I'm reading documentations of SQL Server. On this page at https://learn.microsoft.com/en-us/sql/sql-server/failover-clusters/windows/windows-server-failover-clustering-wsfc-with-sql-server, it says:

Both Always On Availability Groups and Always On Failover Cluster Instances use WSFC as a platform technology, registering components as WSFC cluster resources.

I'm a little bit confused. In previous version of SQL Server (like SQL Server 2008) there is a technology called SQL Server Failover Clustering, is it the same thing as Always On Failover Cluster Instances? Hope someone can clarify on this. Thanks in advance.

1 Answer 1


Yes it is the same thing. See section 3 which explains this term.

First lets get few terms straight.

  1. Windows Server Failover Clustering

A Windows Server Failover Clustering (WSFC) cluster is a group of independent servers that work together to increase the availability of applications and services. SQL Server 2016 takes advantage of WSFC services and capabilities to support Always On availability groups and SQL Server Failover Cluster Instances.

  1. Always On Availability Groups (SQL Server)

The Always On availability groups feature is a high-availability and disaster-recovery solution that provides an enterprise-level alternative to database mirroring. Introduced in SQL Server 2012, Always On availability groups maximizes the availability of a set of user databases for an enterprise. An availability group supports a failover environment for a discrete set of user databases, known as availability databases, that fail over together. An availability group supports a set of read-write primary databases and one to eight sets of corresponding secondary databases. Optionally, secondary databases can be made available for read-only access and/or some backup operations. An availability group fails over at the level of an availability replica.

Failovers are not caused by database issues such as a database becoming suspect due to a loss of a data file, deletion of a database, or corruption of a transaction log.

  1. Always On Failover Cluster Instances (SQL Server)

As part of the SQL Server Always On offering, Always On Failover Cluster Instances leverages Windows Server Failover Clustering (WSFC) functionality to provide local high availability through redundancy at the server-instance level—a failover cluster instance (FCI). An FCI is a single instance of SQL Server that is installed across Windows Server Failover Clustering (WSFC) nodes and, possibly, across multiple subnets. On the network, an FCI appears to be an instance of SQL Server running on a single computer, but the FCI provides failover from one WSFC node to another if the current node becomes unavailable.

Item number 2 and 3 can be combined and item 1 is a prerequisite for implementing any combination. 2 alone, 3 alone, 2 and 3 together.

When you combine 2 and 3 you get: Failover Clustering and Always On Availability Groups (SQL Server)

You can set up a second layer of failover at the server-instance level by implementing SQL Server failover clustering together with the WSFC cluster. An availability replica can be hosted by either a standalone instance of SQL Server or an FCI instance. Only one FCI partner can host a replica for a given availability group. When an availability replica is running on an FCI, the possible owners list for the availability group will contain only the active FCI node. + Always On availability groups does not depend on any form of shared storage. However, if you use a SQL Server failover cluster instance (FCI) to host one or more availability replicas, each of those FCIs will require shared storage as per standard SQL Server failover cluster instance installation.

I would highly recommend reading this FAQ by SQLHA (Allan Hirt) which clarifies many basic questions when you combine these 3 items.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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