1. 12/13/2015rd1 Engineering Economic Analysis Chapter 11  Depreciation

2. 12/13/2015rd2 Depreciation Economic Gradual decrease in utility in an asset with use and time Accounting Systematic allocation of asset's value in portions over its depreciable life Physical Depreciation Functional Depreciation Book Depreciation Tax Depreciation

3. 12/13/2015rd3 Depreciation Property must 1. be used for business to produce income. 2. have a useful life. 3. be an asset that decays, gets consumed, wears out becomes obsolete or loses value to the owner from natural causes. Ergo, land is never depreciable, nor is clearing, grading, planting, and landscaping.

4. 12/13/2015rd4 Cost Basis Total cost to cover the life of the asset to include freight, site preparation, installation, and training. Basic accounting principle is to match cost with revenue.

5. 12/13/2015rd5 Cost Basis Cost of asset$50,000 Freight 1,000 Installation labor 2,000 Site preparation 4,000 Cost basis$57,000

6. 12/13/2015rd6 Cost Basis with Trade-in Old asset $5,000 Less: trade-in 6,000 Unrecognized gain $1,000 Cost of new asset $50,000 Less: Unrecognized gain (1,000) Freight 1,000 Installation labor 2,000 Site preparation 4,000 Cost Basis$56,000

7. 12/13/2015rd7 Property Tangible (can be seen touched and felt) Real (land, buildings, attachments) Personal (equipment, furnishings, machinery) Intangible (has value that can neither be seen nor touched (patents, copyrights, trademarks and franchises)

8. 12/13/2015rd8 Depreciation Methods Straight Line Sum of Years Digits Declining Balance (2, 1.5 and 1.25) Unit of production Percent Depletion MACRS

9. 12/13/2015rd9 Straight Line Depreciation Given P is first cost, S is salvage value at the end of N years, annual depreciation charge (ADC), is computed as ADC = Example: P = 1000, S = 200, N = 4 years ADC = (1000 – 200) / 4 = $200 each year Cannot depreciate below salvage value. (st-line-table 1000 200 4)

10. 12/13/2015rd10 SOYD Depreciation Sum of years digits with P = $1000 and S = $200 and N = 4 years. The sum of the years is 1 + 2 + 3 + 4 = 4*5/2 =10 YearADC 1(4/10) (1000 – 200) = $320 2 (3/10) (1000 – 200) = $240 3(2/10) (1000 – 200) = $160 4(1/10) (1000 – 200) = $ 80 Total ADCs = $800 (soyd-table 1000 200 4)

11. 12/13/2015rd11 Declining Balance Declining balance rates are almost always double the rate of straight-line depreciation. Other rates are 1.25 and 1.5. Still one must never depreciate below salvage value. At double declining balance the ADC is * Book Value. Example: P = $1,000, S = $200; N = 4 years, ddb = 2/N n Bn-1 ADC Bn 1 $1000 $ 500 $ 500 2 $ 500 $ 250 $ 250 3 $ 250 $ 25 $ 225 ST-LINE 4 $ 225 $ 25 $ 200 ST-LINE B n = P(1 -  ) n where  is 2/N for double declining balance. (ddb 1000 200 4 2)

12. 12/13/2015rd12 Optimal time to switch from Declining Balance to ST-Line P = $10,000; N = 5 years; S = 0;  = 40% Year D n B n 1$4000$6000 2 2400 3600 3 1440 2160 4 864 1296 5 518 778 St-Line is 10000/5 = 2000 => switch at year 4 to st-line.

13. 12/13/2015rd13 Declining Balance D 1 =  P D 2 =  (P -  P) =  P(1 -  ) D 3 =  [P -  P -  P(1 -  )] =  P(1 -  ) 2 D n =  P(1 -  ) n-1 TDB = D 1 +… + D n = P[1 – (1 -  ) n ] B n = P – TDB = P(1 -  ) n Example: P = $10K, N = 5 years, Salvage = $778 n B n-1 ADC B n 1 $10000 $4000 $6000 2 6000 2400 3600 3 3600 1440 2160 4 2160 864 1296 5 1296 518 778 ST-LINE

14. 12/13/2015rd14 Book Value Book value = Cost - Depreciation charges made to date.

