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
50,000 < TI < 75,000
75,000 < TI < 100,000
100,000 < TI < 335,000
335,000 < TI < 10,000,000
10,000,000 < TI < 15,000,000
15,000,000 < TI < 18,333,333
TI > 18,333,333
0.15
0.25
0.34
0.39
0.34
0.35
0.38
0.35
.15(TI)
7,500 + .25(TI - 50,000)
13,750 + .34(TI - 75,000)
22.250 + .39(TI - 100,000)
113,900 + .35(TI - 335,000)
3,400,000 + .35(TI - 10,000,000)
5,150,000 + .38(TI - 15,000,000)
.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
Depreciation
Interest
Taxable Income
$ 5,000,000
- 800,000
- 1,500,000
$ 2,700,000
Corporate Tax
Gross Income
Depreciation
Interest
Taxable Income
$ 5,000,000
- 800,000
- 1,500,000
$ 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
$ 5,000,000
Depreciation
800,000
Interest
- 1,500,000
Before Tax Cash Flow $ 2,700,000
After Tax Cash Flow
Gross Income
Interest
Before Tax Cash Flow
Less Tax
After Tax Cash Flow
$ 5,000,000
- 1,500,000
$3,500,000
941,650
$ 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 Cost
n = Useful Life
s = Salvage Value year n
Dt = Depreciation Allowance in year t
Bt = Unrecovered Investment (Book Value) in
year t
Then
Dt = (P - S) / n
Bt = P - [ (P - S) / n ] t
Ex: Straight Line Depr.
Let
P = $100,000
n = 5 years
s = $ 20,000
Then
Dt = (P - S) / n
= $ 16,000
B5 = 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
In declining balance, we write off a constant
% , p, of remaining book value
B2 = B1 - D2 = P(1-p) - pB1
Declining Balance
In declining balance, we write off a constant
% , p, of remaining book value
B2 = B1 - D2 = P(1-p) - pB1
= P(1-p) - pP(1-p)
Declining Balance
In declining balance, we write off a constant
% , p, of remaining book value
B2 = B1 - D2 = P(1-p) - pB1
= P(1-p) - pP(1-p)
= P(1-p)[1 - p]
Declining Balance
In declining balance, we write off a constant
% , p, of remaining book value
B2 = B1 - D2 = P(1-p) - pB1
= P(1-p) - pP(1-p)
= P(1-p)[1 - p]
= P(1-p)2
Declining Balance
In declining balance, we write off a constant
% , p, of remaining book value
B2 = B1 - D2 = P(1-p) - pB1
= 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 = ?
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
B1 = 100,000 - 40,000 = $ 60,000
D5 = ? ,
B5 = ?
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
B1 = 100,000 - 40,000 = $ 60,000
D2 = (2/5)(60,000) = $ 24,000
D5 = ? ,
B5 = ?
