IE463 – Chapter 5
DEPRECIATION AND
INCOME TAXES
Depreciation
Depreciation is the decrease in the value
of physical properties with passage of time
Because, depreciation is a non-cash cost
that affects income taxes we must
consider depreciation properly, when
making After-Tax Engineering Economy
studies
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Depreciable Property
It is a property for depreciation is allowed under
governmental income tax laws and regulations
In general, property is depreciable if it meets the
following basic requirements:
It
must be used in business or held to produce
income.
It must have a determinable useful life, and the life
must be longer than one year.
It must be something that wears out, decays, gets
used up, becomes obsolete, or loses value from
natural causes.
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Basic Terminology
Depreciation = an annual non-cash charge against
income. It represents an estimate of the dollar cost of
fixed assets used in the production of a good or service.
Cost Basis (B) = actual cash cost plus book value of
trade-in (if any) plus costs of making asset serviceable
(e.g., installation).
Book Value (BVk) = value of asset as shown on the
accounting records. Represents amount of money still
invested in the property. BVk = book value at EOY k
SVN = estimated salvage value in year N (used in
depreciation calculations where applicable)
MVN = market (resale) value at EOY N from the disposal
of an asset
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STRAIGHT-LINE (SL) METHOD
A constant amount is depreciated each year over
the asset's life.
N = depreciable life of the asset in years.
dk = annual depreciation deduction in year k
dk = (B - SVN ) / N for k = 1, 2, ..., N
dk* = cumulative depreciation through year k.
dk* = k x dk
BVk = B - dk *
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DECLINING BALANCE (DB) METHOD
Annual depreciation is a constant percentage of
the asset's value at the BOY
R = 2/N
200% declining balance
R = 1.5/N
150% declining balance
d1 = B x R
dk = B(1-R)k-1 (R) = BVk-1 (R)
dk* = B[1 - (1 - R)k ]
BVk = B(1 - R)k
BVN = B(1 - R)N
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SL and DB Example
The La Salle Bus Company has decided to
purchase a new bus for $85,000, with a trade-in
of their old bus. The old bus has a trade-in value
of $10,000. The new bus will be kept for 10
years before being sold. Its estimated salvage
value at that time is expected to be $5,000.
Compute the following quantities using (a) the
straight-line method, (b) the 200% declining
balance method
depreciation deduction in the first year and the fourth
year
cumulative depreciation through year four
book value at the end of the fourth year
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Cost basis:
B = $10,000 + $85,000 = $95,000
trade-in value
SV10
cash-cost
Deduction amounts are fixed for SL:
dk =
95,000 - 5,000
= $9,000 for k = 1 to 10
10
N
Deduction ratios are fixed for DB:
R=
2
= 0.2 , thus, (d k = 0.2 × BVk -1 )
10
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200% DB
8
Straight Line Method (k = 1 to 4)
EOY, k
0
1
2
3
4
dk
0
9,000
9,000
9,000
9,000
BVk
95,000
86,000
77,000
68,000
59,000
= 95,000 – 9,000
= 86,000 – 9,000
= 77,000 – 9,000
= 68,000 – 9,000
d4 = d1
d4* = 4 x 9000 = 36,000
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BV4 = 95,000 – 36,000
9
Straight Line Method (k = 5 to 10)
EOY, k
5
6
7
8
9
10
dk
9,000
9,000
9,000
9,000
9,000
9,000
BVk
50,000
41,000
32,000
23,000
14,000
5,000
= 59,000 – 9,000
= 50,000 – 9,000
= 41,000 – 9,000
= 32,000 – 9,000
= 23,000 – 9,000
= 14,000 – 9,000
SV10 = B – N x dk
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200% Declining Balance Method
(k = 1 to 4)
EOY, k
0
1
2
3
4
dk
0
19,000
15,200
12,160
9,728
d1 = BV0 x R = B x 0.2
= 95,000 x 0.2 = 19,000
BVk
95,000
76,000
60,800 BV1 = BV0 – d1 = B – d1
48,640 = 95,000 – 19,000
38,912
BV4 = B – d4*
d4* = 19,000 + ... + 9,728
= 56,088
d4 = 48,640 x 0.2
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200% Declining Balance Method
(k = 5 to 10)
EOY, k
5
6
7
8
9
10
dk
7,782
6,226
4,981
3,985
3,188
2,550
BVk
31,130
24,904
19,923
15,938
12,750
10,200
BV10 = B – d10* = BV9 – d10
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SL vs. DB
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Consideration of Income Taxes in EE
Income tax represents a significant
cash outflow that we cannot ignore
Notation:
Rk = gross revenues in year k
Ek = operating expenses in year k plus
interest paid on borrowed capital
dk = depreciation allowance for year k
t = effective income tax rate used for
computing income taxes
Tk = income tax liability for year k
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General Tax Procedure
BTCF = Before tax cash flow = R – E
NIBT = Net income before tax = R – E – d
T = tax liability = t ( NIBT ) = t (R – E – d)
NIAT = Net income after tax = NIBT – T
= (R – E – d) – t (R – E – d) = (1 – t)(R – E – d)
ATCF = After tax cash flow = NIAT + d
ATCF = BTCF – t (R – E – d)
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General Tax Procedure - Example
a)
b)
c)
d)
You invested $113,028 on an asset with the
depreciable life of 10 years. You can earn $30,000 per
year from this investment for 10 years. Asset has a
negligible or zero MV at the end of its useful life.
