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Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 16
Depreciation Methods
16.1 Depreciation increases the company’s after–tax cash flow, because depreciation reduces the
16.2 Book value is established on the basis of accepted accounting procedures. Market value is
16.3 Book depreciation is used on internal financial records to reflect current capital investment
16.4 Unadjusted basis refers to the first cost plus any other depreciable costs that make the asset
16.5 MACRS has set n values for depreciation by property class. These are commonly different,
16.6 Quoting Publication 946, 2010 version:
(a) “Depreciation is an annual income tax deduction that allows you to recover the cost or
(b) “An estimated value of property at the end of its useful life. Not used under
MACRS.”
Depreciating stops when property is retired from service, even if its cost is not fully
recovered .
(f) A taxpayer can elect to recover all or part of the cost of certain qualifying property,
up to a limit, by deducting it in the year the property is placed in service. The
16.7 B = 580,000 + 4300 + 6400 = $590,700
16.8 (a) B = $350,000 + 50,000 = $400,000
16.9 Write the cell equations to determine depreciation of $10,000 per year for book purpose
and $5000 per year for tax purposes. Develop the scatter chart to plot book values.
16.11 (a) D3 = (40,000 – 10,000)/10 = $3000
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16.12 (a) D3 = 26,000
BV3 = 62,000 = B – 3(26,000)
16.13 (a) If the machine will have BV = 0 at the end of 5 years, the SL book depreciation
charge for each of the last 2 years will have to be
16.14 BV5 = 200,000 – 5*SLN(200000,10000,7)
16.15 Use the spreadsheet below.
(a) In 2012, BV4 = $450,000
16.16 (a) B = $50,000, n = 4, S = 0, d = 0.25
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Accumulated
Year, t Dt depreciation BVt___
0 — — $50,000
1 $12,500 $12,500 37,500
(b) S = $16,000; d = 0.25; B – S = $34,000
(c) Spreadsheet chart showing S = 0 and S = $16,000 book values are the same as above.
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16.17 Develop difference relations (US minus EU) for (a) depreciation and (b) book value in
year 5 with the SLN function.
16.18 d is decimal amount of BV removed each year.
16.19 (a) d = 2/15 = 0.133
16.20 (a) D for all years = (600,000 – 0)/30 = $20,000
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(d) Hand solution used 3-decimal accuracy and spreadsheet accuracy has more decimal
16.21 D = 2/5 = 0.40
16.22 (a) DDB: d = 2/12 = 0.167
16.23 (a) SL: BV10 = $10,000 by definition
DDB: Determine if the implied S < $10,000 with d = 2/7 = 0.2857
16.24 Select any first cost value to use for B. The spreadsheet below uses $10,000.
16.25 SL is the classic non–accelerated method. Anything that has a BV curve below the SL BV
16.26 A primary intent was economic growth through capital investment and the tax advantages
16.27 (a) D2 = 80,000(0.32) = $25,600
16.28 (a) From MACRS depreciation rate table, d2 = 0.32
B = 24,320/0.32 = $76,000
16.29 Straight line: D = [80,000 – 0.25(80,000)]/5
= $12,000 per year
16.30 MACRS: BV3 = 300,000 – 300,000(0.20 + 0.32 + 0.192)
= 300,000 – 213,600
16.31 Recovery period is 7 years from Table 16-4. Book values are close for both ways.
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16.32 (a) SL: Dt = (320,000-75,000)/7 = $35,000 per year
Straight line
MACRS
Year
Depr
BV
Rate
Depr
BV
0
320,000
320,000
1
35,000
285,000
0.1429
45,728
274,272
2
35,000
250,000
0.2449
78,368
195,904
3
35,000
215,000
0.1749
55,968
139,936
4
35,000
180,000
0.1249
39,968
99,968
5
35,000
145,000
0.0893
28,576
71,392
6
35,000
110,000
0.0892
28,544
42,848
7
35,000
75,000
0.0893
28,576
14,272
8
0
75,000
0.0446
14,272
0
Spreadsheet solution with BV plots follow.
(b) MACRS neglects the salvage value; it always depreciates to zero.
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16.33 (a) MACRS: rate for year 3 is 0.1440; sum of rates for 3 years is 0.4240
D3 = 0.1440(800,000) = $115,200
Spreadsheet solution for all parts follows. The relations used to determine
the values (row 50 are indicated first (row 3).
16.34 (a) MACRS: n = 5, B = $100,000
Hand solution
_______MACRS SL __________________
Year d Depr BV d Depr BV___
0 – – $100,000 – – $100,000
1 0.2000 $20,000 80,000 0.05 $ 5,000 95,000
2 0.3200 32,000 48,000 0.10 10,000 85,000