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Compare the following alternatives based on the
| Project A | Project B | |
| Initial Cost | $60.000 | $90.000 |
| Annual cost of operation | 15.000 | 8.000 |
| Annual cost of reparation | 5.000 | 2.000 |
| Annual increase of the repair | 1.000 | 1.500 |
| Salvage value | 8.000 | 12.000 |
| Life, years | 15 | 15 |
Step by step
Solved in 2 steps
- Consider the following projects. The incremental ROR of the difference is most nearly. Most nearly number for answer. First Cost Life Annual benefit Annual Maintenance cost 1,000 Salvage Value 0.9% O 13% O 12% O 11% O No correct answer A O 10% 10,000 4,000 1.000 10 years B 16,593.4 6,000 2,000 3,000 10 years 1hm1. Determine the B/C ratio for the following project First Cost P100, 000 Project life, years 5 Salvage value Annual benefits P10, 000 P60, 000 Annual O and M Interest rate, % P22, 000 15 Ans: B/C = 1.16 2. Data for two alternatives are as follows: Alternatives A В Investment Р35, 000 P50, 000 Annual benefits P20, 000 P25, 000 Annual O and M Estimated life, years Net salvage value Ре, 450 P13, 830 4 8. Р3, 500 Using an interest rate of 20%, which alternative should be chosen? Ans: Alternative A is referred over Alternative BDetermine the B/C ratio for the following project. First Cost P100, 000 Project life, years 5 Salvage value P10, 000 Annual benefits Р60, 000 Annual O and M Р22, 000 Interest rate, % 15
- From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine A B Initial Cost (Php) Annual Operating Cost 146,000 15,000 80,000 10,000 3 220,000 10,000 75,000 25,000 Annual Revenue Salvage Value Useful Life (years) 61. Determine the B/C ratio for the following project. First Cost P100, 000 Project life, years 5 Salvage value P10, 000 Annual benefits P60, 000 Annual O and M P22, 000 Interest rate, % 15 Ans: B/C = 1.16From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine Initial Cost (Php) Annual Operating Cost A В 146,000 15,000 220,000 10,000 75,000 25,000 Annual Revenue 80,000 10,000 Salvage Value Useful Life (years) 6
- Check whether the project is accepted or rejected and recommend the best project, MARR=6% and n=8 years, using Payback period, ROI, NPV, Aw, IRR methods. ID Project I Project II increase Annual Initial Annual cost revenues initial Annual Salvage value annual salvage value running in cost revenues maintenance revnues cost 1910095 280000 60000 2000 28000 400000 72000 2000 40000 12000Compare the machines below using present worth analysis at i10% per year and find which one should be selected Machine Y First Cost Operating Cost Savage Lfe cycle Machine X 20.000 3.000 3.000 3years Firat Cost 30.000 .000 Annuel Operating Cost Salvage Life cycle 3.000 6 yoarsCompute the benefit cost ratio of the following project: Project cost = P80 000 Gross Income = P25000/year %3D Operating Cost = P6000/year Salvage value = 0 Life of project = 10 years %3D Rate of interest = 12%
- Two investment opportunities are as follows: A B First cost $150 $115 Uniform annual benefit $26 $20 End-of-useful-life salvage value $O SO Useful life, in years 10 10 If the MARR is 13%, which alternative should be selected?Given the following project alternatives, determine the most economical investment given an MARR of 10% using IRR method. Project A B C D 6.12% -2.8067% -3.394% 2699% Investment 13.69% $ 100,000 $ Annual Profit Useful Life Salvage Value $ 140,500 $ 148,200 $ 27,500 38,600 38,500 $ 98,000 $ 22,100 What is the incremental IRR of the second comparison? 5 5 5 5 $ 10,000 $ $ 14,000 25,600 14,000The 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. 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. O 10% O 10-12% O 12-15% O > 15%