Chapter 7 - Depreciation and Income Taxes
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Depreciation Concepts and Terminology
Taxes are important for engineering economic analysis
because they represent significant cash outflow that may
alter the final decision (decision to invest or not to invest,
selection of the best alternative etc.).
One important concept for tax calculations is “depreciation"
which is the reduction in value of an asset (or a property)
over time and usage.
In the accounting terms, depreciation is defined as the
allocation of an asset’s cost over its useful (or depreciable)
life.
It is important to notice that the depreciation is a noncash
cost. Therefore, the depreciation does not a part of actual
cash flow and is only used for tax calculation.
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Depreciation is not allowed for all properties. A depreciable
property must:
be used in business or produce income,
have a determinable useful life,
lose value over its useful life,
not be inventory or investment.
Classification of properties:
Tangible (can be seen or touched)
real property (land, buildings, manufacturing facilities, etc. )
personal property (machinery, vehicles, equipment, etc.)
Intangible (patents, copyrights, trademarks, etc.)
Almost all tangible and intangible property can be
depreciated, except land (determinable life?, lose in value
over time?).
We will focus on tangible properties.
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A company can begin to depreciate property it owns when
the property is placed in service for use in the business
and for the production of income. Property is considered to
be placed in service when it is ready and available for a
specific use, even if it is not actually used yet.
Depreciation stops when the cost of placing an asset in
service has been recovered or when the asset is sold,
whichever occurs first.
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Depreciation Methods and Related Time Periods:
Before 1981: Several methods could be used for
depreciating property placed in service before 1981. The
primary methods used were Straight-Line (SL),
Declining-Balance (DB), and Sum of the Year Digits (SYD).
Those methods are collectively referred as the classical or
historical methods of depreciation.
After 1980 and before 1987: For federal income taxes,
tangible property placed in service during this period must
be depreciated by using the Accelerated Cost Recovery
System (ACRS).
After 1986: Modified Accelerated Cost Recovery System
(MACRS) is required for the depreciation of tangible
property placed in service after 1986.
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Some more terminology:
Unadjusted cost basis: The initial cost of acquiring the asset
(includes the purchase cost, delivery and installation fees).
Adjusted cost basis: Cost basis adjusted with any
improvements and losses.
Book value (BV): The value of an asset on the accounting
records of a company after the total amount of depreciation
deduction to date has been subtracted from its
adjusted cost basis.
BVk = adjusted cost basis −
k
X
(annual depreciation deduction in year j)
j=1
BVk = adjusted cost basis −
k
X
dj
j=1
where BVk is the BV at the end of year k.
In other words, BV is the asset’s remaining unrecovered cost.
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Market value (MV): The amount that will be earned if the asset
is sold in an open market. MV can be different than BV. For
example, IT equipment usually has a MV much lower than its
BV due to rapidly changing technology.
Useful life: The expected (estimated) period that a property
will be used in a trade or business to produce income. It is not
how long the property will last but how long the owner expects
to productively use it.
Salvage value (SV): The estimated value of a property at the
end of its useful life.
Recovery period: The number of years over which the cost
basis of a property is recovered through the accounting
process. For the classical methods of depreciation, recovery
period = useful life. Under MACRS, the way to determine the
recovery period will be discussed later in this chapter.
Recovery rate: A percentage (expressed in decimal form) for
each year of the MACRS recovery period that is utilized to
compute an annual depreciation deduction.
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Classical depreciation methods
Straight-Line (SL) method:
Depreciation deduction is constant each year over the
depreciable (useful) life of an asset.
Notation:
N = depreciable life of the asset in years (useful life);
B = adjusted cost basis;
SVN
dk∗
= estimated salvage value at the end of year N;
= cumulative depreciation through year k.
dk = d = (B − SVN )/N,
dk∗ = k · d
BVk = B −
BVN
for 1 ≤ k ≤ N,
dk∗
= B − k · d,
= SVN .
The asset is fully depreciated under the SL method.
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Example 7.1
An asset has an adjusted cost basis of $25,000 with a $2,000
salvage value at the end of its depreciable life of 8 years.
