Hi, On 02/06/2017 04:08 AM, Jaze Lee wrote: > It is more complicated than i have expected..... > I viewed http://tracker.ceph.com/issues/15653, and know that if the > replica number is > bigger than the host we choose, we may meet the problem. > > That is > if we have > host: a b c d > host: e f g h > host: i j k l > > we only choose one from each host for replica three, and the distribution > is as we expected? Right ? > > > The problem described in http://tracker.ceph.com/issues/15653, may happen > when > 1) > host: a b c d e f g > > and we choose all three replica from this host. But this is few happen > in production. Right? > > > May be i do not understand the problem correctly ? The problem also happens with host: a b c d e f g when you try to get three replicas that are not on the same disk. You can experiment with Dan's script https://gist.github.com/anonymous/929d799d5f80794b293783acb9108992 Cheers > > > > > > > > > > > 2017-02-04 2:54 GMT+08:00 Loic Dachary <loic@xxxxxxxxxxx>: >> >> >> On 02/03/2017 04:08 PM, Loic Dachary wrote: >>> >>> >>> On 02/03/2017 03:47 PM, Sage Weil wrote: >>>> On Fri, 3 Feb 2017, Loic Dachary wrote: >>>>> On 01/26/2017 12:13 PM, Loic Dachary wrote: >>>>>> Hi Sage, >>>>>> >>>>>> Still trying to understand what you did :-) I have one question below. >>>>>> >>>>>> On 01/26/2017 04:05 AM, Sage Weil wrote: >>>>>>> This is a longstanding bug, >>>>>>> >>>>>>> http://tracker.ceph.com/issues/15653 >>>>>>> >>>>>>> that causes low-weighted devices to get more data than they should. Loic's >>>>>>> recent activity resurrected discussion on the original PR >>>>>>> >>>>>>> https://github.com/ceph/ceph/pull/10218 >>>>>>> >>>>>>> but since it's closed and almost nobody will see it I'm moving the >>>>>>> discussion here. >>>>>>> >>>>>>> The main news is that I have a simple adjustment for the weights that >>>>>>> works (almost perfectly) for the 2nd round of placements. The solution is >>>>>>> pretty simple, although as with most probabilities it tends to make my >>>>>>> brain hurt. >>>>>>> >>>>>>> The idea is that, on the second round, the original weight for the small >>>>>>> OSD (call it P(pick small)) isn't what we should use. Instead, we want >>>>>>> P(pick small | first pick not small). Since P(a|b) (the probability of a >>>>>>> given b) is P(a && b) / P(b), >>>>>> >>>>>> >From the record this is explained at https://en.wikipedia.org/wiki/Conditional_probability#Kolmogorov_definition >>>>>> >>>>>>> >>>>>>> P(pick small | first pick not small) >>>>>>> = P(pick small && first pick not small) / P(first pick not small) >>>>>>> >>>>>>> The last term is easy to calculate, >>>>>>> >>>>>>> P(first pick not small) = (total_weight - small_weight) / total_weight >>>>>>> >>>>>>> and the && term is the distribution we're trying to produce. >>>>>> >>>>>> https://en.wikipedia.org/wiki/Conditional_probability describs A && B (using a non ascii symbol...) as the "probability of the joint of events A and B". I don't understand what that means. Is there a definition somewhere ? >>>>>> >>>>>>> For exmaple, >>>>>>> if small has 1/10 the weight, then we should see 1/10th of the PGs have >>>>>>> their second replica be the small OSD. So >>>>>>> >>>>>>> P(pick small && first pick not small) = small_weight / total_weight >>>>>>> >>>>>>> Putting those together, >>>>>>> >>>>>>> P(pick small | first pick not small) >>>>>>> = P(pick small && first pick not small) / P(first pick not small) >>>>>>> = (small_weight / total_weight) / ((total_weight - small_weight) / total_weight) >>>>>>> = small_weight / (total_weight - small_weight) >>>>>>> >>>>>>> This is, on the second round, we should adjust the weights by the above so >>>>>>> that we get the right distribution of second choices. It turns out it >>>>>>> works to adjust *all* weights like this to get hte conditional probability >>>>>>> that they weren't already chosen. >>>>>>> >>>>>>> I have a branch that hacks this into straw2 and it appears to work >>>>>> >>>>>> This is https://github.com/liewegas/ceph/commit/wip-crush-multipick >>>>> >>>>> In >>>>> >>>>> https://github.com/liewegas/ceph/commit/wip-crush-multipick#diff-0df13ad294f6585c322588cfe026d701R316 >>>>> >>>>> double neww = oldw / (bucketw - oldw) * bucketw; >>>>> >>>>> I don't get why we need "* bucketw" at the end ? >>>> >>>> It's just to keep the values within a reasonable range so that we don't >>>> lose precision by dropping down into small integers. >>>> >>>> I futzed around with this