Question

the following describes the first of two equipment being considered for purchase for a new factory. it requires an initial investment of $80,000 and annual maintenance costs of $4,000. it has a salvage value of $9,000 at the end of its 10 years of service life. calculate the present worth of this equipment at marr of 10% per year
a. -$51,952
b. -$108,048
c. -$104,578
d. -$101,108
if the second alternative has a present worth of -$118,853, which alternative should be preferred based on the present worth method?

Explanation:

Step1: Calculate the present worth formula

The present worth ($PW$) formula for a project with initial investment ($I$), annual cost ($A$), salvage value ($S$), and $n$ years at a minimum - attractive - rate - of - return (MARR) $i$ is $PW=-I - A(P/A,i,n)+S(P/F,i,n)$. Here, $I = 80000$, $A = 4000$, $S=9000$, $i = 0.1$, and $n = 10$.

The present - worth factor for an annuity $(P/A,i,n)=\frac{(1 + i)^{n}-1}{i(1 + i)^{n}}$ and the present - worth factor for a single sum $(P/F,i,n)=\frac{1}{(1 + i)^{n}}$.

Step2: Calculate $(P/A,0.1,10)$ and $(P/F,0.1,10)$

For $(P/A,0.1,10)$:

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For $(P/F,0.1,10)$:

$$ (P/F,0.1,10)=\frac{1}{(1 + 0.1)^{10}}=\frac{1}{2.59374246}\approx0.3855 $$

Step3: Calculate the present worth

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Answer:

D. - $101,108$