Total opportunity cost of investment A = 4000(1 + 0.06)20 = 3.2071 × 4000 = 12828
Net benefit = 7330 - 12828 = -5498
Total opportunity cost of investment B = 2000(1 + 0.06)20= 6414.2
Net benefit = 47000 - 6414.2 = 40585.8
Total opportunity cost of investment C = 6000 (1 + 0.06)20= 19242.6
Net benefit = 8730 - 19242.6 = -10512.6
Total opportunity cost of investment D = 1000(1 +0.06)20 = 3207.1
Net benefit = 1340 - 3207.1 = - 1867.1
Total opportunity cost of investment E = 9000(1+0.06)20= 28863.9
Net benefit = 9000- 28863.9 = -19863
Total opportunity cost of investment F = 10000(1+0.06)20 = 32071
Net benefit = 9500 - 32071 = - 22571
Since the net benefit is negative in all investments except B, B is the best alternative.
16. Use incremental analysis to select the best alternatives using Benefit-Cost ration 4000 2000 6000 7330...
16. Use incremental analysis to select the best alternatives using Benefit-Cost ration 4000 2000 6000 7330 470008730 GO0D 1000 9000 10000 Cost Pw(benefits) 90009500 Useful life is 20years Interest 6%
Use incremental analysis to select the best alternatives using
Benefit – Cost ration
16. Use incremental analysis to select the best alternatives using Benefit - Cost ration Cost 4000 2000 7330 47000 8730 1000 1340 9000 9000 10000 9500 6000 Pw(benefits) Useful life is 20years Interest 6%
16. Use incremental analysis to select the best alternatives using Benefit - Cost ration Cost Pw(benefits) 4000 2000 6000 1000 9000 10000 7330 47000 8730 1340 9000 9600 Useful life is 20years Interest 6%
16. Use incremental analysis to select the best alternatives using Benefit- Cost ration 4000 2000 60001000 9000 10000 7330 Cost Pw(benefits) 47000 1340 9500 Useful life is 20years Interest 6%
3. Assuming a MARR of 20%, use incremental analysis (defender vs challenger approach) to select the best choice among the four alternatives: A Initial cost 2500 4800 4200 3600 Annual benefit 850 700 850 1300 Salvage value 2500 1750 1250 3000 Useful life (yrs) 5 4. Use IRR and incremental analysis, assuming a MARR of 20%, to solve problem 83. 5. Use Benefit to Cost (B/C) ratio and incremental analysis to solve problem #3.
brief explanation Solve this problem using the incremental Benefit - Cost ration with, expected life of 10 years and rate of return of 10% Alternative A Initial cost $50,000 Annual maintenance cost $4,000 Estimated annual benefit $10,000 Alternative B Initial cost $30,000 Annual maintenance cost $3,000 Estimated annual benefit $9,000 a. Select A with B/C =1.14 b. Select B with B/C = 1.14 c. Reject A with B/C = 1.14 d. Select B with B/C = 0.14
Using the incremental B-C analysis, B-C ratio with PW and a MARR of 10%, choose the best alternative from the following three mutually exclusive alternatives given below. State your assumptions. Option B Option C Option A $30000 $50000 $70000 5 10 15 Initial investment Life in years Salvage value Annual benefits 0 0 $2000 $8000 $8000 $8000
Q1: How to use incremental rate of return analysis given two alternatives and choosing which one is the best? Q2: Case 1: Alternative 1 Alternative 2 Cost 500 700 Annual Cost . 600 800 Annual Benefit 700 . 900 Case 1: Alternative 1 Alternative 2 Cost 700 500 Annual Cost . 800 600 Annual Benefit 900 700 Based on these two cases, how to I apply incremental return analysis to this, and instead of using incremental rate of return...
international genetic technologies inc. (InGen) is examining
the following three mutually exclusive alternatives.
3) Using benefit-cost ratio analysis, a 10-year useful life and a MARR of 25%, determine which of the following mutually exclusive models should be selected. А в C D E Initial Cost $100 $200 $300 $400 $500 $37 $60 $83 $137 $150 Annual Benefits 4) A big box company is using a benefit-cost ratio analysis to select which one of the 3 alternatives shown below should be...
3. Use a spreadsheet for evaluation of the multiple alternatives provided below. Use incremental B/C analysis. These alternatives are relative to the application of nanotechnology and the use of thin-film solar panels applied to houses to reduce the dependency on fossil-fuel generated electrical energy. A community of 400 new all- electric public housing units will utilize the technology as anticipated proof that significant reductions in overall utility costs can be attained over the expected 15-year life of the housing. The...