Hi, On 22/01/2015 16:37, Chad William Seys wrote: > Hi Loic, >> The size of each chunk is object size / K. If you have K=1 and M=2 it will >> be the same as 3 replicas with none of the advantages ;-) > > Interesting! I did not see this explained so explicitly. > > So is the general explanation of k and m something like: > k, m: fault tolerance of m+1 replicas, space of 1/k*(m+k) replicas, plus > slowness > ? I'm not sure to understand the space formula but it looks like you got the idea. > So one should never bother with k=1 b/c: > k=1, m: fault tolerance of m+1, space of m+1 replicas, plus slowness. > (therefore, just use m+1 replicas!) > > but > k=2, m=1: > might be useful instead of 2 replicas b/c it has fault tolerance of 2 > replicas, space of 1/2*(1+2) = 3/2 = 1.5 replicas, plus slowness. > > And > k=2, m=2: > which should be as tolerant as 3 replicas, but take up as much space as > (1/2)*(2+2)=2 replicas (right?). That's also how I understand it :-) Cheers > Thanks again! > Chad. > -- Loïc Dachary, Artisan Logiciel Libre
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