In the given case, Useful life is different for given project. We will use replacement chain method. Project B can be assumed to repeated with same cash flows once.
Project A
Initial Cost=Co=-$8200
Annual benefit=R=$1850
Useful life=n=8 years
Salvage=S=$500
MARR=i=7%
PW of project A=Co+R*(P/A,7%,8)+S*(P/F,7%,8)
Let us calculate interest factors.
(P/F,0.07,10)=1/(1+0.07)^8=0.582009
PW of project of project A=-8200+1850*5.971299+500*0.582009=$3137.91
Now we consider project B
Project B
Initial Cost=Co=-$5000
Annual benefit=R=$1750
Useful life=n=4 years
Salvage=S=$700
MARR=i=7%
PW of project A=Co+R*(P/A,7%,4)+S*(P/F,7%,4)
Let us calculate interest factors.
(P/F,0.07,10)=1/(1+0.07)^4=0.762895
PW of project of project B=PWs=-5000+1750*3.387211+700*0.762895=$1461.65
PW of project after repeating once=PWs+PWs*(P/F,7%,4)=1461.65+1461.65*0.762895=$2576.74
In a study period of 8 years, PW of project A is higher. It should be selected.
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