ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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- For the cash flows shown, determine the incremental cash flow between machines B and A for (a) year 0, (b) in year 3, and (c) in year 6. Machine First cost, $ AOC, $ per year Salvage value, $ Life, years A -15,000 -1,600 3.000 (a) = = -10000 (b) = 13200 (c) =6000 (a) = -10000 (b) = -1800 (c) = 4200 3 (a) = -10000 (b) = 13200 (c) = 4200 (a) = -10000 (b) = -1800 (c) = 3000 B -25,000 -400 6.000 6arrow_forwardb) National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. The details of the 2 alternative options are summarized in the table below. The interest rate is 10% per year. First Cost, S Annual Operating Cost, S Salvage Value, S Equipment Life (years) Consider only machine A now. Alternative A 15,000 3,500 1,000 4 Alternative B 18,000 3,100 2,000 8 a) Calculate the AW of Machine A for one life cycle. b) Calculate the AW of Machine A for LCM = 8 years (2 life cycles). c) Demonstrate the equivalence of AW over 2 life cycles and AW over one life cycle.arrow_forwardA 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.arrow_forward
- 6arrow_forwardCompare the alternatives C and D on the basis of a present worth analysis using an interest rate of 13% per year and a study period of 10 years. Alternative C $-46,000 $4,000 $-1,300 $6,000 10 First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life. Years The present worth of alternative C is $1 (Click to select) offers the lower present worth. HELIME BIEN D $-20,000 $-7,500 $-100 $1,300 5 and that of alternative D is $ Aarrow_forwardPlease no written by handarrow_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_forwardA consulting engineering firm is considering two models of SUVs for the company principals. A GM model will have a first cost of $36,000, an operating cost of $4000, and a salvage value of $15,000 after 3 years. A Ford model will have a first cost of $32,000, an operating cost of $3100, and also have a $15,000 resale value, but after 4 years. (a) At an interest rate of 15% per year, which model should the consulting firm buy? Conduct an annual worth analysis. (b) What are the PW values for each vehicle?arrow_forwardCompare 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 $arrow_forward
- 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 toarrow_forward5.24 For the cash flows below, use an annual worth comparison to determine which alternative is best at an interest rate of 1% per month. First cost, $ M&O costs, $/month Overhaul every 10 years, $ Salvage value, $ Life, years X -90,000 -400,000 -30,000 -20,000 - Y 7000 3 - 25,000 10 Z -900,000 -13,000 -80,000 200,000 8arrow_forwardMechanical engineer at Company B is considering five equivalent projects, some of which have different life expectations. Salvage value is nil for all alternatives. Assuming that the company’s MARR is 13% per year, determine which should be selected (a) if they are independent, and (b) if they are mutually (c) Explain why your selection in part (b) is correct. First Cost, $ Net Annual Income, $/Year Life, Years A -20,000 +5,500 4 B −10,000 +2,000 6 C −15,000 3,800 6 D −60,000 +11,000 12 E −80,000 +9,000 12arrow_forward
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