ENGINEERING ECO ANALYSIS W/STUDY GUIDE
13th Edition
ISBN: 9780190693053
Author: NEWNAN
Publisher: Oxford University Press
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Question
Chapter 11, Problem 49P
To determine
i.
MACRS
Expert Solution
Answer to Problem 49P
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 49P
Straight Line Depreciation is $20000 per annum.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM depreciation =
Cost / Useful life=$20000010 = $20,000 per year
| Year | Beginning-of-year Book Value ($) | 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)
=1+2+3+4+5+6+7+8+9+10=55
Depreciation base = $200,000
Annual depreciation in year N = Depreciation base×(Number of years remaining at beginning of year N)SOYD
| SOYD | ||||
| Year | Depreciation Base ($) | 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 = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
1Useful Life =110 = 0.1
DDB Rate =
0.1×2=0.2
| DDB | ||||
| Year | Depreciation Base ($) | 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 49P
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)
=1+2+3+4+5+6+7+8+9+10=55
Depreciation base = $200,000
Annual depreciation in year N = Depreciation base×(Number of years remaining at beginning of year N)SOYD
| SOYD | ||||
| Year | Depreciation Base ($) | 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 = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
1Useful Life =110 = 0.1
DDB Rate =
0.1×2=0.2
| DDB | ||||
| Year | Depreciation Base ($) | 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 49P
Double declining balance depreciation is $ 97600.
Explanation of Solution
Given:
Cost of machine = $200000
Useful Life = 10 years.
Calculation:
SLM Depreciation rate =
1Useful Life =110 = 0.1
DDB Rate =
0.1×2=0.2
| DDB | ||||
| Year | Depreciation Base ($) | 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 49P
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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Question 6 of 18
>
The graph shows the average total cost (ATC) curve, the
marginal cost (MC) curve, the average variable cost (AVC)
curve, and the marginal revenue (MR) curve (which is also
the market price) for a perfectly competitive firm that
produces terrible towels. Answer the three questions,
assuming that the firm is profit-maximizing and does not shut
down in the short run.
What is the firm's total revenue?
S
What is the firm's total cost?
$
What is the firm's profit? (Enter a negative number for a loss.)
$
Price
$320
$300
$200
$150
205 260
336
365
Quantity
MC
ATC
AVC
MR=P
1. Suppose that the two nations face the following benefits of pollution, B, and costs of
abatement, C: BN = 10, Bs = 7; CN = 5, Cs = 4. Further assume that if the nation chooses to
abate pollution, it still receives the benefits of pollution but now must pay the cost of
abatement as well.
a. Identify the payoffs that accrue to each nation under the four different possible
outcomes of the game and present these payoffs in the normal form of the game.
b. Recall that the term dominant strategy defines the condition that a player in a game
would prefer to play that strategy (in this case either pollute or abate) regardless of
the strategy chosen by the other player in the game. Does either nation have a
dominant strategy in this game? If so, what is it?
c. Identify the Nash equilibria, or non-cooperative equilibria, of this game.
Chapter 11 Solutions
ENGINEERING ECO ANALYSIS W/STUDY GUIDE
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. 64PCh. 11 - Prob. 65PCh. 11 - Prob. 66PCh. 11 - Prob. 67PCh. 11 - Prob. 68PCh. 11 - Prob. 69PCh. 11 - Prob. 70PCh. 11 - Prob. 71PCh. 11 - Prob. 72P
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