Server Clusters

The cluster configuration I am attempting to construct is not predicated upon throughput.  It is predicated upon fault tollerance.  Want to setup two independent Redhat Linux Servers that are updated simultaneously by the applications.  When one of the Servers goes down the other Server can continue to operate transparently.  Once the failed Server is repaired it is returned to the "cluster".  The Redhat cluster software implies that all servers must share a common external storage - this not what we want.  Have seen some software "Clusterworx" which appears to accommodate this level of independence.  Has anyone out there managed to setup two Linux Redhat Servers in such a configuration.  Thanks in advance.

Respectfully,
Gordon Haas
Network Engineer
saahgAsked:
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pjedmondCommented:
If you haven't been to this site on high availability clusters, then it may be of interest:

http://www.linux-ha.org/
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pjedmondCommented:
The question that you need to ask, is 'How fault tolerant do you want the system to be?'

A web server with 'non dynamic' content is fine for this type of fault tolerance, but when you start looking at databases etc, then there starts to be issues if you don't wish to use a common external storage mechanism (obviously with RAID for fault tolerance).

For example - both systems must have the database associated with them upddated before a true return of success can be given. As a result, the application code must be written specifically for the system concerned, with roll backs if someone puts an axe through one of the systems partway through an upgrade. By writing the software this way, you can effectively carry out RAID mirroring accross all the systems in the cluster, but it would require a lot of thought to make sure that the process was reliable under all circumstances!

Sorry - I've dreamed of this type of setup, but never achieved it:(

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Karl Heinz KremerCommented:
No comment has been added lately, so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area that this question is:
Answered by pjedmond
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