Question

Which of the alternatives presented in the following table would you recommend ding the analysis based on annual worth? 2. Initial cost- I Annual benefit $800 $200 $180 Salvage valueL$500 | $900 | O Life in years 2 years 3 Years Infinity MARR $1500 $900 $1200 10%

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Answer #1

AWa = -1500*(A/P,10%,2)+800+500*(A/F,10%,3)

= -1500*(0.1*(1+0.1)^2)/((1+0.1)^2-1)+800+500*(0.1/((1+0.1)^2-1))

= -1500*0.58+800+500*0.48

= 170

AWb = -900*(A/P,10%,3)+200+900*(A/F,10%,3)

= -900*(0.1*(1+0.1)^3)/((1+0.1)^3-1)+200+900*(0.1/((1+0.1)^3-1))

= 110

AWc = P*i= -1200*0.1 = -120

So looking at the annual worth values, AWc is less which needs to be selected

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