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 тАФ cost minus salvage value тАФ 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=CтИТSLD = \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=BtтИТ1├Чrate,Bt=BtтИТ1тИТDt\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=CтИТStotal┬аestimated┬аunits├Чunits┬аthis┬аperiodD = \frac{C - S}{\text{total estimated units}} \times \text{units this period}

All three converge on the same total, CтИТSC - 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=тИСj=1tDj,Bt=CтИТaccumulatedt\text{accumulated}_t = \sum_{j=1}^{t} D_j, \qquad B_t = C - \text{accumulated}_t

Under straight-line this collapses to Bt=CтИТtDB_t = C - tD; under declining balance it is Bt=C(1тИТrate)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=1тИТ(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 тАФ and any prescribed rate tables or conventions тАФ 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 CтИТSC - S. Salvage acts only as a floor.
  • Depreciating past salvage: once BtB_t would fall below SS, the final charge is trimmed to BtтИТ1тИТSB_{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 CтИТSC - S over the full life. Only the timing changes.

Examples

Step 1: Depreciable base: 24000тИТ4000=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├Ч2500=7,5003 \times 2500 = 7{,}500
Step 4: Book value: 24000тИТ7500=16,50024000 - 7500 = 16{,}500
Step 5: Rate as a percent of cost: 2500/24000тЙИ10.42%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├Ч0.40=12,00030000 \times 0.40 = 12{,}000, book =18,000= 18{,}000
Step 3: Year 2: 18000├Ч0.40=7,20018000 \times 0.40 = 7{,}200, book =10,800= 10{,}800
Step 4: Year 3: 10800├Ч0.40=4,32010800 \times 0.40 = 4{,}320, book =6,480= 6{,}480
Step 5: Year 4: 6480├Ч0.40=2,5926480 \times 0.40 = 2{,}592, book =3,888= 3{,}888
Step 6: Year 5: 3888├Ч0.40=1,555.203888 \times 0.40 = 1{,}555.20 would drop book to 2,332.802{,}332.80, below salvage тАФ so charge only 3888тИТ3000=8883888 - 3000 = 888
Answer: 12,00012{,}000, 7,2007{,}200, 4,3204{,}320, 2,5922{,}592, 888888 тАФ totalling \27{,}000 = C - S,endingatbookvalue, ending at book value $3{,}000$

Step 1: Depreciable base: 90000тИТ10000=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├Ч55000=11,0000.20 \times 55000 = 11{,}000
Step 4: Book value: 90000тИТ11000=79,00090000 - 11000 = 79{,}000
Step 5: If the next year uses 30,000 units: 0.20├Ч30000=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 тИТ Salvage) / Useful life. An asset costing $24,000 with a $4,000 salvage value and an 8-year life depreciates by (24,000 тИТ 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 тАФ 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 тИТ 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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