MARR is 8.4 %35004000A5005000Year O550060006500700010001500600022000200056002500300035005200480044004000800040004500Find the present worth?40005500161752004900180046002000220043004000240026003000400050006000280030003200salvage200034003600

Cash flow diagram:-A diagram that shows the value of future money. In a project's financial records,…

MARR is 8.4 % 3500 4000 4500 Year 5000 5500 6000 6500 7000 1000 1500 22000 6000 2000 5600 2500 5200 3000 3500 4800 4400 4000 4000 8000 4500 Find the present worth? 4000 5500 5200 1800 4900 2000 4600 4300 2200 4000 2400 2600 2800 3000 4000 5000 6000 3000 3200 salvage 2000 3400 3600

MARR is 8.4 % 3500 4000 4500 Year 5000 5500 6000 6500 7000 1000 1500 22000 6000 2000 5600 2500 5200 3000 3500 4800 4400 4000 4000 8000 4500 Find the present worth? 4000 5500 5200 1800 4900 2000 4600 4300 2200 4000 2400 2600 2800 3000 4000 5000 6000 3000 3200 salvage 2000 3400 3600

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Question

MARR is 8.4 %
3500
4000
A500
5000
Year O
5500
6000
6500
7000
1000
1500
6000
22000
2000
5600
2500
3000
3500
5200
4800
4400
4000
8000
4000
4500
Find the present worth?
4000
5500
16
17
5200
4900
1800
4600
2000
2200
4300
4000
2400
2600
3000
4000
5000
6000
2800
3000
3200
salvage
2000
3400
3600

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question 10 Based on the following cash flows and an interest rate of 10% per year Alternative X -300,000 -65,000 25,000 5 First cost, $ Annual cost, $/year Salvage value, $ Life, years -850,000 -10,000 160,000 % 00 1) What is the Annual Worth of Alternative X? $ 2) What is the Annual Worth of Alternative Y? $ 3) What is the annual worth of the first cost and salvage value for project X? $ 4) What is the annual worth of the first cost and salvage value for project Y? $ 5) What is the interest rate? 6) What is the project life for Alternative X? 7) What is the project life for Alternative Y? (Click to select)✓ years years
Graph on right, shows AW of costs versus number of service years of project X. Based on the graph, economic service life of the project is Larger costs Total AW of costs AW of AOC Capital recovery 3.5 10 Years О 3.5 years O 10 years O 1 year О З уears AW of costs, S/year
Question 2 Full explain this question and text typing work only thanks
Marmara Processing Technologies Company wants to buy milling cutters in order to meet its increasing production needs. As a result of the studies, the data in the table below were found. According to these data, what will be the current cost of the annual operating expense parameter of alternative B? A Purchase Cost 10.000 12.000 Annual Operating Expense 1.500 1.700 Scrap Values 2.000 3.000 Economic Life 4 4 Capital Cost 10% 10%
For the cash flows shown, determine the incremental cash flow between machines B and A for (a) year 0, (b) in year 3, and (c) in year 6. Machine First Cost, $ АОС, $/year Salvage value, $ 3,000 Life, years B -15,000 -25,000 -1,600 -400 6,000 3 4 O (a) = -10000 (b) = -1800 (c) = 1200 %3D !3! O (a) = -10000 (b) = 13200 (c) = 4200 !3! %3! O (a) = -10000 (b) = 13200 (c) =6000 %3! %3D O (a) = -10000 (b) = 13200 (c) = 13200 %3D
An industrial plant has hired a contractor to develop a solar farm nearby for the industrial plant's electricity consumption. During the 3-year construction period, the contractor will need to run a generator to power the construction. Cost analysis from previous projects indicates: Generator Brand Installed Cost Cost per Hour A $22M $500 B $23M $400 C $25M $250 D $30M $150 At the end of 3 years, the generator will have a salvage value equal to the cost of removing them. The generator will operate 6,000 hours per year. The lowest interest rate at which the contractor is willing to invest money is 7%. (The minimum required interest rate for invested money is called the minimum attractive rate of return, or MARR.) Select the alternative with the least present worth of cost. O Choice "C" with $29,872,900 O Choice "B" with $29,298,320 O Choice C with $28,936,450 O Choice "A" with $29,872,900
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Solve using future worth cost method
A company buys equipment for $30000. Annual net savings is 1000 per year for 8 years. The equipment is sold for a salvage value of $40,000 at end of 8 years. The company wants 8% return on any investment they make. Is this a good investment for 4) them? Use the Annual Worth method.
i=10%