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 ? 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 -- 谦谦君子 -- 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
