $0, while RB still at $13,446, flat falls below RB between year 4 (flat $16,000 vs RB $19,208) and year 5 (flat $0 vs RB $13,446). They cross at roughly year 4.4.
(iii) Reducing balance better models technology because the largest loss in value is in the first year (a 1-year-old laptop is worth far less than a new one), exactly the steep early decline of exponential decay.
Problem 5, Equipment with salvage cap
Set up. We are valuing an asset under RB and finding when the unadjusted curve crosses the salvage floor.
(i) S = 50,000(0.85)ⁿ. yr 5 = 50,000 × 0.4437 = $22,184.99; yr 10 = 50,000 × 0.1969 = $9,843.74; yr 15 = 50,000 × 0.0874 = $4,367.86.
(ii) Solve 50,000(0.85)ⁿ < 5,000 ⇒ (0.85)ⁿ < 0.10 ⇒ n > ln 0.10 / ln 0.85 = −2.3026 / −0.16252 = 14.17. First year below salvage is year 15 (unadjusted $4,367.86); after that the books are capped at the $5,000 salvage value.
(iii) Reducing balance approaches but never reaches zero because each year multiplies by a factor (1 − r) < 1, the curve is asymptotic; in real accounting a non-zero salvage value is imposed to reflect the practical floor (scrap or resale value) and to stop the books carrying assets at unrealistically small positive values forever.