The Cooper Company is considering investing in a recuperator. The recuperator will have an initial cost of $20,000 and a service life of 10 years. Operating and maintenance costs for the first year are estimated to be $1,500, increasing by $100 every year thereafter. It is estimated that the salvage value of the recuperator will be 20 percent of its initial cost. The recuperator will result in equal annual fuel savings throughout its service life. Assuming MARR is 12 percent per year compounded yearly, what is the minimum value of fuel savings for which the recuperator is attractive

Minimum value of Annual fuel savings = $ 5170.21

Explanation:

PV of Total Cost = Initial Cost + PV of Operating and maintenance costs

PV of Total Cost = 20000 + 1500/1.12 + (1500+100) / 1.12^2 + (1500+100*2) / 1.12^3 + (1500+100*3) / 1.12^4 + (1500+100*4) / 1.12^5 + (1500+100*5) / 1.12^6 + (1500+100*6) / 1.12^7 + (1500+100*7) / 1.12^8 + (1500+100*8) / 1.12^9 + (1500+100*9) / 1.12^10

PV of Total Cost = 30500.74

PV of salvage value = 20000*20%/1.12^10

PV of salvage value = 1287.89

Minimum value of Annual fuel savings = (PV of Total Cost - PV of salvage value) / PVA (rate, nper)

Minimum value of Annual fuel savings = (30500.74 - 1287.89) / PVA (12%,10)

Minimum value of Annual fuel savings = 29212.85/5.650223

Minimum value of Annual fuel savings = $ 5170.21