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

Two processes are being considered for the production of a machine part. Process 1 requires an equipment cost of P180 000.00, a yearly operating cost of P90 000.00, and has zero salvage value at the end of its 6-year life. Process 2 has corresponding figures of P320 000.00, P72 000.00, P45 000.00 and 9-year life. If the applied MARR is 20%, compare the two processes using PW analysis. (Ans. PT1 = -P693 567.36; PT2= - P718 084.62)- Please use and actual formula NOT AN EXCEL PLEASE. We are students not office workers.
You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 12% per year? Why is yours the correct choice? Alternative X Y First cost, $ −45,000 −58,000 Maintenance cost, $/year −8,000 −4,000 Salvage value, $ 2,000 12,000 Life, years 5 5
You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 15% per year? Why is yours the correct choice? Alternative X Y First Cost $-25,000 $-80,000 Maintenance cost, per Year $-13000 $-7000 Salvage Value $3,500 $4,500 Life 5 years 5 years The present wort
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.
2. Two methods can be used for injection molding insulin injection pens. Method A Molding costs $120,000 initially and will have a $30,000 salvage value after 5 years. The operating cost with this method will be $20,000 per year. Method B Molding will have a first cost of $150,000, an operating cost of $15,000 per year, and a $40,000 salvage value after its 6 - year life. Determine which method should be selected at a MARR of 10%. - Use Present Worth Analysis Solve in Excel
Alternate x has first cost of$20000 an operating cost of$9000 per year and $5000salvage value after 8 year .alternate Y will cost $12,183 with an operating cost of$ 7,627 per year and a salvage value of$7,104 after4 year .at MARR of 0.12% per year find the pw of machine y?
Compare two alternatives for a physical security system surrounding a power distribution substation using annual worth analysis and a MARR of 10% per year. System Condi Torro First cost, $ −25,000 −130,000 M&O cost, $ per year −9,000 −2,500 Salvage value, $ 3,000 100,000 Life, years 3 ∞
Two machines can be used to produce a part from titanium. The costs and other cash flows associated with each alternative are estimated. The salvage values are constant regardless of when the machines are replaced. Determine which alternative(s) should be selected for further analysis if alternatives must have a payback of 5 years or less. Perform the analysis with (a) i = 0%, and (b) i = 10% per year. Machine Semiautomatic Automatic First cost, $ −40,000 −90,000 Net annual income, $ per year 10,000 15,000 Maximum life, years 10 10 Salvage value, $ 0 0
A company that makes food-friendly silicone (for use in cooking and baking pan coatings) is considering the independent projects shown, all of which can be considered to be viable for only 10 years. If the company’s MARR is 15% per year, determine which should be selected on the basis of a present worth analysis. Financial values are in $1000 units. A B C D First cost, $ −1,200 −2,000 −5,000 −7,000 Annual net income, $/year 200 400 1100 1300 Salvage value, $ 5 6 8 7
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 20% per year, which alternative has the lower present worth? Method DDM LS First cost, $ −164,000 −370,000 M&O cost, $/year −55,000 −21,000 Salvage value, $ 0 30,000 Life, years 2 4
1. Two methods can be used for producing expansion anchors. Method A costs $70,000 initially and will have a $15,000 salvage value after 3 years. The operating cost with this method will be $30,000 per year. Method B will have a first cost of $135,000, an operating cost of $8,000 per year, and a $40,000 salvage value after its 3-year life. At the MARR of 12% per year, which method should be used on the basis of a present worth analysis?
An upgraded version of a CNC machine has a first cost of $200,000, an annual operating cost of $60,000, and a salvage value of $50,000 after its 8-year life. At an interest rate of 10% per year, the capitalized cost is closest to: (a) $−93,116 (b) $−100,060 (c) $−931,160 (d) $−1,000,600