Timeline for How does variance in task completion time affect makespan?
Current License: CC BY-SA 3.0
8 events
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May 7, 2013 at 16:21 | vote | accept | Patrick87 | ||
May 7, 2013 at 9:21 | comment | added | András Salamon | I don't know where the "hunch" came from; it is not consistent with the rest of the heuristic reasoning. | |
Apr 19, 2012 at 20:15 | vote | accept | Patrick87 | ||
May 7, 2013 at 16:21 | |||||
Apr 13, 2012 at 22:13 | history | edited | Gilles 'SO- stop being evil' | CC BY-SA 3.0 |
formatting
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Apr 13, 2012 at 14:36 | comment | added | Patrick87 | the first is justified by the fact that $1 - (1 - P(X = 5)^k)^n$ tends to $1$ for fixed $k$ and increasing $n$; the latter by the fact that $Var[X + X] = Var[X] + Var[X] = 2\sigma^2 \leq 4\sigma^2 = 4Var[X] = Var[2X]$... so the variance doesn't increase linearly as a function of $k$. Is that compatible with your thinking (that's how I'm interpreting what you have so far)? | |
Apr 13, 2012 at 14:34 | comment | added | Patrick87 | +1 Excellent intuition, and this helps to clarify my thinking as well. So increasing processor counts tends to increase makespan under a weak scaling assumption; and increasing task counts tends to decrease makespan under a strong scaling assumption (of course it takes longer; I mean the work/makespan ratio improves). These are interesting observations, and they seem true; | |
Apr 13, 2012 at 14:22 | history | edited | svinja | CC BY-SA 3.0 |
added 228 characters in body
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Apr 13, 2012 at 14:14 | history | answered | svinja | CC BY-SA 3.0 |