How to improve query performance?


(Dong Aihua) #1

Hi,
Currently, I use 2 servers(2 shards+1 replica) to query 0.46 billion
documents. The average query response time is 1.5s.
I wish I can simply add a new server into the cluster to improve the
query performance. (still 2 shards + 1 replica)
But after adding a new server into the cluster, I found the query
performance didn't improve much. The average query response time still is
1.5s.
Did I miss anything??
In addition, I'm wondering is there any difference of querying only one
server or query 3 servers roundly?

-Regards-
-Jackie-


(Ævar Arnfjörð Bjarmason) #2

This really doesn't tell us much of anything

  • What's your HW setup like, is this data in RAM, on disk, what does
    the CPU/disk use etc. look like when you do queries
  • What does your data look like, do you have an example
  • How does the mapping look like
  • How do the queries that you've tried look like
  • Have you tried increasing the number of shards, if you have 2 shards
    in total with X number of replicas and 3 machines you'll only
    distribute your queries to 2 shards, maybe that's your bottleneck, so
    increasing replicas/boxes wouldn't help.

(Otis Gospodnetić) #3

Jackie,

It doesn't help to add a new server if no shards/replicas are migrated to
it. Check manually via the API or look at SPM.
Also, by adding more servers you are adding more CPUs, so resharding may
make sense.

Otis

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On Friday, June 1, 2012 4:01:33 AM UTC-4, jackiedong wrote:

Hi,
Currently, I use 2 servers(2 shards+1 replica) to query 0.46 billion
documents. The average query response time is 1.5s.
I wish I can simply add a new server into the cluster to improve the
query performance. (still 2 shards + 1 replica)
But after adding a new server into the cluster, I found the query
performance didn't improve much. The average query response time still is
1.5s.
Did I miss anything??
In addition, I'm wondering is there any difference of querying only one
server or query 3 servers roundly?

-Regards-
-Jackie-


(system) #4