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With Hyper-V 2012, it is possible to configure the primary server to replicate to a slave server, to allow virtual guests to be live migrated to the slave server, when maintenance is required on the primary server.

My question is as follows:

Is it possible using Hyper-V 2012, to configure replication from the primary server to two (or more) slave servers?

I'm not having much look finding the answer using Google, so I was hoping somebody might know the answer already.


Correction to the above

Without shared storage, it isn't possible to replicate a VM and live migrate it. To perform a live migration, you first have to remove the replica. The question still stands however.

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I'm curious, maybe I can lab this for you, I wonder if the clones would have to be in sequence, or single source. –  SpacemanSpiff Nov 2 '12 at 19:51

2 Answers 2

I'm assuming you mean replicating the same VM to multiple targets—I'm pretty sure you can have different targets for different VMs.

This blog appears to assert that you can have multiple targets, but it is not exactly clear that is what he is saying:

Hyper-V allows to replicate to multiple locations/sites.

You may also me interested to know that

  • replication does not replicate pre-existing snapshots on the 'master'
  • you can take hourly snapshots as part of the replication process, but only a maximum of 15 (older snapshots are merged in as new ones are taken
  • 'slave' and 'master' servers can have their roles reversed (eg with a planned failover), which further leads me to think it unlikely that multiple slaves are allowed: this would be much more complicated for Hyper-V to manage
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The article below discusses replication. In the section "What Hyper-V Replica Is Not Intended To Do" is says "It is not a cluster where you fail stuff back and forth for maintenance windows"

http://www.aidanfinn.com/?p=12147

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-1; It isn't for clustering, granted, but Microsoft claim it is for maintenance windows, (page 22 of this powerpoint presentation). Regardless, your answer would be better served as a comment, as it doesn't actually address my question. –  Bryan Oct 24 '12 at 17:39

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