AWa = -1500*(A/P,10%,2)+800+500*(A/F,10%,3)
= -1500*(0.1*(1+0.1)^2)/((1+0.1)^2-1)+800+500*(0.1/((1+0.1)^2-1))
= -1500*0.58+800+500*0.48
= 170
AWb = -900*(A/P,10%,3)+200+900*(A/F,10%,3)
= -900*(0.1*(1+0.1)^3)/((1+0.1)^3-1)+200+900*(0.1/((1+0.1)^3-1))
= 110
AWc = P*i= -1200*0.1 = -120
So looking at the annual worth values, AWc is less which needs to be selected
Which of the alternatives presented in the following table would you recommend ding the analysis based...
9. (Ch6) Determine the best altermative using the annual cash fNow analysis from the data given in table below. nitial cost nnual benefit alvage value ife in years 1500 800 500 years 3 Years Infinity 900 200 900 1200 300 ARR 7%
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%
Problem 2 viable investment options for your company. At an MARR of 10%, which ones (if There are three viable investment options for any would you recommend using Annual Worth Analysis B 180 200 150 20 Initial Cost Annual Benefit Annual O&M Salvage Life
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...
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
9-54 Three mutually exclusive alternatives are beine A considered: $500 $400 $300 200 100 Initial cost Benefit at end of the first 200 200 year Uniform benefit at end of 100 125 subsequent years Useful life, in years 4 At the end of its useful life, an alternative is not replaced. If the MARR is 10%, which alternative should be selected (a) Based on the payback period? (b) Based on benefit-cost ratio analysis? 9-54 Three mutually exclusive alternatives are beine...
Problem (2): Consider the following three mutually exclusive alternatives. MARR is 10%. Alternative 1 10,000 Alternative 2 14,500 Alternative 3 20,000 $3,000 increasing by 500 each year thereafter negligible $5,000 Initial investment Annual yielded returns Salvage Value Service life $5,000 $5,000 negligible 6 a) Compute the payback (PB) period and discounted PB period of each alternative. Based on the PB period, which alternative do you recommend? b) Using Annual-worth analysis, which alternative do you recommend?
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...
QUESTION 3For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $26,5383000010000Annual cost, $/year8,0606,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:A- AW for machine A=QUESTION 4For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $1500021,66710000Annual cost, $/year8,8706,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:B- AW for machine B=
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...