Additional Problems and FE Exam Review Questions
16.43 Answer is (b)
16.44 Depreciation is same for all years in straight line method.
16.45 D = 50,000 – 10,000 = $8000 per year
16.48 DDB: d = 2/n = 2/8 = 0.25
16.49 BV = 150,000 – 150,000(0.10 + 0.18 + 0.144) = $86,400
16.54 Factor = (70,000 – 20,000)/25,000 = 2.00 per tree
16.55 Percentage: GI = 65,000(40) = $2.6 million
APPENDIX PROBLEMS
Sum-of–Years–Digits Depreciation
16A.1 By hand: SUM = 36; use SYD rates for (B – S) = €10,000
t dt Dt, € BVt, €__
Spreadsheet:
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16A.2 (a) B = $150,000; n = 10; S = $15,000 and SUM = 55
(b) Same results as in part (a) for years 2 and 7
16A.3 B = $400,000; n = 6 and S = 0.15(400,000) = $60,000
Unit–of–Production Depreciation
16A.4 Dt = (tests per year t/10,000)(70,000)
Year, t
Number
of tests
Dt, $ BVt, $
1
3810
26,670
43,330
2
2720
19,040
24,290
3
5390
24,290*
0
*D3 = 5390/10,000(70,000) = $37,730 is too large; only the remaining BV = $24,290
can be charged in year 3.
16A.5 DDB method does depreciate faster, but UOP, in this case, did depreciate slightly more of
the first cost ($33,600 vs. $32,278).
Switching Methods
16A.6 B = $45,000 n = 5 S = $3000 i = 18%
Switching to
Year, DDB Method SL method Larger
t D, Eq. [16A.7] BV D, Eq. [16A.8] Depreciation____
*BV5 will be $3000 exactly when SL depreciation of $2832 is applied in year 5.
16A.7 Develop a spreadsheet for the DDB–to–SL switch using the VDB function (column B)
16A.8 175% DB: d = 1.75/10 = 0.175 for t = 1 to 5
16A.9 (a) Use Equation [16A.6] for DDB with d = 2/25 = 0.08
(b) 255,000(1-d)25 > 50,000
MACRS Rates
16A.10 By hand: Verify that the rates are the following with d = 0.40
t 1 2 3 4 5 6____
d5: Use the SL rate, n = 5
d6: dSL, 6 is the remainder or 1/2 the d5 rate.
5
By spreadsheet: Function and value spreadsheet are shown. Use Equation [16A.16] to
calculate SL depreciation each year
16A.11 B = $30,000 n = 5 years d = 0.40
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t = 2: For DDB depreciation, use Eq. [16A.12]
d = 0.4
DDB = 0.4(24,000) = $9600
BV2 = 24,000 – 9600 = $14,400
For SL, if switch is better, in year 2, by Eq. [16A.13].
DSL = 24,000 = $5333
5–2+1.5
Select DDB; it is larger.
t = 3: For DDB, apply Eq. [16A.12] again.
Select DDB.
NOTE: If Table 16.2 rates are used, cumulative depreciation in % for 3 years is:
16A.12 Determine MACRS depreciation for n = 7 using Equations [16A.11] through
[16A.13] and apply them to B = $50,000. (S) indicates the selected method and
amount.
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BV2 = $30,595
t = 3: d = 0.286 DSL = 30,595 = $5563
The depreciation amounts sum to $50,000
Year Depr., $ Year Depr., $__
16A.13 (a) The SL rates with the half–year convention for n = 3 are:
Year d rate Formula
The MACRS PWD is larger by $4338