Basics Of Engineering Economy
2nd Edition
ISBN: 9780073376356
Author: Leland Blank, Anthony Tarquin
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 4, Problem 70APQ
To determine
Correct equation.
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Students have asked these similar questions
The PW-based relation for the incremental cash flow series to find 4/"between the lower first-cost alternative X and alternative Y has
been developed.
0=-22,000+ 9000(P/A,4,10)+(-4000(P/F,A*,10))
Determine the highest MARR value for which Y is preferred over X.
Any MARR value greater than
% favors Y.
A company that manufactures brushless blowers invested $700,000 in an automated
quality control system for blower housings. The resultant savings was $180,000 per
year for 5 years. If the equipment had a salvage value of $100,000, what rate of
return per year did the company make and should the company invest in the blower
if MARR is 10%?
7.31% per year (invest in the blower)
None of the above
7.31% per year (do not invest in the blower)
12.30% per year (invest in the blower)
12.30% per year (do not invest in the blower)
perform Breakeven Analysis for each cash flow element and determine the value at which you would breakeven for the expected case scenario (the data provided). It is not necessary to provide a breakeven graph for each cash flow.
Chapter 4 Solutions
Basics Of Engineering Economy
Ch. 4 - State two conditions under which the do-nothing...Ch. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - The costs associated with manufacturing a...
Ch. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Two mutually exclusive projects have the estimated...Ch. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - The manager of engineering at the 900-megawatt...Ch. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Three different plans were presented to the GAO by...Ch. 4 - The U.S. Army received two proposals for a turnkey...Ch. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63APQCh. 4 - Prob. 64APQCh. 4 - Prob. 65APQCh. 4 - Prob. 66APQCh. 4 - Prob. 67APQCh. 4 - Prob. 68APQCh. 4 - Prob. 69APQCh. 4 - Prob. 70APQCh. 4 - Prob. 71APQ
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Similar questions
- The PW-based relation for the incremental cash flow series to find A/* between the lower first-cost alternative X and alternative Y has been developed. 0= -38,000 + 9000 (P/A,Ai *,10) + (-5000(P/F,Ai*,10)) Determine the highest MARR value for which Y is preferred over X. Any MARR value less than % favors Y.arrow_forwardThe ore of a gold mine in the Mountain Province contains, on the average, 0.5 gram of gold per ton. One method of processing costs P1,650 per ton and recovers 93% of the gold, while another method costs only P1,500 per ton and recovers 81% of the gold. If gold can be sold at P8,500 per gram, which method is better and by how much?arrow_forwardTrue or False If breakeven analysis is conducted with PW analysis for different MEAs, it doesnāt guarantee that the fastestalternative to reach its breakeven point has also the highest equivalent worth.arrow_forward
- Hannifin CNG Fuel Dispensers needs to purchase replacement equipment to streamline one of its production lines for a new contract, but may sell the equipment before its expected life is reached at an estimated market value for used equipment. At MARR = 15% per year, select the better option using a future worth analysis over (a) the expected usage period, and (b) the maximum life, when the salvage values are expected to be 50% of the market values for used equipment. Are the selections the same for both plans? Option D E First cost, $ ā62,000 ā77,000 AOC, $ per year ā15,000 ā21,000 Expected market value, $ 8,000 10,000 Expected use, years 3 6 Maximum life, years 4 8arrow_forwardAn industrial machine costing $10,000 will produce net cash savings of $4,000 per year. The machine has a five-year useful life but must be returned to the factory for major repairs after three years of operation. These repairs cost $5,000. The company's MARR is 10% per year. What IRR will be earned on the purchase of this machine? Analyze the sensitivity of IRR to $2,000 changes in the repair cost. Perform the sensitivity analysis. Fill-in the table below. (Round to one decimal place.) Change in the repair cost - $2,000 $0 IRR 21.05% % Is the project acceptable? Yesarrow_forwardThe ore of a gold mine in the mountain province contains, on the average, 0.5 Gram of gold per ton. One method of processing cost P1,650 per ton recovers 93% of the gold, while another method cost only P1,500 per ton and recovers 81% of the gold. If gold can be sold at P8,500 per gram, which method is better and by how much?arrow_forward
- An irrigation return flow drain has sampling equipment that can be powered by solar cells or by running an electric line to the site and using conventional power. Solar cells will cost $14,000 to install with a useful life of 10 years. Annual costs for inspection, cleaning, etc. are expected to be $1500. A new power line will cost $12,000 to install and the power costs are estimated at $600 per year. The salvage value of the solar cells is expected to be 25% of the first cost when the sampling project ends in 4 years. The electric line will stay in place, so its salvage value is considered to be zero. At an interest rate of 10% per year, which alternative should be selected?arrow_forward4. Find the equilibrium in the following single market models (equilibrium consists of ordered pair(s) of (P*,Q)). Use fractions rather than decimals.arrow_forwardDexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with low friction properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 16% per year, which alternative has the better present worth and what is that value (select the closest value)? Method First Cost AOC, per Year Salvage Value Life DDM $170,000 $65,000 $4,000 2 years LS $350,000 $40,000 $29,000 4 years DDM with a PW--$473,000 LS with a PW --$445,900 LS with a PW --$222,055 DDM with a PW--$109,300arrow_forward
- Q1 - With these new estimates, What is the AW of the Investment and Salvage value for YORK AC unit? Q2 - With these new estimates, What is the FW of the Investment and Salvage value for YORK AC unit?arrow_forwardA remotely located air sampling station can be powered by solar cells or by running an electric line to the site and using conventional power. Solar cells will cost $12,600 to install and will have a useful life of 4 years with no salvage value. Annual costs for inspection, cleaning, etc. are expected to be $1400. A new power line will cost $11,000 to install, with power costs expected to be $800 per year. Since the air sampling project will end in 4 years, the salvage value of the line is considered to be zero. At an interest rate of 10% per year, which alternative should be selected on the basis of a future worth analysis?arrow_forwardTwo small engines are designed to help cleaning operation in a petrochemical company. Both seem to be very promising and we need to decide which is better or conditions to make the choice indifferent. COSTS ENGINE 1 Installation/First Costs $ 1,500,000 Operating Costs $/year Salvage Value $ Life in years ENGINE 2 2,250,000 700,000 600,000 100,000 50,000 18 8 If a MARR of 10% per year is provided, then compare Annual Worth and select: [Select] Determine the SALVAGE VALUE for Engine 2 that would make Any choice INDIFFERENT: [Select]arrow_forward
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