engm 661 engineering economics depreciation & taxes

ENGM 661 Engineering Economics

Depreciation&

Taxes

Taxable Income

+ Gross Income - Depreciation Allowance - Interest on Borrowed Money - Other Tax Exemptions = Taxable Income

Corporate Tax Rate

Taxable Income Tax Rate Income Tax 0 < TI < 50,000 0.15 .15(TI)50,000 < TI < 75,000 0.25 7,500 + .25(TI - 50,000)75,000 < TI < 100,000 0.34 13,750 + .34(TI - 75,000)100,000 < TI < 335,000 0.39 22.250 + .39(TI - 100,000)

335,000 < TI < 10,000,000 0.34 113,900 + .35(TI - 335,000)10,000,000 < TI < 15,000,000 0.35 3,400,000 + .35(TI - 10,000,000)15,000,000 < TI < 18,333,333 0.38 5,150,000 + .38(TI - 15,000,000)

TI > 18,333,333 0.35 .35(TI)

Corporate Tax

Ex: Suppose K-Corp earns $5,000,000 in revenue above manufacturing and operations cost. Suppose further that depreciation costs total $800,000 and interest paid on short and long term debt totals $1,500,000. Compute the tax paid.

Corporate Tax

Gross Income $ 5,000,000Depreciation - 800,000Interest - 1,500,000Taxable Income $ 2,700,000

Corporate Tax

Gross Income $ 5,000,000Depreciation - 800,000Interest - 1,500,000Taxable Income $ 2,700,000

Tax = $ 113,900 + .35(2,700,000 - 335,000)

= $ 941,650

After Tax Cash Flow

+ Gross Income - Interest

= Before Tax Cash Flow

After Tax Cash Flow

+ Gross Income - Interest

= Before Tax Cash Flow - Tax

= After Tax Cash Flow

After Tax Cash Flow

Ex: Suppose K-Corp earns $5,000,000 in revenue above manufacturing and operations cost. Suppose further that depreciation costs total $800,000 and interest paid on short and long term debt totals $1,500,000. Compute the after tax cash flow.

After Tax Cash Flow

Gross Income $ 5,000,000Depreciation - 800,000Interest - 1,500,000Before Tax Cash Flow $ 2,700,000

After Tax Cash Flow

Gross Income $ 5,000,000Interest - 1,500,000Before Tax Cash Flow $3,500,000Less Tax 941,650After Tax Cash Flow $ 2,558,350

Methods of Depreciation

Straight Line (SL) Sum-of-Years Digits (SYD) Declining Balance (DB)

Prior to 1981 Accelerated Cost Recovery System (ACRS)

1981-86 Modified Accelerated Cost Recovery

(MACRS) 1986 on

Straight Line (SLD)Let

P = Initial Costn = Useful Lifes = Salvage Value year nDt = Depreciation Allowance in year t

Bt = Unrecovered Investment (Book Value) in year t

ThenDt = (P - S) / n

Bt = P - [ (P - S) / n ] t

Ex: Straight Line Depr.

LetP = $100,000n = 5 yearss = $ 20,000

ThenDt = (P - S) / n

= $ 16,000B5 = P - [ (P - S) / n ] 5

= $ 20,000

Declining Balance

In declining balance, we write off a constant

% , p, of remaining book value D1 = pP , P = initial cost

B1 = P - D1 = P - pP

= P(1-p)

D2 = pB1

= pP(1-p)

Declining Balance


% , p, of remaining book valueB2 = B1 - D2 = P(1-p) - pB1

Declining Balance



= P(1-p) - pP(1-p)

Declining Balance



= P(1-p) - pP(1-p)= P(1-p)[1 - p]

Declining Balance



= P(1-p) - pP(1-p)= P(1-p)[1 - p]= P(1-p)2

Declining Balance



= P(1-p) - pP(1-p)= P(1-p)2

Dt = p [ P (1 - p) t - 1]

Bt = P (1 - p) t

Ex: Declining Balance P = $100,000 n = 5 years S = $20,000 p = 2/5 (200% declining balance)

Then D1 = (2/5)(100,000) = $40,000

D5 = ? , B5 = ?


Then D1 = (2/5)(100,000) = $40,000

B1 = 100,000 - 40,000 = $ 60,000

D5 = ? , B5 = ?


Then D1 = (2/5)(100,000) = $ 40,000

B1 = 100,000 - 40,000 = $ 60,000

D2 = (2/5)(60,000) = $ 24,000

D5 = ? , B5 = ?

