Homework Answers
Part 1
The PW of Alternative A is $(108236)
Present worth = present value of all cash flows
Present worth of alternative A = -55000-(5000/(1.11^1))-(5500/(1.11^2))-(6000/(1.11^3))-(6500/(1.11^4))-(7000/(1.11^5))-(7500/(1.11^6))-(8000/(1.11^7))-(8500/(1.11^8))-(9000/(1.11^9))-(9500/(1.11^10))-(10000/(1.11^11))-(10500/(1.11^12))-(5000/(1.11^3))-(5000/(1.11^6))-(5000/(1.11^9))-(5000/(1.11^2))+(10000/(1.11^12)) = -108236
Part 2
The PW of Alternative B is $(68895)
Present worth of alternative B = -25000-(10000/(1.11^1))-(11000/(1.11^2))-(12000/(1.11^3))-(1300/(1.11^4))-(14000/(1.11^5))-(15000/(1.11^6)) = -68895
Alternative B is recommended and should be considered as the present worth is higher than alternative A
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Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of y...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Inve...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...
Solve for A and B, Engineering Economy please solve it right! Question Help %) Problem 6-52 (algorithmic) Compare al...
Solve for A and B, Engineering Economy
please solve it right!
Question Help %) Problem 6-52 (algorithmic) Compare alternatives A and B with the present worth method if the MARR is 15% per year. Which one would you recommend? Assume repeatability and a study period of 20 years. $40,000 $7,000 at end of year 1 and increasing by $700 per year thereafter $7,000 every 5 years $15,000 $14,000 at end of year 1 and increasing by $1,400 per year thereafter...
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 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.
For the following table, assume a MARR of 12% per year and a useful life for...
For the following table, assume a MARR of 12% per year and a useful life for each alternative of eight years which equals the study period. The rank-order of alternatives from least capital investment to greatest capital investment is Z ·Y? W? X Complete the incremental analysis by selecting the prefer ed altemative. Do nothing" is not an option. $250 $400 $100 Capital investment ? Annual cost savings ? Market value ? PW (12%) 70 100 138 90 50 67...
Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen). The...
Consider the following EOY cash flows for two mutually
exclusive alternatives (one must be chosen). The MARR is 5% per
year.
I need the PW of the Lead Acid and Lithium Ion.
Problem 6-28 (algorithmic) EQuestion Help Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen) The MARR is 5% per year ead Acid $7,000 thium lon Capital investment Annual expenses Useful life Market value at end of useful life $13,000 $2.500 $2,750 12...
determine which alternative should be selected if the analysis period is 18 years, the repeatability assumption...
determine which alternative should be selected if the analysis
period is 18 years, the repeatability assumption does not apply,
and a battery system can be leased for 8,000$ per year after the
useful life of either battery is over.
e chosen) The MARS the owing tot cash rows of two mutually exclusive alle Capital investment Annual expenses Useful life Market value at end of useful in Lead Acid $6.000 $2.500 12 years Lithium Ion $14.000 $2.400 18 years $2.800 Chap...
Can someone help me with part b? I can not seem to find the correct answer for present worth of B1 with 10 year plannin...
Can someone help me with part b? I can not seem to find the
correct answer for present worth of B1 with 10 year planning
horizon or the present worth of B2 with 10 year planning
horizon.
Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the end of each year. Both projects B1 and B2 will be available (or can...
Question 12 For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine...
Question 12 For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine A Machine B 240,000 First cost (AED) 20,000 Annual maintenance cost per year, AED 5,000 2.300 Periodic cost every 10 years, AED 10,000 Salvage cost 2000 Life. vears Match the closest correct answers for the below questions: Calculate the present value of the maintenance costs for...
Toll lanes on a section of the l-40 freeway are being considered in order to reduce traffic congestion and travel times. Since this is a government project, the B-C ratio method must be applied in th...
Toll lanes on a section of the l-40 freeway are being considered in order to reduce traffic congestion and travel times. Since this is a government project, the B-C ratio method must be applied in the evaluation. Construction costs of the project are estimated to be $18,600,000, and $321,000 per year in operating and maintenance costs are anticipated. In addition, the lanes must be resurfaced every fourth year of their 20-year projected life at a cost of $1,110,000 per occurrence...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...
Solve for A and B, Engineering Economy
please solve it right!
Question Help %) Problem 6-52 (algorithmic) Compare alternatives A and B with the present worth method if the MARR is 15% per year. Which one would you recommend? Assume repeatability and a study period of 20 years. $40,000 $7,000 at end of year 1 and increasing by $700 per year thereafter $7,000 every 5 years $15,000 $14,000 at end of year 1 and increasing by $1,400 per year thereafter...
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 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.
For the following table, assume a MARR of 12% per year and a useful life for each alternative of eight years which equals the study period. The rank-order of alternatives from least capital investment to greatest capital investment is Z ·Y? W? X Complete the incremental analysis by selecting the prefer ed altemative. Do nothing" is not an option. $250 $400 $100 Capital investment ? Annual cost savings ? Market value ? PW (12%) 70 100 138 90 50 67...
Consider the following EOY cash flows for two mutually
exclusive alternatives (one must be chosen). The MARR is 5% per
year.
I need the PW of the Lead Acid and Lithium Ion.
Problem 6-28 (algorithmic) EQuestion Help Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen) The MARR is 5% per year ead Acid $7,000 thium lon Capital investment Annual expenses Useful life Market value at end of useful life $13,000 $2.500 $2,750 12...
determine which alternative should be selected if the analysis
period is 18 years, the repeatability assumption does not apply,
and a battery system can be leased for 8,000$ per year after the
useful life of either battery is over.
e chosen) The MARS the owing tot cash rows of two mutually exclusive alle Capital investment Annual expenses Useful life Market value at end of useful in Lead Acid $6.000 $2.500 12 years Lithium Ion $14.000 $2.400 18 years $2.800 Chap...
Can someone help me with part b? I can not seem to find the
correct answer for present worth of B1 with 10 year planning
horizon or the present worth of B2 with 10 year planning
horizon.
Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the end of each year. Both projects B1 and B2 will be available (or can...
Question 12 For alternatives shown in the table below you are trying to decide which alternative you should choose based on their capitalized costs (CC). Use an interest rate of 10% per year. Machine A Machine B 240,000 First cost (AED) 20,000 Annual maintenance cost per year, AED 5,000 2.300 Periodic cost every 10 years, AED 10,000 Salvage cost 2000 Life. vears Match the closest correct answers for the below questions: Calculate the present value of the maintenance costs for...
Toll lanes on a section of the l-40 freeway are being considered in order to reduce traffic congestion and travel times. Since this is a government project, the B-C ratio method must be applied in the evaluation. Construction costs of the project are estimated to be $18,600,000, and $321,000 per year in operating and maintenance costs are anticipated. In addition, the lanes must be resurfaced every fourth year of their 20-year projected life at a cost of $1,110,000 per occurrence...
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