[Hadoop-common-user] Mathematics behind Hadoop-based systems

Just thinking out loud: runtime generally is not equal to the "hours of
data". In your processing model it's your next iteration that will have this
amount of data. By equating them, you're assuming an equilibrium case. But
if you're in an equilibrium, what would increasing the capacity mean? - I
think it'd mean that you process your so-far accumulated data faster, so your
next iteration will have less data to process and so on until you get to a new
equilibrium (but how fast will you get there?). Lesson learned: increasing
capacity shifts equilibrium. It's kind of like a reaction time of your
system... Sometimes, I imagine, as long as you're keeping up with the arrival
rate you don't care if it takes a week's full or a day's. In fact there may
be some constraints on the minimum size of the input.

Another comment: you're assuming that processing time is linear in the input
size. Of course it depends on the processing you're doing, but even if YOUR
processing is linear, Hadoop needs to sort keys, and that is at best
O(n*log(n)), so there is inherent non-linearity present.

Similarly about the number of nodes: doubling your nodes will not double your
service rate for a variety of reasons.

-Yuri

That's a great way of putting it: "increasing capacity shifts the
equilibrium". If you work some examples you'll find that it doesn't take
many iterations for a workflow to converge to its stable runtime. There is
some minimum capacity you need for there to be an equilibrium though, or
else the runtime doesn't stabilize.

Regarding the sorting, I believe the complexity in Hadoop is more like n / R
* log(n/R), where R is in the number of reducers. And in practice, I've
found sorting to not take very long. But you're right - fundamentally this
model is an approximation.