ENGINEERING ECONOMIC ENHANCED EBOOK
14th Edition
ISBN: 9780190931940
Author: NEWNAN
Publisher: OXF
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Question
Chapter 9, Problem 54P
(a)
To determine
To select: The best alternative based on payback period
(b)
To determine
To select: The best alternative based on benefit-cost ratio.
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Two types of robots (Cartesian and Articulated) with the following estimates are under consideration for a dishwasher assembly process. Using an interest rate of 10% per year, determine which one should be selected on the basis of an annual worth analysis. Robot Cartesian Articulated First cost, $ −300,000 −430,000 AOC, $/year −60,000 −40,000 Salvage value, $ 70,000 95,000 Life, years 4 6
Please see attachment and type out the correct answer ASAP with proper step by step explanation of the each option given below within 40 50 minutes.
Determine the present worth, future worth, and annual worth of the following engineering project when the MARR is 15% per year. Is the project acceptable?
Investment cost $10,000Expected life 5 yearsMarket (salvage) value $1,000Annual receipts $8,000Annual expenses $4,000
Chapter 9 Solutions
ENGINEERING ECONOMIC ENHANCED EBOOK
Ch. 9 - Prob. 2QTCCh. 9 - Prob. 3QTCCh. 9 - Prob. 1PCh. 9 - Prob. 2PCh. 9 - Prob. 3PCh. 9 - Prob. 4PCh. 9 - Prob. 5PCh. 9 - Prob. 6PCh. 9 - Prob. 7PCh. 9 - Prob. 8P
Ch. 9 - Prob. 9PCh. 9 - Prob. 10PCh. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Prob. 13PCh. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 27PCh. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - Prob. 31PCh. 9 - Prob. 32PCh. 9 - Prob. 33PCh. 9 - Prob. 34PCh. 9 - Prob. 35PCh. 9 - Prob. 36PCh. 9 - Prob. 37PCh. 9 - Prob. 38PCh. 9 - Prob. 39PCh. 9 - Prob. 40PCh. 9 - Prob. 41PCh. 9 - Prob. 42PCh. 9 - Prob. 43PCh. 9 - Prob. 44PCh. 9 - Prob. 45PCh. 9 - Prob. 46PCh. 9 - Prob. 47PCh. 9 - Prob. 48PCh. 9 - Prob. 49PCh. 9 - Prob. 50PCh. 9 - Prob. 51PCh. 9 - Prob. 52PCh. 9 - Prob. 53PCh. 9 - Prob. 54PCh. 9 - Prob. 55PCh. 9 - Prob. 56PCh. 9 - Prob. 57PCh. 9 - Prob. 58PCh. 9 - Prob. 59PCh. 9 - Prob. 60PCh. 9 - Prob. 61PCh. 9 - Prob. 62PCh. 9 - Prob. 63PCh. 9 - Prob. 64PCh. 9 - Prob. 65PCh. 9 - Prob. 66PCh. 9 - Prob. 67PCh. 9 - Prob. 68PCh. 9 - Prob. 69PCh. 9 - Prob. 70PCh. 9 - Prob. 71PCh. 9 - Prob. 72PCh. 9 - Prob. 73PCh. 9 - Prob. 74PCh. 9 - Prob. 75PCh. 9 - Prob. 77PCh. 9 - Prob. 78PCh. 9 - Prob. 79PCh. 9 - Prob. 80PCh. 9 - Prob. 81PCh. 9 - Prob. 83PCh. 9 - Prob. 84PCh. 9 - Prob. 85PCh. 9 - Prob. 86PCh. 9 - Prob. 87PCh. 9 - Prob. 88P
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- 7arrow_forwardA road resurfacing project costs $250,000, lasts 5 years and saves $150,000 annually in patching costs. MARR is 12%. Determine the annual worth (in $)arrow_forwardYou are being asked to select a piece of machinery for the company you work for. Operating costs are an important criteria when selecting what machinery would be the best choice. What is the EUAW (Equivalent Uniform Annual Worth) of this investment that costs $59,900, has annual benefits of $8,343/year, annual costs of $2,450/year, and a disposal cost of $4,300 at the end of its useful life? It has a useful life of 12 years. Use a 10% MARR.arrow_forward
- DO NOT GIVE SOLUTION IN IMAGEarrow_forwardNASA is considering two materials for use in a space vehicle tracking station in Australia. The estimates are shown below. Which should be selected on an economic basis of present worth values at an interest rate of 10% per year? Material M FF First cost, $ −205,000 −235,000 Maintenance cost, $/year −29,000 −27,000 Salvage value, $ 2,000 20,000 Life, years 2 4arrow_forwardRequired information A remotely located air sampling station can be powered by solar cells or by running an above ground electric line to the site and using conventional power. Solar cells will cost $16,000 to install and will have a useful life of 5 years with no salvage value. Annual costs for inspection, cleaning, and other maintenance issues are expected to be $2,500. A new power line will cost $22,500 to install, with power costs expected to be $1,000 per year. Since the air sampling project will end in 5 years, the salvage value of the line is considered to be zero. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. At an interest rate of 10% per year and using an AW analysis, which alternative should be selected? The annual worth of installing solar cells is $- 25477 The alternative to be selected is installing solar cells and the annual worth of installing a new power line is $- 26291 Ⓡarrow_forward
- A large textile company is trying to decide among three alternatives of sludge dewatering processes. The costs associated with these alternatives are shown below. Alternative Y will need an upgrade of $9700 at the end of year 2. At the end of year 2, alternative Z would be replaced with another alternative Z having the same installed and operating costs. If the MARR is 12% per year, which alternative should be chosen? Alternative Y Installed costs S68500 S48500 $33500 Annual operating costs S6000 S4000 S5000 Overhaul cost in year 2 $9700 Salvage value (S) [your student ID/10) ID +10 ID- 10 ID- 10 Useful life, years 2.arrow_forwardFrom the solution, where did the 670,000 come from?arrow_forwardNikularrow_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_forward7arrow_forwardDetermine the FW of the following engineering project when the MARR is 15% per year. Is the project acceptable? Proposal A Investment cost Expected life Market (salvage) value Annual receipts $11,000 5 years -$1,000 S7,000 $4,000 Annual expenses "A negative market value means that there is a net cost to dispose of an asset Click the icon to view the interest and annuity table for discrete compounding when the MARR is 15% per year The FW of the follawing engineering froject is $ (Round to the nearest dollar.)arrow_forward
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