ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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Question
Chapter 11, Problem 43P
To determine
i.
MACRS
Expert Solution
Answer to Problem 43P
MACRS depreciation is $ 84800.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
A 10-year property class with half-yearly convention has been assumed.
| MACRS | ||||
| Year | Depreciation Base ($) | Depreciation Rate (%) | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 10 | 20,000 | 1,80,000 |
| 2 | 2,00,000 | 18 | 36,000 | 1,44,000 |
| 3 | 2,00,000 | 14.4 | 28,800 | 1,15,200 |
| TOTAL | 84,800 |
Conclusion:
MACRS depreciation is $ 84800.
To determine
ii.
Straight Line Depreciation.
Expert Solution
Answer to Problem 43P
Straight Line Depreciation is $20000 per annum.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM depreciation =
| Year | Beginning-of-year Book Value ($) | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 20,000 | 1,80,000 |
| 2 | 1,80,000 | 20,000 | 1,60,000 |
| 3 | 1,60,000 | 20,000 | 1,40,000 |
| TOTAL | 60,000 |
Conclusion:
Straight Line Depreciation is $20000 per annum
iii.
Sum-of-years-digit depreciation.
Sum-of-years-digit depreciation is $98182.
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
Sum-of-years-digit depreciation (SOYD)
Depreciation base = $200,000
Annual depreciation in year N =
| SOYD | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 10/55 | 36,364 | 1,63,636 |
| 2 | 2,00,000 | 9/55 | 32,727 | 1,30,909 |
| 3 | 2,00,000 | 8/55 | 29,091 | 1,01,818 |
| TOTAL | 98,182 |
Conclusion:
Sum-of-years-digit depreciation is $98182.
iv.
Double declining balance depreciation.
Double declining balance depreciation is $ 97600.
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
DDB Rate =
| DDB | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.2 | 40,000 | 1,60,000 |
| 2 | 1,60,000 | 0.2 | 32,000 | 1,28,000 |
| 3 | 1,28,000 | 0.2 | 25,600 | 1,02,400 |
| TOTAL | 97,600 |
Conclusion:
Double declining balance depreciation is $ 97600
(e).
v.
150% Declining balance depreciation.
150% Declining balance depreciation is $77175.
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
150% Declining balance depreciation rate = 0.1 x 1.5 = 0.15
| 150% DECLINING BALANCE METHOD | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.15 | 30,000 | 1,70,000 |
| 2 | 1,70,000 | 0.15 | 25,500 | 1,44,500 |
| 3 | 1,44,500 | 0.15 | 21,675 | 1,22,825 |
| TOTAL | 77,175 |
Conclusion:
150% Declining balance depreciation is $77175.
To determine
iii.
Sum-of-years-digit depreciation.
Expert Solution
Answer to Problem 43P
Sum-of-years-digit depreciation is $98182.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
Sum-of-years-digit depreciation (SOYD)
Depreciation base = $200,000
Annual depreciation in year N =
| SOYD | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 10/55 | 36,364 | 1,63,636 |
| 2 | 2,00,000 | 9/55 | 32,727 | 1,30,909 |
| 3 | 2,00,000 | 8/55 | 29,091 | 1,01,818 |
| TOTAL | 98,182 |
Conclusion:
Sum-of-years-digit depreciation is $98182.
iv.
Double declining balance depreciation.
Double declining balance depreciation is $ 97600.
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
DDB Rate =
| DDB | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.2 | 40,000 | 1,60,000 |
| 2 | 1,60,000 | 0.2 | 32,000 | 1,28,000 |
| 3 | 1,28,000 | 0.2 | 25,600 | 1,02,400 |
| TOTAL | 97,600 |
Conclusion:
Double declining balance depreciation is $ 97600
(e).
v.
150% Declining balance depreciation.
150% Declining balance depreciation is $77175.
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
150% Declining balance depreciation rate = 0.1 x 1.5 = 0.15
| 150% DECLINING BALANCE METHOD | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.15 | 30,000 | 1,70,000 |
| 2 | 1,70,000 | 0.15 | 25,500 | 1,44,500 |
| 3 | 1,44,500 | 0.15 | 21,675 | 1,22,825 |
| TOTAL | 77,175 |
Conclusion:
150% Declining balance depreciation is $77175.
To determine
iv.
Double declining balance depreciation.
Expert Solution
Answer to Problem 43P
Double declining balance depreciation is $ 97600.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
DDB Rate =
| DDB | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.2 | 40,000 | 1,60,000 |
| 2 | 1,60,000 | 0.2 | 32,000 | 1,28,000 |
| 3 | 1,28,000 | 0.2 | 25,600 | 1,02,400 |
| TOTAL | 97,600 |
Conclusion:
Double declining balance depreciation is $ 97600
(e).
To determine
v.
150% Declining balance depreciation.
Expert Solution
Answer to Problem 43P
150% Declining balance depreciation is $77175.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
150% Declining balance depreciation rate = 0.1 x 1.5 = 0.15
| 150% DECLINING BALANCE METHOD | ||||
| Year | Depreciation Base ($) | Depreciation Rate | Annual Depreciation ($) | Year-end Book Value ($) |
| 1 | 2,00,000 | 0.15 | 30,000 | 1,70,000 |
| 2 | 1,70,000 | 0.15 | 25,500 | 1,44,500 |
| 3 | 1,44,500 | 0.15 | 21,675 | 1,22,825 |
| TOTAL | 77,175 |
Conclusion:
150% Declining balance depreciation is $77175.
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Chapter 11 Solutions
ENGR.ECONOMIC ANALYSIS
Ch. 11 - Prob. 1QTCCh. 11 - Prob. 2QTCCh. 11 - Prob. 3QTCCh. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - Prob. 5PCh. 11 - Prob. 6PCh. 11 - Prob. 7P
Ch. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10PCh. 11 - Prob. 11PCh. 11 - Prob. 12PCh. 11 - Prob. 13PCh. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64P
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