I apologize having to reply to my own message. 
But please replace all CID w/ MID below. I started editing the text and 
inadvertently hit Send.
Thanks,
-- Chris 

On Friday, June 6, 2014 1:03:20 PM UTC-5, Chris Berry wrote:
>
> Greetings,
>
> I have a couple of questions about keys and routing.
>
> Let’s imagine that I have  
>
>    1. A set of time-based Indexes. The indexes are time-based because the 
>    overall set is unbounded (it grows by over 1M/day) . Thus, we'll have, 
> say, 
>    an Index per Quarter to keep them bounded individually and therefore 
>    maintain predictable performance. 
>    2. The documents stored in these Indexes are “keyed” by MID and 
>    "routed to” by PID. 
>    3. Every document has a unique MID (it’s id_) and contains a PID field 
>    that is unique to the document but not across all documents.  
>    4. The common use case is to want all the MIDs for a given PID. 
>
> So no problem there. (Hopefully that makes sense…)
>
> My first question; when I lookup by MID in a given Index, and I do not 
> have a PID (routing key), is that an inefficient lookup? 
>  I.e. will it have to scan all of the Shards to find it??
>
> And my second question is really more about design.
> I also need to be able to lookup a document by it’s CID and I don’t 
> necessarily know which time-based Index I will find it in. (E.g. it may 
> have been inactive and then be resurrected) 
> I was hoping to avoid having some sort of metadata Index that could yield 
> this info for me (I.e. given a CID, return it’s PID and current Index) 
> because it will end up being unbounded (we are talking billions of entries 
> eventually). 
> Perhaps this won’t really matter because it is only looked-up by key, 
> which should be fast as it would go to the correct Shard.   
> And the data set is only a few bytes?? But still, it seems like a I would 
> be creating an eventual problem.
> Although the alternative, looking in every time-based index seems much 
> worse.
>
> Any advise would be greatly appreciated.
>
> Thanks,
> — Chris 
>
>
>

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