AW of A = -350000 * (A/P, 3%,12) + 55000 + 50000 * (A/F, 3%,12)
= -350000 *0.100462 + 55000 + 50000 *0.070462
= 23361.37
CC of A = 23361.37 / 0.03 = 778712.47
CC of B = -700000 + 45000 / 0.03 = 800000
We should select option B, as it has more positive worth
Second option is correct answer
Compare the alternatives shown below on the basis of their capitalized costs, using an effective rate...
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
I cant get alternative y Problem 14.030: Compare alternatives by calculating their capitalized cost Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use 3.2% per year. 10% per year and Alternative First cost,$ -19,000,000 13,500,000 -10,000 82,000 10 AOC, $per year25,000 Salvage value, $ Life, years 105,000 The capitalized cost for alternative X is s -19199107 The capitalized cost for alternative Y is $ 135762 。 Select alternative「-Y- O Problem 14.030: Compare...
Y 0.83 points Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use i=12% per year and f= 3.8% per year. Alternative х First cost, $ -16,000,000 -14,500,000 AOC, $ per year -25,000 -10.000 Salvage value, $ 105,000 82,000 Life, years 10 eBook Hint Print The capitalized cost for alternative X is $ References The capitalized Click to select) IS Select alternativex
Two alternatives to incorporate improved techniques to manufacture computer drivers to play HD DVD optical disc formats have been developed and costed. Compare the alternatives below using capitalized cost and interest rate of 8% per year compounded quarterly. Select the CC of best alternative Alternative x Alternative Y First Cost, $ -240000 -168000 Net income, Sper 5000 9000 quarter Upgrade cost, every 10 - 13000 years Salvage value, $ 30000 5 Life, years 00
Question 12 For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine A Machine B 240,000 First cost (AED) 20,000 Annual maintenance cost per year, AED 5,000 2.300 Periodic cost every 10 years, AED 10,000 Salvage cost 2000 Life. vears Match the closest correct answers for the below questions: Calculate the present value of the maintenance costs for...
Question 12 15 points For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine A Machine B First cost (AED) 20,000 240,000 Annual maintenance cost per year, AED5,000 2,300 Periodi e cost every 10 years, AED 10,000 Salvage cost 2000 Life, years Match the closest correct answers for the below questions: A. [Alternative A] B. -44,483.50] C. I-269,275]...
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC) at i=10% per year, which is the best alternative in this scenario? • Alternative 1, AW = $87,500 and n = (forever) • Alternative 2, PW = -$895,000 and n = (forever) • Alternative 3, First cost (FC) of $900,000, annual operating savings of 3,000 per year, salvage = $200,000, and n = (forever) Alternative 2 Alternative 3 None of them Alternative 1 QUESTION...
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 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
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
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.