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