Question

# Consider the three mutually exclusive alternatives below. At the end of their useful lives, Alternatives X...

Consider the three mutually exclusive alternatives below. At the end of their useful lives, Alternatives X and Z will be replaced with identical replacements so that a 10-year service requirement is met. If the MARR is 3% per year, which alternative (if any) should be chosen based on the annual worth method?

 Alt X Alt Y Alt Z Capital investment \$300,000 \$425,000. \$500,000. Annual savings \$68,750 \$108,750. \$188,750. Salvage value \$90,000 \$125,000. \$140,000. Life, years 10 20 5

Annual worth=CR+A of AOC

CR=P(A/P,i,n)+S(A/F,i,n)

Here,

P=Initial investment cost

S=Salvage value

i=3%

n=10 years

A of AOC=Annual cost of operation

Annual worth of Alt X =-\$300000(A/P,3%,10)-\$90000(A/F,3%,10)+\$68750=-\$300000*0.1172-\$90000*0.0872+\$68750=-35160-7848+68750=\$25742.

Annual worth of Alt Y =-\$425000(A/P,3%,20)-\$125000(A/F,3%,20)+\$108750=-\$425000*0.0672-\$125000*0.0372+\$108750=-28560-4650+108750=\$75540.

Annual worth of Alt Z =-\$500000(A/P,3%,5)-\$140000(A/F,3%,5)+\$188750=-\$500000*0.2184-\$140000*0.1884+\$188750=-109200-26376+188750=\$53174.

Annaul worth of Alt Y is maximum therefore Alt Y is to be chosen.

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