Question

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?

Answer 1

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

Step-by-step explanation:

Machine 1

Yearly savings = $5,000

Time = 6 years

Rate of Interest = 10%

Calculating the Future value:

`FV = \$ 5,000 ( (F)/(A), i, n)\\\\`

`= \$ 5,000 ((F)/(A), 10 \%, 6)\\\\= \$ 5,000 * 7.7156\\\\= \$ 38,578\\\\`

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:

`FV = \$ 6,000 ((F)/(A), i, n) \\\\`

`= \$ 6,000((F)/(A), 10 \%, 5)\\\\= \$ 6,000 * 6.1051\\\\= \$ 36,630.6\\\\`

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

The future value in Year 6 calculate-

`FV = \$ 36,630.6 ((F)/(P), i, n)\\\\`

`= \$ 36,630.6 ((F)/(P), 10 \%, 1)\\\\= \$ 36,630.6 * 1.100 \\\\= \$ 40,293.66 \ \ or \ \ \$ 40,294\\\\`

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

Machine 2 's potential saving value is higher.

Thus, You can pick Machine 2.