15. 12/13/2015rd15 Units-of-Production Example: You operate a truck with an estimated cost of $60,000 for 200,000 miles of life with a $7K salvage value. During the year you used the truck for 25,000 miles. ADC = (60 – 7)K * 25K/200K = $6625.

16. 12/13/2015rd16 Activity Depreciation Example: Cost Depletion -- You buy a small forest for $500,000; the land is worth $100,000. The lumber is estimated to contain 2 million board feet (mbf). If you cut 0.4 mbf, your depletion allowance is (500 – 100)K * 0.4 / 2 = $80,000 Activity depreciation is based on level of activity, not on time.

17. 12/13/2015rd17 Percent Depletion A gold mine has 400,000 ounces of gold on a basis of $50M (cost – land). Gross annual revenue is $20M from selling 50,000 ounces. Mining expenses before depletion is $15M. Gold has 15% depletion percentage. 20M * 0.15 = $3M depreciation. Check taxes. Allowance is computed percentage or 50% of taxable income. Taxable Income = $20M – $15M = $5M 0.5 * 5M = 2.5M Compare 2.5M with $3M You must take the lesser of the two percentage depletions: $2.5M. Based on cost depletion: 50M * 50K/400K = $6.25M. Best here to depreciate on cost depletion.

18. 12/13/2015rd18 Year 3-year 5-year 7-year 10-year 15-year 20-year 1 33.33 20.00 14.29 10.00 5.00 3.750 2 44.45 32.00 24.49 18.00 9.50 7.219 3 14.81* 19.20 17.49 14.40 8.55 6.677 4 7.41 11.52* 12.49 11.52 7.70 6.177 5 --- 11.52 8.93* 9.22 6.93 5.713 6 --- 5.76 8.92 7.37 6.23 5.285 7 --- --- 8.93 6.55* 5.90* 4.888 8 --- --- 4.46 6.55 5.90 4.522 9 --- --- --- 6.56 5.91 4.462* 10 --- --- --- 6.55 5.90 4.461 11 --- --- --- 3.28 5.91 4.462 12 --- --- --- --- 5.90 4.461 13 --- --- --- --- 5.91 4.462 14 --- --- --- --- 5.90 4.461 15 --- --- --- --- 5.91 4.462 16 --- --- --- --- 2.95 4.461 17 --- --- --- --- --- 4.462 18 --- --- --- --- ---4.461 19 --- --- --- --- --- 4.462 20 --- --- --- --- --- 4.461 21 --- --- --- --- --- 2.231

19. MACRS 3-Class SL = 1/3 DDB = 2/3 S = 0 Year CalculationMACRS Percentage 1 ½ * 2/3 = 1/3 or 33.33% 2 2/3 * 2/3 = 4/9 or 44.45% SL dep = (1 – 1/3)/2.5 = 26.67% 3 2/3 * (1 – 0.7778) = 14.81% SL dep (1 – 0.7788)/1.5 = 14.81% 4 ½ * 14.81% = 7.41% In Year 3 SL depreciation  DDB => Switch to SL 12/13/2015rd19

20. 12/13/2015rd20 MACRS A taxpayer buys a $10K 5-year MACRS class asset. Compute the ADCs. If the asset were disposed in year 3, find the ADC for that year. (dmacrs 10000 5) $2000 3200 1920 1152 1152 576 In year of disposal, take ½ of ADC = ½ * 1920 = $960

21. 12/13/2015rd21 A 120-room hotel is bought for $2.5M. A 25-year loan is available for 12%. Study data shows OccupancyProbability 65% full0.4 70%0.3 75%0.2 80%0.1 The operating hotel costs are: Taxes and insurance $20K annually Maintenance $50K annually Operating $200K annually The life of the hotel is 25 years when operating 365 days per year with salvage value $500K. Neglect tax credit and income taxes. Determine the average rate to charge per room per night to return 15% of the initial cost per year.

