Server Fault is a question and answer site for system and network administrators. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a netapp fast 2xxx with iscsi support, with 2 san switches. Is it possible to use one iscsi lun with cluster purposes?

The idea is to create a clustered service which needs storage, the storage is provided by iscsi. The distribution should be red hat.

Has anyone succed on something like this or similar?

share|improve this question
You might want to edit your question to say "Can a Red Hat cluster utilize shared iSCSI storage?" – Jeremy Jan 20 '12 at 16:04
up vote 1 down vote accepted

I've done with Fiber Channel, but I think you could do it with iSCSI the same way. You should show the disks to the two servers (two, three or n servers), after you should install lucci and ricci to configure the RedHat cluster and follow the steps here:

share|improve this answer
I've been told that iscsi is more sensitive to fencing than FC. With windows on top of vmware when the DRS moves the machine, even without cluster, it can become corrupted quite easily. Do you know anything about this? thanks. – marc.riera Jan 8 '12 at 12:44
if a server goes down unexpectly, the data inside could get corrupted. If the data is related to databases (MySQL/MSSQL) the probabilites arise. I think is not related to iSCSI or FC. For try to avoiding you should have a good disk controller in the SAN solution and a good backup solution to recover the data if it gets corrupted. You cant be protected against corrupted data completely, i think it's not very common, and database engines have some tools/procedures to handle these situations, so in my opinion and experience you should not be too worried about this. Regards! – Adrian Perez Jan 8 '12 at 17:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.