11–41
PROBLEM 11-3
(a)
Depreciation Expense ……………………………………………..
3,900
Accumulated Depreciation—Machinery (A)
(5/55 X [$46,000 – $3,100]) …………………………..
3,900
Accumulated Depreciation—Machinery (A) ………………
35,100
Machinery (A) ($46,000 – $13,000) ……………………
33,000
Gain on Disposal of Machinery ………………………..
2,100
(b)
Depreciation Expense ……………………………………………..
Accumulated Depreciation—Machinery (B)
([$51,000 – $3,000] ÷ 15,000 X 2,100) ……………..
6,720
(c)
Depreciation Expense ……………………………………………..
6,000
Accumulated Depreciation—Machinery (C)
([$80,000 – $15,000 – $5,000] ÷ 10) …………………
6,000
(d)
Machinery (E) ……………………………………………………….
28,000
Retained Earnings ………………………………………….
28,000
Accumulated Depreciation—Machinery (E) ………
5,600
*($28,000 X .20)
11–42
PROBLEM 11-4
Net
Income
(5,300)
(3,400)
7,100
(8,250)
4
5
6
As Adjusted
Retained
17,200
40,100
6,400
16,800
Acc. Dep.,
(17,200)
(43,800)
14,400
(16,800)
_______
86,000
42,000
(24,000)
_______
Per Company Books
Retained
22,500
43,500
(700)
25,050
(22,500)
(73,700)
(25,050)
________
112,500
42,000
(2,500)
________
Depreciation
Balances
Purchase of Truck #6
Disposal of Truck #4
Depreciation
(a)
12/31/11
12/31/11
7/1/12
7/1/12
12/31/12
Income effect
PROBLEM 11-4 (Continued)
3Book value of Truck #1 [$18,000 – ($18,000/5 X 4 yrs.)] =
$18,000 – $14,400 ……………………………………………………
= $3,600
Cash received on sale ……………………………………………………..
= 3,500
Loss on sale …………………………………………………………..
$ 100
$22,000/5
=
Truck #4:
$24,000/5
=
Truck #5:
$40,000/5
=
Total
5Book value of Truck #4 $24,000 – [($24,000/5 X 3 yrs.)] …….
= $9,600
Cash received ($700 + $2,500) ………………………………………….
= 3,200
Loss on disposal …………………………………………………….
$6,400
6Truck #2:
$22,000/5 X 1/2
=
$ 2,200
Truck #4:
$24,000/5 X 1/2
=
2,400
Truck #5:
$40,000/5
8,000
Truck #6:
$42,000/5 X 1/2
=
4,200
Total
$16,800
7Truck #2:
(fully dep.)
=
Truck #5:
$40,000/5
=
Truck #6:
$42,000/5
=
Total
(b)
Compound journal entry December 31, 2013:
Accumulated Depreciation—Trucks ………………………
Trucks …………………………………………………………
Retained Earnings ……………………………………….
Depreciation Expense …………………………………
11–44
PROBLEM 11-4 (Continued)
Summary of Adjustments:
Per
Books
As
Adjusted
Adjustment
Dr. or (Cr.)
Trucks
$152,000
$104,000
$(48,000)
Accumulated Depreciation
$129,150
$ 62,600
Totals
$ 67,850
$ 63,300
$ (4,550)
Depreciation Expense, 2013
$ 30,400
$ 16,400
$(14,000)
11–45
PROBLEM 11-5
(a) Estimated depletion:
Estimated Depletion
Depletion
Base
Estimated
Yield
Per
Ton
1ST & 11th
Yrs.
Each of Yrs.
2-10 Incl.
$870,000*
120,000 tons
$7.25
$43,500**
$87,000***
Estimated depreciation:
Asset
Cost
Per ton
Mined
1st
Yr.
Yrs.
2–5
6th
Yr.
Yrs.
7–10
11th
Yr.
