LVM Performance vs. # of disks

[Uday_S_Ankolekar]

Hello,

There is no restriction as such.. but more disks will improve performance when you spread across your big database. Your friend might have confused with controller and disks.

-USA..

[Reston Tech Support]

Yes, spreading across mulitple SCSI controllers I agree with, and have in fact implemented that on our SCSI based machines, for the reason you all have indicated. And thanks a lot for your detailed responses.

But, as you all have confirmed, across fibre channel cards, especially when managed by EMC load balancing software (Power Path), performance is not hindered by the LVM vs. physical disks.

I'm 1 for 1 today!

[Deshpande Prashant]

HI
Do not restrict the volume group for such less number.
With more disk in VG, chanses are better for higher performance and also you have chanse to strip the lvols.
I have some VGs with 35 disks( 9 gb from EMC frame)

Thanks.
Prashant.

[Tim D Fulford]

Hi

I know this is an old thread but I THINK I know what may have gone on here. The EMC guy probably said something like "4/5 disks can flood the buss". This is actually true as a disk 15,000 rpm disks has a burst transfer rate of 50-75MB/s, thus 4 or 5 could THEORETICALLY flood a 2Gbit/s buss. BUT practically speaking the chances of 4/5 disks doing a burst transfere at the same time are slim. This is why volume groups with 20 thimes this number of disks are possible.

OK fluffy cloud stuuf done from here on in it is mathematics. If you hate maths of are adversed to techish then stop reading now...... Now I assume the audience is down to one or two Quantum physicists or mathematicieans. If so I applologise for the quality of the calculations, it is from memory....

p is the probability of a single disk doing a transfer. it is the ratio of the time it takkes to transfer one IO BLOCK to the total time seeking & transfereing.
q = 1-p
Nd = total number of disks on buss
Also asume all the disks are in one stripped group.
Assume that just over 4 disks wil flood the buss
(p+q)^Nd = binomial expansion of the chances of the disks transfereing data at the same time.

P= probability of flooding (1-P)
Q= probability of not flooding (1-Q)

Q0 = q^Nd
Qn+1 = Qn*p(Nd-n)/(n+1)/q
Sum Qn from 0, 1, 2 .. 4 (say)

so IF there is flooding then only 4 disks worth of IO can go ahead, if not then Nd worth of IO can go ahead...

IOrate for buss = IO for disk * ( P*4 + Q*Nd)

You can code this up & pritty soon you will see that at about 35 - 70 disks worth you start to see buss flooding... I'll attach a more detailed paper tomorrow & Excel sprad for the 1-2 of you who may be interested!!!

Regards

Tim

[Tim D Fulford]

Yer tiz

[Tim D Fulford]

Yer tiz

[Tim D Fulford]

Try again (3rd times a charm)