You are ordering new equipment for the boss. Machine 1 will save $5,000 per year for 6 years; Machine 2 will save $6,000 per year for 5 years. Assume an interest rate of 10% compounded annually. What is the future value of savings from Machine 1 and Machine 2 in Year 6? Which machine should you select?

Respuesta :

Answer:

The answer is "$38,578, $40,294, choose Machine 2".

Explanation:

Machine 1

Yearly savings = $5,000

Time = 6 years

Rate of Interest  = 10%

Calculating the Future value:

[tex]FV = \$ 5,000 ( \frac{F}{A}, i, n)\\\\[/tex]

      [tex]= \$ 5,000 (\frac{F}{A}, 10 \%, 6)\\\\= \$ 5,000 \times 7.7156\\\\= \$ 38,578\\\\[/tex]

Machine 1 savings would have a potential value of $38,578  

Machine two  

Yearly savings = 6,000  

Time = five years  

Rate of interest = 10%;  

At the end of 5 years compute Potential value:

[tex]FV = \$ 6,000 (\frac{F}{A}, i, n) \\\\[/tex]

      [tex]= \$ 6,000(\frac{F}{A}, 10 \%, 5)\\\\= \$ 6,000 \times 6.1051\\\\= \$ 36,630.6\\\\[/tex]

At the end of  5 years the potential value is $36,630.6.

The future value in Year 6 calculate-  

[tex]FV = \$ 36,630.6 (\frac{F}{P}, i, n)\\\\[/tex]

      [tex]= \$ 36,630.6 (\frac{F}{P}, 10 \%, 1)\\\\= \$ 36,630.6 \times 1.100 \\\\= \$ 40,293.66 \ \ or \ \ \$ 40,294\\\\[/tex]

Machine 2 saving will be worth $40,294 in future  

Machine 2 's potential saving value is higher.  

Thus,   You can pick Machine 2.