What is off-by-one on the year?

When finding the first year below a threshold, round the comparison up to the next whole year, then state that year.

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How do we model an asset that loses a fixed percentage of its value each year?

Model reducing-balance depreciation with a recurrence relation, compare it with flat-rate depreciation, and find book value and scrap-value timing.

How to model reducing-balance depreciation with a recurrence relation, contrast it with flat-rate depreciation, find an asset's book value after n years, and work out when it reaches a scrap value.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The reducing-balance model
Reducing-balance versus flat-rate
Finding the rate from two values
Choosing a depreciation method
Unit-cost (kilometre) depreciation

What this dot point is asking

You must model reducing-balance depreciation with a recurrence, contrast it with flat-rate, find a book value, and find when an asset reaches a scrap value.

The reducing-balance model

Each year the asset loses a percentage of whatever it is currently worth, so it multiplies by a fixed factor.

A $12\%$ annual depreciation gives $R = 0.88$ . Because $0 < R < 1$ , the value decays towards (but never reaches) zero, flattening as it falls.

Reducing-balance versus flat-rate

The two methods produce very different value curves.

Feature	Flat-rate (straight-line)	Reducing-balance
Loss each year	Same dollar amount	Same percentage
Model	Arithmetic, $V_n = V_0 - dn$	Geometric, $V_n = V_0 R^n$
Graph	Straight line	Decaying curve
Early years	Loses less	Loses more

Worked example

Book value and scrap timing

A printing machine costs $\$120\,000$ and depreciates by $25\%$ of its value each year. Find its book value after $3$ years and the first year it falls below a scrap value of $\$30\,000$ .

Step 1: Set up the depreciation model

A $25\%$ annual loss means $75\%$ of the value survives each year, so the common ratio is $R = 0.75$ and the purchase price gives $V_0 = 120000$ . The recurrence is $V_{n+1} = 0.75 V_n$ and the explicit rule is $V_n = 120000 \times 0.75^n$ .

Step 2: Calculate the book value after 3 years

Substituting $n = 3$ into the explicit rule gives the value at that point in time.

$V_3 = 120000 \times 0.75^3 = 120000 \times 0.421875 = 50625$ , so $\$50\,625$ .

Step 3: Find the first year the value falls below the scrap value

To find when $V_n$ first drops below $\$30\,000$ , rearrange to $0.75^n < 0.25$ and test successive whole-number years. Testing, $0.75^4 = 0.3164$ and $0.75^5 = 0.2373$ , so the value first falls below $\$30\,000$ in year $5$ .

Step 4: Interpret the results

After $3$ years the machine is worth $\$50\,625$ on the books, and it reaches its scrap value early in year $5$ , the point at which it would typically be sold or replaced.

Finding the rate from two values

If you know the value at two times, find the ratio. If $V_0 = 40000$ and $V_3 = 23000$ , then $40000 R^3 = 23000$ , so $R^3 = 0.575$ and $R = \sqrt[3]{0.575} = 0.8315$ , giving an annual depreciation rate of about $16.9\%$ .

Choosing a depreciation method

Businesses choose between the two methods for tax and accounting reasons, and SCSA questions often ask you to compare them. Flat-rate depreciation writes off the same dollar amount each year, so its straight-line graph is simple to budget for and the asset reaches zero in a fixed number of years ( $\dfrac{\text{cost}}{\text{annual loss}}$ ). Reducing-balance depreciation writes off more in the early years, which better matches assets that lose value fastest when new (cars, technology), and it never quite reaches zero because each year removes a percentage of a shrinking value.

A useful check is the total depreciation. Under flat-rate, after $n$ years the total lost is $n \times d$ dollars. Under reducing-balance, the total lost is $V_0 - V_0 R^n = V_0(1 - R^n)$ . Comparing these for the same asset shows reducing-balance losing more early and flat-rate catching up later, the trade-off that decides which method suits a given asset. SCSA comparison questions reward computing both book values at the same year and stating which method leaves the higher value and why (the percentage loss shrinks each year under reducing-balance).

Unit-cost (kilometre) depreciation

A related model charges depreciation per unit of use rather than per year, for example cents per kilometre driven. The value after travelling $k$ kilometres is $V = V_0 - (\text{rate per km}) \times k$ , an arithmetic (flat-rate) model in $k$ rather than in time. Read whether the question depreciates by time or by usage, because the variable in the explicit rule changes accordingly.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksEquipment is bought for

\

80\,000

and depreciates by

18\%

of its value each year (reducing balance). (a) Write a recurrence relation for the book value

V_n

. (b) Find the book value after

years. (c) Find the first year the value falls below the scrap value of

25\,000

Show worked answer →

Reducing-balance loss of $18\%$ keeps $82\%$ each year.

(a) Common ratio $R = 1 - 0.18 = 0.82$ , so $V_{n+1} = 0.82 V_n$ , with $V_0 = 80000$ . (2 marks)

(b) $V_n = 80000 \times 0.82^n$ , so $V_5 = 80000 \times 0.82^5 = 80000 \times 0.37074 = 29659$ , that is $\$29\,659$ . (2 marks)

(c) Solve $80000 \times 0.82^n < 25000$ , so $0.82^n < 0.3125$ . Testing, $0.82^5 = 0.3707$ and $0.82^6 = 0.3040$ , so the value first falls below $\$25\,000$ in year $6$ . (2 marks)

Markers reward the ratio $0.82$ , the explicit substitution, and finding the first whole year below the scrap value.

WACE 20245 marksA van costs

\

48\,000

. Under flat-rate depreciation it loses

6000

per year; under reducing-balance it loses

15\%

per year. Compare the two book values after

4

years and state which method gives the higher value.

Show worked answer →

Compute each method's value at year $4$ .

Flat-rate: $V_4 = 48000 - 6000 \times 4 = 48000 - 24000 = 24000$ , that is $\$24\,000$ . (2 marks)

Reducing-balance: $R = 0.85$ , $V_4 = 48000 \times 0.85^4 = 48000 \times 0.52200 = 25056$ , that is $\$25\,056$ . (2 marks)

Reducing-balance gives the higher value after $4$ years ( $\$25\,056$ versus $\$24\,000$ ), because the percentage loss shrinks each year as the value falls. (1 mark)

Markers reward both correct book values and the comparison.