Engineering Economics

Depreciation and after-tax analysis

Use this for “A machine costs $84,000 installed and is worth $7,000 after 7 years. What is its straight-line depreciation, and its book value after year 4?”

Chart

BV(j) = −11,000j + 84,00001234567010,00020,00030,00040,00050,00060,00070,00090,000100,000Year j (yr)Book value BV ($)(4, 40,000)S
BV(4) = 40,000
This is an example. Type your own values to start a new problem.
$
Dⱼ
=
$
−
$
$
BVⱼ
=
$
C
−
j
×
$
Dⱼ

Formulas

Dⱼ
=
C
−
S
n
11,000
=
84,000 − 7,000
7
BVⱼ
=
C
−
j
×
Dⱼ
40,000 = 84,000 − 4 × 11,000

Assumptions

•
Straight line: the same depreciation Dⱼ every year of the recovery period n.
•
The cost basis C is the first cost with delivery and installation.
•
The book value reaches the salvage value S at year n and stays there after.

Step-by-step

WE KNOW
C = $84,000, S = $7,000, n = 7 yr, j = 4 yr
FIND
yearly depreciation (Dⱼ), book value (BVⱼ)
Step 1 · Find yearly depreciation (Dⱼ)
Use Dⱼ = (C − S) ÷ n
The
cost
less
the
salvage
is
what
wears
away;
spread
it
evenly
over
the
n
years.
Dⱼ
=
84,000 − 7,000
7
Dⱼ
=
77,000
7
Dⱼ = $11,000
Step 2 · Find book value (BVⱼ)
Use BVⱼ = C − j × Dⱼ
After
year
j
the
asset
has
lost
j
years
of
depreciation
from
its
cost
basis.
BVⱼ = 84,000 − 4 × 11,000
BVⱼ = 84,000 − 44,000
BVⱼ = $40,000
Check
Put the numbers back in. Both sides should match.
11,000
=
84,000 − 7,000
7
✓
40,000 = 84,000 − 4 × 11,000 ✓
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