ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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- ! Required information Lego Group in Bellund, Denmark, manufactures Lego toy construction blocks. The company is considering two methods for producing special-purpose Lego parts. Method 1 will have an initial cost of $370,000, an annual operating cost of $135,000, and a life of 3 years. Method 2 will have an initial cost of $770,000, an operating cost of $140,000 per year, and a 6-year life. Assume 12% salvage values for both methods. Lego uses an MARR of 14% per year. Which method should it select on the basis of a present worth analysis? The present worth of method 1 is $ -1,094,517 and that of method 2 is $ -782,345 Method 1 is selected.arrow_forwardCompare three alternatives on the basis of their capitalized costs at /= 11.00% per year and select the best alternative. (Include a minus sign if necessary.) Alternative First Cost AOC, per Year Salvage Value Life, Years E $75000 $-55000 $19000 2 The capitalized cost of alternative E is $ The best alternative is (Click to select) F $-315000 $16000 $69000 4 . alternative F is $ G $ 815000 $-4000 $400000 90 and alternative G is $arrow_forwardb) National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. The details of the 2 alternative options are summarized in the table below. The interest rate is 10% per year. First Cost, S Annual Operating Cost, S Salvage Value, S Equipment Life (years) Consider only machine A now. Alternative A 15,000 3,500 1,000 4 Alternative B 18,000 3,100 2,000 8 a) Calculate the AW of Machine A for one life cycle. b) Calculate the AW of Machine A for LCM = 8 years (2 life cycles). c) Demonstrate the equivalence of AW over 2 life cycles and AW over one life cycle.arrow_forward
- National homebuilders Inc plans to purchase new rain gutter forming equipment. Two manufacturers offered the following estimates, First Cost, $ Annual Operating Cost, $/year Salvage Value, $ Life, years Vendor A 15,000 3500 1000 6 Vendor B 18,000 3100 2000 9 a. Determine which vendor should be selected on the basis of a PW comparison, if the MARR is 15% per year. b. National Homebuilders has a standard practice of evaluating all options over a 5-year period. If a study period of 5 years is used and the salvage values are not expected to change, which vendor should be selected?arrow_forwardCompare the alternatives C and D on the basis of a present worth analysis using an interest rate of 13% per year and a study period of 10 years. Alternative C $-46,000 $4,000 $-1,300 $6,000 10 First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life. Years The present worth of alternative C is $1 (Click to select) offers the lower present worth. HELIME BIEN D $-20,000 $-7,500 $-100 $1,300 5 and that of alternative D is $ Aarrow_forwardQuestion 3 For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A Answer the below questions: A- PW for machine A= 23,979 8,679 4,000 Machine B 30000 6,000 5,000 Machine C 10000 4,000 1,000arrow_forward
- I need help only with the second part of inputting this data into excel calculating the Annual Worth for both options. Then using Goal Seek to find out the new salvage value that will equal AW equations, thank you.arrow_forwardCompare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. Alternative First Cost $-40,000 $-25,000 AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-9,000 $-10,000 $-200 $-600 $11,000 $200 10 5 The present worth of alternative C is $ -33518.05 and that of alternative D is $-102608.3 Alternative D voffers the lower present worth.arrow_forwardProblem 05.017 Alternative Comparison - Different Lives Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (i.e., low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 19% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-160,000 $-40,000 $10,000 2 years The present worth for the DDM method is $ The present worth for the LS method is $ The (Click to select) method is selected. LS $-500,000 $-10,000 $33,000 4 yearsarrow_forward
- 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 $arrow_forward5arrow_forwardCompare the alternatives C and D on the basis of a present worth analysis using an interest rate of 14% per year and a study period of 10 years. Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years с $-48,000 $-9,000 $-1,400 $8,000 10 The present worth of alternative C is $ -117974.14 Alternative D offers the lower present worth. D $-34,000 $-9,000 $-1,500 $1,400 5 and that of alternative D is $ -73129.34 xarrow_forward
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