What is the expected behaviour for this new node?

A pre-vSAN 6.6 node is added to a vSAN cluster that was recently upgraded to vSAN 6.7. What is the expected behaviour for this new node?
A. The node will communicate via unicast with all cluster members.
B. The node will fail to join the vSphere cluster.
C. The node will communicate via multicast with all cluster members.
D. The node will be network partitioned from the other cluster members.

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6 thoughts on “What is the expected behaviour for this new node?

  1. This is an especially vague question. The cluster was recently upgraded to vSAN 6.7, but it doesn’t say from which prior version. Unicast was introduced in vSAN 6.6. Prior to 6.6, hosts used multicast to communicate. Since we don’t know from what version it was upgraded, we don’t know if the cluster is communicating via unicast or multicast.
    According to the documentation, “If some hosts in your vSAN cluster are running earlier versions of software, a multicast network is still required.” So, are we to assume that the administrator has configured multicast? If so, the answer is C. If not, the answer is D.

    In a real test scenario, I would assume no changes were made, and I would answer D.

    https://docs.vmware.com/en/VMware-vSphere/6.7/com.vmware.vsphere.vsan-planning.doc/GUID-031F9637-EE29-4684-8644-7A93B9FD8D7B.html?hWord=N4IghgNiBcIK4DsCWBjMBnALiAvkA

    The answer is probably not B, which reads “vSphere cluster” (as opposed to vSAN cluster). And it is probably not A, since the host would not be configured for unicast (unless, of course, we’re to assume that the cluster is using unicast and the administrator reconfigured the host knowing that it wouldn’t work).

    At the heart of the matter, VMware wants you to demonstrate that you know that unicast was introduced with 6.6. The assh0le who wrote this just did a poor job of formulating a question.

  2. I think the correct answer will be D.
    Any new pre vSAN 6.6 nodes added to cluster in this state will be partitioned away. (e.g.vSAN 6.2 nodes)

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