Ex: Declining Balance (cont)

Dt = p [ P (1 - p) t - 1]D5 = .4(100,000)(.6) 4

= $ 5,184

Bt = P (1 - p) t

B5 = 100,000(.6) 5

= $ 7,776

Ex: Declining Balance (cont)

Dt = p [ P (1 - p) t - 1]D5 = .4(100,000)(.6) 4

= $ 5,184

Bt = P (1 - p) t

B5 = 100,000(.6) 5

= $ 7,776

Note that DecliningBalance will never depreciate book valueto $0. It will, however,depreciate past the salvage value

Time Value of Tax Savings(Tax Rate = 40%)

Time Value of Tax Savings (40%)SL Tax DDB Tax

t Dt Save Dt Save

01 20,000 8,000 40,000 16,0002 20,000 8,000 24,000 9,6003 20,000 8,000 14,400 5,7604 20,000 8,000 8,640 3,4565 20,000 8,000 4,320 1,7286 0 0 8,640 3456

Sum = 100,000 40,000 100,000 40,000Present Value = 30,326 32,191

DDB/SL Conversion(Salvage = $0)

DDB/Straight Line ConversionSL DDB

t Bt-1 Dt Dt Bt

0 100,0001 100,000 20,000 40,000 60,00023456



t Bt-1 Dt Dt Bt

0 100,0001 100,000 20,000 40,000 60,0002 60,000 15,000 24,000 36,0003456



t Bt-1 Dt Dt Bt

0 100,0001 100,000 20,000 40,000 60,0002 60,000 15,000 24,000 36,0003 36,000 12,000 14,400 21,600456



t Bt-1 Dt Dt Bt

0 100,0001 100,000 20,000 40,000 60,0002 60,000 15,000 24,000 36,0003 36,000 12,000 14,400 21,6004 21,600 10,800 8,640 10,80056



t Bt-1 Dt Dt Bt

0 100,0001 100,000 20,000 40,000 60,0002 60,000 15,000 24,000 36,0003 36,000 12,000 14,400 21,6004 21,600 10,800 8,640 10,8005 10,800 10,800 4,320 06 0

Class Problem

Ex: Suppose K-Corp is interested in purchasing a new conveyor system. The cost of the conveyor is $180,000 and may be depreciated over a 5 year period. K-Corp uses 150% declining balance method with a conversion to straight line. Compute the depreciation schedule over the 5 year period.

Class Problem


t Bt-1 Dt Dt Bt

0 180,0001 180,00023456

Class Problem (p = 1.5/5 = .3)


t Bt-1 Dt Dt Bt

0 180,0001 180,000 36,000 54,000 126,00023456

Class Problem


t Bt-1 Dt Dt Bt

0 180,0001 180,000 36,000 54,000 126,0002 126,000 31,500 37,800 88,2003 88,200 29,400 26,460 58,8004 58,800 29,400 17,640 29,4005 29,400 29,400 8,820 06 0

DDB/SL Conversion(Half-Year Convention)

DDB/Straight Line Conversion/Half-YearSL DDB

t Bt-1 Dt Dt Bt

0 100,0001 100,000 10,000 20,000 80,00023456



t Bt-1 Dt Dt Bt

0 100,0001 100,000 10,000 20,000 80,0002 80,000 17,778 32,000 48,0003 48,000 13,714 19,200 28,8004 28,800 11,520 11,520 17,2805 17,280 11,520 6,912 5,7606



t Bt-1 Dt Dt Bt

0 100,0001 100,000 10,000 20,000 80,0002 80,000 17,778 32,000 48,0003 48,000 13,714 19,200 28,8004 28,800 11,520 11,520 17,2805 17,280 11,520 6,912 5,7606 5,760 5,760 2,304 0

Class ProblemA $180,000 piece of machinery is installed and is

to be depreciated over 5 years. You may assume that the salvage value at the end of 5 years is $ 0. The method of depreciation is to be double declining balance with conversion to straight line using the half-year convention (you may only deduct 1/2 year of depreciation in year 1). Establish a table showing the depreciation and the end of year book value for each year.

Class Problem


t Bt-1 Dt Dt Bt

0 180,0001 180,00023456

Solution


t Bt-1 Dt Dt Bt

0 180,0001 180,000 18,000 36,000 144,0002 144,000 32,000 57,600 86,4003 86,400 24,686 34,560 51,8404 51,840 20,736 20,736 31,1045 31,104 20,736 12,442 10,3686 10,368 10,368 4,147 0

MACRS Percentages 3,5,7, & 10 are 200% DB/SL15 & 20 are 150% DB/SL

t 3-Yr. 5-Yr. 7-Yr. 10-Yr. 15-Yr. 20-Yr.

