Which alternative in the table below should be selected when the MARR = 6% per year? The life of each alternative is 10 years. Increment Considered A(A-DN) A Investment cost $700 A(B-A) $600 A(C-B) A(D-C) $1,100 $1,700 A (Annual Revenues $143 $115 $162 $170 less Costs) IRR on A Investment Cost 15.7% 14.0% ? ? The IRR on A(C - B) is %. (Round to one decimal place.)
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- Which alternative in the table below should be selected when the MARR = 7% per year? The life of each alternative is 10 years. Increment Considered A(A - DN) A(B-A) A(C-B) A(D-C) A Investment cost A (Annual Revenues less Costs) $700 $600 $1,100 $1,600 $155 $126 $165 $160 IRR on A Investment Cost 17.9% 16.4% ? ..... The IRR on A(C-B) is %. (Round to one decimal place.)Which alternative in the table below should be selected when the MARR = 9% per year? The life of each alternative is 10 years. Increment Considered A(A – DN) A(B- A) A(C - B) A(D - C) A Investment cost A (Annual Revenues less Costs) $900 $600 $1,000 $1,600 $160 $147 $130 $161 IRR on A Investment Cost 10.1% 17.3% ? ..... The IRR on A(C-B) is %. (Round to one decimal place.)HW-Chp-8-1: A company is considering the following projects. If the MARR is 1 % / month/compounded quarterly and 10 years life, which should be selected based on the Rate-of-Return evaluation for multiple alternatives: Project H D 131,000 First Cont Monthly cont annual Income 111.000 6,300 $4.000 5.000 89,000 4,100 4,200 54,800 53,600 $1,900 Income every years 12.500 13,100 11,000 10.200 (a) PW for project A: Q.C. is every (b) PW comparison If they are Independent: Take Project First Cost OC- Income Income every Compare to Equation: PW Take project PW- HW-Chp-8-1-continued.:(b) PW comparison If they are mutually exclusive: Take only Project First Cast O.C Income Income every Compare to Incr.F.C. Inc.O.C. Incr. Income Incr.lac. every Equation: PW- Take project
- Which alternative in the table below should be selected when the MARR = 4% per year? The life of each alternative is 10 years. Increment Considered A(B-A) A Investment cost $500 $124 A (Annual Revenues less Costs) IRR on A Investment Cost The IRR on A(C-B) is 6.6 %. (Round to one decimal place.) The IRR on A(D-C) is %. (Round to one decimal place.) A(A-DN) $900 $157 11.6% C 21.2% A(C-B) $1,200 $168 ? A(D-C) $1,700 $170 ? DSuppose the MARR is 12%. Use the following table to answer the question-The IRR on the CMS Investment is FMS Initial Investment Annual Revenue Useful Life (Years) A. 17.0 % - 18.0% OB. 15.0 % - 16.0% OC. 11.0% - 12.0% O D.0.5% -1.0% O E.20.0% -21.0% CMS $20,000 6,688 $29,000 9,102Alternatives A, B associated with a project have data as in the following figure, MARR=12% /year, Study period is 6 years. Find: Which alternatives should be selected? A B Capital investment $2800 $5000 Annual cash flow 1100 1400 Useful life (years) 3 6 Market value at end of useful life 0 0
- For a project with: F.C. = JD 4000 Extra cost at the end of the 4 year = 2500 Annual income = 2000 n= 5 years a) Calculate the IRR (i*) b) is it acceptable, if MARR=10% per year?Assume $100,000 is available for investment and MARR =10% per year. If alternative A would earn 25% per year on inveatment of 60,000 and B would be earn 20% per year on investment of 75000 the weighted average(ROR) of AFind the NPV: Financial Calculator: Investment that pays $4,880 at the end of each of the next 3 years. Initial cost of $1,220 and a cost at the end of the 2nd year of $6,100. Interest rate is 1.7%. My numbers in the FC are: N: 3 I/Y: 1.7% PV: CPT; Initial Investment cost of -7,320 PMT: 4,880 each year (3x) CF1/CF2/CF3 FV: 0 My benefit was $14,155.99 - my cost of $7,320.00 = $6,835.99. My answer is incorrect. What am i doing wrong?
- For the following table, assume a MARR of 9% per year and a useful life for each alternative of six years that equals the study period the ior from least capital investment to greatest capital investment is Do nothing A C. B.Complete the IRR analysis by selecting the preferred Do nothing vs A A vs C A vs B Capital investment -$15,000 -%$2,000 -$3,000 O aB Annual Revenues $4,000 $900 $460 Annual costs -S1,000 Ob. Do nothing -S160 $100 Morket value $6.000 -%$2,200 $3,350 Occ RR 12.7% 10.9% OdA1. Compare the following alternatives using the Net Present Worth (NPW) method. Rate 5% per year. Construction $ Benefits $/yr Life years A 400,000 200,000 5 700,000 220,000 10 2. Solve the same problem using the Net Equivalent Uniform Annual (NEUA) method. What do you conclude?Determine the ROR, AW, and PP of the following engineering project when the MARR is 20% per year. Is the project acceptable? Engineering Project Details Investment cost $60,000 Expected life 7 years Annual receipts $15,000 Annual expenses $2,675.40 a. The Rate of Return of the project is ? Write the answer in percentage value, up to 2 decimal places. b. How much is the cash flow excess? Roundoff answer to the whole number. c. The payback Period is Write the answer in two decimal places.