EUAW of Alternative 1 = -45000 * (A/P, 12%,6) + 8000 + 4000 * (A/G, 12%,6) + 6500 * (A/F, 12%,6)
= -45000 * 0.243226 + 8000 + 4000 * 2.172047 + 6500 * 0.123226
= 6543.99
= 6544
EUAW of Alternative 2 = -60000 * (A/P, 12%,5) + 12000 + 9000 * (A/F, 12%,5)
= -60000 * 0.277410 + 12000 + 9000 * 0.157410
= -3227.90
As the EUAW of Alternative 1 is positive, it should be selected
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability ass...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...
Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
Please DO NOT use excel. Show all steps please. You have 2 mutually exclusive alternatives and a MARR of 9%. Which alternative is preferred, based on repeatability assumption? Alternative E F Capital Investment $14,000 $65,000 Annual Expenses $14,000 $9,000 Useful Life (years) 4 20 Market Value at end of useful life $8,000 $13,000
6-52. Compare alternatives A and B with the equivalent worth method of your choice if the MARR is 15% per year. Which one would you recommend? State al assumptions. (6.5) Capital investment $50,000 Operating costs S5,000 at end of year 1 and increasing by S500 per year thereafter S20,000 S10,000 at end of year I and increasing by $1,000 per year thereafter Overhaul costs $5,000 every 5 years None Life Salvage value S10,000 if just 20 years 10 years negligible...
A telecommunications firm is considering a product expansion of a popular cell phone. Two alternatives for the cell phone expansion are summarized below. The company uses an MARR of 9% per year for decisions of this type, and repeatability may be assumed. L Initial Cost (S) Annual Benefit ($) Salvage Value (S) Useful Life (yrs) Expansion A 148,650.000 1,090,000 10.725,000 Expansion B 63.750,000 2,860,000 12,380,000 1. Which Expansion Option should be taken? Use the Equivalent Uniform Annual Worth Method. You...
Question 1 The cash flows given in table below are for two different alternatives. MARR =10% Data IN Initial Cost Annual Benefits Salvage Value Useful Life in years M $20,000 $6,000 $5,000 $80,000 $10,000 $20,000 a) Determine the annual worth of alternative M b) Determine the annual worth of alternative N
QUESTION 6 Data for two mutually exclusive alternatives are given below. Alternatives B $4,000 $800 А Initial Cost $5,000 Annual Benefits (beginning at end of $1,500 year 1) Annual Costs (beginning at end of year $500 1) Salvage Value $500 Useful Life (years) 5 $200 $0 10 Compute the net present worth for each alternative and choose the better alternative. MARR = 8%
Given the following two alternatives, the present worth (PW) of E2 is closest to: Hint: Use positive sign for cash outflow and negative sign for cash inflow. Alt. E1 Alt. E2 Capital Investment, $ 6,000 12,000 Annual Expenses, $ 150 175 Useful life, years Salvage value, $ none 4,000 Less than $10,000 Between $10,000 - $10,500 Between $10,500 - $11,000 O Greater than $11,000
1) Consider these two machines (alternatives): (12 Points) B A $5000 $1750 $700 $8200 $1850 $500 First Cost Uniform annual benefit Salvage Value Useful Life, in Years 4 If the MARR (minimum attractive rate of return) -7 % , which alternative should be selected? Use the Present worth Analysis method. 1) Consider these two machines (alternatives): (12 Points) B A $5000 $1750 $700 $8200 $1850 $500 First Cost Uniform annual benefit Salvage Value Useful Life, in Years 4 If the...
can you solve it by using Excel You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for man- ufacturing high-pressure hydraulic hoses. The pa- 5.7 rameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an inter- est rate of 12% per year? Why is yours the correct choice? Y X Alternative -58,000 -45,000 First cost, $ -4,000 -8,000 Maintenance...