Ex: Declining Balance (cont)
= p [ P (1 - p) t - 1]
D5 = .4(100,000)(.6) 4
= $ 5,184
Dt
Bt = P (1 - p) t
B5 = 100,000(.6) 5
= $ 7,776
Ex: Declining Balance (cont)
= p [ P (1 - p) t - 1]
D5 = .4(100,000)(.6) 4
= $ 5,184
Dt
Bt = P (1 - p) t
B5 = 100,000(.6) 5
= $ 7,776
Note that Declining
Balance will never
depreciate book value
to $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
t
Dt
Save
0
1
20,000
2
20,000
3
20,000
4
20,000
5
20,000
6
0
Sum =
100,000
Present Value =
8,000
8,000
8,000
8,000
8,000
0
40,000
30,326
DDB
Dt
Tax
Save
40,000
24,000
14,400
8,640
4,320
8,640
100,000
16,000
9,600
5,760
3,456
1,728
3456
40,000
32,191
DDB/SL Conversion
(Salvage = $0)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
100,000
20,000
2
3
4
5
6
DDB
Dt
Bt
40,000
100,000
60,000
DDB/SL Conversion
(Salvage = $0)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
100,000
20,000
2
60,000
15,000
3
4
5
6
DDB
Dt
Bt
40,000
24,000
100,000
60,000
36,000
DDB/SL Conversion
(Salvage = $0)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
100,000
20,000
2
60,000
15,000
3
36,000
12,000
4
5
6
DDB
Dt
Bt
40,000
24,000
14,400
100,000
60,000
36,000
21,600
DDB/SL Conversion
(Salvage = $0)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
100,000
20,000
2
60,000
15,000
3
36,000
12,000
4
21,600
10,800
5
6
DDB
Dt
Bt
40,000
24,000
14,400
8,640
100,000
60,000
36,000
21,600
10,800
DDB/SL Conversion
(Salvage = $0)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
100,000
20,000
2
60,000
15,000
3
36,000
12,000
4
21,600
10,800
5
10,800
10,800
6
0
DDB
Dt
Bt
40,000
24,000
14,400
8,640
4,320
100,000
60,000
36,000
21,600
10,800
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
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
180,000
2
3
4
5
6
DDB
Dt
Bt
180,000
Class Problem (p = 1.5/5 =
.3)
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
180,000
36,000
2
3
4
5
6
DDB
Dt
Bt
54,000
180,000
126,000
Class Problem
DDB/Straight Line Conversion
SL
Bt-1
Dt
t
0
1
180,000
36,000
2
126,000
31,500
3
88,200
29,400
4
58,800
29,400
5
29,400
29,400
6
0
DDB
Dt
Bt
54,000
37,800
26,460
17,640
8,820
180,000
126,000
88,200
58,800
29,400
0
DDB/SL Conversion
(Half-Year Convention)
DDB/Straight Line Conversion/Half-Year
SL
DDB
Bt-1
Dt
Dt
t
0
1
100,000
10,000
20,000
2
3
4
5
6
Bt
100,000
80,000
DDB/SL Conversion
(Half-Year Convention)
DDB/Straight Line Conversion/Half-Year
SL
DDB
Bt-1
Dt
Dt
t
0
1
100,000
10,000
20,000
2
80,000
17,778
32,000
3
48,000
13,714
19,200
4
28,800
11,520
11,520
5
17,280
11,520
6,912
6
Bt
100,000
80,000
48,000
28,800
17,280
5,760
DDB/SL Conversion
(Half-Year Convention)
DDB/Straight Line Conversion/Half-Year
SL
DDB
Bt-1
Dt
Dt
t
0
1
100,000
10,000
20,000
2
80,000
17,778
32,000
3
48,000
13,714
19,200
4
28,800
11,520
11,520
5
17,280
11,520
6,912
6
5,760
5,760
2,304
Bt
100,000
80,000
48,000
28,800
17,280
5,760
0
Class Problem
A $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
DDB/Straight Line Conversion/Half-Year
SL
DDB
Bt-1
Dt
Dt
t
0
1
2
3
4
5
6
Bt
180,000
180,000
Solution
DDB/Straight Line Conversion/Half-Year
SL
DDB
Bt-1
Dt
Dt
t
0
1
2
3
4
5
6
180,000
144,000
86,400
51,840
31,104
10,368
18,000
32,000
24,686
20,736
20,736
10,368
36,000
57,600
34,560
20,736
12,442
4,147
Bt
180,000
144,000
86,400
51,840
31,104
10,368
0
MACRS Tables
MACRS Percentages 3, 5, 7, & 10 are 200% DB/SL
15 & 20 are 150% DB/SL
t
3-Yr.
5-Yr.
7-Yr.
10-Yr.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
33. 33%
44. 45%
14. 81%
74. 10%
20. 00%
32. 00%
19. 20%
11. 52%
11. 52%
5. 76%
14. 29%
24. 49%
17. 49%
12. 49%
8. 93%
8. 92%
8. 93%
4. 46%
10. 00%
18. 00%
14. 40%
11. 52%
9. 22%
7. 37%
6. 55%
6. 55%
6. 56%
6. 55%
3. 28%
15-Yr.
