[Beowulf] NFS share - IO rate

Bogdan Costescu bcostescu at gmail.com
Wed Apr 21 02:07:26 PDT 2010

On Tue, Apr 20, 2010 at 9:36 PM, Henning Fehrmann
<henning.fehrmann at aei.mpg.de> wrote:
> Client A says I got the IO-rate Ra which is twice as big as the IO-rate of B:
> Ra = 2 Rb.  The test on B took twice as long as on A.

I look at this differently: the overall rate that the server has dealt
with is given by the total amount of data transferred in the time
taken by the slowest node. So if

Ra=Da/Ta and Rb=Db/Tb then I consider Rt=(Da+Db)/max(Ta, Tb)

It's a similar view to the one I have about a parallel program: the
real time (wallclock) of giving me the solution is what matters, not
whatever built-in counters report. And this real time is the time
taken by the slowest node (=the one which finished last, I'm not
referring to the CPU speed...)

> But one can also interpret the result in a different way.
> Client A was doing its IO test and Client B got no bandwidth left at all.
> Only after A finished the test, B has been served. This results in a twice as small
> average rate on B.

This shows a different point of view: you mention the average rate on
B, I talk about what the server sees. So what are you actually
interested in ? Do you have some rate specified by the manufacturer
for the server that you want to compare with ? Or do you have some
requirement of rate per node ?


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