Depreciation
- Straight-line
- (cost − salvage)/life
- Double-declining balance
- 2/life × book value at the start of the year
- Sum of the years’ digits
- (cost − salvage) × remaining life/SYD
- SYD
- n(n + 1)/2
- Book value
- cost − accumulated depreciation
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
Budgeted sales are 1,500 units and the break-even point is 1,000 units. What is the margin of safety?
Under IFRS 15, revenue is recognised:
A machine costs €80 000, has a useful life of 10 years and a salvage value of €4 000. Using the double-declining-balance method (without switching to straight-line), find the depreciation charge D in year 6.
A computer system costs €12 000, has a useful life of 5 years and a salvage value of €0. Using the sum-of-the-years’-digits method, find the depreciation charge D in year 2.
Answer: A · 500 units
Margin of safety = budgeted sales − break-even sales = 500 units, or one third of budgeted sales.
Answer: D · When (or as) the entity satisfies a performance obligation by transferring control to the customer
The steps: identify the contract, identify performance obligations, determine the price, allocate it, and recognise revenue as each obligation is satisfied.
Answer: D = 5243 €
Depreciation spreads the cost of an asset over its useful life. Straight-line: (cost − salvage value)/life every year. Double-declining balance: 2/life times the book value at the start of each year, so the charge falls each year (salvage is not subtracted, but the book value never goes below it). Sum-of-the-years’-digits: (cost − salvage) × remaining life/(1 + 2 + … + n).
Answer: D = 3200 €
Depreciation spreads the cost of an asset over its useful life. Straight-line: (cost − salvage value)/life every year. Double-declining balance: 2/life times the book value at the start of each year, so the charge falls each year (salvage is not subtracted, but the book value never goes below it). Sum-of-the-years’-digits: (cost − salvage) × remaining life/(1 + 2 + … + n).
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