20-Yr.
5. 00%
9. 50%
8. 55%
7. 70%
6. 93%
6. 23%
5. 90%
5. 90%
5. 91%
5. 90%
5. 91%
5. 90%
5. 91%
5. 90%
5. 91%
2. 95%
3. 75%
7. 22%
6. 68%
6. 18%
5. 71%
5. 29%
4. 88%
4. 52%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
4. 46%
2. 23%
MACRS Percentages 3,5,7, & 10 are 200% DB/SL
15 & 20 are 150% DB/SL
t
3-Yr.
5-Yr.
7-Yr.
10-Yr.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
33.33%
44.45%
14.81%
74.10%
20.00%
32.00%
19.20%
11.52%
11.52%
5.76%
14.29%
24.49%
17.49%
12.49%
8.93%
8.92%
8.93%
4.46%
10.00%
18.00%
14.40%
11.52%
9.22%
7.37%
6.55%
6.55%
6.56%
6.55%
3.28%
15-Yr.
20-Yr.
5.00%
9.50%
8.55%
7.70%
6.93%
6.23%
5.90%
5.90%
5.91%
5.90%
5.91%
5.90%
5.91%
5.90%
5.91%
2.95%
3.75%
7.22%
6.68%
6.18%
5.71%
5.29%
4.88%
4.52%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
4.46%
2.23%
Modified Accelerated Cost
Property Classes
3 yr. - useful life < 4 yrs.
autos, tools
5 yr. - 4 yrs. < useful life < 10 yrs.
office epuipment, computers, machinery
7 yr. - 10 < UL < 16
office furniture, fixtures, exploration
10 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 generation
27.5 - residential rental property
31.5 - non-residential real property
Depreciation
class (3, 5, 7, 10 yr.) uses 200% declining balance
switching to straight-line @ optimal year
class (15, 20) 150% DB switch to SLD
class (27.5, 31.5) use straight-line
After Tax Cash Flow
Formulas
BTCF = Before Tax Cash Flow
= Revenues - Expenses
After Tax Cash Flow
Formulas
BTCF = Before Tax Cash Flow
= Revenues - Expenses
TI = Taxable Income
= Cash Flow - Interest - Depreciation
After Tax Cash Flow
Formulas
BTCF = Before Tax Cash Flow
= Revenues - Expenses
TI = Taxable Income
= Cash Flow - Interest - Depreciation
Tax = TI * Tax Rate
After Tax Cash Flow
Formulas
BTCF = Before Tax Cash Flow
= Revenues - Expenses
TI = Taxable Income
= Cash Flow - Interest - Depreciation
Tax = TI * Tax Rate
ATCF = After Tax Cash Flow
= BTCF - Tax
Ex: After Tax Cash Flow
Tax Rate =
MARR =
t
0
1
2
3
4
5
6
7
NPV =
34%
20%
BTCF MACRS % Depr.
(82,000)
23,500
20.0%
16,400
23,500
32.0%
26,240
23,500
19.2%
15,744
23,500
11.5%
9,446
23,500
11.5%
9,446
23,500
5.8%
4,723
28,500
0
$4,103
Taxable
Income
Tax
7,100
(2,740)
7,756
14,054
14,054
18,777
28,500
2,414
(932)
2,637
4,778
4,778
6,384
9,690
ATCF
(82,000)
21,086
24,432
20,863
18,722
18,722
17,116
18,810
($7,854)
Borrowed Money
Amt. Financed
Interest
Period of Loan
Payment
Tax rate
34%
MARR
20%
t
0
1
2
3
4
5
6
Project
Cash Flow
(500,000)
150,000
150,000
150,000
150,000
150,000
175,000
$210,000
16%
3
#NUM!
