Depreciation Methods
Compare straight-line and diminishing value depreciation, model asset values over time, and solve real-world depreciation problems.
Straight-Line Depreciation
Straight-line depreciation reduces the value of an asset by a fixed amount each period. This creates an arithmetic sequence of values.
Vn = V0 - Dn
where V0 = initial value, D = depreciation per period, n = number of periods
The depreciation amount per year can be calculated as:
D = V0 - Sn
where S = salvage (scrap) value and n = useful life in years
Diminishing Value (Reducing Balance) Depreciation
Diminishing value depreciation reduces the value by a fixed percentage each period. This creates a geometric sequence of values, similar to compound interest but decreasing.
Vn = V0(1 - r)n
where r = depreciation rate per period (as a decimal)
The key difference: straight-line depreciates by a constant amount, while diminishing value depreciates by a constant rate. Diminishing value loses more in early years and less in later years.
Comparing Depreciation Methods
Consider a $20,000 car depreciating over 5 years. Straight-line at $3,200/yr vs diminishing value at 25% p.a.:
Asset Value Over Time
| Year | Straight-Line | Diminishing Value |
|---|---|---|
| 0 | $20,000 | $20,000 |
| 1 | $16,800 | $15,000 |
| 2 | $13,600 | $11,250 |
| 3 | $10,400 | $8,438 |
| 4 | $7,200 | $6,328 |
| 5 | $4,000 | $4,746 |
Key Vocabulary
Depreciation
The decrease in value of an asset over time due to wear, age, or obsolescence.
Salvage Value
The estimated residual value of an asset at the end of its useful life (also called scrap value).
Straight-Line
A depreciation method where the asset loses a constant dollar amount each period, forming an arithmetic sequence.
Diminishing Value
A depreciation method where the asset loses a constant percentage of its current value each period, forming a geometric sequence.
Worked Examples
A machine costs $50,000 and has a salvage value of $5,000 after 10 years. Find the annual straight-line depreciation and the value after 6 years.
Step 1: D = (V0 - S) / n = (50000 - 5000) / 10 = $4,500 per year.
Step 2: V6 = V0 - D x 6 = 50000 - 4500 x 6 = 50000 - 27000 = $23,000.
Answer: Annual depreciation is $4,500 and the value after 6 years is $23,000.
A car worth $35,000 depreciates at 20% p.a. (diminishing value). Find its value after 3 years.
Step 1: Vn = V0(1 - r)n = 35000(1 - 0.20)3 = 35000(0.80)3.
Step 2: (0.80)3 = 0.512.
Step 3: V3 = 35000 x 0.512 = $17,920.
Answer: The car is worth $17,920 after 3 years.
Equipment purchased for $12,000 is worth $4,500 after 4 years using diminishing value depreciation. Find the annual depreciation rate.
Step 1: Use Vn = V0(1 - r)n, so 4500 = 12000(1 - r)4.
Step 2: (1 - r)4 = 4500/12000 = 0.375.
Step 3: 1 - r = (0.375)1/4 = 0.7825..., so r = 1 - 0.7825 = 0.2175.
Answer: The annual depreciation rate is approximately 21.75%.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
A laptop costs $2,400 and depreciates using straight-line depreciation to a salvage value of $0 over 4 years. What is the annual depreciation?
Question 2
A truck worth $80,000 depreciates at 15% p.a. (diminishing value). What is its value after 2 years?
Question 3
Which depreciation method creates values that form a geometric sequence?
Question 4
Equipment worth $40,000 is now worth $10,000 after 5 years using straight-line depreciation. What was the annual depreciation?
Question 5
A printer costs $1,000 and depreciates at 30% p.a. (diminishing value). After how many complete years will its value first fall below $200?
Key Concepts Summary
- ●Straight-line depreciation: Vn = V0 - Dn (arithmetic sequence, constant amount deducted).
- ●Diminishing value depreciation: Vn = V0(1 - r)n (geometric sequence, constant rate applied).
- ●Diminishing value loses more in early years; straight-line loses the same each year.
- ●Diminishing value never reaches zero; straight-line reaches the salvage value exactly at the end of useful life.
- ●To find the depreciation rate from known values, rearrange Vn = V0(1 - r)n and solve for r.