財務(wù)會計概念與應(yīng)用英文版lectu-閱讀頁
2025-01-21 21:28本頁面
【正文】 ipment . . . . . . 1,000 Building . . . . . . . . . . . . . . . . . . . . . . . . . . 5,000 To record impairment of building. 19 Discarding and Selling LongTerm Assets ? An asset’s cost and accumulated depreciation must be removed from the accounting records. ? Assets can be: – Discarded (scrapped) – Sold – Exchanged 20 Discarding Property, Plant, and Equipment Accumulated Depreciation, Conveyor . . . 15,000 Loss on Disposal of Conveyor . . . . . . . . . 300 Conveyor. . . . . . . . . . . . . . . . . . . . . . . . 15,000 Cash. . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Scrapped $15,000 copier and paid $300 disposal costs. Dahl amp。 Sons purchased an advanced copier system for $15,000. It has a 5year life, no salvage value, and is depreciated on a straightline basis. If the copier is sold for $600 after only four years of service, Dahl amp。 Sons purchased Clark amp。 Associates for $1,200,000. 27 Measuring the Management of LongTerm Assets ? Fixed Asset Turnover – The amount of dollars in sales generated by each dollar of fixed assets. – Standard values for this ratio differ from industry to industry. Sales Average PPamp。 Sons purchased a truck for $12,000. The truck has a salvage value of $2,000 and a useful life of 4 years. Compute depreciation using the DDB depreciation method for the first 2 years. 30 SumofYears’Digits Method Sum of the years of the asset’s life SumoftheYear’sDigits (Cost – Salvage Value) Current Year / (4+3+2+1) Year 1 ($12,000 – $2,000) 4/(4+3+2+1) = $4,000 Year 2 ($12,000 – $2,000) 3/(4+3+2+1) = $3,000 Dahl amp。 Sons purchased a truck for $24,000 with a $2,000 salvage value and a 4year useful life. After 3 years, better information reveals the truck has a 6year useful life and a $3,000 salvage value. Calculate a new depreciation expense for the next three years. F o r m u l a C a l c u l a t i o n Tot a l D e pr e c i a t i onA n n u a l d e p r e ci a t i o n C o st S a l v a g e v a l u e = D e p r e ci a t i o n $ 2 4 , 0 0 0 $ 2 , 0 0 0 = $ 5 , 5 0 0 $ 1 6 , 5 0 0f o r f i r st 3 y e a r s E st i m a t e d u sef u l l i f e e x p e n se 4 Y e a r sB o o k v a l u e a f t e r 3 y e a r s C o st A cc u m u l a t e d = B o o k V a l u e $ 2 4 , 0 0 0 $ 1 6 , 5 0 0 = $ 7 , 5 0 0d e p r e ci a t i o n t o d a t eA n n u a l d e p r e ci a t i o n f o r l a st3 y e a r s ( b a sed o n n e w B o o k S a l v a g e v a l u e = D e p r e ci a t i o n $ 7 , 5 0 0 $ 3 , 0 0 0 = $ 1 , 5 0 0 4 , 5 0 0t o t a l l i f e o f 6 y e a r s a n d n e w R e m a i n i n g u sef u l l i f e e x p e n se 3 y e a r ssal v a g e v a l u e o f $ 3 , 0 0 0 ) $ 2 1 , 0 0
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