A Splunk indexer cluster is configured with a replication factor of 3 and a search factor of 2. The cluster has five peer nodes. A bucket is currently searchable on two peers. The cluster master detects that one of those peers has failed. What immediate action does the cluster master take to maintain the search factor?
When the search factor is at risk, the cluster master immediately promotes an existing non-searchable copy to searchable status to restore the search factor. This ensures that the required number of searchable copies is maintained without waiting for full replication. The master selects a peer that has a non-searchable copy of the bucket and instructs it to make that copy searchable. This is a rapid operation compared to replicating a new copy.
Why this answer
The search factor defines the number of searchable copies required. When a peer hosting a searchable copy fails, the cluster master acts to restore the search factor. Its first choice is to promote an existing non-searchable copy on another peer to searchable status, which is faster than replicating a new copy.
This immediate promotion ensures that searches can still access the required number of searchable copies. Only if no non-searchable copies are available will the master replicate a new copy.
Exam trap
The trap here is thinking that the cluster master always replicates a new bucket copy to fix a search factor violation, when it typically promotes an existing non-searchable copy first.