(A/F, I%, infinite) = 0
(A/P, I%, infinite) = i
EUAW of alternative 1 = -160000 * (A/P, 12%,infinite) - 15000 + 1000000 * (A/F, 12%, infinite)
= -160000 * 0.12 - 15000 + 1000000 * 0
= -34200
EUAW of alternative 2 = -25000 * (A/P, 12%,7) - 3000 + 4000 * (A/F, 12%,7)
= -25000 * 0.219118 - 3000 + 4000 * 0.099118
= -8081.47
As the annual cost of alternative 2 is less, it should be selected
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
spreadsheet solution? 5.35 Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per year. Alternative M -150,000 -50.000 Alternative N -800,000 -12.000 First cost, S Annual operating cost, $ per year Salvage value, $ Life, years 8,000 1,000,000
Please dont use excel,show me the formula used 7. Compare the alternatives shown below on the basis of a future worth analysis, using an interest rate of 8% per year. Р First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years -23.000 -4,000 3,000 -30,000 -2.500 1.000
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
Problem 05.023 Alternative Comparison - Different Lives 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. $ 40,000 $-6,000 D $-22.000 $-3,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-200 $-300 $7,000 10 $200 5 The present worth of alternative C is $ and that of alternative D is $...
show the formula Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 12% per year and a study period of 10 years. с $-44,000 $-12,000 $-34,000 $-7,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-1,500 $-1,200 $5,000 10 $1,200 5 The present worth of alternative C is $ -134497.32 and that of alternative D is $...
Compare 10 years 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 Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
Question 3 (20 points) Compare the two mutually exclusive alternatives on the basis of their capitalized costs at i= 10% per year. First cost, $ Annual cost, $/year Salvage value, $ Life, years -110,000 –54,000 9,000 - 700,000 -15.000 2,000,000
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability assumption is acceptable. Aternative I: Initial investment of $45,000, net revenue the first year of $8,000, increasing $4,000 per year for the six year useful life. Salvage value is estimated to be $6500. Alternative II: Initial investment of $60,000, uniform annual revenue of $12,000 for the five year useful life. Slavage value is estimated to be $9,000.