Compare three alternatives on the basis of their capitalized costs at /= 11.00% per year and select the best alternative. (Include a minussign if necessary.)AlternativeFirst CostAOC, per YearSalvage ValueLife, YearsE$75000$-55000$190002The capitalized cost of alternative E is $The best alternative is (Click to select)F$-315000$16000$690004. alternative F is $G$ 815000$-4000$40000090and alternative G is $

Capitalized cost is an expenses that is added to the cost basis of the fixed asset on a company…

Answered: Compare three alternatives on the basis…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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5
Select the best alternative through internal rate of return analysis. Assume that the minimum attractive rate of return is 10% per year. Machine X |Machine YInitial cost $ 400,000.00 $ 500,000.00Annual operating cost $ 15,000.00 |$ 25,000.00 Annual profit $ 20,000.00 |$ 40,000.00Residual value $ 85,000.00 |$ 100,000.00Project life 15 years | none Engineering Economic Analysis
Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 14% per year and a study period of 10 years. Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years с $-48,000 $-9,000 $-1,400 $8,000 10 The present worth of alternative C is $ -117974.14 Alternative D offers the lower present worth. D $-34,000 $-9,000 $-1,500 $1,400 5 and that of alternative D is $ -73129.34 x
Using Incremental with EUAW analysis find the best alternative, MARR = %10. should use Excel and show your equations seperately, see below example: [A Benefit - [IC (A/P, i%, n) - Salvage (A/F, i, n)] + A Cost+ G Cost (A/G, i, n)] Data B C A First Cost $2,300,000 $2,780,000 $2,540,000 Salvage Value $82,000 $118,000 $97,000 Annual Benefit $580,000 $670,000 $650,000 M&O $65,000 $78,000 $71,000 M&O Gradient $11,000 $15,000 $12,500 Useful Life, Years 9 9 9 You - ZOOM
Q1 Table Q1 shows the engineering project with minimum attractive rate of return (MARR) at 15% per year. Table Q1 Proposal A Investment Cost RM 10,000 Expected Life 5 years Market (Salvage Value) -RM 1,000 Annual Receipts RM 8,000 Annual Expenses RM 4,000 (a) Determine the present worth (PW), future worth (FW) and annual worth (AW) based on details given in Table Q1.
Can some one please help me to answer each question correctly? please and thank you.
Given the following pertinent data for four alternatives, what is the best alternative using the incremental ROR method given MARR=10% Initial Cost Annual CF, $ Life, years 30 O Alt. A Alt. B O Alt. C A O Alt. D B +22,000 +35,000 с -200,000 -275,000 -190,000 -331,350 30 D +19500 +42,000 30 30
If the MARR=10%, compute the value of X that makes the alternatives equally desirable. Do not use spreadsheets. Alternatives Machine A Machine B First cost $12,000 $20,000 Annual Operating cost $1,400/year X Salvage value $2,000 $3,000 Life 4 years 8 years
A dyeing and finishing plant is interested in acquiring a dyeing machine for the production of a new product. Three alternatives are being considered assummarized below. Which alternative should be recommended if the plant’sMARR (hurdle rate) is 15% per year using (a) the IRR method and (b) the annual worth method?
Please help.
Complete the following analysis of cost alternatives and select the preferred alternative. The s A D Сapital investment $15,000 S16,000 $13,000 $18,000 Annual costs Market value 250 300 500 100 1,000 1,300 1,750 2,000 at EOY 10 FW(12%) -$49,975 –$53,658 ??? –$55,660 tudy period is 10 years and the MARR = 12% per %3D year.
Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (ie.. low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 10% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-190,000 $-55,000 $4,000 2 years The present worth for the DDM method is $ The present worth for the LS method is $ The LS method is selected. LS $-430,000 $-25,000 $39,000 4 years

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