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Genrish500 [490]
3 years ago
15

An injection-molding machine has a first cost of $1,050,000 and a salvage value of $240,000 in any year. The maintenance and ope

rating cost is $235,000 with an annual gradient of $75,000. The MARR is 10%. What is the most economic life?
a. years (EUAC - $759.376)
b. 7 years (EUAC - $672,616)
c. 3 years (EUAC - $904,384)
d. 5 years (EUAC - $608,428)
Engineering
1 answer:
Ilya [14]3 years ago
4 0

Explanation:

First cost =$ 1,050,000

Salvage Value =$ 225,000 Maintenance & Operating cost =$ 235,000 Maintenance & Operating Gradient =$ 75,000 MARR =10 %

EUAB - EAUC $=\$ 1,050,000(\mathrm{~A} / \mathrm{P}, 10 \%, \mathrm{n})+\$ 225,000(\mathrm{~A} / \mathrm{F}, 10 \%, \mathrm{n})$

$$

-\$ 235,000-\$ 75,000(\mathrm{~A} / \mathrm{G}, 10 \%, \mathrm{n})

$$

Try $\mathrm{n}=4$ years:

$$

\text { EUAB - EAUC }=\$ 331,275+\$ 48,488-\$ 235,000-\$ 103,575=-\$ 621,362

$$

Try $\mathrm{n}=5$ years:

EUAB - EUAC=-\$ 276,990+\$ 36,855-\$ 235,000-\$ 135,750=-$ 610,885

Try n=6 years

{ EUAB - EUAC }=-$ 241,080+$ 29,160-$ 235,000-$ 166,800=-$ 613,720

Thus, year 5 has the minimum EUAB - EUAC, hence the most economic life is 5 years.

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Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

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Standard deviation : \sigma=0.07

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Using the standard normal table for z, we have

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Sonja [21]

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Explanation:

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