OK, I first posted on some of the limitations to indexes in SQL Server 2005 and 2008 in part one here. Now, I want to dive into index internals for a post (or two). And, I often get the question “who is the best audience for your blog – or, for this post” and well, that’s a bit hard to answer. At SQL Connections in Orlando, I delivered a session titled: Index Internals & Usage and while we (fyi – Paul and I co-chair the SQL Connections portion of “DevConnections”) put it in the “developer-focused track,” it was more of a Dev/DBA “hybrid” session with the emphasis on database development and best practices in creating and managing indexes (rather than management/maintenace/operational tuning – which is more for DBAs). Here at TechEd this week, I’m going to focus more on the management/maintenace/operational tuning side with a session called Are your Indexing Strategies Working? I’ll also do a complementary blog post for that as well…
Having said that thought, indexes are definitely in a group of topics – very much so related to performance and scalability (index internals, indexing strategies, log maintenance, general database maintenance) which really needs to cross almost all database-related disciplines (dev, admin, ops, etc…). If you work with SQL Server in almost any capacity, you need to get a feel for at least some aspect of indexing for performance.
So, for this post, I’m continuing with some internals. In the first post (in this series), I wrote about limits. Limits/boundaries are interesting to discuss but it’s also important to remember that good performance takes a lot more than just staying within the bounds of what’s possible. Creating indexes solely because you can – without reason and only with upper limits in mind – can be even worse than under indexing. So, if you find that you’re wanting more about indexes (I have many blog posts that are solely Q&A posts), check out my Indexing category here. Now that you know how many indexes you can create, a better question would be when is it appropriate to create indexes at all?
So, what is “finding the right balance” in indexing? In my opinion, there are three requirements/pre-requisites:
knowing the data
knowing how the users use the data
knowing how the underlying structures and database stores/manipulates and uses indexes
Bringing all of these things together is what I try to do in my workshops, seminars and lectures – in this post, I’ll start with a smaller more digestible piece – internals.
Indexes have 2 components: a leaf level and a non-leaf level (or b-tree). The non-leaf level is interesting to understand and discuss (in terms of internals) but simply put, it’s used for navigation to the leaf level (more than anything else). So, we’ll start with the leaf level (as does SQL Server – the leaf level is always built first). The leaf level of an index contains something (I’ll explain more coming up) for every row of the table in indexed order (note: I am focusing on traditional indexes in every release from SQL Server 2000 up to and including SQL Server 2008 – with the exception of filtered indexes which I will write about in a later post). Once the leaf level is built, non-leaf level(s) can be built to help navigate to the leaf level but the architecture is rather straightforward. The non-leaf level stores something for every page of the level below – and levels are added (each smaller than the previous because each level only contains one the first entry from every page) until the index gets to a root of one page. While it sounds like this could result in a lot of levels (ie. a tall tree), the limitation on the size of the key (which has a maximum of 900 bytes or 16 columns) helps to keep index trees relatively small. In fact, in the example I’ll show coming up – which has a fairly large (large meaning WIDE) index and has a key definition which is at the maximum size – even the tree size of this example index (at the time the index is created) is only 8 levels high/deep…
To see this tree (and the math used to create it – which is the same thing that SQL Server would go through to create it), we’ll use an example where the leaf level of the index contains 1,000,000 “rows.” I put quotes around “rows” because I don’t want to imply that these have to be data rows – these are really just leaf level rows and I’ll explain more on what leaf level rows can be… The leaf level rows are 4,000 bytes per row (therefore only 2 rows per page) or 500,000 pages. This is not ideal but at least the pages are almost full and we’re not wasting a lot of space – if we had two 3000 byte rows we’d still only fit 2 per page and then we’d have 2,000 bytes of wasted space. Now, as for why these are just “rows” and not specifically data rows is because this leaf level could be the leaf level for a clustered index (therefore data rows) OR these leaf level rows could be rows in a non-clustered index that uses INCLUDE (which was new to SQL Server 2005) to add non-key columns to the leaf level of the index (which therefore creates wider leaf rows (wider than the 900 bytes or 16 column maximum). Again, while this doesn’t currently sound interesting, I’ll explain why this can be beneficial coming up (possibly in another post depending on how long this particular post becomes… J).
