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1. Compare the ME alternatives below using an MARR of 10% per year. Assume that "do...
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%
QUESTION 3For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $26,5383000010000Annual cost, $/year8,0606,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:A- AW for machine A=QUESTION 4For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $1500021,66710000Annual cost, $/year8,8706,0004,000Salvage value, $4,0005,0001,000Life, years362Answer the below questions:B- AW for machine B=
Determine whether either of the alternatives below should be selected. Use an MARR of 15% per...
Determine whether either of the alternatives below should be
selected. Use an MARR of 15% per year. Incremental Rate of Return
Analysis must be used. (PLEASE DO NOT USE EXCEL).
First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15
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 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
Question 1 The cash flows given in table below are for two different alternatives. MARR =10%...
Question 1 The cash flows given in table below are for two different alternatives. MARR =10% Data IN Initial Cost Annual Benefits Salvage Value Useful Life in years M $20,000 $6,000 $5,000 $80,000 $10,000 $20,000 a) Determine the annual worth of alternative M b) Determine the annual worth of alternative N
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.
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year....
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 First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC)...
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC) at i=10% per year, which is the best alternative in this scenario? • Alternative 1, AW = $87,500 and n = (forever) • Alternative 2, PW = -$895,000 and n = (forever) • Alternative 3, First cost (FC) of $900,000, annual operating savings of 3,000 per year, salvage = $200,000, and n = (forever) Alternative 2 Alternative 3 None of them Alternative 1 QUESTION...
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%.
For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10%.Machine AMachine BMachine CFirst cost, $ 15000 30000 10,360Annual cost, $/year 8,320 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :C- PW for machine C =
For the following table, assume a MARR of 9% per year and a useful life for...
For the following table, assume a MARR of 9% per year and a useful life for each alternative of six years that equals the study period. The rank-order of alternatives from least capital investment to greatest capital investment is Do Nothing AC B. Complete the IRR analysis by selecting the preferred alternative. A C CB Do Nothing A - $3,500 A Capital investment A Annual revenues A Annual costs A Market value -$2,000 -$15,000 4,000 - 1,000 900 450 -150...
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...
Determine whether either of the alternatives below should be
selected. Use an MARR of 15% per year. Incremental Rate of Return
Analysis must be used. (PLEASE DO NOT USE EXCEL).
First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15
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 year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
Question 1 The cash flows given in table below are for two different alternatives. MARR =10% Data IN Initial Cost Annual Benefits Salvage Value Useful Life in years M $20,000 $6,000 $5,000 $80,000 $10,000 $20,000 a) Determine the annual worth of alternative M b) Determine the annual worth of alternative N
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.
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 First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC) at i=10% per year, which is the best alternative in this scenario? • Alternative 1, AW = $87,500 and n = (forever) • Alternative 2, PW = -$895,000 and n = (forever) • Alternative 3, First cost (FC) of $900,000, annual operating savings of 3,000 per year, salvage = $200,000, and n = (forever) Alternative 2 Alternative 3 None of them Alternative 1 QUESTION...
For the following table, assume a MARR of 9% per year and a useful life for each alternative of six years that equals the study period. The rank-order of alternatives from least capital investment to greatest capital investment is Do Nothing AC B. Complete the IRR analysis by selecting the preferred alternative. A C CB Do Nothing A - $3,500 A Capital investment A Annual revenues A Annual costs A Market value -$2,000 -$15,000 4,000 - 1,000 900 450 -150...
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...
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