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How do we model the loss in value of an asset over time?

Model depreciation using flat-rate, reducing-balance and unit-cost methods, and find the value of an asset over time.

How to model an asset losing value by flat-rate (straight-line), reducing-balance and unit-cost depreciation, find its value after a number of years, and choose the right method.

Generated by Claude Opus 4.88 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
Flat-rate (straight-line) depreciation
Reducing-balance depreciation
Unit-cost depreciation
Depreciation as negative growth
Finding the time to a target value
Choosing and comparing methods

What this dot point is asking

You must apply all three depreciation methods, find an asset's value over time, and recognise which method suits a situation.

Flat-rate (straight-line) depreciation

The asset loses the same amount each year, often a fixed percentage of the original cost. If $V_0$ is the purchase price and $D$ is the annual loss, the value after $n$ years is:

V_n = V_0 - Dn

Worked example

A van costs $\$40\,000$ and depreciates by 15% of its original value each year (flat-rate). Find its value after 4 years, and determine in which year the value first falls below $\$10\,000$ .

Step 1: Calculate the fixed annual loss in dollars

In flat-rate depreciation the percentage is applied to the original cost every year, so the dollar loss is the same each year. Multiplying the purchase price by the depreciation rate gives:

D = 0.15 \times 40\,000 = 6000.

The van loses $\$6000$ of value every year regardless of its current worth.

Step 2: Find the value after 4 years using the flat-rate formula

Because the same $\$6000$ is subtracted each year, after $n$ years the value is $V_n = V_0 - Dn$ . Substituting $n = 4$ :

V_4 = 40\,000 - 6000 \times 4 = 40\,000 - 24\,000 = 16\,000.

Step 3: Find when the value first falls below $\$10\,000$

We set up the inequality $V_n < 10\,000$ and solve for $n$ :

40\,000 - 6000n < 10\,000 \implies 6000n > 30\,000 \implies n > 5.

At $n = 5$ the value is exactly $\$10\,000$ , and at $n = 6$ it falls to $\$4000$ , so the value first drops below $\$10\,000$ during year 6.

Final answer: After 4 years the van is worth $\$16\,000$ . The value first falls below $\$10\,000$ in year 6.

Reducing-balance depreciation

The asset loses a fixed percentage of its current value each year, so the dollar loss shrinks over time. With depreciation rate $r$ (decimal), the value after $n$ years is:

V_n = V_0(1 - r)^n

Worked example

A laptop costs $\$2000$ and depreciates at 25% per year on a reducing-balance basis. Find its value after 3 years.

Step 1: Recognise the depreciation method and write down the formula

The phrase "25% of its current value" signals reducing-balance depreciation: each year the laptop keeps a fixed fraction of its value at the start of that year. This is different from flat-rate, where the dollar loss is always the same. The formula is:

V_n = V_0(1 - r)^n,

where $V_0$ is the purchase price, $r$ is the annual depreciation rate as a decimal, and $n$ is the number of years.

Step 2: Identify the values and substitute

Here $V_0 = 2000$ , $r = 0.25$ (so each year the laptop retains $1 - 0.25 = 0.75$ of the previous year's value), and $n = 3$ :

V_3 = 2000(1 - 0.25)^3 = 2000(0.75)^3 = 2000 \times 0.421875 = 843.75.

Step 3: Interpret the result

The laptop has lost more than half its value in three years, which is typical of reducing-balance at a high rate like 25%. Each year the dollar loss is smaller than the last: $\$500$ in year 1, $\$375$ in year 2, $\$281.25$ in year 3.

Final answer: After 3 years the laptop is worth $\$843.75$ .

Unit-cost depreciation

Here the value falls by a fixed amount for each unit of use (kilometres driven, items produced, hours run), rather than per year. If the asset loses $c$ dollars per unit and has been used $u$ units:

V = V_0 - c \times u

Depreciation as negative growth

Reducing-balance depreciation is mathematically the compound-interest formula with a negative rate: where an investment multiplies by $(1 + i)$ each period, a depreciating asset multiplies by $(1 - r)$ . This is why a reducing-balance asset never reaches exactly zero - each year it keeps a fixed fraction of its value, so the curve approaches the axis without touching it. Flat-rate depreciation, by contrast, is a linear model with a negative gradient and will eventually reach zero (and would go negative if not stopped at the scrap value). Recognising these as the linear and exponential models from earlier topics ties the financial work back to the modelling strand.

Finding the time to a target value

A common task is to find how long until an asset reaches a given value. For flat-rate, set $V_0 - Dn$ equal to the target and solve the linear equation for $n$ . For reducing-balance, set $V_0(1 - r)^n$ equal to the target; solving for $n$ needs logarithms or a financial solver. For unit-cost, set $V_0 - cu$ equal to the target and solve for the number of units $u$ . In every case, round to a sensible whole number of years or units and check the answer makes practical sense in the asset's context.

Choosing and comparing methods

Flat-rate gives a straight-line decline and never quite reflects how assets lose value fastest when new. Reducing-balance drops quickly at first then levels off, matching items like cars and electronics, and never reaches exactly zero. Unit-cost suits assets whose wear depends on use rather than time.

Exam-style practice questions

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

SACE 20233 marksCalculator-assumed. A van costs 40000 dollars and depreciates by 15% of its original value per year (flat-rate). (a) Find its value after 4 years. (b) Find when its value first falls below 10000 dollars.

Show worked answer →

(a) Annual loss $D = 0.15 \times 40000 = 6000$ . Value after 4 years: $V_4 = 40000 - 6000(4) = 16000$ dollars. (2 marks)

(b) Set $40000 - 6000n < 10000$ , so $30000 < 6000n$ , giving $n > 5$ . The value first falls below 10000 dollars during year 6 (at $n = 6$ , $V = 40000 - 36000 = 4000$ ; at $n = 5$ , $V = 10000$ ). (1 mark)

Marks: two for the value after 4 years, one for solving the inequality.

SACE 20222 marksCalculator-assumed. A laptop costs 2000 dollars and depreciates at 25% per year on a reducing-balance basis. Find its value after 3 years.

Show worked answer →

Use $V_n = V_0(1 - r)^n$ with $V_0 = 2000$ , $r = 0.25$ , $n = 3$ :

V_3 = 2000(0.75)^3 = 2000 \times 0.421875 = 843.75 \text{ dollars}.

Marks: one for the reducing-balance formula, one for the value of about 843.75 dollars.

SACE 20212 marksCalculator-assumed. A machine costs 50000 dollars and depreciates by 0.40 dollars for each item it produces (unit-cost). Find its value after producing 60000 items.

Show worked answer →

Use $V = V_0 - c \times u$ with $V_0 = 50000$ , $c = 0.40$ , $u = 60000$ :

V = 50000 - 0.40 \times 60000 = 50000 - 24000 = 26000 \text{ dollars}.

Marks: one for the unit-cost set-up, one for the value of 26000 dollars.