IfMorethan2alternativesyouareCompanyandtodecideofistryinginvestmentA > Incremental analysis (AlternativAlternative: A > Incremental analysisBCiftheadditionalincrementWorth while, compare.pairswinsthenB vs CACompanyConsidering2differentmachinesatMARRat12%Both lifespans=10yearsM/C XM/CYInitial Cost160000285000Annual Operating Cost4500045000Benefits per yearSalvage Value900001050002000040000

Present value refers to the value of a certain sum of money in the future or cash flows in the…

If are you and to Company decide More than 2 alternatives of is frying investment Worth while, compare Alternative: A > Incremental analysis (Alternative) B C if the additional increment pairs then B vs C MARR A Company at Considering 2 different machines at 12% Both life spans = 10 years M/C ✗ м/с у Initial Cost 160000 285000 Annual Operating Cost 45000 45000 Benefits per year Salvage Value 90000 105000 20000 40000

If are you and to Company decide More than 2 alternatives of is frying investment Worth while, compare Alternative: A > Incremental analysis (Alternative) B C if the additional increment pairs then B vs C MARR A Company at Considering 2 different machines at 12% Both life spans = 10 years M/C ✗ м/с у Initial Cost 160000 285000 Annual Operating Cost 45000 45000 Benefits per year Salvage Value 90000 105000 20000 40000

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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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
Engineering Economics . I hope For the solution as soon as possible
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M2
QUESTION 3 For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10% Machine A Machine B Machine C First cost, $ 15,828 30000 10000 Annual cost, $/year Salvage value, s Life, years 8,753 6,000 4,000 4,000 5,000 1,000 Answer the below questions: A- AW for machine A= QUESTION 4 For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10% Machine A Machine B Machine C First cost, $ 15000 21,344 10000 Annual cost, $/year 8,314 6,000 4,000 Salvage value, $ Life, years 4,000 5,000 1,000 Answer the below questions: B- AW for machine B=
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Solve using future worth cost method
There are 5 national projects with infinite life time listed below. Select the best two projects if MARR is 8% per year using rate of return analysis? Project First Cost A 2000 B 1000 C 1500 D 7000 E 5000 ORDER: Annual income 200 130 150 600 260 Calculate i* for each alternative
For the project with cash flow given below find the most sensitive factor among annual benefit, annual cost and salvage value if the changes by ±20%. Initial investment Annual revenue Annual Cost Salvage Value MARR Life in years Project Rs.2,00,000 Rs.1,00,000 Rs. 40,000 Rs. 50,000 10% 5
A sportswear factory has two alternative machines for producing jerseys. Costs are shown below. Minimum acceptable rate of return is 8%.What is the annual cost for machine Y?
Solve using Excel. Show your formula.
FIRST COST ANNUAL COST SALVAGE VALUE ANNUAL INTEREST RATE STUDY PERIOD Power Line 20,000 1,950.90/month × 12= 23,410.8/year Solar 25,000 X 7= 125,000+10,000= 135,000 3,000/year 0 3.5% 25 ALTERNATIVE A: A solar panel costs 25,000. 7 solar panels are needed in a company. The installation fee is 10,000 and will have a useful 25 years with no salvage value. The maintenance per year is 5,000. The annual interest rate for solar panel is 1.5%. 0 1.5% 25 ALTERNATIVE B: A new power line will cost 20,000, with power costs expected to be 1,950.90 per month. The annual increase rate of electric bill is about 3.5%. Which alternative should be selected on the basis of a future worth analysis? Illustrate the cash flow diagram. USE MS EXCEL IN COMPUTATION and SHOW THE SOLUTION PROPERLY.