some more last week trying to get the third >>>> replica to work and ended up doubting that this piece is correct. The >>>> ratio between the big and small OSDs in my [99 99 99 99 4] example varies >>>> slightly from what I would expect from first principles and what I get out >>>> of this derivation by about 1%.. which would explain the bias I as seeing. >>>> >>>> I'm hoping we can find someone with a strong stats/probability background >>>> and loads of free time who can tackle this... >>>> >>> >>> It would help to formulate the problem into a self contained puzzle to present a mathematician. I tried to do it last week but failed. I'll give it another shot and submit a draft, hoping something bad could be the start of something better ;-) >> >> Here is what I have. I realize this is not good but I'm hoping someone more knowledgeable will pity me and provide something sensible. Otherwise I'm happy to keep making a fool of myself :-) In the following a bin is the device, the ball is a replica and the color is the object id. >> >> We have D bins and each bin can hold D(B) balls. All balls have the >> same size. There is exactly X balls of the same color. Each ball must >> be placed in a bin that does not already contain a ball of the same >> color. >> >> What distribution guarantees that, for all X, the bins are filled in >> the same proportion ? >> >> Details >> ======= >> >> * One placement: all balls are the same color and we place each of them >> in a bin with a probability of: >> >> P(BIN) = BIN(B) / SUM(BINi(B) for i in [1..D]) >> >> so that bins are equally filled regardless of their capacity. >> >> * Two placements: for each ball there is exactly one other ball of the >> same color. A ball is placed as in experience 1 and the chosen bin >> is set aside. The other ball of the same color is placed as in >> experience 1 with the remaining bins. The probability for a ball >> to be placed in a given BIN is: >> >> P(BIN) + P(all bins but BIN | BIN) >> >> Examples >> ======== >> >> For instance we have 5 bins, a, b, c, d, e and they can hold: >> >> a = 10 million balls >> b = 10 million balls >> c = 10 million balls >> d = 10 million balls >> e = 1 million balls >> >> In the first experience with place each ball in >> >> a with a probability of 10 / ( 10 + 10 + 10 + 10 + 1 ) = 10 / 41 >> same for b, c, d >> e with a probability of 1 / 41 >> >> after 100,000 placements, the bins have >> >> a = 243456 >> b = 243624 >> c = 244486 >> d = 243881 >> e = 24553 >> >> they are >> >> a = 2.43 % full >> b = 2.43 % full >> c = 2.44 % full >> d = 2.43 % full >> e = 0.24 % full >> >> In the second experience >> >> >>>> sage >>>> >>>> >>>>> >>>>>> >>>>>>> properly for num_rep = 2. With a test bucket of [99 99 99 99 4], and the >>>>>>> current code, you get >>>>>>> >>>>>>> $ bin/crushtool -c cm.txt --test --show-utilization --min-x 0 --max-x 40000000 --num-rep 2 >>>>>>> rule 0 (data), x = 0..40000000, numrep = 2..2 >>>>>>> rule 0 (data) num_rep 2 result size == 2: 40000001/40000001 >>>>>>> device 0: 19765965 [9899364,9866601] >>>>>>> device 1: 19768033 [9899444,9868589] >>>>>>> device 2: 19769938 [9901770,9868168] >>>>>>> device 3: 19766918 [9898851,9868067] >>>>>>> device 6: 929148 [400572,528576] >>>>>>> >>>>>>> which is very close for the first replica (primary), but way off for the >>>>>>> second. With my hacky change, >>>>>>> >>>>>>> rule 0 (data), x = 0..40000000, numrep = 2..2 >>>>>>> rule 0 (data) num_rep 2 result size == 2: 40000001/40000001 >>>>>>> device 0: 19797315 [9899364,9897951] >>>>>>> device 1: 19799199 [9899444,9899755] >>>>>>> device 2: 19801016 [9901770,9899246] >>>>>>> device 3: 19797906 [9898851,9899055] >>>>>>> device 6: 804566 [400572,403994] >>>>>>> >>>>>>> which is quite close, but still skewing slightly high (by a big less than >>>>>>> 1%). >>>>>>> >>>>>>> Next steps: >>>>>>> >>>>>>> 1- generalize this for >2 replicas >>>>>>> 2- figure out why it skews high >>>>>>> 3- make this work for multi-level hierarchical descent >>>>>>> >>>>>>> sage >>>>>>> >>>>>>> -- >>>>>>> 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 >>>>>>> >>>>>> >>>>> >>>>> -- >>>>> Loïc Dachary, Artisan Logiciel Libre >>>>> -- >>>>> 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 >>>>> >>> >> >> -- >> Loïc Dachary, Artisan Logiciel Libre >> -- >> 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 > > > -- Loïc Dachary, Artisan Logiciel Libre -- 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
