Question

ILLUSTRATIVE PROBLEM 4: 4.) In 1992, Silicon Chemical Company purchased a special purpose machine that had a fair price of $120,000. This new machine is expected to produce 100,000 units throughout its estimated useful life of 10 years. Up to the end of the third year it had produced 45,000 units and during the fourth year it produced 15,000 units. If the estimated market value is $12,000 at the end of 10 years, compute for the depreciation amount in the third year and the book value at the end of the fourth year by each of these methods: straight-line method sum-of-the-years digit method 200% declining balance method sinking fund method (i=10%) service output or units-of-production method

Answer 1

	STRAIGHT LINE METHOD:
	Annual Depreciation=(Cost of asset-Salvage Value)/Useful life in years
	Cost of asset	$120,000
	Salvage Salvage Value	$12,000
	Useful Life in years	10
	Annual Depreciation=(120000-12000)/10	$10,800
	Depreciation in 3 rd year	$10,800
	Accumulated Depreciation at end of 4 years	$43,200	(10800*4)
	Book Value at end of 4th year	$76,800	(120000-43200)
	SUM OF YEARS DIGIT METHOD
	Amount to be depreciated=120000-12000=	$108,000
	Sum of digits =1+2+3+….......+10=10*(10+1)/2	55
			N	DF=N/55	X=108000*DF
		Year	Remaining Useful life at beginning of year	Depreciation Factor	Annual Depreciation
		1	10	0.181818182	$19,636
		2	9	0.163636364	$17,673
		3	8	0.145454545	$15,709
		4	7	0.127272727	$13,745
	Depreciation in 3 rd year	$15,709
	Accumulated Depreciation at end of 4 years	$66,764	(19636+17673+15709+13745)
	Book Value at end of 4th year	$53,236	(120000-66764)
	200% DECLINING BALANCE METHOD
	Depreciation Rate =2*(1/Useful Life)=2*(1/10)	20.00%
			A	B	C=A*B	D=A-C
		Year	Beginning Book Value	Depreciation Rate	Annual Depreciation	Ending Book Value
		1	$120,000	20.00%	$24,000	$96,000
		2	$96,000	20.00%	$19,200	$76,800
		3	$76,800	20.00%	$15,360	$61,440
		4	$61,440	20.00%	$12,288	$49,152
	Depreciation in 3 rd year	$15,360
	Book Value at end of 4th year	$49,152
	SINKING FUND METHOD (i-=10%)
	Sinking Fund Formula:
	(Cost-Present Value of Salvage Value)/(1-((1+i)^(-n))/i
	i=10%=0.1
	n=Number of years of useful life=10
	Cost=120000
	Present Value of Salvage Value=12000/(1.1^10)=	$4,627
	Cost-PV of salvage value=120000-4627=	$115,373
	(1-((1+i)^(-n))/i=(1-(1.1^(-10)))/0.1=	6.14456711
	(Cost-PVofSalvage Value)/(1-((1+i)^(-n))/i=115373/6.14456711	$18,777
	This amount represents combined annual depreciation and interest
	DEPRECIATION SCHEDULE		A	B	C=A*10%	D=B-C	E=A-D
		Year	Book Value at Beginning	Combined Depreciation +Interest	Interest	Annual Depreciation	Book Value at end of year
		1	$120,000	$18,777	$12,000	$6,777	$113,223
		2	$113,223	$18,777	$11,322	$7,454	$105,769
		3	$105,769	$18,777	$10,577	$8,200	$97,570
		4	$97,570	$18,777	$9,757	$9,020	$88,550
	Depreciation in 3 rd year	$8,200
	Book Value at end of 4 years	$88,550
	SERVICE OUTPUT OR UNIT OF PRODUCTION METHOD
	Amount of depreciation in 10 years =120000-12000	$108,000
	Expected units of production	100000
	Depreciation rate per unit=108000/100000=	$1.08
	Depreciation in 3 years =45000*1.08	$48,600
	Depreciation in 3 rd year =48600/3=	$16,200
	Depreciation during 4th year=15000*1.08=	$16,200
	Accumulated Depreciation at end of 4 years=48600+16200	$64,800
	Book Value at end of 4th year=120000-64800	$55,200