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Unit 4 · Investing & Planning

📉 Reducing Balance Depreciation

Instead of the same dollar amount each year, this method takes the same percentage of the current value. A car losing 25% each year loses more money in year 1 than year 5, because it's worth less by then. Same structure as compound interest, but going down.

Exponential decay curve
Formula: V = P(1 − r)ⁿ
Mirror of compound interest
Do straight-line first if you haven't. Reducing balance uses the same idea as compound interest from Unit 3, just with a multiplier less than 1 instead of greater than 1. If that's solid, this page will make instant sense.

Compare: Compound interest uses (1 + r)ⁿ  →  value grows.
Reducing balance uses (1 − r)ⁿ  →  value shrinks.
If you only do one thing Value drops by the same percentage each year: V = P(1 − r)n. It is compound interest running in reverse.

The 10-minute version

Low energy? Do just this and you have still won the day. Zero is the only fail.

The one move: read the first formula box and the first worked example above, then do Practice Question 1. Screenshot the win for Nat and stop there.

📺 Watch it explained

Four short ways in. The 🎬 cards are waiting on Nat's videos, and each has a ready-to-read film script tucked underneath.

1

How reducing balance depreciation works

Each year, the asset loses a fixed percentage of its current value. Because the value is smaller each year, the actual dollar loss also gets smaller, giving the characteristic curved graph rather than a straight line.


Value after n years, Reducing Balance V = P(1 − r)ⁿ
(P = purchase price, r = depreciation rate as a decimal, n = years)

The multiplier (1 − r) is called the depreciation factor. For a 25% rate: multiplier = 0.75. Each year you multiply by 0.75 again.

$20,000 $15,000 $11,250 0 1 2 3 4 Years Value ($) Reducing balance Straight-line (ref)

$20,000 car at 25% p.a., the curve shows more dollar loss early on, flattening over time

2

Calculating the value step by step

Worked Example A · full table

A car is purchased for $20,000 and depreciates at 25% p.a. (reducing balance). Find the value at the end of each year for 4 years.

Multiplier:1 − 0.25 = 0.75 (each year × 0.75)
Formula:V = 20,000 × 0.75ⁿ
Year (n)CalculationValue (V)$ Lost this year
020,000 × 0.75⁰$20,000,
120,000 × 0.75¹$15,000−$5,000
220,000 × 0.75²$11,250−$3,750
320,000 × 0.75³$8,437.50−$2,812.50
420,000 × 0.75⁴$6,328.13−$2,109.37

The dollar loss gets smaller each year, that's the key feature of reducing balance depreciation.

Worked Example B · direct formula

A laptop costs $4,000 and depreciates at 30% p.a. Find its value after 2 years.

Multiplier:1 − 0.30 = 0.70
Formula:V = 4,000 × (0.70)²
Calculate:V = 4,000 × 0.49 = $1,960
3

Recurrence relation form

Instead of the formula directly, you can write it as a recurrence relation, especially useful for constructing tables with the Ans key.

Recurrence Relation, Reducing Balance Depreciation V₀ = P
Vₙ = (1 − r) × Vₙ₋₁
(Each term = previous term × depreciation factor)

This is identical to the compound interest recurrence Aₙ = R × Aₙ₋₁ from Unit 3, just with R = (1 − r) instead of R = (1 + r).

Worked Example C · recurrence form

Write the recurrence relation for a $10,000 asset depreciating at 20% p.a. Use it to find the value after 3 years.

Relation:V₀ = 10,000  |  Vₙ = 0.80 × Vₙ₋₁
Year 1:V₁ = 0.80 × 10,000 = 8,000
Year 2:V₂ = 0.80 × 8,000 = 6,400
Year 3:V₃ = 0.80 × 6,400 = 5,120
Check:V = 10,000 × 0.8³ = 10,000 × 0.512 = 5,120 ✓
4

Comparing straight-line vs reducing balance

📉 Straight-Line

  • Same dollar amount each year
  • Linear graph (straight line)
  • Formula: V = P − Dn
  • Simple to calculate
  • Used for: furniture, equipment

📉 Reducing Balance

  • Same percentage each year
  • Curved graph (exponential)
  • Formula: V = P(1 − r)ⁿ
  • More realistic for technology
  • Used for: cars, computers, phones
Exam tip: When comparing methods on the same asset, the reducing balance method gives higher depreciation early on and lower later. The straight-line value will eventually cross above the reducing balance curve.

Practice Questions

1. A phone costs $800 and depreciates at 20% p.a. (reducing balance). Find its value after 2 years. ▶
Multiplier = 1 − 0.20 = 0.80
V = 800 × (0.80)² = 800 × 0.64 = $512
2. A computer costs $10,000 and depreciates at 30% p.a. Find its value after 1 year. ▶
V = 10,000 × (1 − 0.30)¹ = 10,000 × 0.70 = $7,000
3. Write the recurrence relation for a $5,000 machine at 40% p.a. depreciation. Use it to find the value after 2 years. ▶
V₀ = 5,000  |  Vₙ = 0.60 × Vₙ₋₁
Year 1: V₁ = 0.60 × 5,000 = 3,000
Year 2: V₂ = 0.60 × 3,000 = $1,800

Check: V = 5,000 × 0.6² = 5,000 × 0.36 = 1,800 ✓
4. A $6,000 machine uses straight-line depreciation at $1,000 per year. Another $6,000 machine uses reducing balance at 20% p.a. Which method gives a higher value after 3 years? By how much? ▶
Straight-line: V = 6,000 − 1,000 × 3 = 6,000 − 3,000 = $3,000

Reducing balance: V = 6,000 × (0.80)³ = 6,000 × 0.512 = $3,072

Reducing balance gives a higher value by $72. (Straight-line depreciates faster in later years for this asset.)
5. A vehicle costs $25,000 and depreciates at 20% p.a. (reducing balance). Find its value after 3 years. ▶
V = 25,000 × (0.80)³ = 25,000 × 0.512 = $12,800

💻 Ready to test yourself?

Tech Drop escape room, 6 challenges to restore the depreciation system.

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