lock disk or lun in cluster service guard

[sokolova]

what will happens if cluster lock disk, or lun
will corrupts?
is cluster will halts?
or no impact on cluster?

[Duncan Edmonstone]

No the cluster will not halt, in fact the cluster will continue to perform as normal until a cluster reformation event occurs. At that point, if a cluster lock operation is required (i.e quorum cannot be reached - the cluster is split exactly 50/50) then cluster reformation will fail.

You should get messages in syslog though telling you the cluster lock is damaged.

HTH

Duncan

[P Arumugavel]

No, the cluster keeps the functioning until the running cluster fails and is split into two sub-clusters of equal size and reforms.

Serviceguard periodically checks the health of the lock disk and writes messages to the syslog file when a lock disk fails the health check. This file should be monitored for early detection of lock disk problems.

If the lock disk is not available when its arbitration services are needed, the cluster will halt.

Rgds....

[Stephen Doud]

Cluster lock disk/lun devices are used to arbitrate which system(s) remain in a cluster when heartbeat transmission ceases between exactly 50% of the nodes in the cluster. Serviceguard does not use the lock device when a different ratio of nodes lose heartbeat communication. Only when an unexpected loss of heartbeat occurs will Serviceguard seek the arbritration device. If it is not available, the node will TOC (save a memory dump and reboot).

Serviceguard reports problems with the arbitration device every hour in syslog.log, so the admin will have plenty of reminders to resolve the issue.

[Prashanth.D.S]

Hi

As everyone said when a cluster is up and running and cluster lock/LUN goes corrupt nothing happens, you would see messages in syslog about cluster lock not available.

However assume that Node A goes down, now Node2 will try to reform cluster, it wouldn't be able to as cluster lock is missing and Node2 would do a TOC/crash and you may have to start cluster manually on Node B.

Best Regards,
Prashanth