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Home ā€ŗ Investing & Planning ā€ŗ Straight-Line Depreciation
Unit 4 Ā· Investing & Planning

šŸ“‰ Straight-Line Depreciation

When an asset loses the same dollar amount every year, that's straight-line depreciation. A forklift dropping $5000 in value each year. A machine going from $20,000 to $0 over 10 years. The formula is simple, but the exam loves to throw it in unexpected ways.

Linear (equal drops each year)
Formula: V = P āˆ’ Dn
Links to Reducing Balance next
Two types of depreciation are in Unit 4. This page covers straight-line (same $ each year). The next page covers reducing balance depreciation (same % each year, the car/computer one). Start here, it's simpler, and understanding both makes exam comparisons easy.
1

What is straight-line depreciation?

Depreciation is how assets lose value over time. Straight-line depreciation means the asset loses exactly the same dollar amount every single year, like a straight line declining on a graph.


Key terms:
  • P, purchase price (original cost)
  • S, scrap value (residual value at end of life)
  • D, annual depreciation ($ lost per year)
  • n, number of years
  • V, value at year n
Examples of straight-line assets:
  • Office furniture
  • Industrial machines
  • Buildings (sometimes)
  • Equipment with fixed lifespan
$12,000 $2,000 0 1 2 3 4 Years Value ($) equal drops each year $2,000 ā†“

A $12,000 machine depreciating by $2,000 each year, perfectly straight line

2

Finding the annual depreciation (D)

Before you can find the value at any point, you need to know how much the asset loses each year. That comes from the total drop spread evenly over its useful life.

Annual Depreciation D = (P āˆ’ S) / n
(P = purchase price, S = scrap value, n = useful life in years)

Think of it as: total value lost Ć· years of life = equal annual drop.

Worked Example A Ā· finding D

A machine is purchased for $12,000. Its useful life is 5 years, after which it has a scrap value of $2,000. Find the annual depreciation.

Given:P = $12,000  |  S = $2,000  |  n = 5 years
Formula:D = (P āˆ’ S) Ć· n
Substitute:D = (12,000 āˆ’ 2,000) Ć· 5
Simplify:D = 10,000 Ć· 5 = $2,000 per year
3

Finding the value at any point in time

Value after n years V = P āˆ’ D Ɨ n
(P = purchase price, D = annual depreciation, n = years elapsed)

Start at the purchase price and subtract the annual depreciation for each year that passes. Notice this is a linear (arithmetic) sequence, just like the arithmetic sequences from Unit 3!

Recurrence relation version:
Vā‚€ = P     Vā‚™ = Vā‚™ā‚‹ā‚ āˆ’ D   (subtract the same D each step)
Worked Example B Ā· value at year t

Using the same machine (P = $12,000, D = $2,000/year), find the value after 3 years.

Formula:V = P āˆ’ D Ɨ n
Substitute:V = 12,000 āˆ’ 2,000 Ɨ 3
Calculate:V = 12,000 āˆ’ 6,000 = $6,000
Year (n)Value V = 12,000 āˆ’ 2,000nAnnual drop
0$12,000,
1$10,000āˆ’$2,000
2$8,000āˆ’$2,000
3$6,000āˆ’$2,000
4$4,000āˆ’$2,000
5$2,000 (scrap)āˆ’$2,000
4

Working backwards, finding n or P

The formula V = P āˆ’ D Ɨ n can be rearranged to find any missing variable. Two common exam questions:

Find n (years) n = (P āˆ’ V) / D

Use when you know the current or target value and want to know when it reaches that point.

Find P (purchase price) P = V + D Ɨ n

Use when you know the current value and how many years have passed.

Worked Example C Ā· finding n

A piece of equipment costs $1,500 and depreciates by $200 per year. When will it reach its scrap value of $300?

Set V = S:1,500 āˆ’ 200n = 300
Rearrange:200n = 1,500 āˆ’ 300 = 1,200
Solve:n = 1,200 Ć· 200 = 6 years
Worked Example D Ā· finding purchase price

After 4 years a vehicle is worth $11,000. It depreciated by $3,500 per year. What was the purchase price?

Formula:P = V + D Ɨ n
Substitute:P = 11,000 + 3,500 Ɨ 4
Calculate:P = 11,000 + 14,000 = $25,000

Practice Questions

1. A printer costs $8,000, has a scrap value of $2,000, and a useful life of 3 years. Find (a) the annual depreciation D, and (b) its value after 2 years. ā–¶
(a) Annual depreciation:
D = (P āˆ’ S) Ć· n = (8,000 āˆ’ 2,000) Ć· 3 = 6,000 Ć· 3 = $2,000 per year

(b) Value after 2 years:
V = P āˆ’ D Ɨ n = 8,000 āˆ’ 2,000 Ɨ 2 = 8,000 āˆ’ 4,000 = $4,000
2. A laptop is purchased for $2,500 and depreciates by $400 per year. What is its value after 4 years? ā–¶
V = P āˆ’ D Ɨ n = 2,500 āˆ’ 400 Ɨ 4 = 2,500 āˆ’ 1,600 = $900
3. A machine costs $15,000, has a scrap value of $3,000, and a useful life of 6 years. (a) Find D. (b) Find the value after 5 years. ā–¶
(a) D = (15,000 āˆ’ 3,000) Ć· 6 = 12,000 Ć· 6 = $2,000 per year

(b) V = 15,000 āˆ’ 2,000 Ɨ 5 = 15,000 āˆ’ 10,000 = $5,000
4. Using the machine from Q3 (P = $15,000, D = $2,000, S = $3,000), in which year does it reach its scrap value? ā–¶
Set V = S:  15,000 āˆ’ 2,000n = 3,000
2,000n = 12,000
n = 6 years (end of its useful life, makes sense!)
5. A construction vehicle is purchased for $45,000 and has a scrap value of $9,000 after 9 years. Write down D, then find its value after 3 years. ā–¶
D = (45,000 āˆ’ 9,000) Ć· 9 = 36,000 Ć· 9 = $4,000 per year

V after 3 years = 45,000 āˆ’ 4,000 Ɨ 3 = 45,000 āˆ’ 12,000 = $33,000

šŸš— Ready to test yourself?

The Car Yard escape room, 6 depreciation challenges to unlock the lot.

Open Escape Room ā†’