which two conditions is a master elected by vSphere HA?

Under which two conditions is a master elected by vSphere HA? (Choose two.)

A.
When a network partition occurs

B.
When a failure of vCenter Server occurs

C.
When vSphere HA is enabled

D.
When a host is added to an HA enabled cluster

Explanation:

Page 12 from vsphere-esxi-vcenter-server-50-availability-guide.pdf

 

4 Comments on “which two conditions is a master elected by vSphere HA?

  1. Anitha says:

    Believe the answer is C & D – because when network partition occurs the existing master only does the regular HS tasks , no new master is elected.

  2. Mark Dean says:

    I’m not sure what “HS tasks” are but A and C are the correct answers. Here’s why:

    For “A”, in the event of a partition (which is very different than a host isolation event), the slaves that are partitioned off (i.e. in a second site in a metro or geo cluster or across blade enclosures or whatever) will force an election since they will have lost contact with the Master. From the perspective of a segment of the cluster, the Master has failed. The existing Master from before the Network Partition event will remain Master on its ‘side’ of the partition and a new Master will be elected on the other ‘side’ of the partition. Only when the network partition event is resolved, will an election be forced again which will resolve the two Master situation caused by the partition.

    For “C”, the explanation speaks for itself.

  3. CiscoJedi says:

    The reason D is incorrect is because there are TWO types of HA clusters: host and….drum rolls….DATASTORE. So you have to read as “any” cluster, to which there is no master for a datastore cluster.

  4. Robert says:

    To determine which host will be the master, an election process takes place. The election process occurs when:

    -vSphere HA is enabled.
    -Master is placed in maintenance mode, standby mode, or vSphere is reconfigured.
    -When slave host(s) cannot communicate with the master due to a network problem.

    So Mark Dean is correct: answers are A and C.


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