Homework Answers
Annual Worth
As no interest rate is given, let's take 10% for calculation
For Alternative A
= - Construction Cost/PVIFA(10%,40years)
= -$30 million/9.7790507
= -$3,067,782
For Alternative B
(For infinite period)
= - Construction Cost x 0.10
= -$35,000,000 x 0.10
= -$3,500,000
Since the negative amount should be minimum, we prefer alternative A
Note :
PVIFA = Present Value Interest Factor Annuity
We get this from online tables available or we can calculate using calculator
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6. Compre the following alternatives using the Net Equivalent Uniform Annual Worth method. Alt. Construction cost...
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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
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
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.
Find the equivalent uniform annual cost 3. Using an interest rate of 12%, find the equivalent miform annual cost for a piece of construction equipment that has an initial purchase cost o $30,000,...
Find the equivalent uniform annual cost
3. Using an interest rate of 12%, find the equivalent miform annual cost for a piece of construction equipment that has an initial purchase cost o $30,000, an estiinat nomic life of 8 years, and an estimated salvage value of $10,000. Annual mainte- nance will anount to $600 per year and periodic overhauls costing $1.,000 each will occur at the end of the second, fourth, and sixth years.
3. Using an interest rate of...
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
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...
6-52. Compare alternatives A and B with the equivalent worth method of your choice if the...
6-52. Compare alternatives A and B with the equivalent worth method of your choice if the MARR is 15% per year. Which one would you recommend? State al assumptions. (6.5) Capital investment $50,000 Operating costs S5,000 at end of year 1 and increasing by S500 per year thereafter S20,000 S10,000 at end of year I and increasing by $1,000 per year thereafter Overhaul costs $5,000 every 5 years None Life Salvage value S10,000 if just 20 years 10 years negligible...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
2. Consider the following two mutually exclusive alternatives: Cost, $ Uniform annual benefit, $ Useful life,...
2. Consider the following two mutually exclusive alternatives: Cost, $ Uniform annual benefit, $ Useful life, years 100,000 16,000 150,000 24,000 Using a 10% interest rate, and an annual cash flow analysis, determine which alternative should be selected. Draw the CFD.
A telecommunications firm is considering a product expansion of a popular cell phone. Two alternatives for...
A telecommunications firm is considering a product expansion of a popular cell phone. Two alternatives for the cell phone expansion are summarized below. The company uses an MARR of 9% per year for decisions of this type, and repeatability may be assumed. L Initial Cost (S) Annual Benefit ($) Salvage Value (S) Useful Life (yrs) Expansion A 148,650.000 1,090,000 10.725,000 Expansion B 63.750,000 2,860,000 12,380,000 1. Which Expansion Option should be taken? Use the Equivalent Uniform Annual Worth Method. You...
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
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.
Find the equivalent uniform annual cost
3. Using an interest rate of 12%, find the equivalent miform annual cost for a piece of construction equipment that has an initial purchase cost o $30,000, an estiinat nomic life of 8 years, and an estimated salvage value of $10,000. Annual mainte- nance will anount to $600 per year and periodic overhauls costing $1.,000 each will occur at the end of the second, fourth, and sixth years.
3. Using an interest rate of...
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
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...
6-52. Compare alternatives A and B with the equivalent worth method of your choice if the MARR is 15% per year. Which one would you recommend? State al assumptions. (6.5) Capital investment $50,000 Operating costs S5,000 at end of year 1 and increasing by S500 per year thereafter S20,000 S10,000 at end of year I and increasing by $1,000 per year thereafter Overhaul costs $5,000 every 5 years None Life Salvage value S10,000 if just 20 years 10 years negligible...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
2. Consider the following two mutually exclusive alternatives: Cost, $ Uniform annual benefit, $ Useful life, years 100,000 16,000 150,000 24,000 Using a 10% interest rate, and an annual cash flow analysis, determine which alternative should be selected. Draw the CFD.
A telecommunications firm is considering a product expansion of a popular cell phone. Two alternatives for the cell phone expansion are summarized below. The company uses an MARR of 9% per year for decisions of this type, and repeatability may be assumed. L Initial Cost (S) Annual Benefit ($) Salvage Value (S) Useful Life (yrs) Expansion A 148,650.000 1,090,000 10.725,000 Expansion B 63.750,000 2,860,000 12,380,000 1. Which Expansion Option should be taken? Use the Equivalent Uniform Annual Worth Method. You...
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