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 15% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-240,000 $-30,000 $2,000 2 years LS $-480,000 $-15,000 $23,000 4 years The present worth for the DDM method is $ The present worth for the LS method is $[ The (Click to select) method is selected.

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 15% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-240,000 $-30,000 $2,000 2 years LS $-480,000 $-15,000 $23,000 4 years The present worth for the DDM method is $ The present worth for the LS method is $[ The (Click to select) method is selected.

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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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 $
For th below two machines and based on CC analysis which machine we should select? MARR=10% Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years 3 Answer the below question: B- the CC for machine B= 20,538 11,820 7,655 infinite 113,798 8,894
A large textile company is trying to decide which sludge dewatering process it should use ahead of its sludge drying operation. The costs associated with centrifuge and belt press systems are shown. Compare them on the basis of their annual worths using an interest rate of 10% per year. System First cost, $ Centrifuge -250,000 AOC, $/year -31,000 Overhaul in year 2, $ Salvage value, $ 40,000 Life, years 6 Belt Press -170,000 -35,000 -26,000 10,000 4 The annual worth of the centrifuge system is $- 76216 ✪, and the annual worth of the belt press system is $- 82283.7 The system selected on the basis of the annual worth analysis is the centrifuge system.
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
Standby power for pumps at water distribution booster stations can be provided by either gasoline-or diesel-powered engines. The costs for the gasoline engines are as follows: Gasoline First cost, $ Annual M&O, $ per year Salvage value, $ Life, years - 150,000 -41,000 23,000 15 The incremental PW cash flow equation associated with (diesel – gasoline) is 0 = -40,000 + 11,000 (P/A, i, 15) + 16,000(P/F, i, 15) Determine the following: (a) First cost of the diesel engines (b) Annual M&O cost of the diesel engines (c) Salvage value of the diesel engines
Evaluate a combined cycle power plant on the basis of the present worth method when the MARR is 12% per year. Pertinent cost data are as follows: Power Plant (in hundred thousand of P) P 13,000 15 years Investment cost Useful life Market Value (end P 3,000 of year 15) Annual operating P 1,000 expenses Overhaul cost – end P 200 of 5th year Overhaul cost – end P 550 of 10th year
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Subject : Economy and Engineering Management
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.
Mid-Valley Industrial Extension Service, a state-sponsoredagency, provides water quality sampling services toall business and industrial firms in a 10-county region.Last month, the service purchased all necessary lab equipmentfor full in-house testing and analysis. Now, an outsourcingcompany has offered to take over this function ona per-sample basis. Data and quotes for the two optionshave been collected. The MARR for government projectsis 5% per year and a study period of 8 years is chosen.In-house: Equipment and supplies initially cost$125,000 for a life of 8 years, an AOCof $15,000, and annual salaries of$175,000. Sample costs average $25.There is no significant salvage valuefor the equipment and supplies currentlyowned.Outsourced: Cost averages $100 per sample for thefirst 5 years, increasing to $125 persample for years 6 through 8. Determine the breakeven number of tests between the two options.
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 $
Following cash flows for alternatives X and Y at an interest rate of 10% per year. Machine X Machine Y -146,000 -220,000 Initial cost, $ AOC, $/year Annual revenue, $/year Salvage value, $ Life, years -15,000 80,000 10,000 3 -10,000 75,000 25,000 6 In comparing the alternatives on a present worth basis, the PW of Machine X is closest to