[Beowulf] NFS share - IO rate
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Bogdan Costescu bcostescu at gmail.comWed Apr 21 02:07:26 PDT 2010
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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 ? Cheers, Bogdan
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