Basics Of Engineering Economy
2nd Edition
ISBN: 9780073376356
Author: Leland Blank, Anthony Tarquin
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Textbook Question
Chapter 5, Problem 11P
Two machines with the following cost estimates are under consideration for a dishwasher assembly process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of an annual worth analysis.
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Chapter 5 Solutions
Basics Of Engineering Economy
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10P
Ch. 5 - Two machines with the following cost estimates are...Ch. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Estimates have been presented to Holly Farms,...Ch. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - A major repair on the suspension system of Janes...Ch. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32APQCh. 5 - Prob. 33APQCh. 5 - Prob. 34APQCh. 5 - Prob. 35APQCh. 5 - Prob. 36APQCh. 5 - The AW values of three revenue alternatives are ...Ch. 5 - Prob. 38APQCh. 5 - Prob. 39APQCh. 5 - Use an interest rate of 10% per year. The...Ch. 5 - Prob. 41APQCh. 5 - Prob. 42APQ
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- System A costs $150,000 to purchase and install and has a life of 8 years, with no salvage value. It will require an overhaul in four years at a cost of $35,000. Annual operating and maintenance costs (O&M) will be $10,000 per year for the first four years and will increase by $6000 each year for the remaining life (i.e., O&M will cost $16,000 in year 5, $22,000 in year 6, $28,000 in year 7, and $34,000 in year 8). System B costs $200,000 to purchase and install and has a life of 15 years, and a salvage value of $15,000. It will require an overhaul every 5 years at a cost of $30,000 for each overhaul. The O&M costs will be a constant $15,000 per year. Assuming an effective interest rate of 6% per year, use an annual worth analysis to determine which alternative would be recommended.arrow_forwardCtrl 24 Calculate the perpetual equivalent annual cost (years 1 through infinity) of $5 million in year 0, $2 million in year 10, and $100,000 in years 11 through infinity. Use an interest rate of 10% per year. 27 A new bridge across the Allegheny River in Pittsburgh is expected to be permanent and will have an initial cost of $30 million. This bridge must be resurfaced every 5 years at a cost of $1 million.arrow_forwardTwo 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 0arrow_forward
- J. Doe must choose between two different models. The analysis period considered is 6 years. Model 1 has a life of four years with a first cost of $13,500 and maintenance costs of $1,250 per year in years 2, 3, and 4 (no maintenance costs in year 1). The salvage value for this model at the end of its life is $4,271 (For year 2 the salvage value is $7,594). Model 2 has a life of three years with a first cost of $15,000 and maintenance costs of $700 per year. Its salvage value at the end of its life is $9,300 (year 3). Which of the two models should be chosen by J. Doe if considering a MARR of 12%.arrow_forwardEvaluate a combined cycle power plant on the basis of the FW method when the MARR is 18% per year. Pertinent cost data are as follows:arrow_forwardEcology group wishes to purchase a piece of equipment for recycling of various metals. Machine 1 costs $300000, has a tife of 10 years, an annual cost of $30000, and requires one operator at a cost of $72 per hour. It can process 10 tons per hour. Machine 2 costs $180000, has a life of 6 years, an annual cost of $15000, and requires two operators at a cost of $72 per hour each to process 6 tons per hour. Determine the annual breakeven tonnage of scrap metal at i=10% per year.arrow_forward
- Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 10% per year. Q R First cost, $ −42,000 −80,000 AOC, $ per year −6,000 −7,000 in year 1, increasing by $1,000 per year thereafter Salvage value, $ 0 4,000 Life, years 2 4arrow_forwardFabco, Inc., is considering the purchase of flow valves that will reduce annual operating costs by $10,000 per year for the next 12 years. Fabco’s MARR is 7%/year. Using an annual worth approach, determine the maximum amount Fabco should be willing to pay for the valves.arrow_forwardAn engineer is trying to decide which process to use to reduce sludge volume prior to disposal. Belt filter presses (BFP) will cost $203,000 to buy and $85,000 per year to operate. Belts will be replaced one time per year at a cost of $5500. Centrifuges (Cent) will cost $396,000 to buy and $119,000 per year to operate, but because the centrifuge will produce a thicker “cake”, the sludge hauling cost to the monofill will be $37,000 per year less than for the belt presses. The useful lives are 5 and 10 years for alternatives BFP and Cent, respectively, and the salvage values are assumed to be 10% of the first cost of each process whenever they are closed down or replaced. Use PW evaluation to select the more economical process at an interest rate of 6% per year over (a) the LCM of lives, and (b) a study period of 8 years. Are the decisions the same?arrow_forward
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