Straight-Line Depreciation
Learn straight-line depreciation for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-03). Assets such as cars, tools and machines lose value as they get older, and with the straight-line method they lose the same dollar amount every year, so the graph of their value is a straight line sloping down.
You will use the formula \(S = V_0 - Dn\) to find the salvage value, the yearly depreciation, the asset's initial value or the number of years, using a formula, a table or a graph — useful when buying a car or valuing business equipment.
Theory
Straight-line depreciation is part of the Depreciation and loans topic in Year 12 Standard 1 (NSW). An asset loses the same dollar amount every year, so its salvage value after \(n\) years is \(S=V_0-Dn\) and its graph is a straight line sloping down. This guide shows how to find the value, the yearly depreciation, the purchase price and the number of years.
Depreciation is the loss in value of an asset, such as a car, a tool or a machine, as it gets older. Straight-line depreciation takes the same dollar amount off the value every year, so the value falls in equal steps.
The salvage value \(S\) after \(n\) years is \(S=V_0-Dn\), where \(V_0\) is the initial value (purchase price) and \(D\) is the depreciation each year. The total depreciation over \(n\) years is \(Dn=V_0-S\).
In this Year 12 Standard 1 (NSW) topic you use the formula, a table or a graph to find the value after a number of years, the yearly depreciation, the purchase price, or when the asset reaches a given value.
With initial value \(V_0\), depreciation \(D\) dollars per year and \(n\) years, the salvage value is:
The total depreciation is the amount lost:
Rearranging gives the yearly depreciation and the purchase price:
| Year \(n\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Value \(S\) ($) | 20 000 | 17 500 | 15 000 | 12 500 | 10 000 | 7500 |
How to solve a straight-line depreciation problem
- Identify the initial value \(V_0\), the depreciation each year \(D\), the number of years \(n\) and the salvage value \(S\). One of them is unknown.
- Substitute the known values into \(S=V_0-Dn\), multiplying \(D\times n\) before subtracting.
- Solve for the unknown: use \(D=\dfrac{V_0-S}{n}\), \(V_0=S+Dn\), or solve the equation for \(n\).
- Answer the question asked: the value left (\(S\)) or the amount lost (\(Dn\)).
Use \(S=V_0-Dn\) with \(V_0=9600\), \(D=1200\), \(n=5\).
| \(S\) | \(=\) | \(V_0 - Dn\) |
| \(S\) | \(=\) | \(9600 - 1200 \times 5\) |
| \(S\) | \(=\) | \(9600 - 6000\) |
| \(S\) | \(=\) | \(3600\) |
The mower is worth \(\$3600\) after \(5\) years.
The total loss is shared equally over the \(4\) years.
| \(\text{Total depreciation}\) | \(=\) | \(V_0 - S\) |
| \(\) | \(=\) | \(36000 - 20400 = 15600\) |
| \(D\) | \(=\) | \(\dfrac{V_0 - S}{n} = \dfrac{15600}{4}\) |
| \(D\) | \(=\) | \(3900\) |
The car depreciates by \(\$3900\) each year (\(\$15\,600\) in total).
Rearrange \(S=V_0-Dn\) to make \(V_0\) the subject.
| \(V_0\) | \(=\) | \(S + Dn\) |
| \(V_0\) | \(=\) | \(11200 + 2600 \times 3\) |
| \(V_0\) | \(=\) | \(11200 + 7800\) |
| \(V_0\) | \(=\) | \(19000\) |
The food truck cost \(\$19\,000\) when new.
Substitute \(S=12000\) into \(S=V_0-Dn\) and solve for \(n\).
| \(12000\) | \(=\) | \(54000 - 6000n\) |
| \(6000n\) | \(=\) | \(54000 - 12000\) |
| \(6000n\) | \(=\) | \(42000\) |
| \(n\) | \(=\) | \(7\) |
The truck is worth \(\$12\,000\) after \(7\) years.
Common pitfalls
Frequently asked questions
What is straight-line depreciation?
Straight-line depreciation reduces the value of an asset by the same dollar amount every year. The value falls in equal steps, so a graph of value against time is a straight line sloping down.
What is the formula for straight-line depreciation?
The salvage value after n years is S = V0 minus D times n, where V0 is the initial value or purchase price and D is the depreciation each year in dollars.
How do you find the depreciation per year?
Subtract the salvage value from the purchase price and divide by the number of years. A car bought for $36000 and sold 4 years later for $20400 depreciates by 15600 divided by 4, which is $3900 a year.
How do you find the original price of an asset?
Add the total depreciation back on to the current value: V0 = S + Dn. A food truck worth $11200 after 3 years of $2600 depreciation cost 11200 plus 7800, which is $19000.
What is the difference between salvage value and total depreciation?
The salvage value is what the asset is still worth. The total depreciation is how much value it has lost. Together they add up to the purchase price.