The leaf level of this index would result in a 4 GB structure (and this is only at the time it’s created – if a lot of rows are added and the key is not ever increasing then this structure could become heavily fragmented and therefore much larger/taller). In this case, it’s relatively large (again because of “row” width) and with an index key of 900 bytes you can even see that in this case, the tree would be relatively small and only result in 8 levels – as shown below.
Root page of non-leaf level (Level 7) = 2 rows = 1 page
Intermediate non-leaf level (Level 6) = 15 rows = 2 pages (8 rows per page at 900 bytes)
Intermediate non-leaf level (Level 5) = 122 rows = 15 pages (8 rows per page at 900 bytes)
Intermediate non-leaf level (Level 4) = 977 rows = 122 pages (8 rows per page at 900 bytes)
Intermediate non-leaf level (Level 3) = 7,813 rows = 977 pages (8 rows per page at 900 bytes)
Intermediate non-leaf level (Level 2) = 62,500 rows = 7,813 pages (8 rows per page at 900 bytes)
Intermediate non-leaf level (Level 1) = 500,000 rows = 62,500 pages (8 rows per page at 900 bytes)
Leaf level (Level 0) = 1,000,000 rows = 500,000 pages (2 rows per page)
Having said that though, this is NOT a goal. :) In more realistic scenarios [where the key is much smaller and] even when there are more rows, there are fewer levels (3-4 is quite normal). Most importantly, the size of an index (and the number of levels) depends on two things – the width of the key (in terms of the number of bytes) and the number of pages in the leaf level of the indexes. The number of pages in the leaf level of an index depends on the number of rows and the size of the rows (again, in terms of bytes) of the rows in the leaf level.
You can see the size of your index by using one of the following commands:
In SQL Server 2000: DBCC SHOWCONTIG … WITH ALL_LEVELS
In SQL Server 2005/2008: querying the dmv: sys.dm_db_index_physical_levels
To see the syntax of these commands and their output, we’ll use some structures created in the credit sample database. Using credit, you can see exactly how these commands work and how they return the details about every level.
NOTE: you can download a zip of a SQL Server 2000 backup of this database here – and since this is a SQL Server 2000 backup, you can restore this to SQL Server 2000, SQL Server 2005 or SQL Server 2008.
(db_id(), object_id(‘Charge’), 1, NULL, ‘DETAILED’)
DBCCSHOWCONTIG(‘charge’, 1) WITH ALL_LEVELS, TABLERESULTS
Using the DMV or DBCC SHOWCONTIG you can get the same picture of the charge table. Using the detailed (or ALL_LEVELS) parameter, you get the entire structure (all levels) for the clustered index (index_id = 1 is always the clustered index, IF the table is clustered). The reason it returns all levels is that the ‘DETAILED’ mode has been specified.
