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Straight Line Depreciation Calculator

Annual charge, depreciation rate and full book-value schedule, worked step by step
Straight line depreciation on a $45,000 asset with $5,000 salvage over 8 years
Book value after 3 years of a $45,000 asset depreciated to $5,000 over 8 years
First-year charge for an asset placed in service on 1 October
Full schedule for a $12,800 asset with $800 salvage over 5 years

The Straight Line Depreciation Equation

Straight line depreciation spreads an asset's cost evenly across its useful life. The annual charge is

D=Cโˆ’SnD = \frac{C - S}{n}

  • CC โ€” cost, the full amount capitalised (purchase price plus the costs of getting it into service)
  • SS โ€” salvage or residual value, what you expect to recover at the end
  • nn โ€” useful life in years
  • Cโˆ’SC - S โ€” the depreciable base, the part that is actually written off

Two figures follow immediately. The depreciation rate is 1/n1/n of the base per year, and the book value after kk full years is

BVk=Cโˆ’kDBV_k = C - kD

which is a straight line from CC down to exactly SS at k=nk = n - hence the name. The method applies when an asset delivers roughly even service each year; it is the default for buildings, fixtures and most general equipment.

Partial Years and Other Methods

An asset rarely enters service on 1 January. Prorate the first year by the fraction of the year it is in use:

Dfirst=Dร—m12D_{\text{first}} = D \times \frac{m}{12}

with mm the months of service. The remainder, Dโˆ’DfirstD - D_{\text{first}}, falls into an extra final year, so the schedule spans n+1n+1 years. Some entities instead adopt a convention - half-year, mid-month, mid-quarter - that fixes mm by rule.

Compared with accelerated methods. Declining balance charges BVkโˆ’1ร—dBV_{k-1} \times d each year, front-loading the expense; units-of-production charges (Cโˆ’S)ร—unitsk/totalย units(C-S) \times \text{units}_k / \text{total units}. Both write off the same total base, only on a different timetable.

Which method, life and convention you may use for tax purposes is set by your jurisdiction and may differ from the figures in your books. This page computes the arithmetic on the inputs you supply.

Common Mistakes to Avoid

  • Depreciating the full cost: subtract salvage first. Only Cโˆ’SC - S is written off.
  • Letting book value fall below salvage: stop at SS; the schedule ends there even if the asset stays in use.
  • Charging a full year in the year of purchase: prorate by months unless a convention says otherwise.
  • Including land in the cost: land is not depreciated, so split it out of a property's price.
  • Confusing book value with market value: BVkBV_k is an accounting figure, not a resale estimate.
  • Reusing a tax life for book purposes: the allowed life and method for tax depend on your jurisdiction and asset class, and need not match your accounting life.

Examples

Step 1: Depreciable base: Cโˆ’S=45,000โˆ’5,000=40,000C - S = 45{,}000 - 5{,}000 = 40{,}000
Step 2: D=40,000/8=5,000D = 40{,}000 / 8 = 5{,}000 a year
Step 3: Rate: 1/8=12.5%1/8 = 12.5\% of the base each year
Step 4: BV3=45,000โˆ’3ร—5,000=30,000BV_3 = 45{,}000 - 3 \times 5{,}000 = 30{,}000
Step 5: Check the end: BV8=45,000โˆ’8ร—5,000=5,000=SBV_8 = 45{,}000 - 8 \times 5{,}000 = 5{,}000 = S
Answer: \5{,}000ayear,12.5a year, 12.5% of the base, book value$30{,}000$ after 3 years

Step 1: Months in service in the first calendar year: October to December =3= 3
Step 2: Dfirst=5,000ร—3/12=1,250D_{\text{first}} = 5{,}000 \times 3/12 = 1{,}250
Step 3: Years 2 through 8 take the full \5{,}000:: 7 \times 5{,}000 = 35{,}000$
Step 4: Remaining: 40,000โˆ’1,250โˆ’35,000=3,75040{,}000 - 1{,}250 - 35{,}000 = 3{,}750, charged in a ninth partial year
Step 5: Total: 1,250+35,000+3,750=40,0001{,}250 + 35{,}000 + 3{,}750 = 40{,}000
Answer: \1{,}250$ in the first year, and the schedule runs across 9 calendar years

Step 1: D=(12,800โˆ’800)/5=12,000/5=2,400D = (12{,}800 - 800)/5 = 12{,}000/5 = 2{,}400 a year
Step 2: Year 1: BV=12,800โˆ’2,400=10,400BV = 12{,}800 - 2{,}400 = 10{,}400
Step 3: Year 2: 8,0008{,}000; Year 3: 5,6005{,}600
Step 4: Year 4: 3,2003{,}200; Year 5: 800800
Step 5: Accumulated depreciation: 5ร—2,400=12,0005 \times 2{,}400 = 12{,}000, and 12,800โˆ’12,000=800=S12{,}800 - 12{,}000 = 800 = S
Answer: \2{,}400ayear,endingatthea year, ending at the$800$ salvage value

Frequently Asked Questions

D = (cost โˆ’ salvage value) / useful life in years. The same amount is charged every full year, and book value after k years is BV = cost โˆ’ kD, falling in a straight line to the salvage value.

Multiply the annual charge by the months in service over 12. An asset placed in service on 1 October with a $5,000 annual charge takes 5,000 ร— 3/12 = $1,250 in year one, and the leftover $3,750 falls into an extra final year.

It is 1/n of the depreciable base per year, where n is the useful life โ€” 12.5% a year over 8 years, 20% over 5. Note this is a percentage of cost minus salvage, not of cost, unlike declining-balance rates.

Not necessarily. Tax rules in each jurisdiction prescribe their own methods, asset lives and conventions, which may differ from the life you use in your books. Compute the accounting figure here, then apply whatever your local rules require for a tax return.

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