A newly proposed project has a first cost of $ 211962 and estimated annual income of $46410 per year for 10 years. Determine the PW value if the MARR is 10.4% per year.
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A newly proposed project has a first cost of $ 211962 and estimated annual income of $46410 per year for 10 years. Determine the PW value if the MARR is 10.4% per year.
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- Biomet Implants is planning new online patient diagnostics for surgeons while they operate. The new system will cost $300,000 to install in an operating room, $5000 annually for maintenance, and have an expected life of 4 years. The revenue per system is estimated to be $80,000 in year 1 and to increase by $10,000 per year through year 4. Determine if the project is economically justified using PW analysis and an MARR of 10% per year. $-18,470 not justified $-18,475 not justified $-18,477 not justified $-18,479 not justified O No correct answerTwo processes can be used for producing a polymer that reduces friction loss in engines. Process T will have a first cost of $660,000, an operating cost of $90,000 per year, and a salvage value of $80,000 after its 2-year life. Process W will have a first cost of $1,100,000, an operating cost of $25,000 per year, and a $120,000 salvage value after its 4-year life. Process W will also require updating at the end of year 2 at a cost of $90,000. Which process should be selected on the basis of a present worth analysis at a MARR of 12% per year? The present worth of process T is $- and the present worth of process W is $- The process selected on the basis of the present worth analysis is process (Click to select) ✓Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with low friction properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 16% per year, which alternative has the better present worth and what is that value (select the closest value)? Method First Cost AOC, per Year Salvage Value Life DDM $170,000 $65,000 $4,000 2 years LS $350,000 $40,000 $29,000 4 years DDM with a PW--$473,000 LS with a PW --$445,900 LS with a PW --$222,055 DDM with a PW--$109,300
- A professional mechanics who specializes in truck engines paid $46,000 for equipment that will have a $4800 salvage value after 5 years. The costs with each usage amount to $60 per day. The income is $290 per day for his services, how many days per year must he be worked in order to break even at an interest rate of 7% per 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 yearsThe cost of the extending a certain road at Yellowstone National Park is $1.7 million. Resurfacing and other maintenance are expected to cost $350,000 every 3 years with an interest rate of 6% per year. a) What is the Annual Worth based capitalized cost of the road? b) How will the answer (a) change if its Salvage is expected to be $3 million at the end of its useful Life?
- Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The operating cost with this method will be $29,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $130,000, an operating cost of $6000 in year 1, increasing by $6000 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 14% per year, what are the present worth of Method A and Method B? a. PWA = $-133,655 PWB = $-129,647 PWB = $-125,178 PWB = $-129,647 PWB = $-125,178 O b. PWA $-133,655 = O C.PWA $-150,260 O d. PWA = $-150,260 =Concurris Prototyping is committed to using the newest and finest equipment in its labs. Accordingly, Wilma, a senior engineer, has recommended that a 2-year-old piece of precision measurement equipment be replaced immediately. She believes it can be demonstrated that the proposed equipment is economically advantageous at a 15%-per year return and a planning horizon of 5 years. Perform the replacement analysis using the annual worth method, a 5-year study period, and the estimates below. Was Wilma correct Equipment Current Proposed Original purchase price, $ Current market value, $ -30,000 -42,000 15,000 Remaining life, years 5 15 Estimated value in 5 years, $ 7,000 10,000 5,000 -3,000 Salvage value after 15 years, $ AOC, $ per year The AW of the defender is $- Wilma (Click to select) correct. -14,000 and the AW of the challenger is $-1. The annual worth for years 1 through infinity of $75,000 now, $25,000 per year in years 1 through 15 and S40,000 per year in years 16 through infinity at a 10% interest per year is closest to: (a) $27,500 (b) $36,000 (c) $44,000 (d) $ 19,500 1. For the estimates in table below, calculate the equivalent annual cost of the project: First Cost, $ -950,000 Replacement Cost, year 2 AOC, $/ year Salvage value, $ -450,000 -600,000 175,000 Life, years 4 Interst Rate % 10
- Based on company records of similar equipment, a consulting aerospace engineer at Aerospatiale estimated AW values for a presently owned, highly accurate steel rivet inserter as shown. A challenger has ESL = 2 years and AWC= $−41,300 per year. The MARR is 12% per year. If Retained ThisNumber of Years The AW Value Is,$ per Year 1 −62,000 2 −51,000 3 −49,000 4 −53,000 5 −70,000 When should the next replacement evaluation take place, and under what assumption? The next replacement evaluation should take place in (Click to select) 4 2 3 5 years, and under the assumption that the (Click to select) defender challenger estimates do not change.A project has a first cost of $200,000 with annual costs of $50,000 and revenue of $90,000 per year.What is the payback period at (a) no-return, and (b) i = 7% per year?Polypropylene wall caps, used for covering exterior vents for kitchen cooktops. bathroom fans, dryers, and other building air exhausts. can be made by two different methods. Method X will have a first cost of $75,000. an operating cost of $32,000 per year, and a $9000 salvage value after 4 years. Method Y will have a first cost of $140,000. an operating cost of $24,000 per year, and a $19,000 salvage value after its 4-year life. At an interest rate of 10% per year, which method should be used on the basis of an annual worth analysis? 10% TABLE 15 Discrete Cash Flow: Compound interest Factors 10% Single Payments Uniform Series Payments Arithmetic Gradients Compound Amount Capital Recovery Present Sinking Fund Compound Amount Present Gradient Gradient Worth Worth Present Worth Uniform Series F/P P/F A/F F/A A/P P/A P/G A/G L1000 110000 0.9091 09091 08264 10000 0.47619 0.30211 10000 04762 0.9366 12100 21000 0.57619 0.40211 1.7355 13310 0.7513 3.3100 2.4869 23291 14641 06830 0.21547 4,6410…