The clustered index in this table has 1,600,000 rows (DMV column: record_count or SHOWCONTIG column: rows) and these are stored on 9303 pages (DMV column: page_count or SHOWCONTIG column: pages). If you read to the next level which is level 1 because the leaf level is level 0 (remember index levels always start with the leaf level 0 and then go up to the root), you can see that it’s number of “rows” is equal to the number of pages in the leaf level… and this keeps going until you get to a root of 1 page. In this case, the clustered index (which is the widest structure of the table) has a very narrow clustering key (the key is on charge_no which is an int) only has a total of 3 levels even though the table has 1,600,000 rows. Ideally, you should run this on a few of your production tables (in a development/test environment) and you can start to get some insight into how big your structures are. However, a BIG factor that you might see in production is fragmentation. If a particular level (or levels for that matter) are heavily fragmented then each level might be wider and less compact (and therefore less performant). Reviewing the DMV columns avg_fragmentation_in_percent and avg_page_space_used_in_percent, you can get a feel for how full each page is. Poor page density reflects that your pages are not as full as they could be but there are many factors for why this is the case: bad row size, splits due to inserts, splits due to updates of varchar columns or even a poorly chosen fillfactor that has left too much space on the pages. However, page density is only one piece of the puzzle and if your avg_fragmentation_in_percent is very low (0-5%) then I wouldn’t be over worried about your pages not being entirely full unless you have the time to possibly re-design tables (eg. vertically partition them) and then rewrite your applications to direct your statements at only the appropriate base table. But, another factor to consider is the rate at which your fragmentation occurs as well as when you can fix that fragmentation. This is a HUGE discussion that requires time… And, I want to get back to index structures for now. However, both Paul and I have blogged quite a bit about rebuilding v. defragging indexes and what those operations do/how, etc. In fact, just today, Paul has blogged a Q&A about myths and misconceptions about index rebuild operations. So, I’ll get back to internals for now! :)
You can use LIMITED (which is the default mode), SAMPLED, or DETAILED. All three have excellent uses and all use IS locks (to minimize blocking). Limited gives you a quick overview of fragmentation and mostly describes how intact and in order the levels are. Limited is quite clever in that it only scans the first non-leaf level above the leaf to determine how much fragmentation there is… since the non-leaf level always tracks the first entry (and a pointer to the page) then they know EACH and EVERY page in the leaf level by ONLY reading the non-leaf level (which is [typically] a lot smaller and therefore faster). However, because they don’t touch every page and determine page density then they only track how out of order the levels are and not how dense/full the pages are (which is also a form of fragmentation). So, if you want a bit more details, you can use SAMPLED. The SAMPLED mode returns the fragmentation from reading every 100th page of the index (or heap). If the table has less than 80MB used (which is 10,000 pages), every page is read instead (which is a DETAILED scan). The DETAILED mode reads every page of every level to calculate the most accurate picture of your tables fragmentation. This is the best form of analysis but also takes the most time.
If you’re interested in learning a few more of the tips/tricks with using this DMV, check out the following script: Using dm_db_index_physical_stats.zip (2.23 KB)
A favorite tip is that the database in which you want to analyze tables does NOT have to be in 9.0 compatibility mode in order to use this DMV. Don’t get me wrong, you will get errors if you try to use this DMV in a database that’s not in 9.0 compat mode; however, if you are in master (which is set appropriately and cannot be changed) and then use the first parameter to target a non-9.0 compat mode database, then this DMV works great. However, a second “gotcha” is for parameter 2… as long as you don’t use 2-part naming for the objectname (2nd) parameter, everything will work as expected. If you specify object_id(‘tablename’) from master for a table that’s in credit then object_id will return NULL. The query will still run but against all tables in credit rather than the one you thought you were targeting. If you want to use this DMV across databases, you will need to supply the database name in the first parameter and then make sure that you use 3-part naming for the second parameter.
Now that you are getting to know some of the structures (in terms of seeing physical structures and internals), where do we go from here? The best route to start “finding the right balance” for performance is to know the data and as well as get some general insight into usage patterns (this is probably the hardest component to know and sometimes you only know exactly what’s going on if you profile what’s actually happening in production – is that too late? To a certain extent yes and to another extent no…there are still many things for which you can plan and other things you can confirm or test once the application is running (i.e. Profiler). All of those things together are going to help to “find the right balance”.
Having said that, and having discussed the general internals of a b-tree (and therefore an index structure), what’s the difference between a clustered and non-clustered index? Well… stay tuned, that will be part 3 in this series. And, then (finally), we’ll get to appropriate uses for INCLUDE (which was new for SQL Server 2005) and then appropriate uses for Filtered Indexes (a new feature in SQL Server 2008). Also, somewhere in there I’ll post a few tips from my TechEd session so that you can start to determine if your indexing strategies are working??
Thanks for reading!