For the cash flows shown, use an annual worth comparison and an interest rate of 8% per year and detemine which service alternativeshould be selected.BInitial cost, $300000680000Annual cost, S/yearS000030000Annual cost started at year 4, S/year15000Extra cost every 6 years, $40000Salvage value, $50000150000Life, Years8.AW of alternative A is:AW of alternative B is:Which alternative will be selected?

Annual worth is the equivalent & uniform annual value of all the cash flows of an investment.

For the cash flows shown, use an annual worth comparison and an interest rate of 8% per year and deteine which service alternative should be selected. B Initial cost, $ 300000 680000 Annual cost, S/year 50000 30000 Annual cost started at year 4, S/year 15000 Extra cost every 6 years, $ 40000 Salvage value, S 50000 150000 Life, Years 8. 00 AW of alternative A is AW of alternative B is: Which alternative will be selected?

For the cash flows shown, use an annual worth comparison and an interest rate of 8% per year and deteine which service alternative should be selected. B Initial cost, $ 300000 680000 Annual cost, S/year 50000 30000 Annual cost started at year 4, S/year 15000 Extra cost every 6 years, $ 40000 Salvage value, S 50000 150000 Life, Years 8. 00 AW of alternative A is AW of alternative B is: Which alternative will be selected?

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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Using (incremental) benefit-cost ratio analysis OR (incremental) internal rate of return, determine which of the following three mutually exclusive alternatives should be selected. Each alternative has a 6-year useful life. Assume the MARR is 6%. First Cost Uniform Annual Benefit Salvage Value $560 $140 $40 A $340 $100 0 B $125 $40 0 C
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Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. Alternative C First Cost $-40,000 $-11,000 $-24,000 AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-2,000 $-200 $-500 $6,000 $200 10 The present worth of alternative C is $ and that of alternative D is $ (Click to select) voffers the lower present worth.
please detail solution and explnation
Can someone please solve this question using hand calculations only? PLEASE AND THANK YOU!!!
A machine cost is 3000$ and salvage value 850$, if the rate of interest is 9% and number of years are 5 then what willI be the present worth for fixed output condition? Select one: a. 2448$ b. 2418$ c. 2427$
The equivalent annual worth of alternative A over an infinite time period is closest to:a. $-25,000b. $-27,200c. $-31,600d. $-37,100
The equivalent annual worth of alternative A is closest to:a. $-25,130b. $-37,100c. $-41,500d. $-42,900
From the following information calculate the netpresent value of the two projects and suggestwhich of the two projects should be acceptedassuming a discount rate of 10% Project A Project BInitial investment Rs. 20,000 Rs. 30,000Estimated life 5 years 5 yearsScrap value Rs. 1,000 Rs. 2,000 The profit before depreciation and after taxes areas follows:Year Project ARs.Project BRs.1 5,000 20,0002 10,000 10,0003 10,000 5,0004 3,000 3,0005 2,000 3,000
I will rate. Solve correctly and show all works.
Compare three alternatives on the basis of their capitalized costs at /= 8% per year and select the best alternative. Alternative E F G $-460,000 $-30,000 $90,000 4 First Cost AOC, per Year Salvage Value Life, Years $-80,000 $-65,000 $40,000 2 The capitalized cost of alternative E is $ The best alternative is (Click to select) ▼ alternative F is $ $-960,000 $-3,000 $450,000 ∞0 and alternative G is $
Compare three alternatives on the basis of their capitalized costs at i = 13% per year and select the best alternative. Alternative E F G First Cost $-85,000 $-325,000 $-825,000 AOC, per Year $-65,000 $-20,000 $-6,000 Salvage Value $25,000 $75,000 $500,000 Life, Years 2 4 ∞ The capitalized cost of alternative E is $ , alternative F is $ , and alternative G is $ . The best alternative is E G F .

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