Q6 5 Points Determine the economic service life at MARR 10 p

Q.6 (5 Points) Determine the economic service life at MARR 10% per year for a machine that has a first cost of $15,000 and estimated operating costs and yearend market values as shown. MI Market Value ($) rating Cost (S 1500 2200 3000 4000 5000 Year Ope 10000 2 3 4. 5 7000 5000 3500 2500

Solution

P= 15000, F= 10000, i =10%, n =1 , Operating Cost = 1500

CR(i) =(15000-10000)*(A/P,10%,1)+10000*0.1+1500

CR(i) = 5000*1.1+1000+1500=5500+1000+1500=8000

End of year 2,

P=15000, F= 7000, i= 0.1, n=2

OC= 1500+700(A/G,10%,2)=1500+700*0.476=1853.2

CR(i) = (15000-7000)*(A/P,10%,2)+7000*0.1+1853.2=8000*0.5762+700+1853.2=4610+700+1853.2=7163.2

End of year 3

P=15000, F=5000, n=3, i=0.1

OC=1500+800(A/G,10%,3)=1500+800*0.937=1500+749.6=2250

CR(i) =  (15000-5000)*(A/P,10%,3)+5000*0.1+2250=10000*0.4021+700+2250=4021+700+2250=6921

End of Year 4

P= 15000, F=3500, n=4, i=0.1

OC=1500+1000*(A/G,10%,4)1.321=2821.1

CR(i)=(150000-3500)*(A/P,10%4)+3500*0.1+2821.1=3171.1+11500(A/P,0.1,4)=3171,1+115000.3378=3970+3171.1=7141.1

CR(i) for year =3< CR(i) for year =4

Hence CR(i) at 3 years is Minimum therfore economic service life for machine is 3