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Engineering Economic AnalysisEngineering Economic Analysis

Chapter 11 Depreciation

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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

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DepreciationDepreciation

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.

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Cost BasisCost 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.

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Cost BasisCost Basis

Cost of asset $50,000Freight 1,000

Installation labor 2,000Site preparation 4,000

Cost basis $57,000

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Cost Basis with Trade-inCost Basis with Trade-in

Old asset $5,000Less: trade-in 6,000Unrecognized gain $1,000

Cost of new asset $50,000Less: Unrecognized gain (1,000)Freight 1,000Installation labor 2,000Site preparation 4,000Cost Basis $56,000

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PropertyProperty

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)

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Depreciation MethodsDepreciation Methods

Straight Line

Sum of Years Digits

Declining Balance (2, 1.5 and 1.25)

Unit of production

Percent Depletion

MACRS

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Straight Line DepreciationStraight 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)

P-S

N

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SOYD DepreciationSOYD 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

Year ADC 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)

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Declining BalanceDeclining BalanceDeclining 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

Bn = P(1 - )n where is 2/N for double declining balance.

(ddb 1000 200 4 2)

2P

N

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Optimal time to switch from Declining Optimal time to switch from Declining Balance to ST-LineBalance to ST-Line

P = $10,000; N = 5 years; S = 0; = 40%Year Dn Bn 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.

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Declining BalanceDeclining BalanceD1 = PD2 = (P - P) = P(1 - )D3 = [P - P - P(1 - )] = P(1 - )2

Dn = P(1 - )n-1

TDB = D1 +… + Dn = P[1 – (1 - )n]Bn = P – TDB = P(1 - )n

Example: P = $10K, N = 5 years, Salvage = $778

n Bn-1 ADC Bn 1 $10000 $4000 $6000 2 6000 2400 3600 3 3600 1440 2160 4 2160 864 1296 5 1296 518 778 ST-LINE

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Book ValueBook Value

Book value = Cost - Depreciation charges made to date.

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Units-of-ProductionUnits-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.

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Activity DepreciationActivity 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.

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Percent DepletionPercent DepletionA 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.

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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

MACRS 3-ClassMACRS 3-Class

SL = 1/3 DDB = 2/3 S = 0

Year Calculation MACRS 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

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MACRSMACRS

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

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A 120-room hotel is bought for $2.5M. A 25-year loan is available for 12%. Study data shows

Occupancy Probability65% full 0.4 70% 0.375% 0.280% 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.

Problem 11-1Problem 11-1

First cost = $10K; S = $1600; Life 6 years; SOYD

(SOYD-table 10e3 1.6e3 6)

n Bn-1 ADC Bn 1 10000 2400 7600 2 7600 2000 56003 5600 1600 40004 4000 1200 28005 2800 800 2000 6 2000 400 1600

… continued

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Problem 11-1Problem 11-1

First cost = $10K; S = $1600; Life 6 years; DDB

(DDB 10e3 1.6e3 6 2) n Bn-1 ADC Bn

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.

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Problem 11-2Problem 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

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Problem 11-9Problem 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)

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Problem 11-16Problem 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

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Depreciation SchedulesDepreciation Schedules

Year A B C D E1 $323.3 $212.0 $424.0 $194.0 $107.02 258.7 339.2 254.4 194.0 216.03 194.0 203.5 152.6 194.0 324.04 129.3 122.1 91.6 194.0 216.05 64.7 122.1 47.4 194.0 107.06 61.1

970.0 1060.0 970.0 970.0 970.0

_____ _____ _____ _____ _____

First cost = _____ Salvage = _____

(Depreciation 10000 1000 6 2)

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Depreciation ScheduleDepreciation Schedule

n A B C D

1 $8K $22,857 $11,429 $22,857

2 8K 16,327 19,592 16,327

3 8K 11,661 13,994 11,661

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 = ________

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Problem 11-26Problem 11-26

First cost is $20K for a MACRS 3-year class property sold in 2nd 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

MV2 – BV2 = 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

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Example 11-9Example 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)

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ExampleExampleAn 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)

BV3 = 230K - $109,307.50 = $120,692.50

BV3 < 150K < 230K => 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

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