Defender: . Current salvage value = $8,000, decreasing at an annual rate of 1000$ from the previous year's value Required overhaul = $1,500 O&M - $2,000 in year 1, increasing by 30% per year Challenger: Nc.=5 AECc = 5500 MARR-10% Find the loss in PV from mistakenly keeping the defender one year longer than is optimal. So in other words, if you found that the optimal indefinite plan was (10.1). 0.5)inf, then find the PV difference between that and the plan (0,2).0.5)inf. The PV difference is within $10 of: 289.11 269.11 329.11 None of the above 309.11 Defender: . Current salvage value = $8,000, decreasing at an annual rate of 1000$ from the previous year's value Required overhaul = $1,500 O&M - $2,000 in year 1, increasing by 30% per year Challenger: Nc.=5 AECc = 5500 MARR-10% Find the loss in PV from mistakenly keeping the defender one year longer than is optimal. So in other words, if you found that the optimal indefinite plan was (10.1). 0.5)inf, then find the PV difference between that and the plan (0,2).0.5)inf. The PV difference is within $10 of: 289.11 269.11 329.11 None of the above 309.11