EOY k
0
1
2
3
4
5
6
7
8
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25,000−2,000
8
dk
2,875
2,875
2,875
2,875
BVk
$25,000
22,125
19,250
16,375
13,500
= 2,875
2,875
2,875
2,875
(25, 000 − 5(2, 875)) = 10,625
7,750
4,875
2,000
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Declining-Balance (DB) method:
DB method is also known as constant-percentage method or
the Matheson formula.
Annual depreciation deduction is a constant percentage of the
BV at the beginning of the year.
Let R be the ratio of depreciation.
For R = k/N, it is known as (k × 100)% DB method.
For example, R=2/N corresponds to 200% declining balance
(Double-Declining-Balance, DDB), which is twice the SL rate of
1/N.
d1 = B(R),
dk = B(1 − R)k−1 (R),
k
P
dk∗ =
dj = B[1 − (1 − R)k ],
j=1
BVk = B(1 − R)k ,
BVN = B(1 − R)N .
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Back to Example 7.1, use 200% DB method (i.e. R=2/N).
EOY k
0
1
2
3
4
5
6
7
8
dk
6,250.00
4,687.50
3,515.63
2,636.72
1,977.54
1,483.15
1,112.37
834.27
BVk
$25000
18,750.00
14,062.50
10,546.88
7,910.16
5,932.62
4,449.46
3,337.10
2,500.00
Sample calculation:
R=2/8=0.25,
d5 = $25, 000(1 − 0.25)4 (0.25) = $1, 977.54
BV5 = $25, 000(1 − 0.25)5 = $5, 932.62.
SV8 =$2,000 < $2,500. The asset is not fully depreciated.
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DB with switchover to SL:
To fully depreciate the asset and take enough depreciation in
Example 7.1, one modification can be made: whenever the
depreciation deduction in a year by DB method is lower than
that of SL method, switch to SL method.
EOY, k
BOY BV (BVk−1 )
DDB
1
2
3
4
5
6
7
8
25,000
18,750
14,062.50
10,546.87
7,910.15
5,932.61
4,449.46
3,224.73
6,250
4,687.50
3,515.63
2,636.72
1,977.54
1,483.15
1,112.36
834.27
∗ 18,750−2,000
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k−1 −SVN
SL= BV
N−(k−1)
2,875
2,392.86∗
2,010.42
1,709.37
1,477.54
1,310.87
1,224.73
1,224.73
Depreciation
amount
6,250.00
4,687.50
3,515.63
2,636.72
1,977.54
1,483.15
1,224.73
1,224.73
= 2, 392.86.
BV8 = $3,244.73-$1,244.73=$2,000=SV8 .
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Sum-of-years’-digits method:
dk =
dN∗
=
(N − k + 1)
h
i (B − SVN )
N(N+1)
2
N
X
dk
k=1
=
B − SVN X
N(N − k + 1)
N(N+1)
2
k=1
= B − SVN
BVN
= B − dN∗ = SVN
The asset is fully depreciated under sum-of-years’-digits
method.
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Back to Example 7.1. N=8,
sum-of-years’-digits=1+2+3+...+8= 8(9)
2 = 36.
d1 =
d2 =
d3 =
8−1+1
36 (25, 000
8−2+1
36 (25, 000
8−3+1
36 (25, 000
− 2, 000) = $5, 111.1
8−8+1
36 (25, 000
− 2, 000) = 638.8
− 2, 000) = 4, 472.2
− 2, 000) = 3, 833.3
...
d8 =
Depreciation deduction decreases at a constant amount of
$638.8 from one year to the next; arithmetic series.
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Units-of-Production method:
Depreciation is based on activity (number of units produced)
rather than time.
Depreciation per unit of production= (B-SVN )/Estimated lifetime
production units.
Example 7.2: An equipment has a cost basis of $50,000 and a
salvage value of $10,000 after 30,000 hours of use. a) What is
the depreciation rate per hour of use? b) What is the BV of the
equipment after 10,000 hours of operation?
a) ($50,000-$10,000)/30,000=$1.33 per hour.
b) $50,000-10,000(1.33)=$36,700.
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Modified Accelerated Cost Recovery System
(MACRS)
MACRS applies to most tangible depreciable property
placed in service after Dec. 31, 1986. Except those
properties which to be depreciated under a method that is
not based on a term of years (units-of-production method)
and intangible property.
Under MACRS, SVN is assumed to be 0 and useful life
estimates are not used directly in calculating depreciation
amounts.
