Compare the following 2 alternatives using the Net Present Worth (NPS) method – rate is 3%...
Compare the following 2 alternatives using the Net Present Worth (NPS) method – rate is 3% per year. Draw all cash flow diagrams.
Please show work using a formula.
| Alt. | Construction Cost ($) | Benefit ($/yr.) | Service Life (yrs.) |
| A | 380,000 | 200,000 | 7 |
| B | 450,000 | 220,000 | 7 |
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Compare the following 2 alternatives using the Net Present Worth
(NPS) method – rate is 3%...
6. Compre the following alternatives using the Net Equivalent Uniform Annual Worth method. Alt. Construction cost...
6. Compre the following alternatives using the Net Equivalent Uniform Annual Worth method. Alt. Construction cost $ Life (yrs) A $30million 40 B $35million Infinite
The rate is 2.75% 5. Copmare the following 2 altrernatives using the Net Equivalent Uniform Annual...
The rate is 2.75%
5. Copmare the following 2 altrernatives using the Net Equivalent Uniform Annual method Alt. Construction cost $ Benefit ($/yr) Salvage $ Service Life (yrs) A 1,500,000 400,000 40,000 9 B 2,300,000 550,000 80,000 18
5. Compare the following two alternatives by the IRR method, given MARR of 8%/year. Is the...
5. Compare the following two alternatives by the IRR method, given MARR of 8%/year. Is the incement in cost form A to B justified? Alt. Construction cost | Benefits Styr | Salvage Service Life (yrs 510,000 145,000 10,000 775,000 155,000 20,000
Compare two alternatives, A and B. on the basis of a present worth evaluation using /=...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability ass...
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability assumption is acceptable. Aternative I: Initial investment of $45,000, net revenue the first year of $8,000, increasing $4,000 per year for the six year useful life. Salvage value is estimated to be $6500. Alternative II: Initial investment of $60,000, uniform annual revenue of $12,000 for the five year useful life. Slavage value is estimated to be $9,000.
show the formula Compare the following alternatives on the basis of Present worth Analysis at an...
show the formula
Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
Compare the two following alternatives in terms of present worth using MARRz 6% for a study...
Compare the two following alternatives in terms of present worth using MARRz 6% for a study period of 2 years Assume that the salvage value does not change depending on year sold. Alternative 1: First cost: 30,000 Yearly cost 6,000 Salvage value: 9,000 Lifetime: 4 Alternative 2 First cost: 75,000 Yearly cost 5,000 Salvage value: 13,000 Lifetime: 6 Alt 1: $-80,826, Alt 2: S-105,266 Alt 1: S-74,228, Alt 2: S-112,525 Alt 1 S-32,990, Alt 2: S-72,597 Alt 1: S-102,270, Alt...
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate of 3% per year. Use firs...
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate of 3% per year. Use first the B/C method to determine feasibility and then the incremental B/C method to determine the oprimum level of investment. Alt Construction Cost $ Annual Benefits $/yr Life yrs A 105,000 40,000 5 B 230,000 52,000 6 C 350,000 64,000 7 D 600,000 100,000 8
The following data is available for three different alternatives. Assume an interest rate of 9% per...
The following data is available for three different alternatives. Assume an interest rate of 9% per year, compounded annually. Initial Cost Annual Benefit Useful Life (yrs) Alternative A 7,000 1,275 infinite Alternative B 9,400 1,327 4 Alternative 14,000 4,867 17 1 Alternatives B and Care replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative. Choose Alternative A because it lasts the longest Choose Alternative A because its net present worth...
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate...
Compare the following alterntives given a market rate of 5.45% per year and an inflation rate of 3% per year. Use first the B/C method to determine feasibility and then the incremental B/C method to determine the oprimum level of investment. Alt Construction Cost $ Annual Benefits $/yr Life yrs A 105,000 40,000 5 B 230,000 52,000 6 C 350,000 64,000 7 D 600,000 100,000 8
6. Compre the following alternatives using the Net Equivalent Uniform Annual Worth method. Alt. Construction cost $ Life (yrs) A $30million 40 B $35million Infinite
The rate is 2.75%
5. Copmare the following 2 altrernatives using the Net Equivalent Uniform Annual method Alt. Construction cost $ Benefit ($/yr) Salvage $ Service Life (yrs) A 1,500,000 400,000 40,000 9 B 2,300,000 550,000 80,000 18
5. Compare the following two alternatives by the IRR method, given MARR of 8%/year. Is the incement in cost form A to B justified? Alt. Construction cost | Benefits Styr | Salvage Service Life (yrs 510,000 145,000 10,000 775,000 155,000 20,000
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
3. Compare the two following two alternatives using an equivalent worth method and a MARR of 12%. The repeatability assumption is acceptable. Aternative I: Initial investment of $45,000, net revenue the first year of $8,000, increasing $4,000 per year for the six year useful life. Salvage value is estimated to be $6500. Alternative II: Initial investment of $60,000, uniform annual revenue of $12,000 for the five year useful life. Slavage value is estimated to be $9,000.
show the formula
Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
Compare the two following alternatives in terms of present worth using MARRz 6% for a study period of 2 years Assume that the salvage value does not change depending on year sold. Alternative 1: First cost: 30,000 Yearly cost 6,000 Salvage value: 9,000 Lifetime: 4 Alternative 2 First cost: 75,000 Yearly cost 5,000 Salvage value: 13,000 Lifetime: 6 Alt 1: $-80,826, Alt 2: S-105,266 Alt 1: S-74,228, Alt 2: S-112,525 Alt 1 S-32,990, Alt 2: S-72,597 Alt 1: S-102,270, Alt...
The following data is available for three different alternatives. Assume an interest rate of 9% per year, compounded annually. Initial Cost Annual Benefit Useful Life (yrs) Alternative A 7,000 1,275 infinite Alternative B 9,400 1,327 4 Alternative 14,000 4,867 17 1 Alternatives B and Care replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative. Choose Alternative A because it lasts the longest Choose Alternative A because its net present worth...
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