Building
$36,000
$.30*
$1,800
$3,600
$3,600
$3,600
$1,800
Machinery (1/2)
30,000
.25**
1,500
3,000
3,000
3,000
1,500
Machinery (1/2)
30,000
.50***
3,000
6,000
3,000
0
0
Total depreciation
PROBLEM 11-6
(a)
Original cost
$550 X 3,000 =
$1,650,000
Deduct residual value of land
$200 X 3,000 =
600,000
Cost of logging road
150,000
$1,200,000
(b)
Inventory ……………………………………………………….
240,000
Timber ……………………………………………………….
240,000
Depletion, 2012: 20% X 500,000 bd. ft. = 100,000 bd. ft.;
100,000 bd. ft. X $2.40 = $240,000
Cost of salvaging timber …………………………………..
Less recovery ($3 X 400,000 bd. ft.) ……………………
$ 340,000
Loss of land value …………………………………………….
600,000
Logging equipment …………………………………………..
300,000
Extraordinary loss due to the eruption
11–47
PROBLEM 11-7
Instructors should note the changing depletion base in this problem.
2012
Computation of Depletion Base for 2012
Estimated Depletion for 2012
$9,250,000
X 0.08
(540,000/6,750,000)
Depletion Expense for 2012
$ 740,000
Depreciation of Removable Equipment
Cost
$ 225,000
Salvage Value
(9,000)
Depreciable base
$ 216,000
Annual Depreciation using SL ($216,000/15)
$ 14,400
Depreciation Expense for 2012
$ 10,800
(9/12 X $14,400)
Depletion Base for 2013
Base for 2012
$9,250,000
Less Depletion for 2012
Plus Seedling Planting Costs
120,000
Depletion Base for 2013
$8,630,000
Depletion Base for 2013
$8,630,000
Times
(774,000/6,450,000)
Depletion for 2013
Depreciation Expense for 2013
$ 14,400
Cost per acre
$1,700
Land Cost
800
Timber Cost
$ 900
10,000 acres
Road Cost
Total Depletion Base
PROBLEM 11-7 (Continued)
2014
Depletion Base for 2014
Base for 2013
$ 8,630,000
Less: Depletion for 2013
1,035,600
Plus: Seedling Planting Costs
150,000
Depletion Base for 2014
$ 7,744,400
Depletion Base for 2014
$ 7,744,400
Times
(650,000/6,500,000)
Depletion for 2014
$ 774,440
Depreciation Expense for 2014
$ 14,400
11–49
PROBLEM 11-8
(a) The amounts to be recorded on the books of Darby Sporting Goods
Inc. as of December 31, 2012, for each of the properties acquired from
Encino Athletic Equipment Company are calculated as follows:
Cost Allocations to Acquired Properties
Appraisal
Value
Remaining
Purchase
Price
Allocations
Renovations
Capitalized
Interest
Total
Supporting Calculations
1Balance of purchase price to be allocated.
Total purchase price …………………………………………………..
$400,000
Less: Land appraisal ………………………………………………….
Machinery
11–50
PROBLEM 11-8 (Continued)
2Capitalizable interest.
Expenditures
Capitalization
Period
Weighted-Average
Accumulated Expenditures
Date
Amount
1/1
$ 50,000
12/12
$ 50,000
Weighted-Average
Interest
Avoidable
Accumulated Expenditures
Rate
Interest
$175,000
X
12%
=
$21,000
and $6,300 ($21,000 X 45/150) would be allocated to the machinery.
(b) Darby Sporting Goods Inc.’s 2013 depreciation expense, for book
purposes, for each of the properties acquired from Encino Athletic
Equipment Company is as follows:
1.
Land: No depreciation.
2013 depreciation expense
= Cost X Rate X 1/2 year
3.
Machinery: Depreciation rate
= 2.00 X 1/5 = .40
4/1
90,000
$500,000
PROBLEM 11-8 (Continued)
(c) Arguments for the capitalization of interest costs include the following.
1. Diversity of practices among companies and industries called for
standardization in practices.
11–52
PROBLEM 11-9
(a) Carrying value of asset: $10,000,000 – $2,500,000* = $7,500,000.
