Is considering two different pumps. Which of pumps should they purchase? The data on the pumps is given in the following Table. Minimum attractive interest rate is 15%. You can use net present worth (cost) criteria with common multiple life which is 6. Do before tax analysis which means do not account for tax and depreciation. Operating Cost (S/year) 50,000 40,000 TCI ($) Pump Alternatives Life of alternatives (years) 100,000 150,000 23

Is considering two different pumps. Which of pumps should they purchase? The data on the pumps is given in the following Table. Minimum attractive interest rate is 15%. You can use net present worth (cost) criteria with common multiple life which is 6. Do before tax analysis which means do not account for tax and depreciation. Operating Cost (S/year) 50,000 40,000 TCI ($) Pump Alternatives Life of alternatives (years) 100,000 150,000 23

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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1. A company intends to replace equipment that is no longer able to meet its needs. The two best alternatives are to purchase a manual tool (A1) or an automatic tool (A2). The estimated economic consequences of the two alternatives are: A1 A2 Initial investment ($) 28.000 65.000 Annual withdrawal ($) 14.000 9.000 Additional withdrawal every 5 years ($) 14.000 9.000 Lifespan (years) 10 20 Residual value ($) 8.000 13.000 Based on the present worth analysis, with an interest rate of 15% per year, an analysis period of 20 years, and the assumption of recurrence, determine the best alternative that must be chosen using future worth analysis.
Q1).Using AW method compare between the following alternatives: Operating and maintenance Alt. F.C. Income at the Annual nd S.V. Useful life i. end of 2 yearIncome A 35000 3500 2800 7000 8 8% B. 32000 1500 4700 2500 8%
Please do in excel please
1. The data for new and used machines are shown below: Initial cost($) Annual operating cost ($/year) Salvage value ($) Life (years) Used machine 15,000 8,000 5,000 3 New machine 40,000 2,000 10,000 6 Using an interest rate of 10% per year, determine the present worth of the new machine.
A new bridge project is being evaluated at i = 20%. Recommend an alternative based on the capitalized cost for each. AnnualO&M Life(years) $250,000 $500,000 Construction Cost Concrete $50million Steel $40million Solution: 1. Here 2. CC concrete 50 3. CC steel 4. Choose capitalized cost (CC). million dollars; million dollars; 70 50
Fitzgerald, Ivy, Garcia, Nichols, Eudy, Williams, Thomas, Owens, and Nagy (FIGNEWTON) Inc. must replace its fig - crushing equipment. The alternatives under consideration are presented below. Alternative First Cost Net Annual Benefits Useful Life A $170, 500 $14, 675 5 years B 205,000 17,000 7 years C 242,500 16, 350 8 years D 290,000 14,825 10 years a) Which anlaysis method should be used to select the alternative? b) Why should that analysis method be used? c) if FIGNEWTON uses a MARR of 8%, what alternative should be chosen? d) if FIGNEWTON uses a MARR of 18%, what alternative should be chosen? Submit one excel file with answers a), b), c) and d) highlighted clearly.
which process line should be built for a new chemical? the expected market for the chemical is 16 years. an 18% rate is used to evaluate new process facilities, which are compared with present worth. how much does the better choice save? First cost O &M cost/year salvage life A $14 M $2.5 M $2 M 8 years B 22 M 3 M 7.5 M 16 years
A manufacturer with a MARR of 20% is considering the installation of one of three packaging machines. The economic parameters of each machine are as follows: Packaging machine: Initial cost ($) Uniform annual benefit ($) Uniform annual O&M cost ($) X Y 20,000 40,000 120,000 180,000 70,000 50,000 6 Service life (years) 3 Salvage value end of life ($) 2,000 4,000 Z 15,000 100,000 70,000 12 2,000 Which choice below gives the correct PW equation for machine X over the appropriate analysis period? O PW (costs) = $20k + $20k (P/F,20%,6) + $70k (P/A,20%, 12) - $2k (P/F,20%,6) - $2k (P/F,20%,12) NPW = [ $120k (P/A,20%, 12) ] - [ $20k + $20k (P/F,20%,6) + $70k (P/A, 20%, 12) - $2k (P/F,20%,6) - $2k (P/F,20 %,12)] PW (benefits) = $120k (P/A,20%, 12) O NPW = ($120k-$70k) (P/A,20%,12)
A company is considering the purchase of either machine A or machine B. And the interest rate is 10%. The following table shows the cash flows of both machines. Machine Initial investment ($) Other costs ($) Service life (years) Salvage value ($) Cash flow diagram for Machine A: A $80,000 $10,000 for the first 10 years, increasing at a rate of 5% from year 11 onwards 20 20,000 PW of machine A: B $100,000 $15,000 for the first 15 years and $20,000 for the remaining years a) Which alternative should the company choose if it adopts an interest rate of 10%? Use present worth (PW) method to evaluate the two machines and show your calculations in the space given below. 20 25,000