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MACRS consists of two subsystems:
1. General Depreciation System (GDS):
Personal property is classified into six classes: 3-, 5-, 7-,
10-, 15-, 20-year property, which are the GDS recovery
periods.
3-, 5-, 7-, and 10-year: 200% DB method with a switchover
to the SL method when that method provides a greater
deduction.
15- and 20-year: 150% DB method with a switchover to the
SL method when that method provides a greater deduction.
Real property is classified into two classes: nonresidential
and residential. The SL method is used over the GDS
recovery period.
Nonresidential real property: the recovery period is 39 years
(31.5 years if placed in service before May 13, 1993).
Residential real property: the recovery period is 27.5 years.
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2. Alternative Depreciation System (ADS):
The SL method is used over the ADS recovery period.
Property that is placed in any tax-exempt use and property
used predominantly outside the U.S. must be depreciated
under ADS.
Any property that qualifies under GDS can be depreciated
under ADS, if elected. However, once ADS is elected for an
asset, it cannot switch back to the GDS.
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MACRS uses a half-year time convention:
All assets placed in service during the year are treated as
if use began in the middle of the year, and one-half year of
depreciation is allowed.
If the asset is disposed of before the full recovery is used,
then only half of the normal depreciation deduction can be
taken for that year.
The depreciation deduction (dk ) for an asset under GDS is
computed with
dk = B(rk ), k = 1, .., N + 1. (N is the recovery period. Recovery
extends to year N+1 because of the half-year time convention.)
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Example 7.3
Use MACRS(GDS) to depreciate a bus with a cost basis of
$95,000 which will be sold in year 5. The GDS recovery period
is given as 5 years.
EOY k
0
1
2
3
4
5
Factor
0.2000
0.3200
0.1920
0.1152
0.0576
dk
19,000
30,400
18,240
10,944
5,472
BVk
$95,000
76,000
45,600
27,360
16,416
10,944
Note that from Table 7.3, r5 = 0.1152. The asset was sold
before the full recovery period is used, therefore only half of the
normal depreciation deduction can be taken for year 5.
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Income Taxes
Taxable Income (TI) = Gross income − all expenses (except
capital investments) − depreciation deductions.
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Example 7.4
A company generated $200M of gross income from sales and
incurred $140M of operating expenses this year. A capital
investment of $1.5M was made to purchase a specialized
equipment at the beginning of the year, and the company
deducted $0.1M as the annual depreciation charge for this
equipment. What is the taxable income of the company?
TI = $200M − $140M − $0.1M = $59.9M.
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Income tax rates:
Income tax on
$60,000=($50,000)(0.15)+($10,000)(0.25)=$10,000.
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Gain (loss) on the disposal of an asset
When a depreciable asset is disposed (sold), the MV is seldom
equal to its BV. Suppose the asset is disposed at year N.
If MVN > BVN , a gain (MVN − BVN ) is obtained on the disposal
(selling) of the asset. This is called depreciation recapture, or
gain on disposal. Tax must be paid on the depreciation
recapture.
The opposite situation, i.e., MVN < BVN , is called
loss on disposal, and it results in earning tax credit.
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After-tax economic analysis
Similar profitability analysis as before, but use after-tax cash
flows (ATCF) instead of before-tax cash flows (BTCF), and use
an after-tax MARR. Let,
Rk = cash inflow in year k.
Ek = cash outflow in year k.
dk = sum of all noncash, or book, cost in year k, such as
depreciation deduction.
t = annual income tax rate.
Tk = income tax liability (or credit) for year k.
BTCFk = Before-tax cash flows in year k.
ATCFk = After-tax cash flows in year k.
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Taxable income = Rk − Ek − dk
Tk = −t(Rk − Ek − dk ).
Therefore, when Rk > (Ek + dk ), a tax liability (i.e. negative cash
flow) occurs. When Rk < (Ek + dk ), a decrease in the tax
amount (a credit) occurs.
BTCFk = Rk − Ek
ATCFk = BTCFk + Tk
= (Rk − Ek ) − t(Rk − Ek − dk )
= (1 − t)(Rk − Ek ) + tdk
After-tax MARR≈Before-tax MARR × (1-income tax rate).
This approximation is exact if no deduction involved.
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The last row in Figure 7-4 is due to "gain (loss) on disposal" of
the asset.
