Use incremental analysis to select the best alternatives using Benefit – Cost ration
BCR= PV Benefits/ Pv Cost
Where PV Benefits = Present value of benefits
Pv Cost = Present value of cost
A . Pw of benefits = 7330 (p/f, 6%,20 )=7330 * .312 = 2285.53
BCR = PV Benefits/ PV Cost
= 2285.53 / 4000
= 0.5713
B . Pw of benefits = 47000 ( P/F , 6 % ,20 ) = 47000 * 0.312 = 14664
BCR = PV Benefits/ PV Cost
= 14664 / 2000
= 7.332 > 1
C. Pw of benefits = 8730 ( P/F , 6 % ,20 ) = 8730 * 0.312 = 2723.76
BCR = PV Benefits/ PV Cost
= 2723.76 / 6000
= 0.45396
D. Pw of benefits = 1340 ( P/F , 6 % ,20 ) = 1340 * 0.312 = 418.08
BCR = PV Benefits/ PV Cost
= 418.08 / 1000
= 0.41808
E. Pw of benefits = 9000 ( P/F , 6 % ,20 ) = 9000 * 0.312 =2808
BCR = PV Benefits/ PV Cost
= 2808 / 9000
= 0.312
F. Pw of benefits = 9500 ( P/F , 6 % ,20 ) = 9500 * 0.312 =2964
BCR = PV Benefits/ PV Cost
= 2964 / 10000
= 0.2964
Since, B / C of the incremental alternative is greater than one,
choose the higher first cost alternative which is alternative B
Use incremental analysis to select the best alternatives using Benefit – Cost ration 16. Use incremental...
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%
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%
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%
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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
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solve this problem using engineering economy Compare and pick the best alternative based on "Incremental ROR analysis for a MARR of 15%. All alternatives have the same lives with no salvage value. (6 + 1 pts) Alternatives Investment $5000 $4000 $2500 Operating Cost $240 $500 $1000
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...