Published income tax rate is 40% on annual taxable
income (NIBT). Use after-tax MARR of 15% per year,
and straight line depreciation method.
NIBT?
NIAT?
ATCF?
Is it profitable investment after taxes?
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Solution to (a), (b) and (c)
d = 113,028 / 10 = $11,303 (depreciation amount)
(+) Net income
30,000
(-) Deprecation
11,303
(a) NIBT
$18,697 (R – E – d)
(-) Income Tax (0.4) 7,479
(b) NIAT
$11,218 (1 – t)(R – E – d)
(+) Depreciation
11,303
(c) ATCF
$22,521
(1 – t)(R – E – d)+d
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Solution to part (d)
AW (15%) = ATCF – 113,028 (A/P, 15%, 10)
= 22,521 – 22,521 = $0
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Typical Before-Tax Cash Flow Diagram:
Typical After-Tax Cash Flow Diagram:
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After-Tax Cash Flow Analysis
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After-tax MARR
To perform an after-tax evaluation of a project's
after-tax cash flows, we must use an after-tax
MARR.
After - tax MARR
Before - tax MARR =
1 - effective income tax rate, t
Example: Suppose the before-tax MARR = 20% and t =
40%. What is the approximate after-tax MARR?
MARR BT =
MARR AT
⇒ MARR AT = 0.2 (1 - 0.4) = 0.12
1- t
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ATCF Analysis – Example
Investment
$10,000
Net Annual Receipts
$4,000/yr
Study Period
4 years
Market Value at EOY 4
$5,000
After-tax MARR
15%
Effective income tax rate
40%
Depreciable recovery period
5 years
Is this a worthwhile investment after taxes? Use
Straight Line Method for depreciation.
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Step 1: Find depreciation amounts
for the study period of 4 years:
dk =
10,000 - 0
= $2,000 for k = 1 to 5
5
EOY, k
0
1
2
3
4
dk
--2000
2000
2000
2000
BVk
10000
8000
6000
4000
2000
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Step 2: Determine the ATCF with tax
rate of 40%:
dk
TIk
Tk
(t = 0.4)
EOY, k
BTCFk
ATCFk
0
- 10,000
1
4000
2000
2000
-800
3200
2
3
4a
4b
4000
4000
4000
5000
2000
2000
2000
2000
2000
2000
3000
-800
-800
-800
-1200
3200
3200
3200
3800
- 10,000
MV4 – BV4 = 5000 – 2000
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Step 3: Use the ATCF to evaluate
this investment @ MARR = 15%:
After-tax cash flow diagram:
PW(15%) = – 10,000 + 3,200 (P/A, 15%, 3)
+ 7,000 (P/F, 15%, 4) = + $1,309
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Affordable Investment - Example
An asset purchased at the price of (I), and
it is depreciated for 10 years, with straight
line method. Estimated market value @
EOY 10 is zero, how much can you afford
to purchase this asset when it produces
annual $30,000 net income for 10 years.
Effective income tax is 40% and
MARRAT = 15%.
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Solve for I?
d = (I – 0) / 10
ATCF = 30000 – 0.4 (30000 – d)
AW(15%) = – I(A/P,15%,10)
+ 30000(1 – 0.4) + 0.4 (I/10)
AW(15%) = 0
I = 30000(1 – 0.4) / ((A/P,15%,10) – 0.04)
I = $113,028
0.1993
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