Engineer Economic Analys Final Exam

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Please answer the questions in the docx file, about Engineer Economic Analys. Thank you

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Final Exam Total Points: 10 Due date: July 27, 2018 1. (2.5 points) Find the present value, P, for the following cash flows. i = 12% 2. (2.5 points) Star Inc. must purchase a small size milling machine. The following is known about the machine and about possible cash flows. The machine is expected to have a useful life of 8 years. The company has a MARR of 7%. Determine the NPW of the machine. p=.25 First cost Annual savings Annual costs Actual salvage value 3. p=.50 p=.25 $40,000 $40,000 $40,000 2,000 5,000 8,000 12,000 8,000 6,000 4,000 5,000 6,500 (2.5 points) The company accountant is uncertain which of three depreciation methods the firm should use for welding equipment that costs $150,000, and has a zero salvage value at the end 1 of a 10-year depreciable life. Compute the depreciation schedule for the welding equipment using the methods listed: a. Straight line b. Double declining balance c. Sum-of-years’ –digits Based on your analysis, which depreciation method is most profitable for the firm? 4. (2.5 points) A Petroleum company recently completed construction on a large refinery in Louisiana. The final construction cost was $71,000,000. The refinery covers a total of 260 acres. The Expansion and Acquisition Department at the company is currently working on plans for a new refinery in Texas. The anticipated size is approximately 360 acres. If the powersizing exponent for this type of facility is .70, what is the estimated cost of construction? 2 ...
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Tutor Answer

EngAsh
School: Cornell University

Here is the answerπŸ˜€ If you have any question, please don't hesitate to ask

Question1:

The Answer
3

𝑃𝑉 = βˆ‘
𝑑=0

𝐢𝐹
(1 + 𝑖)𝑑

T

Cash flow

Present value

0

30

30

1

20

17,85

2

20

15,94

3

30

21,35

𝑃𝑉 = 85.14
Question2:

The Answer
𝑡𝑷𝑾 = βˆ’π’‡π’Šπ’“π’”π’• 𝒄𝒐𝒔𝒕 + βˆ‘πŸ–π’•=𝟏

𝒂𝒏𝒏𝒖𝒂𝒍 π’”π’‚π’—π’Šπ’π’ˆ
βˆ’
(𝟏+𝟎.πŸŽπŸ•)𝒕

βˆ‘πŸ–π’•=𝟏

𝒂𝒏𝒏𝒖𝒂𝒍 𝒄𝒐𝒔𝒕
(𝟏+𝟎.πŸŽπŸ•)𝒕

+

𝑨𝒄𝒕𝒖𝒂𝒍 π‘Ίπ’‚π’π’—π’‚π’ˆπ’† 𝑽𝒂𝒍𝒖𝒆𝒔
(𝟏+𝟎.πŸŽπŸ•)πŸ–

𝑃
𝑃
𝑝
π‘ƒπ‘Š1 = βˆ’40,000 βˆ’ 12000 ( , 7%, 8) + 2000 ( , 7%, 8) + 4000 ( , 7%, 8)
𝐴
𝐴
𝑓
𝑃
𝑃
π...

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awesome work thanks

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