Is every node share the same metadata ?

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Is every node share the same metadata ?

Alex Feng

Hi Riak Users,

I am a little bit confused about the metadata, it is stored in memory and synced in every node, so bascially every node share the same metadata, right ?
But then from the official document, the formula to calcuate the RAM requirement, the RAM requirement is for whole cluster, for example, based on the formula below, the requirement is 50G, then if we have 5 nodes, then each node only needs 10G RAM. Seems the metadata is divided to different nodes, I must hvae misunderstood something, could someone please to clarify this, thank you very much.



Approximate RAM Needed for Bitcask = (static bitcask per key overhead + estimated average bucket+key length in bytes) * estimate total number of keys * n_val



Br,
Alex

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Re: Is every node share the same metadata ?

Alexander Sicular-2
Hi Alex,

You seem to be referring to bitcask metadata. That metadata is loaded into ram on each node. As you note, the bitcask capacity planner calculates ram required by a cluster to service a certain number of keys. I think where you are confused is that this metadata is not synced across the cluster. It is specific to the data stored specifically on each node. 

-Alexander



Alexander Sicular
Solutions Architect
Basho Technologies
9175130679
@siculars

On Fri, Jan 6, 2017 at 11:31 AM, Alex Feng <[hidden email]> wrote:

Hi Riak Users,

I am a little bit confused about the metadata, it is stored in memory and synced in every node, so bascially every node share the same metadata, right ?
But then from the official document, the formula to calcuate the RAM requirement, the RAM requirement is for whole cluster, for example, based on the formula below, the requirement is 50G, then if we have 5 nodes, then each node only needs 10G RAM. Seems the metadata is divided to different nodes, I must hvae misunderstood something, could someone please to clarify this, thank you very much.



Approximate RAM Needed for Bitcask = (static bitcask per key overhead + estimated average bucket+key length in bytes) * estimate total number of keys * n_val



Br,
Alex

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