The cost is the same (1%) for both the slow and fast cases. Does that
mean the warning can be ignored? Is there a way to show "actual" times
or costs. That would be so much better! Actual row counts are the same
for the operation with the spill.
The cost shown is always the optimizer's estimated cost of the iterator, computed according to its internal model. This model does not reflect your server's particular performance characteristics; it is an abstraction that happens to produce reasonable plan shapes most of the time for most queries on most systems. There is no way to show 'actual' costs/execution times per iterator.
Besides performing a manual text diff of xml execution plans to find
the differences in warnings, how can I tell what the 1500% increase in
runtime is actually due to?
Typically, you can't. Spill warnings (sorts, hashes, exchanges) are new in execution plans for 2012, but they are just an indication of something you should investigate and look to eliminate if possible. The impact of a particular spill is something that needs to be measured - it is not possible to say that a spill of a particular type will always result in an x% performance drop for example.
For slow case, tempdb before/after (select *
sys.fn_virtualfilestats(db_id('tempdb'),null)) (only showing a few
100ms of latency)
Spilling to tempdb and back is certainly undesirable, but the overall impact is hard to assess. For sort and hash spills, the impact is largely due to the I/O and access pattern, which may be small-block synchronous I/O e.g. for sort spills. With ~100ms of latency, you don't need too many synchronous I/Os to introduce a significant delay. The nature of the process and I/O patterns means tempdb spills can still be a problem on very low latency storage systems like fusion-io.
For exchange spills, there is an extra delay. The intra-query deadlock must be detected by the regular deadlock monitor, which by default only wakes up once every 5 seconds (more frequently if a deadlock has been found recently).
The resolver must then choose one or more victims, and spool exchange buffers to tempdb until the deadlock is resolved. The amount of spooling needed and the complexity of the deadlock will largely determine how long this takes.
Ultimately, preserved ordering is a Very Bad Thing for parallelism in general. Ideally, we want multiple concurrent threads operating on data streams with no inter-dependence. Preserving sort order introduces dependencies, so producer and consumer threads in different parallel branches can become deadlocked waiting for order-preserving iterators to receive rows to decide which input sorts next in sequence.
The precise nature of the deadlock depends on data distribution and per-thread sort order at runtime, so it is typically very hard to debug. Hence my recommendation to avoid order-preserving iterators in parallel plans, especially at high DOP. I do explain a very simplified example of an order-preserving parallel deadlock in some talks I do, but real examples are always more complex, though the underlying cause is the same.
In case the concepts are not familiar, it may help to follow the following example, reproduced from the (somewhat epic) 1993 paper Query Evaluation Techniques for Large Databases by Goetz Graefe:
If a different partitioning strategy than range-partitioning is used,
sorting with subsequent partitioning is not guaranteed to be
deadlock-free in all situations. Deadlock will occur if (i) multiple
consumers feed multiple producers, and (ii) each producer produces a
sorted stream and each consumer merges multiple sorted streams, and
(iii) some key-based partitioning rule is used other than range
partitioning, i.e., hash partitioning, and (iv) flow control is
enabled, and (v) the data distribution is particularly unfortunate.
Figure 37 shows a scenario with two producer and two consumer
processes, i.e., both the producer operators and the consumer
operators are executed with a degree of parallelism of two. The
circles in Figure 37 indicate processes, and the arrows indicate data
paths. Presume that the left sort produces the stream 1, 3, 5, 7, ...,
999, 1002, 1004, 1006, 1008, ..., 2000 while the right sort produces
2, 4, 6, 8, ..., 1000, 1001, 1003, 1005, 1007, ..., 1999.
The merge operations in the consumer processes must receive the first
item from each producer process before they can create their first
output item and remove additional items from their input buffers.
However, the producers will need to produce 500 items each (and insert
them into one consumer’s input buffer, all 500 for one consumer)
before they will send their first item to the other consumer. The data
exchange buffer needs to hold 1000 items at one point of time, 500 on
each side of Figure 37. If flow control is enabled and the exchange
buffer (flow control slack) is less than 500 items, deadlock will
occur.
The reason deadlock can occur in this situation is that the producer
processes need to ship data in the order obtained from their input
subplan (the sort in Figure 37) while the consumer processes need to
receive data in sorted order as required by the merge. Thus, there are
two sides which both require absolute control over the order in which
data pass over the process boundary. If the two requirements are
incompatible, an unbounded buffer is required to ensure freedom from
deadlock.
I initially thought you were on to something here. Working assumption was along the lines that perhaps the buffer pool wasn't immediately flushed as it requires "some work" to do so and why bother until the memory was required. But...
Your test is flawed.
What you're seeing in the buffer pool is the pages read as a result of re-attaching the database, not the remains of the previous instance of the database.
And we can see that the buffer pool was not totally blown away by the
detach/attach. Seems like my buddy was wrong. Does anyone disagree or
have a better argument?
Yes. You're interpreting physical reads 0
as meaning there were not any physical reads
Table 'DatabaseLog'. Scan count 1, logical reads 782, physical reads
0, read-ahead reads 768, lob logical reads 94, lob physical reads 4,
lob read-ahead reads 24.
As described on Craig Freedman's blog the sequential read ahead mechanism tries to ensure that pages are in memory before they're requested by the query processor, which is why you see zero or a lower than expected physical read count reported.
When SQL Server performs a sequential scan of a large table, the
storage engine initiates the read ahead mechanism to ensure that pages
are in memory and ready to scan before they are needed by the query
processor. The read ahead mechanism tries to stay 500 pages ahead of
the scan.
None of the pages required to satisfy your query were in memory until read-ahead put them there.
As to why online/offline results in a different buffer pool profile warrants a little more idle investigation. @MarkSRasmussen might be able to help us out with that next time he visits.
Best Answer
Actually, the warning for excessive memory grant uses an hardcoded logic. It raise an alert when a certain % of the allocated memory is not use but it does not considere the amount of wasted memory.
It cannot be "trusted" and you should not worry about it if, like in this case, it's only 1Mb that is allocated and not use.
I've seen cases where I got a warning for something like this (1Mb allocated, 1 Mb non-used) and cases where there was 256Mo allocated and like 136 Mo used (so 120Mo non-used) and there was no warning.