EXERCISE 11-2 (20-25 minutes)

(a) If there is any salvage value and the amount is unknown (as is the case here), the cost would have to be determined by looking at the data for the double-declining balance method.

100%	= 20%; 20% X 2 = 40%
5

Cost X 40% = $20,000

$20,000 ¸ .40 = $50,000 Cost of asset

(b) $50,000 cost [from (a)] – $45,000 total depreciation = $5,000 salvage value.

(c) The highest charge to income for Year 1 will be yielded by the double-declining balance method.

(a) The highest charge to income for Year 4 will be yielded by the straight-line method.

(b) The method that produces the highest book value at the end of Year 3 would be the method that yields the lowest accumulated depreciation at the end of Year 3, which is the straight-line method.

Computations:

St.-line = $50,000 – ($9,000 + $9,000 + $9,000) = $23,000 book value, end of Year 3.

S.Y.D. = $50,000 – ($15,000 + $12,000 + $9,000) = $14,000 book value, end of Year 3.

D.D.B. = $50,000 – ($20,000 + $12,000 + $7,200) = $10,800 book value, end of Year 3.

(c) The method that will yield the highest gain (or lowest loss) if the asset is sold at the end of Year 3 is the method which will yield the lowest book value at the end of Year 3, which is the double-declining balance method in this case.

EXERCISE 11-5

(a)	($117,900 – $12,900)	= $21,000/yr. = $21,000 X 5/12 = $8,750
	5

2004 Depreciation — Straight line = $8,750

(b)	($117,900 – $12,900)	= $5.00/hr.
	21,000

2004 Depreciation — Machine Usage = 800 X $5.00 = $4,000

(c)	Machine			Allocated to
	Year		Total	2004	2005
	1		5/15 X $105,000 = $35,000	$14,583*	$20,417**
	2		4/15 X $105,000 = $28,000	______	11,667***
				$14,583	$32,084
	*	$35,000 X 5/12 = $14,583
	**	$35,000 X 7/12 = $20,417
	***	$28,000 X 5/12 = $11,667

2005 Depreciation — Sum-of-the-Years’-Digits = $32,084

(d) 2004 40% X ($117,900) X 5/12 = $19,650

2005 40% X ($117,900 – $19,650) = $39,300

OR

1^st full year (40% X $117,900) = $47,160

2^nd full year [40% X ($117,900 – $47,160)] = $28,296

2004 Depreciation =	5/12 X $47,160 =	$19,650
2005 Depreciation =	7/12 X $47,160 =	$27,510
	5/12 X $28,296 =	11,790
		$39,300

EXERCISE 11-6 (20-30 minutes)

(a)	2003	Straight-line	$212,000 – $12,000	= $25,000/year
			8
		3 months — Depreciation $6,250 = ($25,000 X 3/12)

(b	2003	Output	$212,000 – $12,000			= $5.00/output unit
			40,000
		1,000 units X $5.00 = $5,000
(c)	2003	Working hours		$212,000 – $12,000			= $10.00/hour
				20,000
		525 hours X $10.00 = $5,250
(d)	8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 OR				n (n + 1)			=	8(9)	= 36
					2				2

			Allocated to
Sum-of-the-years’-digits		Total	2003	2004	2005
Year 1	8/36 X $200,000 =	$44,444	$11,111	$33,333
2	7/36 X $200,000 =	$38,889		9,722	$29,167
3	6/36 X $200,000 =	$33,333	_______	_______	8,333
			$11,111	$43,055	$37,500

2005: $37,500 = (9/12 of 2^nd year of machine’s life plus 3/12 of 3^rd year of machine’s life)

(e) Double-declining balance 2004: 1/8 X 2 = 25%.

