Question

e chosen) The MARS the owing tot cash rows of two mutually exclusive alle Capital investment Annual expenses Useful life Mark
pre Info TUL TUTUR Given UT- TUTT AE TURBOCOWN - Given P FIP 1.0500 1.1025 1.1576 1.2155 1.2763 1.3401 1.4071 1.4775 1.5513 1

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.
atives (one must be chosen). The MARR is 5% per year. Capital investment Annual expenses Useful life Market value at end of u
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Answer #1

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.

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