Straight-Line Depreciation
Master straight-line depreciation for NSW Year 12 Mathematics Standard 2. In this topic an asset loses the same fixed dollar amount of value each year, so its book value falls in a straight line, modelled by \(S=V_0-Dn\) from the purchase price down towards a salvage value.
You will learn to find an asset's value after any number of years, calculate the annual depreciation, find the original purchase price, and work out when an asset reaches a target or scrap value — a practical Standard 2 financial-maths skill for vehicles, machinery and equipment, often set out in a spreadsheet.
Theory
Straight-line depreciation reduces an asset's value by the same fixed dollar amount each year. This Year 12 Standard 2 (NSW) guide uses \(S=V_0-Dn\) to find an asset's value after \(n\) years, work out the annual depreciation, find the original purchase price, and find when an asset reaches a given salvage value.
Straight-line depreciation (also called the prime-cost method) reduces an asset's value by the same fixed dollar amount every year. Because the loss each year is constant, the value falls in a straight line against time.
The value after \(n\) years is \(S=V_0-Dn\), where \(V_0\) is the purchase price and \(D\) is the annual depreciation (the fixed amount lost each year). The value \(S\) is often called the book value or, at the end of the asset's life, the salvage (scrap) value.
On a graph, \(V_0\) is the vertical intercept and \(D\) is the size of the (negative) gradient — the line slopes downwards by \(D\) each year. This is a core Year 12 Standard 2 (NSW) financial-maths skill, frequently modelled in a spreadsheet and compared with declining-balance depreciation.
The value \(S\) of an asset after \(n\) years of straight-line depreciation is:
where \(V_0\) is the purchase price and \(D\) is the constant annual depreciation. Rearranging gives the annual depreciation and the purchase price:
How to solve a straight-line depreciation problem
- Identify the purchase price \(V_0\) and the annual depreciation \(D\) (if \(D\) is not given, find it with \(D=\dfrac{V_0-S}{n}\)).
- Write the model \(S=V_0-Dn\).
- Substitute the known values, then solve for the unknown — the value \(S\), the year \(n\), or (rearranged) \(V_0\).
- Interpret: state the answer with units, and check the value has not fallen below \(\$0\) or the stated salvage value.
Substitute \(V_0=12{,}000\), \(D=1500\), \(n=3\) into \(S=V_0-Dn\).
| \(S\) | \(=\) | \(V_0 - Dn\) |
| \(S\) | \(=\) | \(12000 - 1500 \times 3\) |
| \(S\) | \(=\) | \(12000 - 4500\) |
| \(S\) | \(=\) | \(\$7500\) |
Each year the value drops by the same \(\$1500\), so a spreadsheet counts down in equal steps:
| Year \(n\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Value \(S\) | $12,000 | $10,500 | $9000 | $7500 |
Share the total loss in value equally over the \(5\) years.
| \(D\) | \(=\) | \(\dfrac{V_0 - S}{n}\) |
| \(D\) | \(=\) | \(\dfrac{16000 - 6000}{5}\) |
| \(D\) | \(=\) | \(\dfrac{10000}{5} = \$2000\) |
The trailer depreciates by \(\$2000\) each year.
Set \(S=3000\) in \(S=V_0-Dn\) and solve for \(n\).
| \(3000\) | \(=\) | \(18000 - 3000n\) |
| \(3000n\) | \(=\) | \(15000\) |
| \(n\) | \(=\) | \(5\) |
The printer reaches its \(\$3000\) scrap value after 5 years.
Rearrange \(S=V_0-Dn\) to make \(V_0\) the subject.
| \(V_0\) | \(=\) | \(S + Dn\) |
| \(V_0\) | \(=\) | \(4000 + 1500 \times 4\) |
| \(V_0\) | \(=\) | \(4000 + 6000\) |
| \(V_0\) | \(=\) | \(\$10{,}000\) |
The rowing machine cost \(\$10{,}000\) when new.
Common pitfalls
Frequently asked questions
What is straight-line depreciation?
Straight-line depreciation, also called the prime-cost method, is when an asset loses the same fixed dollar amount of value every year. Because the yearly loss is constant, the value falls in a straight line, from the purchase price down towards a salvage value.
What is the formula for straight-line depreciation?
The value after n years is S equals V nought minus D times n, written S = V0 - Dn. Here V0 is the purchase price, D is the annual depreciation (the fixed amount lost each year), n is the number of years, and S is the value at that time.
How do you find the annual depreciation?
Divide the total loss in value by the number of years: D equals V nought minus S, all over n. For example, an asset bought for $16,000 and worth $6000 after 5 years loses $10,000 over 5 years, so D = 10000 divided by 5 = $2000 a year.
What is the salvage value?
The salvage value (or scrap value) is what an asset is worth at the end of its useful life, once it has finished depreciating. In a straight-line model the value falls steadily towards this amount, and the value should not be modelled below it.
What is the difference between straight-line and declining-balance depreciation?
Straight-line depreciation takes off the same fixed dollar amount each year, giving a straight-line graph. Declining-balance depreciation takes off the same percentage of the current value each year, giving a curve that is steep at first and then flattens out. Straight-line uses S = V0 - Dn; declining balance uses S = V0 times (1 - r) to the power n.
How do you find how long until an asset reaches a certain value?
Substitute the target value for S in S = V0 - Dn and solve for n. For instance, an $18,000 printer depreciating $3000 a year reaches $3000 when 3000 = 18000 - 3000n, giving n = 5 years.