Following cash flows for alternatives X and Y at an interest rate of 10% per year. Initial cost, $ AOC, $/year Annual revenue, $/year Salvage value, $ Life, years Machine X Machine Y -146,000 -220,000 Multiple Choice -15,000 -10,000 80,000 75,000 10,000 25,000 3 6 In comparing the alternatives on a present worth basis, the PW of Machine X is closest to
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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%In engineering economics it is often used present worth analysis to determine best alternative between options. I was wondering how do we know what type of formula to use. Sometimes it is used the formula with P/A, others A/P, as F/A and P/F and so on. How do we know which fraction to use?Please do not give solution in image format thanku 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 operatingand maintenance costs of $600,000, and requires overhauls every 10 years at a cost of $3,000,000. Using a 5 percent/year interest rate, determine the capitalized cost for each design and recommend which should be chosen
- 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% 5Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 10% per year, compounded annually. Alternative A Alternative B Initial Cost 490 1,315 Annual Benefit 80 266 Salvage Value 228 451 Useful Life (yrs) 2 3 O Alternative B. because it costs $142.96 more than Alternative A, in terms of present worth O Alternative A, because it costs $142.96 less than Alternative B, in terms of present worth Alternative A. because its present worth is positive Alternative B, because it only incurs the initial cost once every three years instead of every two yearsA company is considering the following two alternatives: Alt. 1 $42,500 6,000 18,500 12,000 1.5 years Alt. 2 $70,000 First cost Annual maintenance Annual savings Salvage value Useful life 4,000 20,000 25,000 3 years If interest is 20% per year, which alternative is preferred?
- Required information Akash Uni-Safe in Chennai, India, makes Terminator fire extinguishers. The company needs replacement equipment to form the neck at the top of each extinguisher during production. Machine First cost, $ AOC, $ per year Salvage value, $ Life, years D E -70,000 -15,000 10,000 6 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. -60,000 -12,000 8,000 Select between two metal-constricting machines. Use the corporate MARR of 15% per year with future worth analysis using tabulated factors. The future worth of machine D is $- 8008 and the future worth of machine E is $- 20280 The machine selected based on the future worth analysis isEconomics Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Delivery costs are $31,500 per year. Nancy can buy a used truck for $8,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $24,000 per year. At the end of 3 years, the used truck will have an estimated salvage value of $3,200. Nancy’s MARR is 25%/year. What is the present worth of this investment? $5. For the cash flows shown below, use an annual worth comparison to determine which alternative is best at an interest rate of 12% per year compounded monthly. Alternative First Cost, $ Monthly Costs, Smonth 30,000 Cost every 4 years,S Salvage Value,S Life, VrS 60,000 300,000 20,000 900,000 15,000 60,000 200,000 25,000 20 7,000
- A Cost $4000 $2000 $6000 $1000 $9000 Annual benefit 639 410 761 117 750 Useful life, in years 20 20 20 20 20 i* 15% 20% 11.1% 9.9% 5.45% Perform incremental ROR analysis to select best one. MARR=6%; (P/A, 6%, 20) = 11.47 %3DDRAW THE CASH FLOW DIAGRAM Company A is considering two alternatives. Machine A has a first cost of $15,000 and a salvage value of $2,000. The life is 6 years and it has an annual maintenance and operating cost of $1,000. Machine B has a first cost of $18,000, a life of 8 years and no salvage value. The annual operating cost is $800. Which machine should be used to justify the purchase of such machine if money is worth 7% and calculate the difference between the equivalent annual worths.The following data is available for three different alternatives. Assume an interest rate of 2% per year, compounded annually. Alternative A Alternative B 7,000 6,600 975 2,148 infinite 10 Initial Cost Annual Benefit Useful Life (yrs) Alternative C 14,000 6,027 5 Alternatives B and C are replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative. Choose Alternative A because it lasts the longest Choose Alternative A because its net present worth is postive O Choose Alternative C because its net present worth is $82,185.21 more than its nearest competitor Choose Alterntive C because it has the highest annual benefit