Question

$250,000 is invested in equipment having a salvage value equal to $250,000(0.80n) after n years of use. O&M costs equal $60,000 the first year and increase $8,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC?

Answer 1

P = $250,000

EAC Capital Recovery = (P-S) × (A/P, 10%, n) + S*i

EAC Maintenance = 60000+8000(A/G, 10%, n)

(A/P, 10%, n) = Value*(1-(1+i)^-n)/i

(A/G, 10%, n) = (1/i)-n/((1+i)ⁿ-1)

Sinc Total EUAC is minimum for year 6

So the optimum replacement interval is after 6 years

Minimum EUAC= $ 126696.4