Multiple Choice$40,560.$23,160.$72,432.$58,950. Following cash flows for alternatives X and Y at an interest rate of 10% per year.Machine X Machine YInitial cost, $-146,000-220,000AOC, $/year-15,000 -10,000Annual revenue, $/year80,00075,000Salvage value, $10,00025,000Life, years36In comparing the alternatives on a present worth basis, the PW of Machine X is closest to

Step 1: Given Value for Calculation Initial Cost = i = $146,000Annual Operating Cost = aoc =…

Answered: Following cash flows for alternatives X…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

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Buying Equipment 1 from XYZ company and company ABC will give the production similar productivity input of 400,000.00 per year. The equipment from company XYZ has a purchase price of 200,000.00, annual maintenance of 5,000.00, and production life of 10 years, while the equipment from company ABC has a purchase price of 100,000.00, annual maintenance of 2,000.00, and production life of 12. Using present worth method, what is the total present worth of each alternatives? Use MARR of 20%
help please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all working
In comparing the machines on a present worth basis, the present worth of machine X, the cost of money is 10% per year, is closest to Machine X Machine Y Initial cost, $ -66,000 -46,000 Annual cost, $/year Salvage value, $ Life, years -10,000 -15,000 10,000 24,000 3. $-103,908 $-87,840 None of the answers S-65,270
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
QUESTION 8 Two processes can be used for producing a polymer that reduces friction loss in engines. Which process should be selected on the basis of a present worth analysis using the LCM, at an interest rate of 10% per year? Answer the below questions to select the best Option. Process A Process B First cost, $ 1,397,151 534,579 Annual operating cost, $ per year 90,383 32,000 Salvage value, $ 80,000 130,000 Update cost at year 2, $ 15,000 Life, years 4 8 EGN 3212 - Eco PW for Process A-
am. 122.
Required information A company that manufactures magnetic flow meters expects to undertake a project that will have the cash flows estimated. First cost, $ Equipment replacement cost in year 2, $ Annual operating cost, $/year Salvage value, $ Life, years -800,000 -300,000 -850,000 250,000 4 At an interest rate of 10% per year, what is the equivalent annual cost of the project? Find the AW value using tabulated factors. The equivalent annual cost of the project is $-[
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Machine Initial cost, $ Annual cost, $/year Salvage value, $ Life, years O-$87,840 O-$103,910 The estimates for the machines are for two mutually exclusive alternatives. The co of money is 10% per year. The present worth of Machine X is closest to: O $114,310 $65.270 O X -66,000 -10,000 10,000 At 6 Y -46,000 -15,000 24,000 3
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