An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at /= 13% per year on the basis of an annual worth evaluation. First Cost AOC per Year Salvage Value Life Land $-145,000 $-91,000 $23,000 4 years Incineration $-910,000 $-50,000 $310,000 6 years The environmental engineer selects a (Click to select) private disposal contract land application fluidized-bed incineration Contract 0 $-132,000 0 2 years

An environmental engineer wants to evaluate three different methods for disposing of nonhazardous chemical waste: land application, fluidized-bed incineration, and private disposal contract. Use the estimates below to help her determine which has the least cost at /= 13% per year on the basis of an annual worth evaluation. First Cost AOC per Year Salvage Value Life Land $-145,000 $-91,000 $23,000 4 years Incineration $-910,000 $-50,000 $310,000 6 years The environmental engineer selects a (Click to select) private disposal contract land application fluidized-bed incineration Contract 0 $-132,000 0 2 years

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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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 Centrifuge Belt Press First cost, $ −235,000 −150,000 AOC, $/year −48,000 −41,000 Overhaul in year 2, $ -- −26,000 Salvage value, $ 40,000 10,000 Life, years 6 4 The annual worth of the centrifuge system is $− , and the annual worth of the belt press system is $− . The system selected on the basis of the annual worth analysis is the (Click to select) belt press centrifuge system.
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 First Cost Maintenance cost, per Year Y $-45,000 $-80,000 $-14000 $-2000 Salvage Value Life $3,500 5 years $9,000 5 years The present worth of alternative X is $ and that of alternative Y is $ Alternative ((Click to select) + is selected by the company.
Two processes can be used for producing a polymer that reduces friction loss in engines. Process T will have a first cost of $760,000, an operating cost of $100,000 per year, and a salvage value of $80,000 after its 2-year life. Process W will have a first cost of $1,200,000, an operating cost of $25,000 per year, and a $120,000 salvage value after its 4-year life. Process W will also require updating at the end of year 2 at a cost of $90,000. Which process should be selected on the basis of a present worth analysis at a MARR of 12% per year? The present worth of process T is $-[ and the present worth of process W is $- The process selected on the basis of the present worth analysis is process W
QUESTION 8 A professional photographer who specializes in wedding-related activities paid $48,000 for equipment that will have a $2000 salvage value after 5 years. He estimates that his costs associated with each event amount to $65 per day. If he charges $350 per day for his services, how many days per year must he be employed in order to break even at an interest rate of 6% per year? The number of days required for break even, (in days per year) Round to the nearest three (3) decimal places
A professional photographer who specializes in wedding-related activities paid $49,000 for equipment that will have a $2000 salvage value after 5 years. He estimates that his costs associated with each event amount to $65 per day. If he charges $300 per day for his services, how many days per year must he be employed in order to break even at an interest rate of 8% per year? The number of days required for break even is determined to be per year.
ChemCo Enterprises is the manufacturer of Ultra-Dry, a hydrophobic coating that will waterproofanything. Over a 5 year period, the costs associated with the pilot test product line were as follows: first cost of $43,000 and annual costs of $18,000. Annual revenue was $40,000 and used equipment was salvaged for $4,000. What rate of return did the company make on the product?
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.
Two processes can be used for producing a polymer that reduces friction loss in engines. Process T will have a first cost of $660,000, an operating cost of $90,000 per year, and a salvage value of $80,000 after its 2-year life. Process W will have a first cost of $1,100,000, an operating cost of $25,000 per year, and a $120,000 salvage value after its 4-year life. Process W will also require updating at the end of year 2 at a cost of $90,000. Which process should be selected on the basis of a present worth analysis at a MARR of 12% per year? The present worth of process T is $- and the present worth of process W is $- The process selected on the basis of the present worth analysis is process (Click to select) ✓
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1. The annual worth for years 1 through infinity of $75,000 now, $25,000 per year in years 1 through 15 and S40,000 per year in years 16 through infinity at a 10% interest per year is closest to: (a) $27,500 (b) $36,000 (c) $44,000 (d) $ 19,500 1. For the estimates in table below, calculate the equivalent annual cost of the project: First Cost, $ -950,000 Replacement Cost, year 2 AOC, $/ year Salvage value, $ -450,000 -600,000 175,000 Life, years 4 Interst Rate % 10
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