Declining-balance depreciation
The declining-balance formula S = V0(1 - r)^n, depreciation tables and graphs, finding the rate or the time, and comparing declining-balance with straight-line depreciation.
A new car loses value the moment you drive it out of the dealership. Phones, laptops and work tools do the same. Businesses track this loss, called depreciation, because the tax office lets them claim it as a cost. The usual method takes the same percentage off the value every year, which is compound interest running backwards.
The declining-balance formula
With declining-balance depreciation, the value falls by a fixed percentage of the current value each year. Losing $r$ of the value leaves $(1 - r)$ of it, so after $n$ years:
- $S$ is the salvage value (the value after $n$ years).
- $V_0$ is the original value.
- $r$ is the depreciation rate per year, as a decimal.
- $n$ is the number of years.
The total depreciation is $V_0 - S$.
A tradie buys a new ute for $\$42\,000$. It depreciates at $18\%$ p.a. using the declining-balance method. Find its value after $3$ years and the total depreciation.
Declining-balance vs straight-line
In Year 11 you used straight-line depreciation, where the same dollar amount is lost each year: $S = V_0 - Dn$. The two methods behave very differently.
| Year | Declining balance, $18\%$ ($\$$) | Straight line, $\$5000$ a year ($\$$) |
|---|---|---|
| $0$ | $42\,000$ | $42\,000$ |
| $1$ | $34\,440.00$ | $37\,000$ |
| $2$ | $28\,240.80$ | $32\,000$ |
| $3$ | $23\,157.46$ | $27\,000$ |
| $4$ | $18\,989.11$ | $22\,000$ |
| $5$ | $15\,571.07$ | $17\,000$ |
| $6$ | $12\,768.28$ | $12\,000$ |
| $7$ | $10\,469.99$ | $7000$ |
| $8$ | $8585.39$ | $2000$ |
- Declining balance loses the most in the first year, when the value is biggest, then loses less and less. Its graph is a curve that flattens out and never reaches zero.
- Straight line loses the same amount every year. Its graph is a straight line that reaches zero.
- Here the ute is worth less under declining balance for the first $5$ years, and more from year $6$ on. The graphs cross between years $5$ and $6$.
Check the table above, then drag the declining-balance rate up. Does the crossover move earlier or later? Which method suits a laptop that is almost worthless after $3$ years?
Finding the rate or the time
A ute bought for $\$42\,000$ is worth $\$18\,000$ after $4$ years. Find the declining-balance rate, to $2$ decimal places.
To find the number of years, try whole values of $n$ until the value first falls below the target, or use the table or graph.
Your turn
A laptop costs $\$2400$ and depreciates at $30\%$ p.a. using the declining-balance method. Find its value after $2$ years.
Show the working
$S = 2400(0.7)^2 = 2400 \times 0.49 = \$1176$.
A laptop costs $\$2400$ and depreciates at $30\%$ p.a. (declining balance). Find its value after $2$ years.
$S = 2400(0.7)^2 = \$1176$.
A tractor bought for $\$30\,000$ depreciates at $15\%$ p.a. (declining balance). Find the total depreciation over $3$ years, to the nearest cent.
$S = 30\,000(0.85)^3 = \$18\,423.75$. Depreciation $= 30\,000 - 18\,423.75 = \$11\,576.25$.
A machine bought for $\$30\,000$ depreciates by $\$4500$ each year using the straight-line method. Find its value after $4$ years.
$S = 30\,000 - 4500 \times 4 = \$12\,000$.
Which best describes the graph of value against time for declining-balance depreciation?
Each year the value is multiplied by $(1 - r)$, a number between $0$ and $1$. The value shrinks by less each year and never quite reaches zero.
Equipment worth $\$50\,000$ depreciates at $20\%$ p.a. (declining balance). After how many whole years is it first worth less than $\$20\,000$?
$50\,000(0.8)^4 = \$20\,480$, still above. $50\,000(0.8)^5 = \$16\,384$, below. So after $5$ years.
An e-bike is worth $\$12\,750$ after losing $15\%$ of its value in one year. What was it worth a year ago?
$12\,750 = V_0(0.85)$, so $V_0 = \dfrac{12\,750}{0.85} = \$15\,000$. Adding $15\%$ to $12\,750$ would be wrong, because the $15\%$ was of the old value.
Which expression gives the value after $6$ years of a $\$9000$ camera that depreciates at $12\%$ p.a.?
Losing $12\%$ keeps $88\%$ each year, so $S = 9000(0.88)^6$.
A ute bought for $\$42\,000$ is worth $\$18\,000$ after $4$ years of declining-balance depreciation. Find the rate as a percentage, to $2$ decimal places.
$(1 - r)^4 = \dfrac{18\,000}{42\,000}$, so $1 - r = \sqrt[4]{0.428\,57} \approx 0.8091$ and $r \approx 19.09\%$.
Exam style: A delivery van costs $\$58\,000$ and depreciates at $22.5\%$ p.a. using the declining-balance method. Find the total depreciation over $5$ years, to the nearest cent.
$S = 58\,000(0.775)^5 \approx \$16\,215.73$. Depreciation $= 58\,000 - 16\,215.73 = \$41\,784.27$.
Exam style: Equipment costing $\$20\,000$ can be depreciated straight-line by $\$2500$ a year, or by declining balance at $25\%$ p.a. After $3$ years, how much lower is the declining-balance value than the straight-line value?
Straight line: $20\,000 - 2500 \times 3 = \$12\,500$. Declining balance: $20\,000(0.75)^3 = \$8437.50$. Difference: $12\,500 - 8437.50 = \$4062.50$.
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