Alternative X has a first cost of 25000 an annual operating cost of 4700 , and a salvage value of 7125 after 24 rear. Alternative Y has a first cost of 26000 an annual operating cost of 4800, and a salvage value of 10200 after 24 year. If MARR of 24% per year, approximately what is the PW of each alternative? elect one: O a. PW of X= -44874 and PW of Y= -46285 O b. PW of X= -45776 and PW of Y= -47215 OC. PW of X= -44430 and PW of Y= -45827 O d. PW of X= -45323 and PW of Y= -46748

Alternative X has a first cost of 25000 an annual operating cost of 4700 , and a salvage value of 7125 after 24 rear. Alternative Y has a first cost of 26000 an annual operating cost of 4800, and a salvage value of 10200 after 24 year. If MARR of 24% per year, approximately what is the PW of each alternative? elect one: O a. PW of X= -44874 and PW of Y= -46285 O b. PW of X= -45776 and PW of Y= -47215 OC. PW of X= -44430 and PW of Y= -45827 O d. PW of X= -45323 and PW of Y= -46748

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

See similar textbooks

Similar questions

! 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.
Compare 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 $
b) 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.
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?
Alternative X has a first cost of 9000 an annual operating cost of 1500, and a salvage value of 2325 after 8 year. Alternative Y has a first cost of 10000 an annual operating cost of 1600 , and a salvage value of 3800 after 8 year. If MARR of 8% per year, approximately what is the PW of each alternative? Select one: O a. PW of X= -16528 and PW of Y= -17313 O b. PW of X= -16860 and PW of Y= -17661 O c. PW of X= -16693 and PW of Y= -17486 O d. PW of X= -16364 and PW of Y= -17142
The Bureau of Public Highways is considering possible types of road surfacing with cost estimates per kilometer: Type B P650k First Cost Resurfacing period Resurfacing Cost Type A P500k 8 years P200k Average annual maintenance cost P15k The periodic resurfacing do not include the base of the subgrade. Compare these types by calculating the present worth for 24 year service, with no salvage value at the end of this time using interest rate of 10%. What if Capitalized Cost is used, which is better? 12 years P250k P20k
An investment of P270,000 can be made in a project that will produce a uniform annual revenue of P185,400 for 5 years and then have a salvage value of 10% of the investment. Out-of-pocket cost for operation andmaintenance will be P81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. (a) Future Worth Method(b) Present Worth Method What is the future worth of net cash flows
Compare 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 $ A
Question 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 1,000
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.
Compare 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.
Problem 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 years