Reducing Balance Depreciation

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.

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

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)	Calculation	Value (V)	$ Lost this year
0	20,000 × 0.75⁰	$20,000	,
1	20,000 × 0.75¹	$15,000	−$5,000
2	20,000 × 0.75²	$11,250	−$3,750
3	20,000 × 0.75³	$8,437.50	−$2,812.50
4	20,000 × 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

Recurrence relation form

Instead of the formula directly, you can write it as a recurrence relation, especially useful on the CAS and for constructing tables.

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 ✓

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?

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📉 Reducing Balance Depreciation