The following two alternatives are given. Data A B First Cost $8.200 $5,600 Annual Cost $1,000 $800 Annual Benefit $2,700 $2,100 Life, Years 7 7 Salvage Value $2,800 $1,000 Assume that MARR is 15%. Use the incremental rate of return analysis to determine which alternative (A or B) one should choose. Find the AIRR, or a range of AIRR. <10% 10-12% 12-15% 15%

The following two alternatives are given. Data A B First Cost $8.200 $5,600 Annual Cost $1,000 $800 Annual Benefit $2,700 $2,100 Life, Years 7 7 Salvage Value $2,800 $1,000 Assume that MARR is 15%. Use the incremental rate of return analysis to determine which alternative (A or B) one should choose. Find the AIRR, or a range of AIRR. <10% 10-12% 12-15% 15%

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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Use the following alternatives to develop an incremental analysis choice table and answer the following questions. Alternative A Initial Cost - $15,000 Annual Revenue (Year 1-15) - $2500 Salvage Benefit (Year 15) - $2000 Alternative B Initial Cost - $25,000 Annual Revenue (Year 1-15) - $3500 Salvage Benefit (Year 15) - $5000 1. Determine the incremental rate of return IRRHigh-Low. Provide your answer as a percentage and round to the nearest hundredth. 2. Determine the internal rate of return for the higher cost alternative IRRHigh. Provide your answer as a percentage and round to the nearest hundredth. 3. Determine the internal rate of return for the lower cost alternative IRRLow. Provide your answer as a percentage and round to the nearest hundredth. Please answer all 3.
Any suggestions? Thanks for the help! Given the data for three different alternatives in the table below, determine the best alternative using the incremental rate of return (∆RoR) analysis. MARR =9%. A B C First cost $15,000 $25,000 $20,000 O &M Cost/ year 1,600 400 900 Benefit/year 8,000 13,000 9,000 Salvage value 3,000 6,000 4,600 Life in years 4 4 4 1. The better alternative between the first increment is ________________. 2. The better alternative between the second increment is ___________________.
Compare the following two alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of return (AROR). Alt. Construction cost $ Benefits $/yr Salvage $ Service Life (yrs) A 410,000 55,000 20,000 11 B 250,000 35,000 10,000 11
Given the financial data for four mutually exclusive alternatives in the table below, determine the best alternative using the incremental rate of return (AROR) analysis. MARR = 10%. A B C D $15,000 $21,200 $36,000 45,000 1,600 700 400 1,000 First cost O &M Cost/ year Benefit/year 8,000 9,000 13,000 Salvage value 3,000 4,600 6,000 Life in years 4 15,000 10,000
Data for 2 alternatives are given below, determine the cost of alternative B so that the two alternatives will be equally desirable. Assume an interest rate of 15%. Use benefit cost ratio analysis Alternatives A Cost Salvage Value Annual Benefit Life (years) B P100k X 10k 16k 28k 30k 10 12
1. Data for two alternatives are given in the table below. Determine the X cost of alternative B so that the two alternatives will be equally desirable. Assume an interest of 11%. Use BCR analysis. ALTERNATIVE A Cost 9500 Salvage value 1000 1600 Annual benefit 2800 3700 Life (years) 12
Using (incremental) benefit-cost ratio analysis OR (incremental) internal rate of return, determine which of the following three mutually exclusive alternatives should be selected. Each alternative has a 6-year useful life. Assume the MARR is 6%. First Cost Uniform Annual Benefit Salvage Value $560 $140 $40 A $340 $100 0 B $125 $40 0 C
5. Compare the following two alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of return (AROR). Alt. Benefits S/yr Salvage S Service Life (ys) Construction cost $ 410,000 55,000 20,000 11 B 250,000 35,000 10,000 11
Given the financial data for four mutually exclusive alternatives in the table below, determine the best alternative using the incremental rate of return (AROR) analysis. MARR =10%. A B D First cost $15,000 $21,200 $36,000 45,000 O &M Cost/ 1,600 700 400 1,000 year Benefit/year 8,000 9,000 13,000 15,000 Salvage value 3,000 4,600 6,000 10,000 Life in years 5
6. Compare the following three alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of retum method (AROR). Construction cost S Benefits S/yr Salvage S Alt. Service Life (yrs) 510,000 145,000 -10,000 775.000 155,000 15,000 1,075,000 165,000 20,000 6.
For the following alternatives compute the Delta B/C ratio of Alternative C minus Alternative B. Use 11% as MARR. (Remember for our convention, salvage value is a minus cost.) Project Initial Investment Annual Benefit Salvage Value Useful Life 2.13 2.98 1.18 0.99 1.28 A -1500 350 320 5 B -2000 500 610 6 -2500 600 820 7 D -5200 850 2300 9
Two different alternatives are being compared. Using Net Present Worth Analysis which alternative is best? Use an interest rate of 9%. Alternative Initial Cost A В $ 1,425 $ 2,168 $ 1,325 $ $ Annual Benefit 620 Salvage Value Useful Life (years) 465 $ 897 3 9. A is the best alternative Choose neither alternative Choose either alternative B is the best alternative

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