Part-a:
The AW of lead acid =Present Value of Cash Flows from Lead acid/PVIFA(5%,12)
= Present Value of Salvage Value -(Capital Investment + Present Value of Annual Expenses) }/PVIFA(5%,12)
= [0 – {6,000 + 2,500*PVIFA(5%,12)} ]/ PVIFA(5%,12)
= {0 –(6,000 + 2,500*8.8633)}/8.8633
= -28,158.25/8.8633
= -$3,176.95
.
.
The AW of lithium ion =Present Value of Cash Flows from Lithium Ion/PVIFA(5%,18)
= Present Value of Salvage Value -(Capital Investment + Present Value of Annual Expenses) }/PVIFA(5%,12)
= [2,800*0.4155 – {14,000 + 2,400*PVIFA(5%,18)} ]/ PVIFA(5%,18)
= {1,163.40 –(14,000 + 2,400*11.6896)}/11.6896
= -40,891.64/11.6896
= -$3,498.12
.
.
Lead Acid should be selected since it has lower AW i.e. Annual Outflow from Lead Acid is lower as compared to Lithium Ion.
.
.
Part-b:
Present Value of Cash Flows from Lead Acid = -Present Value of Lease Rentals -(Capital Investment + Present Value of Annual Expenses)
= 8,000*PVIFA(5%,6)/1.05^12 – (6,000 + 2,500*PVIFA(5%,12)
= -8,000*5.0757/1.79585 - 28,158.25
= -4,454.70 – 28,158.25
= -$26,612.95
.
Present Value of Cash Flows from Lithium Ion = = Present Value of Salvage Value -(Capital Investment + Present Value of Annual Expenses)
= [2,800*0.4155 – {14,000 + 2,400*PVIFA(5%,18)}
= {1,163.40 –(14,000 + 2,400*11.6896)}
= -40,891.64
.
.
Lead Acid should be selected since it has lower negative NPV.
determine which alternative should be selected if the analysis period is 18 years, the repeatability assumption...
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...
I keep getting the Lithium Ion ones wrong, someone please help. Consider the following EOY cash flows for two mutually exclusive alternatives one must be chosen The MARR is 4% per year Lead Acid $9,000 $2,750 Lithium lon Capital investment Annual expenses Useful life Market value at end of useful life $15,000 $2,200 12 years 18 years 50 $3,000 Click the icon to view the interest and annuity table for discrete compounding when ,24% per year. (a) Determine which alternative...
Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen). The MARR is 12% per year. Capital Investment Annual expenses Useful life Market value at end of useful life Lead Acid $8,000 $2,250 12 years $0 Lithium lon $13,000 $2,300 18 years $2,800 Click the icon to view the interest and annuity table for discrete compounding when /= 12% per year. (a) Determine which altemative should be selected based on the PW method. Assume repeatabllity...
Please DO NOT use excel. Show all steps please. You have 2 mutually exclusive alternatives and a MARR of 9%. Which alternative is preferred, based on repeatability assumption? Alternative E F Capital Investment $14,000 $65,000 Annual Expenses $14,000 $9,000 Useful Life (years) 4 20 Market Value at end of useful life $8,000 $13,000
0.6. Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen): Solar Panel A Solar Panel B Capital investment, $ 7,000 13.000 Annual operating expenses, SL 2.200 2.000 Market value, $ 1.000 2.30) Useful life, years 12 The MARR is 12% per year. Determine (using FW method) which alternative should be selected if the analysis period is 18 years, the repeatability assumption does not apply, and a solar panel can be leased for $6,000 per...
Question 1 A. Using Ms Excel, find out which alternative should be selected on the basis of the Present Worth method, if the rate of interest is 8% per year. • Alternative 1: Initial purchase price = $2500000, Annual operating cost $45000 at the end of 1st year and increasing by $3000 in the subsequent years till the end of useful life, Annual income = $120000, Salvage value = $550000, Useful life = 3 years. Alternative 2: Initial purchase price...
c. If the total capital investment budget is $150,000 , which alternative should be selected ? d.If the total capital investment budget available is $200,000 , which alternative should be selected ( if the alternatives are independent ) ? Four mutually exclusive alternatives are being evaluated, and their costs and revenues are itemized below. a. If the MARR is 15% per year and the analysis period is 12 years, use the PW method to determine which alternatives are economically acceptable...
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...
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 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=