Question

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

Answer 1

Answer:

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.