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R. Baker Kearfott wrote:
Nate, On 5/12/2010 8:07 AM, Nate Hayes wrote:John Pryce wrote:. . .Nate, you point out mid-rad can save memory, hence bandwidth, on large interval calculations. But the saving is always less than 50%, ne c'est pas? because one has to store the "mid" value to full precision. If you had a method that saves an order or two of magnitude, it would be more convincing. But that usually comes from an improved algorithm.John, I think you don't know what you're talking about. Amdahl's Law is nonlinear. When you're already above 99% parallel, reducing the sequential portion of a program by a very tiny amount can mean the difference between 1,000X and 10,000X speedups.Nate, I've very puzzled by this. John's point was that the "mid" in mid-rad usually needs to be stored in full precision, while only the "rad" part could be economized. What does that have to do with Amdahl's law? Please explain or give an example. Where was John incorrect?
Baker,Reading and writing RAM is NOT a parallel activity, so any memory spillage of processor cache increases the sequential part of the program! Worse, if memory is spilled, it must later be read back in so this can double the hit to the sequential portion. Worse again if an algorithm "almost" fits into processor cache but repeatedly spills to RAM because it is just "slightly over" the cache limit. This can be HUGELY detrimental.
If a program is already 99.99% parallel (because its using compact mid-rad intervals, for example, and fits snugly into cache), but repeated memory spills during a long computation with larger inf-sup intervals increase the sequential portion of the program to .06%, the original 10,000X speedup is reduced to 2,000X (almost an order of magnitude).
John says it will always be less than a 50% difference. How can he justify this!? I would like to know. It seems to me he is just drawing a linear correlation between the speedup and the size, in bits, of a single compact mid-rad vs. inf-sup interval!!!
Nate