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Problem 8-2A Depreciation methods LO P1 A machine costing $211,200 with a four-y

ID: 2578387 • Letter: P

Question

Problem 8-2A Depreciation methods LO P1

A machine costing $211,200 with a four-year life and an estimated $18,000 salvage value is installed in Luther Company’s factory on January 1. The factory manager estimates the machine will produce 483,000 units of product during its life. It actually produces the following units: 121,700 in 1st year, 124,100 in 2nd year, 120,200 in 3rd year, 127,000 in 4th year. The total number of units produced by the end of year 4 exceeds the original estimate—this difference was not predicted. (The machine must not be depreciated below its estimated salvage value.)

Required:

Compute depreciation for each year (and total depreciation of all years combined) for the machine under each depreciation method. (Round your per unit depreciation to 2 decimal places. Round your answers to the nearest whole dollar.)

Straight-Line Depreciation Depreciation Expense Year Total

Explanation / Answer

Answer

A

Cost

211200

B

Salvage Value

18000

C=A-B

Depreciable base

193200

D

Life (in years)

4

E=C/D

SLM Annual Depreciation

48300

Year

Depreciation expense

1

48300

2

48300

3

48300

4

48300

TOTAL

193200

A

Cost

211200

B

Salvage Value

18000

C=A-B

Depreciable base

193200

D

Estimated Units

483000

E=C/D

Depreciation per unit

0.4

Year

Depreciable units

Depreciation per unit

Depreciation expense

1

121700

0.4

48680

2

124100

0.4

49640

3

120200

0.4

48080

4

127000

46800 [cannot lead to made total depreciation exceed $193200, hence restricted to $46800]

TOTAL

493000

193200

A

Cost

211200

B

Salvage Value

18000

C=A-B

Depreciable base

193200

D

Life (in years)

4

E=C/D

SLM Annual Depreciation

48300

F=E/C

SLM %

25%

G=F x 2

DDB %

50%

Depreciation for the period

End of Period

Year

Beginning of Period book Value

Depreciation Rate

Depreciation expense

Accumulated Depreciation

Book Value

1

211200

50%

105600

105600

105600

2

105600

50%

52800

158400

52800

3

52800

50%

26400

184800

26400

4

26400

8400

193200

18000

TOTAL

193200

A

Cost

211200

B

Salvage Value

18000

C=A-B

Depreciable base

193200

D

Life (in years)

4

E=C/D

SLM Annual Depreciation

48300