SQL Server – Rank Function Creating Duplicate Partitions

Basically seems to be picking the MAX id from a partition, not max across all partitions

Writing TOP (1) without an ORDER BY clause to define which row is 'top' means the query processor is logically free to return any row from the set. The query plan selected by the optimizer happens to return a particular row (highest id from the first partition) but you cannot rely on this, even if it were a useful result.

Whenever you use TOP you should always specify an ORDER BY at the same scope to produce deterministic behaviour - unless you really do not care which row(s) come back.

Given the table size SELECT MAX(id) FROM tableA will not perform well enough

The optimizer is lacking some logic to transform a scalar MAX or MIN aggregate over a partitioned index to a global aggregate over per-partition aggregates. Itzik Ben-Gan explains the limitation and provides a general workaround in this article.

If the highest partition number is known and guaranteed not to change, the workaround to specify a literal partition using the $partition function will work, though it may fail in a non-obvious way if the partitioning strategy changes in future.

This 'solution' works by eliminating all but one partition, resulting in a simple seek on one partition of the index.

Adding an order by id does not improve performance for some reason

The same optimizer limitation broadly applies to TOP (1) ... ORDER BY. The ORDER BY makes the result deterministic, but does not help produce a more efficient plan in this particular case (but see below).

Implied Index Keys

Your index is on id DESC, timeSampled DESC. In SQL Server 2008 and later, partitioning introduces an extra implied leading key on $partition ASC (it is always ascending, it is not configurable) making the full index key $partition ASC, id DESC, timeSampled DESC.

Since id and timeSampled increase together (though there is nothing in the schema to guarantee this) you could rewrite the query as TOP (1) ... ORDER BY $partition DESC, id DESC. Unfortunately, the DESC keys on your index and ASC implied leading key $partition means the index could not be used to scan just one row from the index in order.

If your index keys were instead id ASC, timeSampled ASC the whole index key would be $partition ASC, id ASC, timeSampled ASC. This all-ASC index could be scanned backward, returning just the first row in key order. This row would be guaranteed to have the highest id value in the highest-numbered partition. Given the (unenforced) relationship between id and partition id, this would produce the correct result with an optimal execution plan that reads just a single row.

This 'solution' lacks integrity because the id-timeSampled relationship is not enforced, and you probably do not want to rebuild the nonclustered primary key anyway. Nevertheless, I mention it because it may enhance your understanding of how partitioning interacts with indexes.

This:

SET SHOWPLAN_XML ON;
GO
SELECT * FROM sys.objects;
GO

Is equivalent to pressing Display Estimated Execution Plan on the toolbar (or hitting Ctrl + L). You'll notice that no rows are returned from the query, like there is when you use Include Actual Execution Plan (Ctrl + M).

The spill warning is only a runtime warning. There is no way that SQL Server can know, when displaying the estimated plan, that a spill will happen at runtime. This is because a spill is caused by factors that might only be present during certain invocations of the query (for example, when there is memory pressure). The estimated plan knows roughly how much memory it's going to ask for, but it can't know until execution that it isn't going to get it.

As an aside, may I recommend* our free tool, SQL Sentry Plan Explorer? I think it provides much more obvious information than Management Studio. I recently wrote a lengthy blog post that can act as a tutorial, and Jonathan Kehayias has a great PluralSight course on it as well.

_{* Disclaimer: I work for SQL Sentry.}