2003: 25% X $212,000 X 3/12 = $13,250

2004: 25% X ($212,000 – $13,250) = $49,688

OR

1^st full year (25% X $212,000) = $53,000

2^nd full year [25% X ($212,000 – $53,000)] = $39,750

2003 Depreciation 3/12 X $53,000 = $13,250

2004 Depreciation 9/12 X $53,000 = $39,750

3/12 X $39,750 = 9,938

$49,688

EXERCISE 11-16

(a)	December 31, 2004
	Loss on Impairment................................................................	3,200,000
	Accumulated Depreciation—Equipment..................		3,200,000

	Cost	$9,000,000
	Accumulated depreciation	1,000,000
	Carrying amount	8,000,000
	Fair value	4,800,000
	Loss in impairment	$3,200,000
(b)	December 31, 2005
	Depreciation Expense...........................................................		1,200,000
	Accumulated Depreciation—Equipment..................			1,200,000
	New carrying amount	$4,800,000
	Useful life	4 years
	Depreciation per year	$1,200,000

(c) No entry necessary. Restoration of any impairment loss is not permitted.

EXERCISE 11-17

(a)	Loss on Impairment................................................................		3,220,000
	Accumulated Depreciation—Equipment..................			3,220,000
	Cost	$9,000,000
	Accumulated depreciation	1,000,000
	Carrying amount	8,000,000
	Less: Fair value	4,800,000
	Plus: Cost of disposal	20,000
	Loss on impairment	$3,220,000

(a) No entry necessary. Depreciation not taken on assets intended to be sold.

(c)	Accumulated Depreciation—Equipment............................	500,000
	Recovery of Loss on Impairment............................		500,000
	Fair value	$5,300,000
	Less: Cost of disposal	20,000	5,280,000
	Carrying amount		4,780,000
	Recovery of impairment loss		$ 500,000

EXERCISE 11-23

(a)	$970,000 + $170,000 + $40,000* – $100,000	= .09 depletion per unit
	12,000,000

*(Note to instructor: The $40,000 should be depleted because it is a cost of the mine. This cost is incurred to get the land back to its original value of $100,000.)

2,500,000 units extracted X $.09 = $225,000 depletion for 2004

(b) 2,100,000 units sold X $.09 = $189,000 charged to cost of goods sold for 2004

EXERCISE 11-24

(a) Asset turnover ratio:

$13,234	= .96 times
$14,212 + $13,362
2

(a) Rate of return on assets:

$76	= .55%
$14,212 + $13,362
2

(b) Profit margin on sales:

$76	= .57%
$13,234

(c) The asset turnover ratio times the profit margin on sales provides the rate of return on assets computed for Eastman Kodak as follows:

Profit margin on sales	X	Asset Turnover		Return on Assets
.57%	X	.96	=	.55%

Note the answer .55% is the same as the rate of return on assets computed in (b) above.

PROBLEM 11-3

(a)	Depreciation Expense—Asset A		2,900
	Accumulated Depreciation—Asset A			2,900
	(5/55 X [$35,000 – $3,100])
	Accumulated Depreciation—Asset A		26,100
	Asset A ($35,000 – $13,000)			22,000
	Gain on Disposal of Plant Assets			4,100
(b)	Depreciation Expense—Asset B		6,720
	Accumulated Depreciation—Asset B			6,720
	([$51,000 – $3,000] ¸ 15,000 X 2,100)
(c)	Depreciation Expense—Asset C		6,000
	Accumulated Depreciation—Asset C			6,000
	([$80,000 – $15,000 – $5,000] ¸ 10)
(d)	Asset E		22,000
	Retained Earnings			22,000
	Depreciation Expense—Asset E		4,400*
	Accumulated Depreciation—Asset E			4,400
	*($22,000 X .20)
		PROBLEM 11-8

(a) The amounts to be recorded on the books of Selig Sporting Goods Inc. as of December 31, 2004, for each of the properties acquired from Starks Athletic Equipment Company are calculated as follows:

Cost Allocations to Acquired Properties

	Appraisal Value	Remaining Purchase Price Allocations	Renovations	Capitalized Interest	Total
(1) Land	$280,000				$280,000
(2) Building		$ 84,000^1	$100,000	$21,600^2	205,600
(3) Machinery	_______	36,000^1	_______	______	36,000
Totals	$280,000	$120,000	$100,000	$21,600	$521,600

Supporting Calculations

^1Balance of purchase price to be allocated.
Total purchase price					$400,000
Less land appraisal					280,000
Balance to be allocated					$120,000
	Appraisal Values	Ratios			Allocated Values
Building	$105,000	105/150 =	.70	X $120,000	$ 84,000
Machinery	45,000	45/150 =	.30	X $120,000	36,000
Totals	$150,000		1.00		$120,000

²Capitalizable interest.