1 33.33% 20.00% 14.29% 10.00% 5.00% 3.75%2 44.45% 32.00% 24.49% 18.00% 9.50% 7.22%3 14.81% 19.20% 17.49% 14.40% 8.55% 6.68%4 74.10% 11.52% 12.49% 11.52% 7.70% 6.18%5 11.52% 8.93% 9.22% 6.93% 5.71%6 5.76% 8.92% 7.37% 6.23% 5.29%7 8.93% 6.55% 5.90% 4.88%

8 4.46% 6.55% 5.90% 4.52%9 6.56% 5.91% 4.46%10 6.55% 5.90% 4.46%11 3.28% 5.91% 4.46%12 5.90% 4.46%13 5.91% 4.46%14 5.90% 4.46%15 5.91% 4.46%16 2.95% 4.46%17 4.46%18 4.46%19 4.46%20 4.46%21 2.23%

MACRS Tables

MACRS Percentages 3,5,7, & 10 are 200% DB/SL15 & 20 are 150% DB/SL

t 3-Yr. 5-Yr. 7-Yr. 10-Yr. 15-Yr. 20-Yr.

1 33.33% 20.00% 14.29% 10.00% 5.00% 3.75%2 44.45% 32.00% 24.49% 18.00% 9.50% 7.22%3 14.81% 19.20% 17.49% 14.40% 8.55% 6.68%4 74.10% 11.52% 12.49% 11.52% 7.70% 6.18%5 11.52% 8.93% 9.22% 6.93% 5.71%6 5.76% 8.92% 7.37% 6.23% 5.29%7 8.93% 6.55% 5.90% 4.88%8 4.46% 6.55% 5.90% 4.52%9 6.56% 5.91% 4.46%10 6.55% 5.90% 4.46%11 3.28% 5.91% 4.46%12 5.90% 4.46%13 5.91% 4.46%14 5.90% 4.46%15 5.91% 4.46%16 2.95% 4.46%17 4.46%18 4.46%19 4.46%20 4.46%21 2.23%

Modified Accelerated Cost

Property Classes3 yr. - useful life < 4 yrs.

autos, tools5 yr. - 4 yrs. < useful life < 10 yrs.

office epuipment, computers, machinery7 yr. - 10 < UL < 16

office furniture, fixtures, exploration10 yr. - 16 < UL < 20

vessels, tugs, elevators (grain)15 yr. - 20 < UL < 25

data communication, sewers, bridges, fencing

MACRS (Cont.)

20 yr. - UL > 25

farm buildings, electric generation27.5 - residential rental property31.5 - non-residential real property

Depreciationclass (3, 5, 7, 10 yr.) uses 200% declining balance switching to straight-line @ optimal yearclass (15, 20) 150% DB switch to SLDclass (27.5, 31.5) use straight-line

After Tax Cash Flow

Formulas BTCF = Before Tax Cash Flow

= Revenues - Expenses

After Tax Cash Flow



TI = Taxable Income = Cash Flow - Interest - Depreciation

After Tax Cash Flow




Tax = TI * Tax Rate

After Tax Cash Flow




Tax = TI * Tax Rate

ATCF = After Tax Cash Flow= BTCF - Tax

Ex: After Tax Cash Flow

Tax Rate = 34%MARR = 20%

Taxablet BTCF MACRS % Depr. Income Tax ATCF

0 (82,000) (82,000)1 23,500 20.0% 16,400 7,100 2,414 21,0862 23,500 32.0% 26,240 (2,740) (932) 24,4323 23,500 19.2% 15,744 7,756 2,637 20,8634 23,500 11.5% 9,446 14,054 4,778 18,7225 23,500 11.5% 9,446 14,054 4,778 18,7226 23,500 5.8% 4,723 18,777 6,384 17,1167 28,500 0 28,500 9,690 18,810

NPV = $4,103 ($7,854)

Borrowed Money

Year Principle Interest Total Pmt. Loan Bal.Amt. Financed $210,000 0 $210,000Interest 16% 1 70000 33600 103600 $140,000Period of Loan 3 2 70000 22400 92400 $70,000Payment #NUM! 3 70000 11200 81200 $0Tax rate 34%MARR 20%

Project Loan Loan Before Tax MACRS MACRS Taxable Tax After Taxt Cash Flow Payment Interest Cash Flow % deduct Income 34% Cash Flow

0 (500,000) 210,000 (290,000) (290,000)1 150,000 70,000 33,600 46,400 20.0% 100,000 16,400 5,576 40,8242 150,000 70,000 22,400 57,600 32.0% 160,000 (32,400) (11,016) 68,6163 150,000 70,000 11,200 68,800 19.0% 95,000 43,800 14,892 53,9084 150,000 150,000 12.0% 60,000 90,000 30,600 119,4005 150,000 150,000 12.0% 60,000 90,000 30,600 119,4006 175,000 175,000 6.0% 30,000 145,000 49,300 125,700

NPW = ($29,471)

Class Problem

A company plans to invest in a water purification system (5 year property) requiring $800,000 capital. The system will last 7 years with a salvage of $100,000. The before-tax cash flow for each of years 1 to 6 is $200,000. Regular MACRS depreciation is used; the applicable tax rate is 34%. Construct a table showing each of the following for each of the 7 years.

