Depreciation
Depreciation is the term used for the loss of value of an item over time. The loss of value is often due to usage and wear and tear on the item, but can also be due to other factors such as a newer edition of the item becoming available. For example, the value of a new car depreciates rapidly in the first few years of usage and continues to depreciate the older it gets.
It should be remembered that the value of anything, in terms of money, depends on how much people are willing to pay for it. For this reason, no mathematical formula can infallibly model the changes in value of an item over time. There are too many complicating factors involved.
However, estimates are possible as guides as to what values can be expected for various purposes. Importantly, the Australian Taxation Office (as well as other tax institutions from around the world) enables people and businesses to estimate the amount of depreciation on pieces of equipment that are used for the purpose of generating income. The depreciation becomes an allowable deduction that reduces the amount of tax payable.
Depreciation can be calculated in either of two ways. One way is to assume that the item loses a fixed percentage of its current value each year. This is known as the declining-balance method of depreciation. The other method is to assume that a fixed amount of value is lost each year. This is called straight-line method of depreciation, and is what we will focus on in this lesson.
Straight-line depreciation
The straight-line method is named so because the graph showing the value at yearly intervals will appear as a downward-sloping straight line. The slope of the line reflects the fixed quantity lost from the value in each period. A steeper slope means a higher annual rate of depreciation.
Similar to the simple interest formula, the straight line method assumes the value of depreciation is constant per period. It can also be referred to as flat-rate or prime cost depreciation.
Below is the graph showing the straight line depreciation of a $\$20000$$20000 car depreciating at a rate of $\$3000$$3000 per year. The equation of the line is $S=20000-3000n$S=20000−3000n, where $S$S is the salvage value of the car, and $n$n is the number of time periods.
The formula to find the salvage value $S$S of an asset after $n$n time periods, using the method of straight-line depreciation, is
$S=V_0-Dn$S=V0−Dn,
where $V_0$V0 is the initial value of the asset, and $D$D is the rate of depreciation.
(This formula can be found on the HSC Reference Sheet)
Note that salvage value may also be called book value, scrap value or written-down value. Also note that the depreciation rate $D$D will be a negative number.
Worked examples
Example 1
A car purchased for $\$20000$$20000 depreciates by $\$2500$$2500 per year. How much is it worth after $6$6 years?
Think: We can use the formula above, with $V_0=\$20000$V0=$20000, $D=-\$2500$D=−$2500 and $n=6$n=6 years.
Do:
| $S$S | $=$= | $V_0-Dn$V0−Dn |
| $=$= | $\$20000-\$2500\times6$$20000−$2500×6 years | |
| $=$= | $\$5000$$5000 |
Example 2
A car purchased for $\$15000$$15000 is worth $\$9000$$9000 after $8$8 years. By how much did it depreciate per year?
Think: We are asked to find the rate of depreciation $D$D. We know the initial value is $V_0=\$15000$V0=$15000 and the final salvage value is $S=\$9000$S=$9000 after $n=8$n=8 years. So we can substitute these known values into the formula and solve for $D$D.
Do:
Substituting, we have that $\$9000=\$15000-D\times8$$9000=$15000−D×8 years. Rearranging for $D$D, we get that $D=\frac{15000-9000}{8}=\$750$D=15000−90008=$750/year.
So, using straight-line depreciation, the car depreciated by $\$750$$750 per year.
Example 3
The following graph shows the declining value of a high volume printer.
Assuming equal amounts of value were lost in each of the eight years, how much depreciation was there per year?
Think: If an equal amount of value was lost each year, then there was a constant rate of depreciation. We can find this by calculating the gradient of the line.
Do: Over the eight years of its life, the value went from $\$50000$$50000 to $\$0$$0. So the gradient of the line is $\frac{0-50000}{8}=\$6250$0−500008=$6250/year. That is, there was $\$6250$$6250 of depreciation each year.
Example 4
A business buys a truck for $\$67100$$67100. The truck is estimated to depreciate at the rate of $\$6100$$6100 per year. How many years will it take for the truck to have no remaining value?
Think: We can divide the original value by the rate of depreciation to find the number of time periods (years) for the value to reach zero.
Do: Dividing, we get that $\frac{67100}{6100}=11$671006100=11 years. So the truck will have no remaining value after $11$11 years.
Practice questions
Question 1
The spectator attendance at an annual sporting event was recorded for four consecutive years from its first year of running: $44500$44500, $43800$43800, $43100$43100, $42400$42400
By how much did the attendance decrease each year?
$\editable{}$ spectators each year
If the attendance continues to decrease at the same annual rate, what will the expected attendance be in its fifth year?
Question 2
The graph shows the depreciation of a car's value over 4 years.
What is the initial value of the car?
By how much did the car depreciate each year ?
After how many years will the car be worth $\$14400$$14400 ?
What is the value of the car after 4 years ?
Question 3
A car was originally valued at $\$33300$$33300 and depreciates by $\$3000$$3000 per year.
What is the salvage value of the car after $4$4 years?
What percentage of the original value is the car worth after $4$4 years?
Round your answer to two decimal places.
What percentage of the original value has been lost after $4$4 years?
Round your answer to two decimal places.