Each of the three mutually exclusive alternatives shown has a 5-year useful life. If the MARR is 10%, which alternative should be selected? Solve the problem by benefit-cost ratio analysis.
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- The cash flows for three mutually exclusive alternatives are given in table below. MARR = 4%. ALB Alt. C 27,000 24,000 7,600 6,500 13% 11% Initial cost Annual benefits ROR Life in years Alt. A $15,000 $4,500 15% 5 Reference: Case Study 8 The best alternative for a MARR of 2.0% using the incremental rate of return analysis is A. Alt. C B. Alt. B C. Alt. A OD.Do-nothingWould love some help on how to approach this - thanks! The cash flows for three different alternatives are given in table below. MARR =10%. Alt. A Alt. B Alt. C Initial cost $5,000 9,000 7,500 Annual benefits $1,457 2,518 2,133 RoR 14% 13% 12.4% Life in years 5 1. ΔRoR for the first increment (Alt. C-Alt. A) is ___________________. A.10.12% B. 9.38% C. 11.85% D. 11.00% 2. ΔRoR for the second increment is ___________________. A. 10.12% B. 9.38% C. 8.94% D. 9.87% 3. The best alternative for a MARR of 10% using the incremental rate of return analysis is ____________. A. Alt. C B. Alt. A C. Alt. B D. Do nothingAny help would be appreciated! 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 ________________. A. Alt. A or Alt. B B. Alt. A C. Alt.C D. Alt. B 2. The better alternative between the second increment is ___________________. A. Alt. B or Alt. C B. Alt. B C. Alt. C D. Alt. A
- The following information is provided for five mutually exclusive alternatives that have 20-year useful lives. If the MARR is 12%, which alternative be selected? Use incremental rate of return Alternatives A B D E Cost $4,000 $2,000 $6,000 $1,000 $9,000 Annual Benefit $639 $410 $761 $117 $785 PW of Benefit $7,330 $4,700 $8,730 $1,340 $9,000 Rate of Return 15% 20% 11% 10% 6%Given the data for two different alternatives in the table below, determine the best alternative using the incremental rate of return (AROR) analysis. MARR =9%. A B First cost $15,000 $20,000 O&M Cost/year 1,600 900 Benefit/year 8,000 9,000 Salvage value 3,000 4,600 Life in years 4 4Using Incremental with EUAW analysis find the best alternative, MARR = %10. should use Excel and show your equations seperately, see below example: [A Benefit - [IC (A/P, i%, n) - Salvage (A/F, i, n)] + A Cost+ G Cost (A/G, i, n)] Data B C A First Cost $2,300,000 $2,780,000 $2,540,000 Salvage Value $82,000 $118,000 $97,000 Annual Benefit $580,000 $670,000 $650,000 M&O $65,000 $78,000 $71,000 M&O Gradient $11,000 $15,000 $12,500 Useful Life, Years 9 9 9 You - ZOOM
- 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 Group of answer choices Cannot determine Alterative A Alterative B Alterative CUsing Incremental with EUAW analysis find the best alternative, MARR = %10. should use Excel and show your equations seperately, see below example: [A Benefit - [IC (A/P, i%, n) - Salvage (A/F, i, n)] + A Cost+ G Cost (A/G, i, n)] A B C $2,300,000 $2,780,000 $2,540,000 $82,000 $118,000 $97,000 $580,000 $670,000 $650,000 $65,000 $78,000 $71,000 $11,000 $15,000 $12,500 9 9 9 Data First Cost Salvage Value Annual Benefit M&O M&O Gradient Useful Life, Years YouGiven 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 The better alternative between the second increment (B – C) is:. Group of answer choices Alt. B Alt. C Alt. B or Alt. C Alt. A
- :[A] { > Incremental analysis ([ B Alternative], [B wins ]): C A company considering 2 different machines at MARR at 12% Both life spans = 10 years Initial Cost Annual Operating lost Benefits per yin ar Salvage Value \table MM If you are and of frying investment to company decide More than 2 alternatives if the additional increment is worth while, compare Alternative: A Incremental analysis (Alternative) pairs then B C A MARR Company at Considering 2 different machines at 12% Both life spans = 10 years. M/C X м/с у Initial Cost 160000 285000 Annual Operating Cost Benefits per year Salvage Value 45 000 90000 45000 105000 20000 40000Given cash flows for two alternatives as shown in table below, choose the most attractive alternative if MARR = 8%. Year 0 1 2 3 through ∞ Alt. A -$42K $3.6K $3.6K $3.6K Alt. B -$54K $4.7K $4.7K $4.7K Group of answer choices Alt. A Alt. B Select neither Select eitherDetermine which alternative, if any, should be chosen based on Annual Worth method using 15% MARR. Use Repeatability Method. Alternative A B First Cost (Investment Cost) $ 5,000 $10,200 Uniform Annual Benefit $1,100 $2,300 Useful Life 5 years 10 years a. The Annual Worth of Alternative A is = $ Blank 1 b. The Annual Worth of Alternative B is = $ Blank 2 c. Choose Alternative (Type only A or B) = Blank 3 Note: Show final answer to the nearest WHOLE NUMBER. No need to write the Unit of Measure. Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer