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
