Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 16
Depreciation Methods
Fundamentals of Depreciation
16.1 Depreciation is a tax-allowed deduction; it reduces the tax burden.
16.2 Unadjusted basis is the installed cost of an asset and includes all costs incurred in preparing
16.3 (a) Book value is the remaining capital investment after depreciation charges have been
16.4 Productive life – Time the asset is actually expected to provide useful service.
16.5 B = 18,000 + 300 + 1200 = $19,500
(c) Percent written off = [(40,000 + 24,000 + 14,000)/100,000](100%)
16.8 Tax depreciation: Dt = Rate*BVt-1
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S = 0.1(300,000) = $30,000
Dt = (400,000 – 30,000)/8
= $46,250 per year t (t = 1, …, 8)
BV4 = 400,000 – 4(46,250) = $215,000
16.12 (a) Dt = 12,000 – 2000 = $1428.57
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16.13 Depreciation charge is the same each year
16.14 (a) Salvage value = book value at end of recovery period, year 4
16.15 (a) Depreciation is constant and determined from the change in book value
(b) (B – 80,000)/5 = 72,000
16.16 (a) D = (50,000 – 5000)/4 = $11,250
16.17 (a) D = (750,000 – 50,000)/10 = $70,000
Declining Balance Depreciation
16.18 (a) By hand:
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D2 = (0.3)(500,000)(0.7)1
= $105,000
SL: D2 = (500,000 – 50,000)/5
= $90,000
(b) By spreadsheet:
16.19 (a) d = 2/5 = 0.40
16.20 d = 2/5 = 0.40
16.21 Substitute 0.25B for BV; d = 2/5 = 0.4; find t
16.22 d = 2/5 = 0.4
16.23 d = 2/5 = 0.40
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= $8640
Difference = 8640 – 4000 = $4640
16.24 Largest accumulated depreciation in year 2 is for DDB method: $37,500
MACRS Depreciation and Recovery Periods
16.25 (a) Personal property recovery periods are 3,5,7,10,15, or 20 years
16.26 (a) The half-year convention assumes that all property is placed in service at the
16.27 Recovery period is 15 years (Table 16-4)
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D3 = 0.0855(770,000) = $65,835
BV3 = B – total depreciation through year 3
= 770,000(1 – 0.050 – 0.095 – 0.0855)
= $592,515
16.28 3 years: BV3 = 400,000 – 400,000(0.20 + 0.32 + 0.192)
16.29 (a) Total depreciation for 3 years = 100% – 57.6%
(b) Function = VDB(140000,0,10,MAX(0,4-1.5),MIN(10,4-0.5),2) displays the value
16.30 (a) MACRS: BV3 = 300,000 – 300,000(0.20 + 0.32 + 0.192)
(b) Schedules show that DDB has a lower BV3 of $64,800. Same difference of $21,600.
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(c) Depreciation is limited by the DDB function to ensure that S = $60,000 is maintained.
16.31 (a) From MACRS depreciation rate table, d3 = 0.192
16.32 Use rates for real property with n = 39 years
Fairfield hopes to sell it for $452,183 more than they paid for it.
16.33 (a) MACRS in United States: n = 5, BV6 = $0
(b) MACRS goes to BV6 = $0 and SL stops at BV10 = $150,000. BV6 values are
highlighted.
16.34 SL: D3 = [80,000 – 0.25(80,000)]/5
16.35 (a) MACRS: d3 = 0.1440; sum of rates for 3 years is 0.4240
BV3 = 1,000,000 – 0.4240(1,000,000) = $576,000
(b) DDB: d = 2/15 = 0.13333
(c) ADS SL: d = 1/15 = 0.06666 years 2 through 15; ½ that for years 1 and 16.
Depletion
16.36 Depreciation applies to assets that can be replaced. Depletion is applicable only to
16.37 (a) CDt = 2,100,000/350,000 = $6.00 per ounce
(b) Remaining investment = 2,100,000 – 1,050,000
16.38 Percentage depletion for coal is 10% of gross income, provided it does not exceed
50% of taxable income (TI).
Gross* % depletion 50% Allowed
Year Income, $ @10% of TI depletion
*tons x $/ton
16.39 CDt = 500,000/200 = $2500 per million board feet
16.40 Percentage depletion = 0.20(GI) = 700,000
Let N = number of barrels
16.41 (a) CDt = 2,900,000/100 = $29,000 per 1000 tons
Year
Volume,
1000 tons
Cost depletion,
$ per year
1
10
290,000
2
9
261,000
3
15
435,000
4
15
435,000
5
18
522,000
Total
67
$1,943,000
(b) No, cost depletion is limited by total first cost. Since $1.943 million < $2.9 million,
16.42 Cost depletion: Remaining investment = 35.0 – 24.8 million = $10.2 million