newnansm11 - Chapter 11: Income, Depreciation, and Cash...

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Unformatted text preview: Chapter 11: Income, Depreciation, and Cash Flow 11-1 Year SOYD DDB 1 $2,400 $3,333 2 $2,000 $2,222 3 $1,600 $1,482 4 $1,200 $988 5 $800 $375* 6 $400 $0 Sum $8,400 $8,400 *Computed $658 must be reduced to $375 to avoid depreciating the asset below its salvage value. 11-2 DDB Schedule is: Year n d(n)=(2/n)[P sum d(n)] DDB Depreciation 1 (2/6) ($1,000,000 - $0) = $333,333 2 (2/6) ($1,000,000 - $333,333) = $222,222 3 (2/6) ($1,000,000 - $555,555) = $148,148 4 (2/6) ($1,000,000 - $703,703) = $98,766 5 (2/6) ($1,000,000 - $802,469) = $65,844 6 See below = $56,687 If switch DDB to SL for year 5: SL = ($1,000,000 - $802,469 - $75,000)/2 = $61,266 Do not switch. If switch DDB to SL for year 6: SL = ($1,000,000 - $868,313 - $75,000)/1 = $56,687 Do switch. Sum-of-Years Digits Schedule is: SOYD in N = [(Remain. useful life at begin. of yr.)/[(N/2)(N+1)]] (P S) 1 st Year: SOYD = (6/21) ($1 mil - $75,000) = $264,286 2 nd Year: = (5/21) ($1 mil - $75,000) = $220,238 3 rd Year: = (4/21) ($1 mil - $75,000) = $176,190 4 th Year: = (3/21) ($1 mil - $75,000) = $132,143 5 rd Year: = (2/21) ($1 mil - $75,000) = $ 88,095 6 th Year: = (1/21) ($1 mil - $75,000) = $ 44,048 Question: Which method is preferred? Answer: It depends, on the MARR%, i% used by the firm (individual) As an example: If i% is= PW of DDB is= PW of SOYD is= Preferred is 0% $925,000 $925,000 Equal, same 2% $881,211 $877,801 DDB 10% $738,331 $724,468 DDB 25% $561,631 $537,130 DDB Thus, if MARR% is > 0%, DDB is best. One can also see this by inspection of the depreciation schedules above. 11-3 DDB Depreciation Year DDB Depreciation 1 (2/5) ($16,000 - $0) = $6,400 2 (2/5) ($16,000 - $6,400) = $3,840 3 (2/5) ($16,000 - $10,240) = $2,304 4 (2/5) ($16,000 - $13,926) = $830 Sum $14,756 Converting to Straight Line Depreciation If Switch for Year Beginning of Yr Book Value Remaining Life SL = (Book Salvage)/ Remaining Life Decision 2 $9,600 4 yrs $2,400 Do not switch 3 $5,760 3 yrs $1,920 Do not switch 4 $3,456 2 yrs $1,728 Switch to SL 5 $2,074 1 yr $2,074 Resulting Depreciation Schedule: Year DDB with Conversion to Straight Line 1 $6,400 2 $3,840 3 $2,304 4 $1,728 5 $1,728 Sum $16,000 11-4 P=$12,000 S=$3,500 N=4 (a) Straight Line Depreciation SL = -(P-S) / N =($12,000-$3,500) / 4 $2,125 (b) Sum-of-Years Digits Depreciation SOYD in yr. N = [(Remain. useful life at begin. of yr.)/[(N/2)(N+1)]] (P S) 1 st Year: SOYD = (4/10) ($12,000 - $3,500)= $3,400 2 nd Year: = (3/10) ($12,000 - $3,500)= $2,550 3 rd Year: = (2/10) ($12,000 - $3,500)= $1,700 4 th Year: = (1/10) ($12,000 - $3,500)= $850 (c) Double Declining Balance Depreciation DDB in any year = 2/N(BookValue) DDB in any year = 2/N (Book Value) Year: DDB = (2/4) ($12,000 - $0) = $6,000 2 nd Year: = (2/4) ($12,000 - $6,000) = $3,000 3 rd Year: = (2/4) ($12,000 - $9,000) = $1,500 4 th Year: = (2/4) ($12,000 - $10,500) = $750 (d) CCA Special handling equipment classifies as a Class 43 asset with a CCA rate of 30% Year UCC at Start of Year CCA Rate...
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newnansm11 - Chapter 11: Income, Depreciation, and Cash...

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