Question

1) Consider these two machines (alternatives): (12 Points) B A $5000 $1750 $700 $8200 $1850 $500 First Cost Uniform annual benefit Salvage Value Useful Life, in Years 4 If the MARR (minimum attractive rate of return) -7 % , which alternative should be selected? Use the Present worth Analysis method.

Answer 1

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.

1 1+i (P/A, i, n)

1 (1+0.07) 0.07 (P/A, 0.07, 10) 5.971299

(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.

1 1+i (P/A, i, n)

1 (1+0.07) (P/A, 0.07, 10) 3.387211 0.07

(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.