I don't have the formula.
I suggest you schedule a test with the storage team, explaining that you are
trying to determine this so you can minimize and standardize space request
frequency in the future. You may also want to ask for their input on the
geometry and/or controller configuration and so forth of the media they are
landing you on. Possibly OCR and voting disks should be on devices that are
immune or resistant to disk farm activity storms driven outside your sphere
of influence. You can explain that while the volume of i/o might be small,
spikey latency and be a real pain in everyone's daily routine.
I'm in Vegas and #C17LV at the moment (all y'all should jump on the next
plane and do on site registration if you aren't here already. Don't you know
Oaktable World has a day here?), so I'll lay 50-50 odds they will say
something like "will 128GB always be enough? can we just carve that off as
even tracks (if some spinning device config) or a chunk of SSD?"
Or they might want you to know exactly what you need. But either way you
have enlisted them as your ally in determining a useful allocation, which
almost certainly should be rounded up to consume an integral number of
tracks if you're on spinning rust that can be allocated that way. If you're
in magic gizmo button pushy allocate space without regard to geometry land,
that probably doesn't matter. Enlisting them as an ally is the key bit. You
want to minimize total effort by both you and them now and in the future. So
round up, 'cause the history is they keep adding little bits here and there
to that repository.
If they agree to the actual experiments, super. Let us know your results
(which will be quite perishable per release, so still round up, even if you
know exactly what you need right now. Disk is cheap. Human time is
expensive.)
mwf
From: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx]
On Behalf Of Herring, David
Sent: Thursday, March 30, 2017 4:45 PM
To: 'oracle-l@xxxxxxxxxxxxx'
Subject: OCR/vd diskgroup sizing under 12c
Folks,
I'm wondering if someone could help me understand the size requirements for
diskgroup CLSTR (for OCR and voting disks) under 12c (only concerned with
RHEL 6/7 for the moment). Why can't I just ask for more space each time we
hit INS-30515 errors during installs? Because management won't like that
and wants reasons for the space request, even though an additional 1 or 2 GB
is insignificant compared to the TB's of disk we're frequently allocating
for DBFLASH/DBDATA per new install.
Back to my question, the space requirement has increased for OCR/voting
disks between 11g and 12c and requires additional space when more hosts are
added to the cluster but I can't quite seem nail down the formula. Here's
what I've found:
1. In 12cR2 Grid Infrastructure guide, Table 8-5 for standalone
clusters says that for normal redundancy (3 disks min) and 2 hosts in the
cluster with an AU of 4, the required size is at least 2.5 GB.
Unfortunately there's no breakdown, in terms of the impact of the AU and
number of hosts in the cluster. If we don't use flex diskgroups and stick
with a default 1 AU and have 6 hosts, how does that affect the required
size?
2. Section 8.1.1 Indentifying Storage Requirements for Oracle
Automatic Storage Management gives a formula of " total = [2 * ausize *
disks] + [redundancy * (ausize * (all_client_instances + nodes + disks + 32)
+ (64 * nodes) + clients + 543)]", This is "additional" space needed for
ASM metadata, in addition to the 2.5 GB from above.
3. Checking 12cR1 I find the following formula in Table 7-5:
4 hosts or less -> (2*5.2 GB + 2*400 MB + 3*300 MB) = 12.1 GB.
5 hosts or more -> (2 * (5.2 GB +2*(500 MB)) + (2*400 MB) + (3*300 MB)) =
14.1 GB.
2*400 MB is for failure group -> "normal" redundancy needs 2 additional
failure groups.
3*300 MB is for the # of disks in the diskgroup -> "normal" redundancy needs
3 disks.
If I stick with the 12cR1 formula in point #3 for a 1 AU, 4 host cluster I'd
allocate 12.1 GB. But if I use the combination of points #1 and #2 I'd get:
2.5 GB + ((2 * 1 * 3) + (2 * (1 * (1 + 4 + 3 + 32) + (64 * 4) + 1 + 543)) =
2.5 GB + (6 + (2 * (1 * (40) + (256) + 544)) = 2.5 GB + 1686 MB = 4.15 GB.
I don't see 4.15 fiting any way with 12.1 (from point #3) and we know from
previous 12c RAC installs that we need, for a 2 host cluster, a total of at
least 10.77 GB.
Does anyone see what I'm missing/confusing?
Regards,
Dave