Dates of loans in 2004	Amounts		Periods Outstanding		Interest At 12%
1/1	$ 50,000	X	12/12	X .12	$ 6,000
4/1	130,000	X	9/12	X .12	11,700
10/1	130,000	X	3/12	X .12	3,900
12/31	190,000	X		X .12	_______
Totals	$500,000				$21,600

Note t: If the interest is allocated between the building and the machinery, $15,120 ($21,600 X 105/150) would be allocated to the building and $6,480 ($21,600 X 45/150) would be allocated to the machinery.

(b) Selig Sporting Goods Inc.’s 2004 depreciation expense, for book purposes, for each of the properties acquired from Starks Athletic Equipment Company is as follows:

1.	Land: No depreciation.
2.	Building: Depreciation rate	= 1.50 X 1/15 = .10
	2004 depreciation expense	= Cost X Rate X 1/2 year
		= $205,600 X .10 X 1/2
		= $10,280
3.	Machinery: Depreciation rate	= 2.00 X 1/5 = .40
	2004 depreciation expense	= Cost X Rate X 1/2
		= $36,000 X .40 X 1/2
		= $7,200

(c) Arguments for the capitalization of interest costs include the following.

(1) Diversity of practices among companies and industries called for standardization in practices.

(2) Total interest costs should be allocated to enterprise assets and operations, just as material, labor, and overhead costs are allocated. That is, under the concept of historical costs, all costs incurred to bring an asset to the condition and location necessary for its intended use should be reflected as a cost of that asset.

Arguments against the capitalization of interest include the following:

(1) Interest capitalized in a period would tend to be offset by amortization of interest capitalized in prior periods.

(2) Interest cost is a cost of financing, not of construction.

PROBLEM 11-9

(a) Carrying value of asset: $8,000,000 – $2,000,000 = $6,000,000.

Future cash flows ($5,300,000) < Carrying value ($6,000,000)

Impairment entry:

Loss on Impairment.. 1,600,000*.....

Accumulated Depreciation 1,600,000

*$6,000,000 – $4,400,000

(b) Depreciation Expense....... 1,100,000**

Accumulated Depreciation 1,100,000

**(4,400,000 ¸ 4)

(c) No depreciation is recorded on impaired assets to be disposed of. Recovery of impairment losses are recorded.

Accumulated Depreciation........... 200,000

Recovery of Impairment Loss 200,000

PROBLEM 11-10

(1)	$82,000	Allocated in proportion to appraised values
		(1/10 X $820,000).
(2)	$738,000	Allocated in proportion to appraised values
		(9/10 X $820,000).
(3)	Forty years	Cost less salvage ($738,000 – $40,000) divided by
		annual depreciation ($17,450).
(4)	$17,450	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)	$30,000	Fair market value.
(8)	$4,500	Cost ($30,000) times percentage (1/10 X 150%).
(9)	$3,825	Cost ($30,000) less prior year’s depreciation ($4,500)
		equals $25,500. Multiply $25,500 times 15%.
(10)	$150,000	Total cost ($164,900) less repairs and maintenance
		($14,900).
(11)	$32,000	Cost less salvage ($150,000 – $6,000) times 8/36.
(12)	$9,333	Cost less salvage ($150,000 – $6,000) times 7/36 times
		one-third of a year.

(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 payments are at the beginning of each year. Alternatively, to convert from an ordinary annuity to an annuity due factor, proceed as follows: For eleven payments use the present value of an ordinary annuity for 11 years (7.139) times 1.08. Multiply this factor (7.710) times $6,000 annual payment to obtain $46,260, and then add the $5,740 down payment.
(14)	$2,600	Cost ($52,000) divided by estimated life (20 years).