5% per year, a machine has a first cost of $8,00 costs and year-end salvage values as shown. The annu operate the machine for 3 years is closest to which va Year 123 Operating Cost, $ -1000 -1200 -1500 Salvage Value, $ 7000 5000 4200
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![At i = 15% per year, a machine has a first cost of $8,000 and estimated operating
costs and year-end salvage values as shown. The annual equivalent cost if you
operate the machine for 3 years is closest to which value?
Year
123 45
Operating
Cost, $
-1000
-1200
-1500
-2000
-3000
= $-3,870
=$-3,505
= $-3,932
= $-3,637
=$-3,154
Salvage
Value, $
7000
5000
4200
3000
2000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d674337-9c39-48f9-a27e-80b963b8f0af%2F24a357a1-f71e-4640-99be-b84802b7489b%2F42wsyps_processed.png&w=3840&q=75)
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- Gardner Denver Company is considering the purchase of a new piece of factory equipment that will cost $420,000 and will generate $95,000 per year for 5 years. Calculate the IRR for this piece of equipment. For further Instructions on internal rate of return in Excel, see Appendix C.At i = 10% per year, a machine has a first cost of $9,000 and estimated operating costs and year-end salvage values as shown. The annual equivalent cost if you operate the machine for 3 years is closest to which value? Operating Salvage Cost, $ Value, $ 7000 Year -1000 -1200 5000 -1500 4200 -2000 3000 -3000 2000 O-$-3,568 = $-3,770 -1500 4200 -2000 3000 -3000 2000 O =$-3,568 O = $-3,770 = 2-3,636 %3D O =$-3,454 O = $-3,832 123 n 345At i = 15% per year, a machine has a first cost of $15,000 and estimated operating costs and year-end salvage values as shown. The annual equivalent cost if you operate the machine for 5 years is closest to which value? Salvage Value, $ 10,000 9,000 Operating Cost, $/year Year 1 -1,000 2 -1,200 3 -1,500 8,000 4 -2,000 3,000 -5,000 1,000
- Compare the machines below using present worth analysis at i10% per year and find which one should be selected Machine Y First Cost Operating Cost Savage Lfe cycle Machine X 20.000 3.000 3.000 3years Firat Cost 30.000 .000 Annuel Operating Cost Salvage Life cycle 3.000 6 yoarsThe AW method is to be used to select the better alternative of the two machines listed below, at 10% per year. Answer the below questions: Machine R Machine S First cost, $ 250,000 497,732 Annual operating cost, $ per year 40,000 50,000 Salvage value, $ Life, years 20,000 30,000 3. 5. The AW of Machine S=Machine X has an intial cost of $10,000. It is expected to last 12 years, to cost $200 per year to maintain and to have a salvage value of $1,000 at the end of its useful life. The equivalent uniform annual cost of the machine at 8% interest is most nearly ___________. A. $1,160 B. $1,507 C. $1,580 D. $1,475 E. None of the above
- A company with a MARR of 10% must install one of two production machines. The economic parameters of each machine are as follows: Machine Initial cost (5) Service life (years) Salvage value end of life (S) $15,000 4 years $22,000 6 years 52,000 $1,000 The Future worth (FW) for the machine Y over the 12 years analysis period is:A manufacturing company is trying to decide between the two machines shown below. Determine which machine should be selected on the basis of rate of return. Assume the MARR is 20% per year. Machine A Machine B Initial Cost, $ -18,000 -35,000 Annual operating cost, $/year -4,000 -3,600 Salvage value, $ 1,000 2,700 Life, years 3 6For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A 15000 B- AW for machine B= 8,342 4,000 Answer the below questions : Machine B 6 20,447 6,000 5,000 Machine C 10000 4,000 1,000
- An engineer must choose between two technically equivalent alternatives, A and B. Based upon the data in the table below over the 12year analysis period and an interest rate of 9%, the PW of machine A is: A B First Cost $3,000 $6,000 Annual Maintenance $500 $300 End of Useful Life Salvage Value $700 $1,000 Useful Life 5 years |15 уears O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/F,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,10)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15)From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine A B Initial Cost (Php) Annual Operating Cost 146,000 15,000 80,000 10,000 3 220,000 10,000 75,000 25,000 Annual Revenue Salvage Value Useful Life (years) 6need help. MARR = 9.00%This machine has a 9 year economic service life after which it will have a $13,098 salvage value. The operatingcosts for this machine are expected to be $6,433 for the first year, increasing by $640/year for each yearthereafter (for example, $6,433 for year 1, $7,073 for year 2, $7,713 for year 3, etc). At what market value ofthe existing machine would make the proposed machine equally economically attractive?
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