Solution

Solution



0 (800,000) (800,000)1 200,0002 200,0003 200,0004 200,0005 200,0006 200,0007 300,000

NPV = ($51,173)

Solution



0 (800,000) (800,000)1 200,000 20.0% 160,000 40,000 13,600 186,4002 200,000 32.0% 256,000 (56,000) (19,040) 219,0403 200,000 19.2% 153,600 46,400 15,776 184,2244 200,000 11.5% 92,160 107,840 36,666 163,3345 200,000 11.5% 92,160 107,840 36,666 163,3346 200,000 5.8% 46,080 153,920 52,333 147,6677 300,000 0 300,000 102,000 198,000

NPV = ($51,173) ($136,824)

Residential Rental

MACRS - ADS Election

Straight Line with either a half-year or half-month convention.

Required for property outside U.S. having tax-exempt status financed by tax-exempt bonds covered by executive order

Example

Ex: A press forming machine is purchased for the manufacture of steel beams for $300,000. The press is considered a 7 year property class (MACRS-GDS = 7). Compute the annual depreciation using the MACRS Alternative Depreciation Election.

Example

Soln: MACRS - ADS has a longer life than does MACRS - GDS. In this case 14 years.

Dn = $300,000/14

= $21,428 n = 2, . . ., 14

= $21,428 / 2= $10,714 n = 1, 15

Units of Production Method

Allows for equal depreciation for each unit of output

whereUt = units produced during the year

U = total units likely to be produced during life

(P-F) = depreciable amount allowed

U

UFPD t

t )(

Operating Day Method


whereQt = total hours used during the year

Q = total hours available during the year(P-F) = depreciable amount allowed

Q

QFPD t

t )(

Income Forecast Method


whereRt = rent income earned during the year

R = total likely rent to be earned during life(P-F) = depreciable amount allowed

R

RFPD t

t )(

Depletion Method


whereVt = volume extracted during the year

V = total volume available in reserve(P-F) = depreciable amount allowed

V

VFPD t

t )(

Example

Ex: NorCo Oil has a 10 year, $27,000,000 lease on a natural gas reservoir in western South Dakota. The reservoir is expected to produce 10 million cubic ft. of gas each year during the period of the lease. Compute the expected depletion allowance for each year.

Example

Ex:

000,700,2$

000,000,1010

000,000,10000,000,27$

x

Dt

Percentage Depletion

Depletion is taken as a constant percentage of gross income

Allowable PercentagesOil/Gas 15%Natural Gas 22%Sulphur/Uranium 22%Gold, silver, … 15%Coal 10%

Example

Ex: NorCo Oil has a 10 year, $27,000,000 lease on a natural gas reservoir in western South Dakota. The reservoir is expected to produce 10 million cubic ft. of gas each year during the period of the lease at $1.50 per cubic ft.

Gross Income = 1.5(10,000,000)= 15,000,000

Depletion = 15,000,000 (0.22)= $3,300,000

Depreciation Recapture

Ex: K-Corp purchases a Loader for $250,000 which has a 7 year property class life. After 3 years, $140,675 has been depreciated and the book value is now $109,325. K-Corp now sells the loader for $150,000.



Recapture = 150,000 - 109,325 = $40,675



Recapture = 150,000 - 109,325 = $40,675

$40,675 taxed as ordinary income


Ex: Suppose K-Corp were able to sell this same loader for $ 275,000.

Capital Gain = 275,000 - 250,000 = $25,000

Depr. Recapture = 250,000 - 109,325 = $140,675


Ex: Suppose K-Corp were able to sell this same loader for $ 275,000.

Capital Gain = 275,000 - 250,000 = $25,000

Depr. Recapture = 250,000 - 109,325 = $140,675

$ 25,000 taxed at 28%$140,675 taxed at 35%


Non residential or commercial real propertyIf ThenFt > P Ft - P is section 1231 capital gain

Bt < Ft < P Ft - Bt recaptured as ordinary income

Ft < Bt Bt - Ft is section 1231 loss


Non residential or commercial real propertyIf ThenFt > P Ft - P is section 1231 capital gain

Bt < Ft < P Ft - Bt recaptured as ordinary income


Residential or Commercial real propertyIf ThenBt < Ft Ft - Bt is section 1231 gain


Investment Tax Credit

Stimulate investment by providing reduced taxation in year in which asset is placed in service.

On-again, off-again Repealed in 1985 with tax rate 46%

35%

Investment Tax Credit

K-Corp purchases a CNC machine for $100,000.

ITC = 100,000(0.10) = 10,000

Initial Cost Basis (for depreciation) is reduced 5%

Padj = 100,000(.95) = 95,000

engm 661 engineering economics depreciation & taxes

Documents