What is the smallest acceptable annual income from a project which has a $ 92,498 investment cost and a $9,936 salvage value is the life is 15 years and the MARR is 15% ?
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- A computerized fabricating system has a first cost of $180,000, and an annual operating cost of $84,000 in years 1 and 2, increasing by $5,000 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within the useful life of 5 years. Using a MARR of 15% per year, determine the annual worth of total cost when keeping the system for 2 years and 4 years.Your firm is currently paying $3,000 a year to a commercial garbage collection agency to haul waste paper to the city dump. The paper could be sold as waste paper if it were baled and strapped. A paper baler is available at the following conditions: Purchase price = $6,500 Labor to operate baler = $3,500/year Strapping material = $300/year Life of baler = 30 years Salvage value = $500 MARR = 10%/year If it is estimated that 500 bales would be produced per year, what would the selling price per bale to a wastepaper dealer have to be to make this project acceptable? Assume no inflation.Apricot Computers is considering replacing its material handling system and either purchasing or leasing a new system. The old system has an annual operating and maintenance cost of $31,000, a remaining life of 8 years, and an estimated salvage value of $5,300 at that time. A new system can be purchased for $259,000; it will be worth $24,000 in 8 years; and it will have annual operating and maintenance costs of $16,000/year. If the new system is purchased, the old system can be traded in for $19,000. Leasing a new system will cost $25,000/year, payable at the beginning of the year, plus operating costs of $7,100/year, payable at the end of the year. If the new system is leased, the old system will be sold for $9,000. MARR is 15%. Compare the annual worths of keeping the old system, buying a new system, and leasing a new system based upon a planning horizon of 8 years. Click here to access the TVM Factor Table Calculator For calculation purposes, use 5 decimal places as displayed in the…
- Engineering economy please show solutionConsider the following mutually exclusive alternatives: Alternative A Alternative B Capital investment $473,000 $1,114,000 Net annual receipts $104,100 $235,000 Both alternatives have a useful life of 20 years and no market value at that time. The MARR is 20 % per year. Determine the annual worth (AW) of the most profitable course of action. (Enter your answer as a number without the dollar sign.)Three independent project R, S, & T. If the PW of these projects at MARR is as follows: PWS= $80,000, PWR = -$50,000, PWT = $70,000. Write down the bundles that you can be immediately excluded from the analysis.
- Answer part A Answer part BWhat is the payback period for a project with the following characteristics, given a minimum attractive rate of return (MARR) of 12%? First Cost $20,000 Annual Benefits $8,000 Annual Maintenance $2,000 in year 1, then increases by $500 per year Salvage value $2,000 Useful Life 10 yearsProblem 05.024 Alternative Comparison - Different Lives You and your partner have become very interested in cross-country motorcycle racing and wish to purchase entry-level equipment. You have identified two alternative sets of equipment and gear. Package K has a first cost of $250,000, an operating cost of $5,500 per quarter, and a salvage value of $30,000 after its 2-year life. Package L has a first cost of $170,000 with a lower operating cost of $3,500 per quarter and an estimated $25,000 salvage value after its 4-year life. Which package offers the lower present worth analysis at an interest rate of 8% per year, compounded quarterly? The present worth of package K is $[ and that of package L is $[ (Click to select) offers the lower present worth.
- A factory needs to increase its facilities and study two alternatives:Alternative 1: Construction of a reinforced concrete shed at a price of $5 million and a useful life of 40 years. The demolition cost $200,000 and its annual maintenance cost is $100,000.Alternative 2: Construction of a masonry shed at a cost of $3 million and a useful life of 40 years with an expected renovation of $2 million at the end of 20 years and a salvage value of $100,000. The annual maintenance cost is $150,000.Considering a TMAR of 20% per year, what is the best alternative?A $45,000 investment in a new conveyor system is projected to improve throughput and increasing revenue by $14,000 per year for five years The conveyor will have an estimated market value of $4,000 at the end of five years. Using an MARR = 12% per year, what is the FW? $-19,329.40 $13,635.70 $16,998.90 none of the above K ?The plant manager at a company would like to perform an analysis for a new $250,000 machine. She estimates benefits of $20,000 in the first year, and benefits are increasing by 10% per year. 1) What is the payback period for the machine? 2)Suppose that the machine life is 15 years and machine has a salvage value of 20% of the initial cost at the end of its useful life. If the MARR of the company is 11 % per year, is this investment acceptable? Why?