22. Problem 11-1 First cost = $10K; S = $1600; Life 6 years; SOYD (SOYD-table 10e3 1.6e3 6)  n B n-1 ADC B n 1 10000 2400 7600 2 7600 2000 5600 3 5600 1600 4000 4 4000 1200 2800 5 2800 800 2000 6 2000 400 1600 … continued  12/13/2015rd22

23. Problem 11-1 First cost = $10K; S = $1600; Life 6 years; DDB (DDB 10e3 1.6e3 6 2)  n B n-1 ADC B n 1 $10000.00 $3333.33 $6666.67 2 $6666.67 $2222.22 $4444.44 3 $4444.44 $1481.48 $2962.96 4 $2962.96 $ 987.65 $1975.31 5 $1975.31 $ 187.65 * $1787.65 ST-LINE 6 $1787.65 $ 187.65 $1600.00 ST-LINE * May take $375.30 in year 5. 12/13/2015rd23

24. Problem 11-2 $1M first cost has salvage value of $75K after 6 years. Determine if SOYD or DDB is better at MARR = 12%. (SOYD 1e6 75e3 6)  (list-pgf '(0 220238.09 176190.49 132142.86 88095.24 44047.62) 12)  $693,233.45 (DDB 1e6 75e3 6 2)  (list-pgf '(0 333333.34 222222.22 148148.14 98765.42 65843.61 43895.74) 12)  $ 702,589.76 12/13/2015rd24

25. Problem 11-9 First cost = $1.5M with a 5-year life and no salvage value. Find the MACRS depreciation. (dmacrs 1.5e  (300000 480000 288000 172800 172800 86400) 12/13/2015rd25

26. Problem 11-16 (Depreciation 100e3 20e3 5 2)  St Line SOYD DDB MACRS *** 16000 26666.67 40000 20000 16000 21333.33 24000 32000 16000 16000 14400 19200 16000 10666.67 8640 11520 16000 5333.33 5184 11520 ---- ---- ---- 5760 12/13/2015rd26

27. Depreciation Schedules Year A B C D E 1$323.3 $212.0 $424.0$194.0 $107.0 2 258.7 339.2 254.4 194.0 216.0 3 194.0 203.5 152.6 194.0 324.0 4 129.3 122.1 91.6 194.0 216.0 5 64.7 122.1 47.4 194.0 107.0 6 61.1 970.0 1060.0 970.0 970.0 970.0 _____ _____ _____ _____ _____ First cost = _____ Salvage = _____ (Depreciation 10000 1000 6 2) 12/13/2015rd27

28. Depreciation Schedule n A B C D 1$8K$22,857$11,429 $22,857 2 8K 16,327 19,59216,327 3 8K 11,661 13,99411,661 4 8K 5,154 9,996 8,330 5 8K 0 7,140 6,942 6 8K 0 7,140 6,942 7 8K 0 7,140 6,942 8 0 0 3,570 0 First cost = _______Salvage value = ________ 12/13/2015rd28

29. Problem 11-26 First cost is $20K for a MACRS 3-year class property sold in 2 nd year for $14K. Find any depreciation recapture, ordinary losses or capital gains. (DMACRS 20e3 3)  (6666 8890 2962 1482) Book value = 20K – 6666 – 0.5(8890) = $8889 MV 2 – BV 2 = 14K – 8889 = $5111 (depreciation recapture) or ordinary gains. Gains (losses) = salvage value – book value Gains = Salvage value – book value = (salvage value – cost basis) + (cost basis – book value) capital gains ordinary gains 12/13/2015rd29

30. Example 11-9 Cost basis is $10K for MACRS 3-year property disposed of after 5 years of operation for a)$7K; 7K - 0 = $7K depreciation recapture (ordinary gains) b)$0;0 – 0 = 0 => neither recapture nor loss c)Cost of $2K; -2K = -$2K loss. (Dmacrs 10e3 3)  (3333 4445 1481 741) 12/13/2015rd30

31. Example An asset costing $230K in MACRS 7-year class is sold at the end of 3 years for $150K and is taxed at 34%. a) Find the gain or loss. b) Repeat if sold for $100K a) (dmacrs 230E3 7)  (32867 56327 40227 28727 20539 20516 20539 10258) 32867 + 56327 + ½ * 40227 = $109,307.50 (depreciation charges) BV 3 = 230K - $109,307.50 = $120,692.50 BV 3 no capital gains b) BV = $120,692.50 – 100K = $20,692.50 loss => Tax savings 20.692.5 * 0.34 = $7035.45 Net = $107,035.50 12/13/2015rd31