Hi,
If FTT=1 RAID 1 requires 3 hosts, why FTT=2 RAID 1 requires 5 hosts (not 4)?
What is the object placement/explanation that justifies this?
Thanks
Hello andvm,
I covered the layout of these recently here:
FTT=X with FTM=RAID1 vs RAID5/6, breakdown of how this is composed:
FTT=0, FTM=RAID0 = 2(0)+1 = 1 component (1 data-replica)
FTT=1, FTM=RAID1 = 2(1)+1 = 3 components (2 data-replicas + 1 witness component)
FTT=2, FTM=RAID1 = 2(2)+1 = 5 components (3 data-replicas + 2 witness components)
FTT=3, FTM=RAID1 = 2(3)+1 = 7 components (4 data-replicas + 3 witness components)
FTT=0, FTM=RAID5 = doesn't exist as RAID5/6 doesn't work like this on vSAN nor on any storage platform
FTT=1, FTM=RAID5 = 4 components (all components contain a combination of data and a single set of distributed parity data)
FTT=2, FTM=RAID6 = 6 components (all components contain a combination of data and two sets of distributed parity data)
So, with 3 data-replicas we need to have 2 witness components instead of 1 as in FTM=RAID1 we need to retain odd-numbers of total components per Object for quorum purposes (e.g. if you only had 4 total components then it could result in a scenario where a cluster is partitioned but both still have 2 components each e.g. a tie instead of one partition having quorum).
And as we need to place components on separate hosts/Fault Domains, if 5 components are needed for the Storage Policy then a minimum of 5-nodes is needed for placements.
Bob
Hello andvm,
I covered the layout of these recently here:
FTT=X with FTM=RAID1 vs RAID5/6, breakdown of how this is composed:
FTT=0, FTM=RAID0 = 2(0)+1 = 1 component (1 data-replica)
FTT=1, FTM=RAID1 = 2(1)+1 = 3 components (2 data-replicas + 1 witness component)
FTT=2, FTM=RAID1 = 2(2)+1 = 5 components (3 data-replicas + 2 witness components)
FTT=3, FTM=RAID1 = 2(3)+1 = 7 components (4 data-replicas + 3 witness components)
FTT=0, FTM=RAID5 = doesn't exist as RAID5/6 doesn't work like this on vSAN nor on any storage platform
FTT=1, FTM=RAID5 = 4 components (all components contain a combination of data and a single set of distributed parity data)
FTT=2, FTM=RAID6 = 6 components (all components contain a combination of data and two sets of distributed parity data)
So, with 3 data-replicas we need to have 2 witness components instead of 1 as in FTM=RAID1 we need to retain odd-numbers of total components per Object for quorum purposes (e.g. if you only had 4 total components then it could result in a scenario where a cluster is partitioned but both still have 2 components each e.g. a tie instead of one partition having quorum).
And as we need to place components on separate hosts/Fault Domains, if 5 components are needed for the Storage Policy then a minimum of 5-nodes is needed for placements.
Bob