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  • $\begingroup$ Maybe I got something completely wrong but please have a look at the code below which uses $N=6$ als calculation reference and on my machine produces: {{0.590009, 1.09306, 1.09899}, {0.770624, 2.43555, 2.45416}, {0.975321, 4.93756, 4.989}, {1.2041, 9.2909, 9.27697}, {1.45696, 16.4594, 16.3822}, {1.7339, 27.7425, 27.5057}} and thus a scaling that is much more like $N^6$ than $N^2$: exp[baseTime_, i_, n_] := baseTime*(i/6)^n; baseTime = AbsoluteTiming[BuildMatrix[6]][[1]]; Table[{exp[baseTime, i, 2], exp[baseTime, i, 6], AbsoluteTiming[BuildMatrix[i]][[1]]}, {i, 7, 12}] $\endgroup$ Commented Dec 23, 2016 at 17:26
  • $\begingroup$ It's a bit difficult to see all this code in a comment. Maybe better to edit the original question to show what it is you are measuring. $\endgroup$ Commented Dec 23, 2016 at 17:36
  • $\begingroup$ I included my thoughts into the initial question. Remark: I use the time for $N=6$ to make predictions according to assumed scaling behaviour. At least on my machine the assumption of a behaviour $\propto N^6$ fits much better. So I am not sure where your $\propto N^2$ scaling arises from. $\endgroup$ Commented Dec 23, 2016 at 17:47
  • 2
    $\begingroup$ Umm, 6^n and n^6 are very different beasts. $\endgroup$ Commented Dec 23, 2016 at 19:04
  • $\begingroup$ I don't understand your comment to be honest. At first, your code definitely won't work this way (your exp awaits two arguments, you provide three), secondly, please have a more thorough look at what I am doing above, the n is simply a parameter for the estimated exponent in my exp function. And I compare scaling according to $N^2$ (n=2) and $N^6$ (n=6) with one another to show you your initial estimation with a quadratic scaling is wrong. The (i/6)^n comes from the fact that I measure the baseTime in the $N=6$ case and has nothing to do with overall scaling. $\endgroup$ Commented Dec 23, 2016 at 22:55