*($10,000,000 ÷ 8) X 2
Future cash flows ($6,300,000) < Carrying value ($7,500,000)
*$7,500,000 – $5,600,000
(b) Depreciation Expense …………………………………….. 1,400,000**
Accumulated Depreciation—Equipment ….. 1,400,000
**($5,600,000 ÷ 4)
11–53
PROBLEM 11-10
(1)
$80,000
Allocated in proportion to appraised values
(1/10 X $800,000).
(3)
Fifty years
Cost less salvage ($720,000 – $40,000) divided by
annual depreciation ($13,600).
(4)
$13,600
Same as prior year since it is straight-line depreciation.
(5)
$91,000
[Number of shares (2,500) times fair value ($30)]
plus demolition cost of existing building ($16,000).
(6)
None
No depreciation before use.
(7)
$40,000
Fair value.
(8)
$6,000
Cost ($40,000) times percentage (1/10 X 150%).
(9)
$5,100
equals $34,000. Multiply $34,000 times 15%.
(10)
$168,000
Total cost ($182,900) less repairs and maintenance
($14,900).
(11)
$36,000
Cost less salvage ($168,000 – $6,000) times 8/36.
(12)
$10,500
Cost less salvage ($168,000 – $6,000) times 7/36 times
one-third of a year.
(2)
$720,000
Allocated in proportion to appraised values
(9/10 X $800,000).
PROBLEM 11-10 (Continued)
(13)
$52,000
Annual payment ($6,000) times present value of annuity
due at 8% for 11 years (7.710) plus down payment ($5,740).
This can be found in an annuity due table since the
(14)
$2,600
Cost ($52,000) divided by estimated life (20 years).
PROBLEM 11-11
(a)
1.
Straight-line Method:
$90,000 – $6,000
= $16,800 a year
5 years
2.
Activity Method:
$90,000 – $6,000
= $.84 per hour
100,000 hours
Year
2010
20,000 hrs. X $.84 =
$16,800
2011
25,000 hrs. X $.84 =
21,000
2012
15,000 hrs. X $.84 =
2013
30,000 hrs. X $.84 =
25,200
2014
10,000 hrs. X $.84 =
Year
2010
2011
4/15 X $84,000 =
22,400
2012
3/15 X $84,000 =
16,800
2013
2/15 X $84,000 =
2014
1/15 X $84,000 =
4. Double-Declining-Balance Method: Each year is 20% of its total
life. Double the rate to 40%.
Year
2010
40% X $90,000 =
$36,000
2011
40% X ($90,000 – $36,000) =
21,600
2012
40% X ($90,000 – $57,600) =
12,960
2013
2014
Enough to reduce to salvage =
(b) 1. Straight-line Method:
Year
2010
$90,000 – $6,000
X 9/12 =
$12,600
5 years
2011
Full year
16,800
2012
Full year
16,800
2013
Full year
11–56
PROBLEM 11-11 (Continued)
2. Sum-of-the-Years’-Digits:
2010
(5/15 X $84,000) X 9/12 =
$21,000
2011
(5/15 X $84,000) X 3/12 =
$ 7,000
(4/15 X $84,000) X 9/12 =
16,800
23,800
3. Double-Declining Balance Method:
Year
Cost
Accum.
Depr. at
beg. of
year
Book
Value at
beg. of
year
Depr.
Expense
2010
$90,000
—
$90,000
$27,000 (1)
2011
25,200 (2)
2014
5,443 (5)
(1) $90,000 X 40% X 9/12
(2) ($90,000 – $27,000) X 40%
(3) ($90,000 – $52,200) X 40%
2012
(4/15 X $84,000) X 3/12 =
(3/15 X $84,000) X 9/12 =
12,600
18,200
2013
(3/15 X $84,000) X 3/12 =
(2/15 X $84,000) X 9/12 =
12,600
2014
(1/15 X $84,000) X 9/12 =
2015
(1/15 X $84,000) X 3/12 =
*PROBLEM 11-12
(a) The straight-line method would provide the highest total net income
for financial reporting over the three years, as it reports the lowest
total depreciation expense. These computations are provided below.
