Comparing Depreciation Methods
Learn to compare depreciation methods for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-03). Straight-line depreciation takes the same dollar amount off each year, while declining balance takes the same percentage, so the two methods give different values for the same asset.
You will set straight-line and declining balance depreciation side by side using numbers, a graph you are given and a spreadsheet, work out both values for the same year and decide which method gives the greater depreciation — a practical skill when choosing how to value a car, a machine or other equipment.
Theory
Comparing depreciation methods is part of the Depreciation and loans topic in Year 12 Standard 1 (NSW). Straight-line depreciation takes off the same dollar amount each year (\(S=V_0-Dn\)); declining-balance takes off the same percentage (\(S=V_0(1-r)^n\)). This guide shows how to work out both values for the same year and decide which is higher and which method gives the greater depreciation.
Straight-line depreciation uses \(S=V_0-Dn\): the same dollar amount \(D\) comes off every year, and the graph is a straight line falling to \(\$0\).
Reading a straight-line graph: \(V_0\) is the value at \(n=0\), and \(D\) is the fall in value divided by the number of years (for example \(\$12\,000\) falling to \(\$0\) in \(8\) years gives \(D=\$1500\)).
Declining-balance depreciation uses \(S=V_0(1-r)^n\): the same percentage of the current value comes off every year, and the graph is a curve that falls steeply at first and then levels off.
To compare the methods, find both values at the same \(n\). The difference is the larger value minus the smaller value, and the total depreciation under either method is \(V_0-S\). In this Year 12 Standard 1 (NSW) topic you compare the methods using formulas, tables, spreadsheets and graphs.
If both methods lose the same amount in the first year, the values are equal after \(1\) year; after that, declining balance stays higher. Over a long time the straight line always ends up lower, because it keeps losing the full \(D\) each year.
The two depreciation formulas, with initial value \(V_0\) and \(n\) years:
For either method:
| Year | Straight-line ($) | Declining-balance ($) |
|---|---|---|
| 0 | 12 000 | 12 000.00 |
| 1 | 10 500 | 9600.00 |
| 2 | 9000 | 7680.00 |
| 3 | 7500 | 6144.00 |
| 6 | 3000 | 3145.73 |
How to compare two depreciation methods
- Identify \(V_0\), the yearly amount \(D\) for straight-line, the rate \(r\) for declining-balance, and the year \(n\) asked about.
- Calculate both values at the same \(n\): \(S=V_0-Dn\) and \(S=V_0(1-r)^n\).
- Compare: the larger \(S\) is the higher value; the difference is larger minus smaller; total depreciation is \(V_0-S\).
- Answer in words, naming the method and stating the dollar amounts.
Work out both values at \(n=2\), then compare.
| \(\text{Straight: } S\) | \(=\) | \(15000 - 2000 \times 2\) |
| \(\) | \(=\) | \(11000\) |
| \(\text{Declining: } S\) | \(=\) | \(15000(0.85)^2\) |
| \(\) | \(=\) | \(15000 \times 0.7225\) |
| \(\) | \(=\) | \(10837.50\) |
| \(11000\) | \(>\) | \(10837.50\) |
Straight-line gives the higher value, \(\$11\,000\) against \(\$10\,837.50\).
Find both values, then subtract the smaller from the larger.
| \(\text{Straight: } S\) | \(=\) | \(24000 - 4000 \times 3\) |
| \(\) | \(=\) | \(12000\) |
| \(\text{Declining: } S\) | \(=\) | \(24000(0.90)^3\) |
| \(\) | \(=\) | \(24000 \times 0.729 = 17496\) |
| \(\text{Difference}\) | \(=\) | \(17496 - 12000\) |
| \(\) | \(=\) | \(5496\) |
The values differ by \(\$5496\) (declining-balance is higher).
Total depreciation \(=V_0-S\) for each method.
| \(\text{Straight: } S\) | \(=\) | \(9000 - 1200 \times 3 = 5400\) |
| \(\text{loss}\) | \(=\) | \(9000 - 5400 = 3600\) |
| \(\text{Declining: } S\) | \(=\) | \(9000(0.80)^3 = 4608\) |
| \(\text{loss}\) | \(=\) | \(9000 - 4608 = 4392\) |
Declining-balance depreciates more: \(\$4392\) against \(\$3600\).
| Year | Straight-line ($) | Declining-balance ($) |
|---|---|---|
| 0 | 30 000 | 30 000 |
| 1 | 25 000 | 24 000 |
| 2 | 20 000 | 19 200 |
| 3 | ? | ? |
Straight-line subtracts \(\$5000\) again; declining balance multiplies by \(0.80\) again.
| \(\text{Straight: } S\) | \(=\) | \(20000 - 5000 = 15000\) |
| \(\text{Declining: } S\) | \(=\) | \(19200 \times 0.80 = 15360\) |
After \(3\) years declining-balance leaves the higher value, \(\$15\,360\) against \(\$15\,000\), even though it was lower in Years \(1\) and \(2\).
Common pitfalls
Frequently asked questions
What is the difference between straight-line and declining-balance depreciation?
Straight-line depreciation takes the same dollar amount off every year, using S = V0 minus Dn, so its graph is a straight line. Declining-balance takes the same percentage of the current value off every year, using S = V0 times (1 minus r) to the power n, so its graph is a curve that levels off.
Which depreciation method gives a higher value?
It depends on the year. Work out both values for the same number of years and compare them. Often declining balance is lower in the early years, but the straight line keeps losing the full amount each year and ends up lower later on.
How do you find the difference between two depreciation values?
Calculate the value under each method for the same year, then subtract the smaller value from the larger one. For a $24000 ute after 3 years, 17496 minus 12000 gives a difference of $5496.
Which method gives the greater total depreciation?
The method that leaves the lower value. Total depreciation is the purchase price minus the value, so a lower value means more has been lost.
Can declining-balance depreciation reach zero?
No. Declining balance always keeps a fraction of the current value, so it never reaches exactly zero. Straight-line depreciation does reach zero, after V0 divided by D years.