nods för disposing of å non-hazardous An environmental engineer is considering three chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. Determine the equivalent present worth value of the LAND APPLICATION using its AW value based on 10% per year. Refer to table 2 for additional information.
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- 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 AnalysisPlease no written by hand solution Two projects are shown below. Interest rate is 10% per year with yearly compounding. Project A Project B Project Life 10 years 10 years Benefits $500 at year = 0 and $100 per year (at year = 1,2,3,…,10) $15 per year starting year 2 (at year = 2,3,4,…,10) Disbenefits $40 per year (at year = 1,2,3,…,10) $1 per year starting year 2 (at year = 2,3,…,10) Costs $600 at year = 0, $100 at year = 1 $50 at year = 0 (1) Find the conventional B/C ratio for Project A. (2) Find the conventional B/C ratio for Project B. (3) Based on B/C analysis (using conventional B/C ratio), which between A and B is more attractive?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 Original purchase price, $ Current market value, $ Remaining life, years Estimated value in 5 years, $ Salvage value after 15 years, $ AOC, $ per year Current -30,000 15,000 5 7,000 -11,000 The AW of the defender is $- challenger is $- Wilma ((Click to select)) correct. Proposed -40,000 -- 15 10,000 5,000 -3,000 and the AW of the
- Equipment needed at Valero Corporation refinery for the conversion of corn stock toethanol, a cleaner burning gasoline additive, will cost $175,000 and have net cashflows of $35,000 the first year, increasing by $10,000 per year over the life of 5 years.Develop a spreadsheet chart that plots AW vs interest rate to show where (what interestrate) the project switches from financially justified to unjustified. Use AW as the verticalaxis of the chart, and interest (i) as the horizontal axis of the chart. Use a range ofinterest rates spanning from 10% to 20%, and use increments of 1%. Label both axesand use a title for the chart.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--$109,300Consider the comparison between alternatives A and B. Using present worth method of analysis, and at an interest rate of 10% per year, the values of n that you should use in the uniform series factors to make a correct comparison are: A B First Cost ₱500,000 ₱900,000 Annual Operating Cost ₱100,000 ₱40,000 Salvage Value ₱130,000 ₱150,000 Life 3 years 6 years a. n=3 years for A and n=3 years for B b. n=3 years for A and n=6 years for B c. n=6 years for A and n=6 years for B d. n=6 years for A and n=3 years for B
- The future worth of a project with initial cost P, positive annual cash flows of A, salvage value S, and interest rate of i over a life of n years can be calculated using which statement? (a) FW = −P(F/P, i%, n) + A(F/A, i%, n) + S(F/P, i%, n) (b) FW = P(F/P, i%, n) + A(F/A, i%, n) + S (c) FW = −P(P/F, i%, n) + A(F/A, i%, n) − A[(P/A, i%, n) + S (d) FW = −P(F/P, i%, n) + A(F/A, i%, n) + S?Seawater contains 2.1 pounds of magnesium per ton. By using the processingmethod A, 85% of the metal can be recovered at a cost of 3.25 per ton of water pumped andprocessed. If process B is used, 70% of the available metal is recovered, at a cost of only 2.60per ton of water pumped and processed. The two processes are substantially equal as toinvestment costs and time requirements. (a) If the extracted metal can be sold for 2.40 perpound, which processing method should be used? (b) At what selling price for the metal wouldbe two processes be equally economical?Answer with complete solution brief explanation. Thank you.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 years
- 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 $-Most likely estimates for a project are as follows. MARR Useful life Initial investment Receipts - Expenses (R-E) 10% per year 9 years $5,000 $1,200/year Click the icon to view the relationship between the PW and the percent change in parameter. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. (b) To which variable is the PW most sensitive to? OA. Receipts - Expenses OB. Usefule life OC. Initial Investment (a) Determine whether the statement "An initial investment of $6,000 keeps the investment economical." is true or false. O False O TrueA remotely located air sampling station can be powered by solar cells or by running an electric line to the site and using conventional power. Solar cells will cost $7000 to install and will have a useful life of 4 years with no salvage value. Annual costs for inspection, cleaning, etc. are expected to be $1600. A new power line will cost $14000 to install, with power costs expected to be $1300 per year. Since the air sampling project will end in 4 years, the salvage value of the line is considered to be zero. At an interest rate of 8.00% per year, which alternative should be selected on the basis of a future worth analysis? (Include a minus sign if necessary.) The future worth of solar cells is $[ and that of electric line is $ (Click to select) :) should be selected on the basis of a future worth analysis.