Question

d. 17.5 marks] A remotely situated fuel cell has an installed cost of $2,000 and will reduce existing surveillance expenses by $350 per year. The border security agency's MARR is 10%. What is the approximate minimum useful life that makes the fuel cell purchase break even? i. 8.52 ii. 8.63 iii. 8.77 iv. 8.89

Answer 1

By break even we mean where total cost is equal to total income. Here total cost is $ 2,000 and annual saving is $ 350. Let us assume in n years total cost will be equal to total income.

$\large 2,000 = 350(P/A,10\%,n)$

$\large \implies \frac{1-1.1^{-n}}{0.1}= \frac{2,000}{350}$

$\large \implies 1-1.1^{-n}= 5.714285 \times 0.1$

1.1"=l-0.5714285

$\large \implies 1.1^{-n}= 0.428571$

$\large \implies 1.1^{n}= \frac{1}{0.428571}$

$\large \implies 1.1^{n}= 2.3333$

Taking log on both sides we get

$\large \implies n\times log(1.1)= log(2.3333)$

$\large \implies n= \frac{log(2.3333)}{log(1.1)}$

$\large \implies n \approx 8.89 \: Years$

The number of years in which the break even will occur is 8.89 years.

Correct option is last.

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