Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
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Question
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Chapter 11, Problem 51P
To determine

i.

Book Value after 2 years.

Expert Solution
Check Mark

Answer to Problem 51P

Book Value after 2 years is $40800.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

MACRS depreciation is calculated by using the following formula=

dt=B×rt

dt is the depreciation charge in any year t. B is the cost of the asset made ready for use.

rt is the appropriate MACRS schedule for the asset by MACRS depreciation.

Recovery Year, t (year) MACRS
1 20%
2 32%
3 19.20%
4 11.52%
5 11.52%
6 5.76%
100%

The asset is a 5-year property with asset depreciation range (ADR) class life of more than 10 years and less than 16 years.

Book Value is calculated of an asset at the end of 1st year of its useful life period.

BV3=$85000(3×$8750)=$85000$26250=$58750

BVt is the Book Value of the asset at any year t, B is the purchase cost of the asset and t=1tdt is sum of depreciation charges till any year t.

T = 1

B = $85000

dt=$17000

BVt=Bt=1tdt=Bdt=$85000$17000=$68000

Years Depreciation charge Book Value
1 17000 68000
2 27200 40800
3 16320 24480
4 9792 14688
5 9792 4896
6 4896 0
7 0 0
8 0 0
85000

Conclusion:

Book Value after 2 years is $40800.

To determine

ii.

Book Value after 4 years.

Expert Solution
Check Mark

Answer to Problem 51P

The book value of the asset after 4 years of its useful life is $14688.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

The book value of the asset after 4 years of its useful life is $14688.

Years Depreciation charge Book Value
1 17000 68000
2 27200 40800
3 16320 24480
4 9792 14688
5 9792 4896
6 4896 0
7 0 0
8 0 0
85000

Conclusion:

The book value of the asset after 4 years of its useful life is $14688.

To determine

iii.

The accumulated depreciation charge of the asset through the 5th year.

Expert Solution
Check Mark

Answer to Problem 51P

The accumulated depreciation charge of the asset through the 5th year is $80104.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

The accumulated depreciation charge of the asset through the 5th year

t=1tdt=d1+d2+d3+.........+dt

Here, t=1tdt is the accumulated depreciation charge of the asset till year t and d1, d2.... Are depreciation charges for different years.

Year depreciation (d)
1 $17,000
2 $27,200
3 $16,320
4 $9,792
5 $9,792
$80,104

Conclusion:

The accumulated depreciation charge of the asset through the 5th year is $80104.

To determine

iv.

Depreciation charge of the asset for the 6th year.

Expert Solution
Check Mark

Answer to Problem 51P

The depreciation charge of the asset for the 6th year is $4896.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

Years Depreciation charge Book Value
1 17000 68000
2 27200 40800
3 16320 24480
4 9792 14688
5 9792 4896
6 4896 0
7 0 0
8 0 0
85000

Conclusion:

The depreciation charge of the asset for the 6th year is $4896.

To determine

v.

The book value of the asset for the 8th year.

Expert Solution
Check Mark

Answer to Problem 51P

The book value of the asset for the 8th year, using MACRS is $0.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

Book Value is calculated of an asset at the end of 1st year of its useful life period.

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

BV3=$85000(3×$8750)=$85000$26250=$58750

BVt is the Book Value of the asset at any year t, B is the purchase cost of the asset and t=1tdt is sum of depreciation charges till any year t.

T = 1

B = $85000

dt=$17000

BVt=Bt=1tdt=Bdt=$85000$17000=$68000

Years Depreciation charge Book Value
1 17000 68000
2 27200 40800
3 16320 24480
4 9792 14688
5 9792 4896
6 4896 0
7 0 0
8 0 0
85000

Conclusion:

The book value of the asset for the 8th year, using MACRS is $0.

To determine

vi.

Book Value at the end of 3rd year using straight Line Method.

Expert Solution
Check Mark

Answer to Problem 51P

The book value of the asset after 3rd year is $58570.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

Straight Line depreciation Method:

dt=BSN

dt is the depreciation charge in any year t, N is the number of years in a depreciable life, B is the cost of the asset made ready for use.

T = 1 year

B = $85000

S = $15000

N = 8

dt=BSNdt=$85000$150008=$8750

Straight Line depreciation is constant throughout the depreciable Life. Thus, the depreciation charge for the asset is $8750 for every year throughout the depreciable life of the asset.

Book Value at the end of 3rd year

BVt=Bt=1tdt=Bndt

BV3=$85000(3×$8750)=$85000$26250=$58750.

Conclusion:

Thus, the book value of the asset after 3rd year is $58570.

To determine

vi.

Book value of the asset after 8th year.

Expert Solution
Check Mark

Answer to Problem 51P

Book value of the asset after 8th year is $15000.

Explanation of Solution

Given:

Useful Life = 5 years

Cost of asset = $85000

Salvage Value = $15000

ADR life = 8 years.

Calculation:

BVt=Bt=1tdt=Bndt

BV8=$85000(8×$8750)=$85000$70000=$15000.

Conclusion:

Thus, the book value of the asset after 8th year is $15000.

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