Add new IP/Subnet to Windows Failover Cluster

Yana Glozman 20 Reputation points
2024-08-07T22:18:20.1566667+00:00

I have a 2 node Windows Cluster. Each node is a VM and has 1 NIC.

This cluster has an IP address 10.1.x.x.

I need to add a second IP address to this cluster to make sure that in a DR scenario, the cluster will still work.

I need to add and new IP on a different subnet - 10.6.y.y.

How do I accomplish this?

I found some old articles on the web that add an IP with PowerShell command and then configure it in Cluster GUI, but I do not have a second Subnet.

What am I missing?

Any help will be appreciated.

Windows for business Windows Server Storage high availability Clustering and high availability
Windows for business Windows Server User experience Other
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  1. Alex Bykovskyi 2,241 Reputation points
    2024-08-08T15:12:37.6633333+00:00

    Hey,

    You can add additional adapter to your VMs and configure new NICs on a new subnet. In this scenario, cluster will see a new network on a new subnet. I've tested it in my lab. I would recommend you to run cluster validation afterwards. https://community.spiceworks.com/t/add-cluster-network-to-hyper-v-cluster/228391

    Might also help with cluster configuration: https://www.starwindsoftware.com/resource-library/starwind-virtual-san-for-hyper-v-2-node-hyperconverged-scenario-with-windows-server-2016/

    Cheers,

    Alex Bykovskyi

    StarWind Software

    Note: Posts are provided “AS IS” without warranty of any kind, either expressed or implied, including but not limited to the implied warranties of merchantability and/or fitness for a particular purpose.

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  1. Anonymous
    2024-08-08T02:51:32.7433333+00:00

    Hi,

    I'm afraid new subnets cannot be added to the failover cluster. You can only add an IP address of the existing subnet to the Failover cluster. The subnets can only be configured when you create the failover cluster.

    Best Regards,

    Ian Xue


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  2. Alex Bykovskyi 2,241 Reputation points
    2024-08-08T15:17:06.4633333+00:00

    duplicate answer

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