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- Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 15% per year and a study period of 10 years. Alternative C First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-50,000 $-9,000 $-28,000 $-5,000 $-700 $-900 $6,000 $700 10 The present worth of alternative C is $ and that of alternative D is $ Alternative D offers the lower present worth.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 D First cost, $ −40,000 −32,000 AOC, $/year −7,000 −3,000 Annual increase in operating cost, $/year −1,000 — Salvage value, $ 9,000 500 Life, years 10 5Two 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. First cost, $ AOC, $ per year Salvage value, $ Life, years Q -42,000 -6,000 0 2 R -80,000 -7,000 in year 1, increasing by $1,000 per year thereafter 4,000 4
- Q1:Compare the alternatives below using future worth analysis at i = 8% per year using LCM. First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years - 23,000 -30,000 -4,000 - 2,500 3,000 1,000 3 6Compare 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.Compare 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 C $-50000 $-9000 $-700 $6000 10 The present worth of alternative C is $ (Click to select) offers the lower present worth. D $-28000 $-5000 $-900 $700 5 and that of alternative D is $
- Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 15% per year. Q R $-44,000 $-11,000 $3,000 $-84,000 $-6,000 in year 1, increasing by $1,000 per year thereafter $10,000 First Cost AOC per Year Salvage Value Life 2 years 4 years Alternative (Click to select) vshould be selected. (Click to select) QTwo 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 $-44,000 R $-84,000 First Cost AOC per Year Salvage Value $-10,000 $-5,000 in year 1, increasing by $1,000 per year thereafter $4,000 $5,000 Life 2 years 4 years Alternative (Click to select) should be selected.Compare the following alternatives based on the equivalent uniform annual value analysis, using a 15% annual interest rate with continuous capitalization. Alternative A Alternative B Initial Cost $18000 $25000 Annual Cost $4000 $3600 Salvage Value $3000 $2500 Life (years) 3 4
- 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 Initial Cost (Php) Annual Operating Cost A В 146,000 15,000 220,000 10,000 75,000 25,000 Annual Revenue 80,000 10,000 Salvage Value Useful Life (years) 68. If the interest rate is 24% per year compounded monthly, compare the alternatives shown below on the basis of a present worth analysis Alternative 1 Alternative 2 First cost Annual operating cost Annual income Salvage value Life, years S18000 $33000 $5600 $27000 $4200 $4600 $20000 $3000 4Calculate the Equivalent X(4) for a project with the following cash flows: First Cost (F.C) = JD 13000, Annual Income for the last 6 years= JD 2400 Operating Cost (O.C.) = JD 30 Income, at the end of the 4th year = JD 5000, Salvage Value (S.V.) = 1800. If n= 9 years and i=7% per year. ( show the CFD and calculations)