You have a failover cluster named Cluster1 that has the following configurations:
-> Number of nodes: 6
-> Quorum: Dynamic quorum
-> Witness: File share, Dynamic witness
What is the maximum number of nodes that can fail simultaneously while maintaining quorum?
Click on the arrows to vote for the correct answer
A. B. C. D. E.C
Note this question is asking about nodes failing 'simultaneously', not nodes failing one after the other.
With six nodes and one witness, there are seven votes. To maintain quorum there needs to be four votes available (four votes is the majority of seven). This means that a minimum of three nodes plus the witness need to remain online for the cluster to function. Therefore, the maximum number of simultaneous failures is three.
https://docs.microsoft.com/en-us/windows-server/storage/storage-spaces/understand-quorumIn a failover cluster, quorum is the process of ensuring that the cluster can continue to function in the event of a failure. Quorum is necessary to avoid a "split-brain" situation where the cluster nodes cannot agree on which node is the active node.
Dynamic quorum is a feature of Windows Server Failover Clustering that adjusts the number of votes needed for quorum as nodes join or leave the cluster. This helps to ensure that the cluster can maintain quorum even if a large number of nodes fail simultaneously.
A witness is an additional component that helps to maintain quorum by providing an additional vote. In this case, the cluster is configured with a file share witness and a dynamic witness.
To determine the maximum number of nodes that can fail simultaneously while maintaining quorum, we need to consider the number of nodes in the cluster and the number of votes needed for quorum.
In a six-node cluster with dynamic quorum and two witnesses, the total number of votes is seven (six votes for the nodes plus one vote for the file share witness).
To maintain quorum, the cluster needs to have at least four votes. This means that up to three nodes can fail simultaneously (since six - three = three nodes remain, which still have four votes).
Therefore, the answer is C. 3.