Computations of depreciation expense and accumulated depreciation under
various assumptions:
(1) Straight-line:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$240,000
$ 240,000
2012
$ 480,000
2013
$ 720,000
$720,000
(2) Double-declining-balance:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$504,000
(40% X $1,260,000)
$ 504,000
2012
(40% X $756,000)
$ 806,400
2013
(40% X $453,600)
$ 987,840
$987,840
(3) Sum-of-the-years’-digits:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$400,000
(5/15 X $1,200,000)
$ 400,000
2012
(4/15 X $1,200,000)
$ 720,000
2013
(3/15 X $1,200,000)
$ 960,000
11–58
*PROBLEM 11-12 (Continued)
(4) Units-of-output:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$288,000
($24* X 12,000)
$288,000
2012
264,000
($24 X 11,000)
$552,000
2013
240,000
($24 X 10,000)
$792,000
$792,000
Total Cost
MACRS
Rates (%)*
Annual
Depreciation
Accumulated
Depreciation
2011
$1,260,000
X
14.29
=
$180,054
$180,054
*Taken from the MACRS rates schedule.
Optional straight-line method:
Total Cost
Depreciation
Rate
Annual
Depreciation
Accumulated
Depreciation
TIME AND PURPOSE OF CONCEPTS FOR ANALYSIS
CA 11-1 (Time 25–35 minutes)
Purpose—to provide the student with an understanding of the basic objective of depreciation accounting.
In addition, the case involves a reverse sum-of-the-years’-digits situation and the student is to comment
on the propriety of such an approach. Finally, the classic issue of whether depreciation provides funds
must be considered. The tax effects of depreciation must be considered when this part of the case is
examined. An excellent case for covering the traditional issues involving depreciation accounting.
CA 11-2 (Time 20–25 minutes)
Purpose—to provide the student with a basic understanding of the difference between the unit and
group or composite depreciation methods. The student is required to indicate the arguments for and
against these methods and to indicate how retirements are handled.
CA 11-3 (Time 25–35 minutes)
Purpose—to provide the student with an understanding of a number of unstructured situations involving
depreciation accounting. The first situation considers whether depreciation should be recorded during a
strike. The second situation involves the propriety of employing the units-of-production method in
certain situations. The third situation involves the step-up of depreciation charges because properties
are to be replaced due to obsolescence. The case is somewhat ambiguous, so cut-and–dried approaches
should be discouraged.
CA 11-4 (Time 25–35 minutes)
Purpose—to provide the student with an understanding of the objectives of depreciation and the
theoretical basis for accelerated depreciation methods.
CA 11-5 (Time 20–25 minutes)
Purpose—to provide the student with the opportunity to examine the ethical dimensions of the depre–
ciation method choice.
11–60
CA 11-1
(a) The purpose of depreciation is to distribute the cost (or other book value) of tangible plant assets,
less salvage, over their useful lives in a systematic and rational manner. Under generally accepted
accounting principles, depreciation accounting is a process of allocation, not of valuation, through
which the productive effort (cost) is to be matched with productive accomplishment (revenue) for
the period. Depreciation accounting, therefore, is concerned with the timing of the expiration of the
cost of tangible plant assets.
On the other hand, if the expected level of operations (including equipment usage) in the early
years of asset life is expected to be low as compared to that of later years because of slack demand
or production policies, the pattern of the depreciation charges of the proposed method approximately
parallels expected benefits (and revenues) and hence is reasonable. Although the units–of–production
depreciation method is the usual selection to fit this case, the proposed method also conforms to
generally accepted accounting principles in this case provided that proper justification is given.
(c) (1) Depreciation charges neither recover nor create funds. Revenue-producing activities are the
sources of funds from operations: if revenues exceed out-of-pocket costs during a fiscal period,
funds are available to cover other than out-of-pocket costs; if revenues do not exceed out-of–
pocket costs, no funds are made available no matter how much, or little, depreciation is charged.