Ceph » RADOS

discussion¶

• crush multipick anomaly

description, with example¶

CRUSH will correctly choose items with relative weights with the right probabilities for each independent choice. However, when choosing multiple replicas, each choice is not indepent, since it
has to be unique. The result is that low-weighted devices get too many items.

Simple example:

maetl:src (master) 03:20 PM $ cat cm.txt
# begin crush map

# devices
device 0 device0
device 1 device1
device 2 device2
device 3 device3
device 4 device4

# types
type 0 osd
type 1 domain
type 2 pool

# buckets
domain root {
    id -1        # do not change unnecessarily
    # weight 5.000
    alg straw2
    hash 0    # rjenkins1
    item device0 weight 10.00
    item device1 weight 10.0
    item device2 weight 10.0
    item device3 weight 10.0
    item device4 weight 1.000
}

# rules
rule data {
    ruleset 0
    type replicated
    min_size 1
    max_size 10
    step take root
    step choose firstn 0 type osd
    step emit
}

# end crush map
maetl:src (master) 03:20 PM $ ./crushtool -c cm.txt  -o cm
maetl:src (master) 03:20 PM $ ./crushtool -i cm --test --show-utilization --num-rep 1 --min-x 1 --max-x 1000000 --num-rep 1
rule 0 (data), x = 1..1000000, numrep = 1..1
rule 0 (data) num_rep 1 result size == 1:    1000000/1000000
  device 0:         stored : 243456     expected : 200000
  device 1:         stored : 243624     expected : 200000
  device 2:         stored : 244486     expected : 200000
  device 3:         stored : 243881     expected : 200000
  device 4:         stored : 24553     expected : 200000
maetl:src (master) 03:20 PM $ ./crushtool -i cm --test --show-utilization --num-rep 1 --min-x 1 --max-x 1000000 --num-rep 3
rule 0 (data), x = 1..1000000, numrep = 3..3
rule 0 (data) num_rep 3 result size == 3:    1000000/1000000
  device 0:         stored : 723984     expected : 600000
  device 1:         stored : 722923     expected : 600000
  device 2:         stored : 723153     expected : 600000
  device 3:         stored : 723394     expected : 600000
  device 4:         stored : 106546     expected : 600000

Note that in the 1x case, we get 1/10th the items on device 4, as expected. For 3x, it grows to 1/7th. For lower weights the amplification is more pronounced.

detailed explanation¶

The chances of getting a particular device during the first draw is the weight of the device divided by the sum of the weight of all devices. For example let say there are 5 devices in a bucket, with the following weights a = 10, b = 10, c = 10, d = 10, e = 1. The chances of getting e is 1/41 and the chances of getting a is 10/41.

Things get more complicated for the second draw because we have to account for a first draw that does not include a given device: it is the sum of the weight of all devices except the one we're interested in, divided by the weight of all devices. So, if we want to know the chances of e showing up in the second draw, the first draw must not include it and this has a 40/41 chance of happening. Also, during the second draw, the chance of getting e is increased because there is one less device to chose from (the one that was picked during the first draw): 1/31 (i.e. 41 - the weight of the device that was chosen). Because the second draw depends on the first draw, the probability must be multiplied: 40/41 * 1/31.

Since the chance of getting the device e in a first draw or getting the device e in a second draw are independent, the chances of getting the device e in both situations is the sum of their probability: 1/41 (first draw) + (40/41 * 1/31) (second draw).

This is a special case because all devices have the same weight except for e. If we are to calculate the probability of a being selected in the second draw, we have to sum the case where e is selected and the case where b, c or d is selected during the first draw, because they do not have the same weight. If e is selected during the first draw, a will be selected during the second draw with a probability of (1/41 * 10/40). If b, c, or d is selected during the first draw, a will be selected during the second draw with a probability of (30/41 * 10/31).

The chances of getting the device a in the first draw and the second draw is therefore: 10/41+(30/41 * 10/31)+(1/41 * 10/40)

To summarize:

• probability of getting e : 1/41 + (40/41 * 1/31) = .05586
• probability of getting a : 10/41+(30/41 * 10/31)+(1/41 * 10/40) = .48603

We are therefore 8.7 ( 0.48603/0.05586 ) more likely to get e than to get a.

From the point of view of the users, this is counter intuitive because they expect that the weight reflects the probability, which is only true for a single draw. With just one draw a is (10/41)/(1/41) = 10 times more likely to be selected than e. With two draws, a is only 8.7 times more likely to be selected than e, as shown above.