On Tue, Apr 05, 2016 at 08:35:43AM -0400, Sage Weil wrote: > On Mon, 4 Apr 2016, Allen Samuels wrote: >> But there's an approximation that gets the job done for us. >> >> When U is VERY SMALL (this will always be true for us :)). >> >> The you can approximate 1-(1-U)^D as D * U. (for even modest values of >> U (say 10-5), this is a very good approximation). >> >> Now the math is easy. >> >> The odds of failure for reading a block of size D is now D * U, with >> checksum correction it becomes (D * U) / (2^C). >> >> It's now clear that if you double the data size, you need to add one bit >> to your checksum to compensate. >> >> (Again, the actual math is less than 1 bit, but in the range we care >> about 1 bit will always do it). >> >> Anyways, that's what we worked out. > > D = block size, U = hw UBER, C = checksum. Let's add N = number of bits > you actually want to read. In that case, we have to read (N / D) blocks > of D bits, and we get > > P(reading N bits and getting some bad data and not knowing it) > = (D * U) / (2^C) * (N / D) > = U * N / 2^C > > and the D term (block size) disappears. IIUC this is what Chris was > originally getting at. The block size affects the probability I get an > error on one block, but if I am a user reading something, you don't care > about block size--you care about how much data you want to read. I think > in that case it doesn't really matter (modulo rounding error, minimum read > size, how precisely we can locate the error, etc.). > > Is that right? Yep, that's pretty much what I was thinking. My probability algebra is more than a little rusty but that looks like the formula I was starting to put together. Chris -- To unsubscribe from this list: send the line "unsubscribe ceph-devel" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
