Midterm(Case6)

Midterm(Case6) - Joseph Murphy Case Study 6 Initial MARR =...

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Joseph Murphy Case Study 6 Initial MARR = 15% PowrUp Lloyd's Investment Annual Repair Investment Annual Repair Year and Salvage Main. Savings and Salvage Main. Savings AW Values $(6,642.00) $(800.00) $25,000.00 $(7,025.00) $(300.00) $35,000.00 0 $(26,000.00) $- $- $(36,000.00) $- $- 1 $- $(800.00) $25,000.00 $- $(300.00) $35,000.00 2 $- $(800.00) $25,000.00 $- $(300.00) $35,000.00 3 $- $(800.00) $25,000.00 $- $(300.00) $35,000.00 4 $- $(800.00) $25,000.00 $- $(300.00) $35,000.00 5 $- $(800.00) $25,000.00 $- $(300.00) $35,000.00 6 $2,000.00 $(800.00) $25,000.00 $- $(300.00) $35,000.00 7 $- $(300.00) $35,000.00 8 $- $(300.00) $35,000.00 9 $- $(300.00) $35,000.00 10 $3,000.00 $(300.00) $35,000.00
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New MARR = 15% Lloyd's Investment Annual Repair AW PowrUp Year and Salvage Main. Savings $17,558.00 AW Values $(7,025.00) $(300.00) $35,000.00 AW Lloyd's 0 $(36,000.00) $- $- $27,675.00 1 $- $(300.00) $35,000.00 2 $- $(300.00) $32,000.00 3 $- $(300.00) $28,000.00 4 $- $(1,200.00) $26,000.00 5 $- $(1,320.00)
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Midterm(Case6) - Joseph Murphy Case Study 6 Initial MARR =...

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