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Example 7.5
Suppose that an asset with a cost basis of $100,000 is
expected to produce net revenues of $30,000 per year during
the six-year period, and its terminal market value is negligible.
If the effective income tax rate is 40%, how much can a firm
afford to spend for this asset and still earn the after-tax MARR
of 10% per year. The asset is being depreciated under the ADS
method over a recovery period of five years.
EOY k
0
1
2-5
6
BTCFk
-$100,000
30,000
30,000
30,000
dk
TI
Tk
10,000
20,000
10,000
20,000
10,000
20,000
-8,000
-4,000
-8,000
ATCFk
-100,000
22,000
26,000
22,000
PW(10%) of
ATCF=-100,000+22,000[(P/F,10%,1)+(P/F,10%,6)]+
26,000(P/A,10%,4)(P/F,10%,1)=$7,344.9.
You can afford at most $107,344.9.
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Suppose that the asset will be sold at a market value of
$25,000 at the end of six years. Modify the ATCF analysis table.
EOY k
0
1
2-5
6
6
BTCFk
-$100,000
30,000
30,000
30,000
25,000
dk
TI
Tk
10,000
20,000
10,000
20,000
10,000
20,000
25,000
-8,000
-4,000
-8,000
-10,000
ATCFk
-100,000
22,000
26,000
22,000
15,000
The asset is fully depreciated at the end six years. BV6 = 0.
Since MV6 = $25, 000 > BV6 = $0, a gain (MV6 - BV6 ) is
obtained on the disposal of the asset. Tax must be paid on the
depreciation recapture.
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Example 7.6
The Ajax Semiconductor company is attempting to evaluate the
profitability of adding another integrated circuit production line
to its present operations. The company would need to
purchase two or more acres of land for $275K. The facility
would cost $60M and have no MV at the end of five years. The
facility could be depreciated, using a GDS recovery period of
five years. An increment of working capital would be required
for adding the new production line, and its estimated amount is
$10M. Both land and working capital are considered as
non-depreciable assets. The new production line is expected to
generate additional $30M gross income each year for five
years. On the other hand, the operating expenses of the new
production line are estimated to be $8M per year for five years.
The firm’s effective income tax rate is 40%.
a) Set up a table and determine the ATCF for this project.
b) Is the investment worthwile when the after-tax MARR is
12% per year?
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Solution:
a)
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b)
BV5 = 60, 000, 000 −
5
X
dk
k=1
= 6, 912, 000 >
selling price = 0.
Therefore, a loss on disposal of $6,912,000 will be claimed at
the end of year 5.
Tax credit = (0.4)(6,912,000)=$2,764,800.
PW(12%) of ATCF = $936,715 > 0
Therefore, this investment is recommended.
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Example 7.7
Two fixtures are being considered for a particular job in a
manufacturing firm:
Capital investment
Annual operating expense
Useful life
Market value
at the end of the useful life
Depreciation method
Fixture X
$30,000
$3,000
6 years
$6,000
Fixture Y
$40,000
$2,500
8 years
$4,000
SL to zero BV, 5-year
GDS, 5-year.
If the tax rate is 50% and the after-tax MARR is 8% per year,
which of the two fixtures should be recommended?
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Solution:
Assume repeatability and find the after-tax AW(8%) of each
alternative.
Fixture X:
EOY k
0
1-5
6
6
BTCF
-$30,000
-3,000
-3,000
6,000
dk
TI
Tk
6,000
-9,000
-3,000
6,000
4,500
1,500
-3,000
ATCF
-$30,000
1,500
-1,500
3,000
After-tax AW(8%) = -$4,989.
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Fixture Y:
EOY k
0
1
2
3
4
5
6
7
8
8
BTCF
-$40,000
-2,500
-2,500
-2,500
-2,500
-2,500
-2,500
-2,500
-2,500
4,000
dk
TI
Tk
8,000
12,800
7,680
4,608
4,608
2,304
-10,500
-15,300
-10,180
-7,108
-7,108
-4,804
-2,500
-2,500
4,000
5,250
7,650
5,090
3,554
3,554
2,402
1,250
1,250
-2,000
ATCF
-$40,000
2,750
5,150
2,590
1,054
1,054
-98
-1,250
-1,250
2,000
After-tax AW(8%) = -$5,199. Select Fixture X.
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