Loan
Payment
210,000
70,000
70,000
70,000
Year
0
1
2
3
Principle
Interest
70000
70000
70000
33600
22400
11200
Loan
Before Tax MACRS
Interest Cash Flow
%
(290,000)
33,600
46,400
20.0%
22,400
57,600
32.0%
11,200
68,800
19.0%
150,000
12.0%
150,000
12.0%
175,000
6.0%
Total Pmt. Loan Bal.
$210,000
103600 $140,000
92400
$70,000
81200
$0
MACRS
deduct
Taxable
Income
100,000
160,000
95,000
60,000
60,000
30,000
16,400
(32,400)
43,800
90,000
90,000
145,000
Tax
34%
After Tax
Cash Flow
(290,000)
5,576
40,824
(11,016) 68,616
14,892
53,908
30,600
119,400
30,600
119,400
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
Tax Rate =
MARR =
t
0
1
2
3
4
5
6
7
NPV =
34%
20%
BTCF MACRS %
(800,000)
200,000
200,000
200,000
200,000
200,000
200,000
300,000
($51,173)
Depr.
Taxable
Income
Tax
ATCF
(800,000)
Solution
Tax Rate =
MARR =
t
0
1
2
3
4
5
6
7
NPV =
34%
20%
BTCF MACRS %
(800,000)
200,000
20.0%
200,000
32.0%
200,000
19.2%
200,000
11.5%
200,000
11.5%
200,000
5.8%
300,000
($51,173)
Depr.
Taxable
Income
Tax
160,000
256,000
153,600
92,160
92,160
46,080
0
40,000
(56,000)
46,400
107,840
107,840
153,920
300,000
13,600
(19,040)
15,776
36,666
36,666
52,333
102,000
ATCF
(800,000)
186,400
219,040
184,224
163,334
163,334
147,667
198,000
($136,824)
Residential Rental
MACRS - ADS Election
Straight Line with either a half-year or halfmonth 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
= $21,428 / 2
= $10,714
n = 2, . . ., 14
n = 1, 15
Units of Production
Method
Allows for equal depreciation for each unit of
output
Ut
Dt  ( P  F )
U
where
Ut = units produced during the year
U = total units likely to be produced during
life
(P-F) = depreciable amount allowed
Operating Day Method
Allows for equal depreciation for each unit of
output
Qt
Dt  ( P  F )
Q
where
Qt = total hours used during the year
Q = total hours available during the year
(P-F) = depreciable amount allowed
Income Forecast Method
Allows for equal depreciation for each unit of
output
Rt
Dt  ( P  F )
R
where
Rt = rent income earned during the year
R = total likely rent to be earned during life
(P-F) = depreciable amount allowed
Depletion Method
Allows for equal depreciation for each unit of
output
Vt
Dt  ( P  F )
V
where
Vt = volume extracted during the year
V = total volume available in reserve
(P-F) = depreciable amount allowed
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:
10,000,000
Dt  $27,000,000
10 x 10,000,000
 $2,700,000
Percentage Depletion
Depletion is taken as a constant percentage of
gross income
Allowable Percentages
Oil/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
Depletion
= 1.5(10,000,000)
= 15,000,000
= 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.
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
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
$40,675 taxed as ordinary income
Depreciation Recapture
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
Depreciation Recapture
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%
Depreciation Recapture
Non residential or commercial real property
If
Then
Ft > P
Bt < Ft < P
Ft < Bt
Ft - P is section 1231 capital gain
Ft - Bt recaptured as ordinary income
Bt - Ft is section 1231 loss
Depreciation Recapture
Non residential or commercial real property
If
Then
Ft > P
Bt < Ft < P
Ft < Bt
Ft - P is section 1231 capital gain
Ft - Bt recaptured as ordinary income
Bt - Ft is section 1231 loss
Residential or Commercial real property
If
Then
Bt < Ft
Ft < Bt
Ft - Bt is section 1231 gain
Bt - Ft is section 1231 loss
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
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