On 15/09/17 03:20, Brassow Jonathan wrote:
There is definitely a difference here. You have 2 stripes with 5 devices in each stripe. If you were writing sequentially, you’d be bouncing between the first 2 devices until they are full, then the next 2, and so on.
When using the -i argument, you are creating 10 stripes. Writing sequentially causes the writes to go from one device to the next until all are written and then starts back at the first. This is a very different pattern.
I think the result of any benchmark on these two very different layouts would be significantly different.
brassow
BTW, I swear at one point that if you did not provide the ‘-i’ it would use all of the devices as a stripe, such that your two examples would result in the same thing. I could be wrong though.
that's what I thought I remembered too.
I guess a big question, from user/admin perspective is: are
those two stripes LVM decides on(when no -i) is the best
possible choice LVM makes after some elaborative
determination so the number of stripes(no -i) would, might
vary depending on raid type, phy devices number and maybe
some other factors or, 2 stripes are simply hard-coded defaults?
On Sep 14, 2017, at 10:49 AM, lejeczek <peljasz@yahoo.co.uk> wrote:
On 14/09/17 15:58, Brassow Jonathan wrote:
Seems strange on the surface. Would you mind posting the layout of each? ‘lvs -a -o +devices’
brassow
here is for LV created without -i, both times with & without I supplied all ten(all that VG has) pvs as arguments to lvcreate.
$ lvs -a -o +devices,stripes,stripe_size chenbro0.1
LV VG Attr LSize Pool Origin Data% Meta% Move Log Cpy%Sync Convert Devices #Str Stripe
raid0.A chenbro0.1 rwi-aor--- 21.18t raid0.A_rimage_0(0),raid0.A_rimage_1(0) 2 16.00k
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdak(0) 1 0
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdam(0) 1 0
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdao(0) 1 0
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdaq(0) 1 0
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdas(0) 1 0
[raid0.A_rimage_0] chenbro0.1 iwi-aor--- 10.59t /dev/sdau(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdal(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdan(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdap(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdar(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdat(0) 1 0
[raid0.A_rimage_1] chenbro0.1 iwi-aor--- 10.59t /dev/sdav(0) 1 0
I cannot remove this LV for a while thus will not be able to recreate with -i for now, sorry.
On Sep 13, 2017, at 8:53 AM, lejeczek <peljasz@yahoo.co.uk> wrote:
hi boys, girls
man page reads: -i ...This is equal to the number of physical volumes to scatter the logical volume data....
I wonder, when I do not use -i while creating an LV with 10 phy devs.
$ lvcreate -n raid0.A --type raid0 -I 16 -l 97%pv
a dbench would show:
$ dbench -t 60 20
...
Throughput 112.309 MB/sec 20 clients 20 procs max_latency=719.409 ms
Yet when I say: this many stripes:
$ lvcreate -n raid0.A --type raid0 -I 16 -i 10 -l 97%pv
dbench:
...
Throughput 83.2822 MB/sec 20 clients 20 procs max_latency=816.027 ms
And though the results would vary, xfs, a dbench for LV with no -i as an argument(which LVM chooses then to be 2) would always look better.
And I thought, as in the manual, always make stripes to go to all phy devices.
Question - is there some "little" magic LVM does? And if yes then how/what it is?
many thanks, L.
.
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