Problem involving Moore's Law

  • #1
Thread originally posted in a non-homework section, so missing the homework template
[Mentor note: Thread title changed to reflect problem content]

I have a problem that I would like to solve with Python, but I don't know the equation for this, here it is:

It has been shown that Moore's law not only applies to semiconductor density, but it also predicts the increase in (reasonable) simulation sizes, and the reduction in computational simulation run-times. First show for a fluid mechanics simulation that takes 4 hours to run on a machine today, that it should only take 1 hour to run on machines built 3 years from now, and only 15 minutes on machines built 6 years from now. Then show that for a large simulation that has an estimated run-time of 5 years that it would complete sooner if we waited 3 years to start the simulation.

What is the formula I would use to solve this?
 
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Answers and Replies

  • #2
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I have a problem that I would like to solve with Python, but I don't know the equation for this, here it is:

It has been shown that Moore's law not only applies to semiconductor density, but it also predicts the increase in (reasonable) simulation sizes, and the reduction in computational simulation run-times. First show for a fluid mechanics simulation that takes 4 hours to run on a machine today, that it should only take 1 hour to run on machines built 3 years from now, and only 15 minutes on machines built 6 years from now. Then show that for a large simulation that has an estimated run-time of 5 years that it would complete sooner if we waited 3 years to start the simulation.

What is the formula I would use to solve this?
A better question is "What is Moore's Law?" Have you done any research on the statement of this law?
 
  • #3
I know that formula is expressed as Pn = Po x 2^n, the only thing I can surmise is that n = 3 / 2 = 1.5 for 3 years, 6 years would n = 6 /2 = 3, but that's all, how do I translate the rest, like how it should only take 1 hour on one and 15 minutes on the other, or for a large simulation with an estimated run-time, how do I work that into the Moore's Law formula?
 
  • #4
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I know that formula is expressed as Pn = Po x 2^n
No. According to Wikipedia,
"Moore's law" is the observation that, over the history of computing hardware, the number of transistors in a dense integrated circuit doubles approximately every two years.
Your formula doesn't take into account that the doubling occurs every two years.
TheQuizmaster said:
, the only thing I can surmise is that n = 3 / 2 = 1.5 for 3 years, 6 years would n = 6 /2 = 3, but that's all, how do I translate the rest, like how it should only take 1 hour on one and 15 minutes on the other, or for a large simulation with an estimated run-time, how do I work that into the Moore's Law formula?
 
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  • #5
Guess I shouldn't trust this source then http://www.umsl.edu/~abdcf/Cs4890/link1.html [Broken]
That is why I'm here, so I can learn how to do these kinds of problems, what is the correct way to do it?
 
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  • #6
NascentOxygen
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Guess I shouldn't trust this source then http://www.umsl.edu/~abdcf/Cs4890/link1.html [Broken]
That article states:

Equation: Pn = Po x 2^n
[....]
n = number of years to develop a new microprocessor divided by 2 (ie. every two years)
 
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  • #7
Right, so it isn't suitable for the one I'm trying to solve, how would you go about solving this?
 
  • #8
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Right, so it isn't suitable for the one I'm trying to solve, how would you go about solving this?
Adjust the formula so that you get doubling every two years instead of each year. It's not a big change. It shouldn't be too hard for a quizmaster...
 
  • #9
Adjust the formula so that you get doubling every two years instead of each year. It's not a big change. It shouldn't be too hard for a quizmaster...
I'm not that good yet, how would I adjust that, I want to learn how to solve this, but I don't know how?
 
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  • #10
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If you wanted doubling every six months (half a year), it would be ##2^{2n}##. How could you change it so that the doubling occurs every two years?
 

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