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I'm building a new SQL 2008 R2 (Enterprise) Cluster that is backed by an EqualLogic iSCSI SAN. Instead of building a failover cluster (which I have done), I would instead like to setup several nodes that are all active simultaneously. I want to spread out the CPU/RAM load across the nodes.

Can anyone point me towards a document on this? I'm only finding info on HA/Failover clusters.

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Unlike Oracle RAC, your SQL Server instances will not be able to distribute their work across multiple nodes. Is that what you mean? You can build a failover cluster where every node has an active instance, but that's different from what you're asking for. – Nick Chammas Sep 2 '11 at 17:15
up vote 5 down vote accepted

You don't get "load balancing" with SQL Server. There are a lot of "High Availability" features including clustering.

That is, you can't have read/write on the same data with multiple SQL Server instances.

You can off load reporting onto a snapshot, or replicate data between Instances, or "shared scalabl" but none is full OLTP on all data on every node.

You scale up, not out, for SQL Sever

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Thanks for the answer. – Christopher Sep 6 '11 at 2:09

SQL Server does not support load balancing at the server level. There is a similar discussion on this at here. You might look at SQL Server and Soft NUMA architecture. If your hardware is compatible you can take multiple servers and make it run like one big server, with CPU and RAM being shared between all. Jonathan Kehayias did a short write-up on it here. Although you do loose the HA in this, cause if one of those NUMA nodes goes down it is treated as if a single server went down (the whole thing is ca-put).

The only ways I have seen in distributing the workload is at the database level. That involves more development work though in what section of the data is stored were and how the app gets to it.

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Thanks for the answer. – Christopher Sep 6 '11 at 2:09

In short you don't. There are ways to do this with replication and load balancers but this is very complex to get setup and requires your schema be designed correctly for this to work correctly.

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Thanks for the answer. – Christopher Sep 6 '11 at 2:09

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