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- When comparing mutually exclusive alternatives that have different lives by the present worth method, it is necessary to: (a) Always compare them over a period equal to the life of the longer-lived alternative. (b) Always compare them over a time period of equal service. (c) Always compare them over a period equal to the life of the shorter-lived alternative. (d) Find the present worth over one life cycle of each alternative. (e) None of the above a C dTwo alternatives for a construction project are being considered. Both projects have a 5-year life. Alternative A's initial cost is $2,260 and yields $355 annually for 5 years. Alternative B initially costs $5500 and yields $1,250 annually for 5 years. The rate of return on the difference between the alternatives is approximately Group of answer choices 3.6% 15% 8% 12%Two incinerators are being considered by a waste management company. Design A has an initial cost of $2,500,000, has annual operating and maintenance costs of $800,000, and requires overhauls every 5 years at a cost of $1,250,000. Design B is more sophisticated, including computer controls; it has an initial cost of $5,750,000, has annual operating and maintenance costs of $600,000, and requires overhauls every 10 years at a cost of $3,000,000. Using a 5%/year interest rate, determine the capitalized cost for each design and recommend which should be chosen.
- 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 AnalysisQ1 - With these new estimates, What is the AW of the Investment and Salvage value for YORK AC unit? Q2 - With these new estimates, What is the FW of the Investment and Salvage value for YORK AC unit?Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $18,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR is 10%/year. Solve, a. What is the future worth of this investment? b. What is the decision rule for judging the attractiveness of investments based on future worth? c. Is the filter economically justified?
- CASE STUDY ANNUAL WORTH ANALYSIS THEN AND NOW Background and Information Harry, owner of an automobile battery distributorship in Atlanta, Georgia, performed an economic analysis 3 years ago when he decided to place surge protectors in-line for all his major pieces of testing equipment. The estimates used and the annual worth analysis at MARR = 15% are summarized below. Two different manufacturers' protectors were compared. Cost and installation, $ Annual maintenance cost, S per year Salvage value, $ Equipment repair savings. $ Useful life, years. O 1 2 3 4 5 6 7 8 9 10 A MARR= Year 0 1 234567 11 12 13 14 15 16 17 AW element 18 Total AW 899 10 PowrUp -26,000 -800 2,000 25.000 B 15% Investment and salvage -26,000 ...... 2,000 -6,642 6 Lloyd's -36,000 -300 C 3,000 35.000 10 PowrUp Annual maintenance 0 -800 -800 -800 -800 -800 -800 -800 D Case Study Exercises 1. Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven…Select between the two options using the corporate MARR of 15% per year and a future worth analysis for the expected use period. Option First Cost AOC, per Year Expected Market Value Expected Use E $-54,000 $-12,000 $5,000 З years $-64,000 $-14,000 $6,000 6 years The future worth of option D is $ The future worth of option E is $ Option (Click to select) vis sefected. (Click to select) DEIn case of high interest rate, the best suitable method of analysis is: Select one: a. future worth analysis b. present worth analysis c. annual worth analysis d. incremental analysis
- Three independent alternatives are given below. If MARR is 18%, what is your decision? A B C Initial Cost $4.50 $1.90 $1.20 Annual Revenues $4.00 $2.50 $3.00 Salvage Value $0.50 $0.90 $0.00 Annual Operating & Maintenance Costs $1.20 $1.90 $2.70 Estimated life, in years 3 InfiniteWhat is the annual worth approach? How is it useful?Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 15.00% per year and a study period of 10 years. (Include a minus sign if necessary.) Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years The present worth of alternative C is $ с $-50000 $-8000 $-1500 $14000 10 $-21000 $-9000 $-200 $1500 5 and that of alternative D is $