Depreciation Calculator

Straight-line, declining balance and units-of-production methods with AI-powered step-by-step solutions
Straight-line depreciation: cost $24,000, salvage $4,000, life 8 years
Double declining balance schedule: cost $30,000, salvage $3,000, life 5 years
Units of production: cost $90,000, salvage $10,000, 400,000 units, 55,000 used
Book value after 3 years of straight-line depreciation on a $24,000 asset

The Depreciation Methods

Depreciation spreads an asset's depreciable base — cost minus salvage value — across the periods that use it. Three inputs drive every method: cost CC, salvage (residual) value SS, and useful life LL.

Straight-line charges the same amount every year:

D=CSLD = \frac{C - S}{L}

Declining balance applies a fixed rate to the book value, which shrinks each year, so the charge front-loads:

rate=kL,Dt=Bt1×rate,Bt=Bt1Dt\text{rate} = \frac{k}{L}, \qquad D_t = B_{t-1} \times \text{rate}, \qquad B_t = B_{t-1} - D_t

with k=2k = 2 for double declining balance and k=1.5k = 1.5 for the 150% variant. Note the crucial asymmetry: declining balance ignores salvage in the rate, so you must stop once book value reaches SS.

Units of production ties the charge to usage rather than time:

D=CStotal estimated units×units this periodD = \frac{C - S}{\text{total estimated units}} \times \text{units this period}

All three converge on the same total, CSC - S; they differ only in how that total is distributed across the years.

Book Value, Rates and Choosing a Method

Book value and accumulated depreciation

accumulatedt=j=1tDj,Bt=Caccumulatedt\text{accumulated}_t = \sum_{j=1}^{t} D_j, \qquad B_t = C - \text{accumulated}_t

Under straight-line this collapses to Bt=CtDB_t = C - tD; under declining balance it is Bt=C(1rate)tB_t = C(1 - \text{rate})^t, until the salvage floor bites.

Solving for a rate

Given a starting value, an ending value and a life, the constant annual rate that connects them is

rate=1(BLC)1/L\text{rate} = 1 - \left(\frac{B_L}{C}\right)^{1/L}

This is the depreciation twin of a CAGR, and it is how you back out the implied rate from a published value table.

Which method applies

Straight-line suits assets that wear evenly with time; declining balance suits assets that lose most value early; units of production suits machinery whose wear tracks output. Which one you are permitted to use for tax purposes — and any prescribed rate tables or conventions — is set by the accounting standards and tax rules of your jurisdiction, and those change. This page computes any method exactly on the figures you give it; it does not select a method for you or provide tax guidance.

Common Mistakes to Avoid

  • Subtracting salvage in declining balance: the rate is applied to full book value, starting from CC, not from CSC - S. Salvage acts only as a floor.
  • Depreciating past salvage: once BtB_t would fall below SS, the final charge is trimmed to Bt1SB_{t-1} - S and depreciation stops.
  • Using the rate on cost every year: declining balance multiplies the current book value. Applying the rate to CC every year is straight-line with an odd rate.
  • Confusing rate with life: 2/L2/L is the DDB rate, so a 5-year life gives 40%40\%, not 20%20\%.
  • Forgetting the salvage estimate entirely: leaving SS out of straight-line overstates the annual charge by S/LS/L.
  • Mixing units and time: units of production takes the units used, not the fraction of the life elapsed.
  • Expecting the totals to differ: every method totals CSC - S over the full life. Only the timing changes.

Examples

Step 1: Depreciable base: 240004000=20,00024000 - 4000 = 20{,}000
Step 2: D=20000/8=2,500D = 20000/8 = 2{,}500 per year
Step 3: Accumulated after 3 years: 3×2500=7,5003 \times 2500 = 7{,}500
Step 4: Book value: 240007500=16,50024000 - 7500 = 16{,}500
Step 5: Rate as a percent of cost: 2500/2400010.42%2500/24000 \approx 10.42\%
Answer: \2{,}500ayear;bookvaluea year; book value$16{,}500$ after 3 years

Step 1: Rate =2/5=0.40= 2/5 = 0.40, applied to book value (salvage is not subtracted first)
Step 2: Year 1: 30000×0.40=12,00030000 \times 0.40 = 12{,}000, book =18,000= 18{,}000
Step 3: Year 2: 18000×0.40=7,20018000 \times 0.40 = 7{,}200, book =10,800= 10{,}800
Step 4: Year 3: 10800×0.40=4,32010800 \times 0.40 = 4{,}320, book =6,480= 6{,}480
Step 5: Year 4: 6480×0.40=2,5926480 \times 0.40 = 2{,}592, book =3,888= 3{,}888
Step 6: Year 5: 3888×0.40=1,555.203888 \times 0.40 = 1{,}555.20 would drop book to 2,332.802{,}332.80, below salvage — so charge only 38883000=8883888 - 3000 = 888
Answer: 12,00012{,}000, 7,2007{,}200, 4,3204{,}320, 2,5922{,}592, 888888 — totalling \27{,}000 = C - S,endingatbookvalue, ending at book value $3{,}000$

Step 1: Depreciable base: 9000010000=80,00090000 - 10000 = 80{,}000
Step 2: Rate per unit: 80000/400000=0.2080000 / 400000 = 0.20 per unit
Step 3: This year: 0.20×55000=11,0000.20 \times 55000 = 11{,}000
Step 4: Book value: 9000011000=79,00090000 - 11000 = 79{,}000
Step 5: If the next year uses 30,000 units: 0.20×30000=6,0000.20 \times 30000 = 6{,}000, cumulative 17,00017{,}000
Answer: \11{,}000fortheyear,leavingabookvalueoffor the year, leaving a book value of$79{,}000$

Frequently Asked Questions

D = (Cost − Salvage) / Useful life. An asset costing $24,000 with a $4,000 salvage value and an 8-year life depreciates by (24,000 − 4,000)/8 = $2,500 every year, until book value reaches the $4,000 salvage figure.

Apply a rate of 2 / useful life to the current book value each year, without subtracting salvage from the base. A 5-year life gives 40%: $30,000 becomes $12,000 in year one, then 40% of the remaining $18,000, and so on. Stop once book value reaches salvage.

Because the shrinking book value already tapers the charge — subtracting salvage first would taper it twice. Salvage instead acts as a floor: the final year's charge is trimmed so book value lands exactly on it.

No. Every method totals Cost − Salvage over the full life. They differ only in timing: declining balance front-loads the expense, straight-line spreads it evenly, and units of production follows actual usage.

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