Monthly Archives: February 2013

Statistics (Part 3) – Histograms

As noted in the previous entry density can be used to predict the number of records that will be returned from a query against a table. This is fine if there is an even distribution of data but what about the cases where there is not.

As demonstrated in the previous post the density for the column AddressLine1 of 0.0003552398 multiplied by the number of rows in the table, 4040008, gives us 1435.17, which is the number of rows the query optimizer estimates will be returned from a query against the table with a single predicate against the column AddressLine1. Given an even distribution of data each unique value in the column would exist in the set roughly 1435 times. SQL Server can use the density vector to estimate the number or rows that will be returned from a query. The caveat here is ‘given and even distribution’. What if the distribution is uneven?

What if you have 10000 rows with 9999 values in a column equal to A and 1 equal to B? This was the example I set up in my first post in this series. In this case the density would be 1/Number of Unique Values or ½ or .5. If we use the technique of multiplying the density by the number of rows in the table we would get an estimate of 5000 rows for each value. This could cause the optimizer to produce a plan that scans the entire table as opposed to the plan that queries an index and does a book mark lookup. If the query is looking for records where the value of AddressLIne1 is A plan produced is OK since it is likely a table scan will be used anyway. But looking for values of B would require a table scan to find a single record. On the flip side using bookmark lookups to retreive 9999 rows when a table scan would require less I/O is not a good idea either. I refer you to my first post in the series where I showed the number of physical reads required from a table scan relative to a bookmark lookup.

SQL Server can use histograms to make more refined predictions.

In my previous post I created a test table with 4040008 rows and an index on the column AddressLine1. The index automatically creates a statisitcs objects. You can see the histogram of the statistics attached to the demo index by using DBCC SHOW_STATISTICS. I use the WITH HISTOGRAM option so that only the histogram is returned.


One thing to note about the histogram is that the values of RANGE_HI_KEY are sorted in ascending order. All values not equal to a RANGE_HI_KEY naturally will be of greater or lesser value than the RANGE_HI_KEYs and, subsequently, can be thought of as existing between two keys that are side by side. If our range hi keys are 5, 10, 20, and 25 the value 7 exists between the entries of 5 and 10. While it also exists between the entries of 5 and 20 the next highest entry to 7 is 10. For query optimization the data in the row of the next highest entry is used to for cardinality estimates. In our example the value of ‘081, boulevard du Montparnasse’ exists between ‘#500-75 O’Connor Street’ and ‘1, place Beaubernard’ when these values are sorted in ascending order.

With this we know then there are two types of estimates that can be made from the histogram. Values equal to the RANGE_HI_KEY and values that are between two RANGE_HI_KEYs which are side by side.

The cardinality estimates for predicates equal to the RANGE_HI_KEY come from the EQ_ROWS column. If we query the table for the value, “1, place Beaubernard” in the column AddressLine1, the optimizer will predict 4040 rows to be returned.

SELECT * FROM dbo.AWStatTest where AddressLine1 = '1, place Beaubernard'

With a full sample the EQ_ROWS value for each RANGE_HI_KEYS show the exact number of times that each of the RANGE_HI_KEYS value is in the index/heap’s column. The values of in the RANGE_ROWS and DISTINCT_RANGE_ROWS are used to calculate AVG_RANGE_ROWS which, in turn, is used to provide cardinality estimates for values that don’t match the RANGE_HI_KEYs.

RANGE_ROWS is the number or records which contain a value that sorts between two side by side range rows, the current row and the next lowest row. In our example range rows for ‘1, place Beaubernard’ is 14147. This means there are 14147 records which contain a value between ‘1, place Beaubernard’ and ‘#500-75 O’Connor Street’.

We can use this query to test this.

WHERE AddressLine1 > '#500-75 O''Connor Street'
AND AddressLine1 < '1, place Beaubernard'

DISTINCT_RANGE_ROWS are the number of distinct values between two range hi keys.

SELECT COUNT(*) [count]
FROM dbo.AWStatTest
WHERE AddressLine1 < '1, place Beaubernard'
AND AddressLine1 > '#500-75 O''Connor Street'
) AA

If you recall from the previous lesson density is calculated as 1/Unique number of values. So we can calculate the density of the range between two RANGE_HI_KEYS. The density of the range between ‘1, place Beaubernard’ and ‘#500-75 O’Connor Street’ is 1/13 or .0769231.

Density times the number of rows gives us the average number of times each value will exist in this range. .0769231 X 14147 = 1088.231 which is the value of the AVG_RANGE_ROWS column for the RANGE_HI_KEY ‘1, place Beaubernard’. If we query AddressLine1 for a value that exists in this range the number of rows estimated to be returned should be 1088.231.

SELECT * FROM dbo.AWStatTest where AddressLine1 = '081, boulevard du Montparnasse'

This density estimation between the RANGE_HI_KEYs works just like the density estimation for the entire column but with more precision. It can be more precise because the range is smaller and because as the histogram is built SQL Server can choose range hi keys so that the steps between them have a more even distribution.

These are examples of single columns statistics being used with queries against individual columns. What happens if the query against the table contains two predicates. That is the focus of the next post.

Statistics (Part 2) – I am your density

In order to show how statistics work I need to first create some data to work with. This table and the one created in the first post of this series will be used in this example.

ID int identity(1,1),
AddressLine1 nvarchar(60),
AddressLine2 nvarchar(60),
City nvarchar(30),
PostalCode nvarchar(15)

Using the following command with some variability of the TOP clause and the number of execution specified on the GO command I created a table with some data to test against.

INSERT INTO AWStatTest (AddressLine1,AddressLine2, City,PostalCode)
SELECT TOP 1000 AddressLine1,AddressLine2,City,PostalCode
FROM AdventureWorks.Person.Address
GO 1000

The content of the table looks something like this. It has 4040008 rows.

I then created this index on it (which also creates statistics of the same name) and look at the stats using DBCC SHOW_STATISTICS.

CREATE INDEX IX_AWStatTest ON dbo.AWStatTest (AddressLine1,AddressLine2,City,PostalCode)

The properties for the statistics IX_AWStatTest looks like this.

The information displayed by DBCC SHOW_STATISTICS is divided into three sections: header, density vector and histogram. There is useful information in the header but the density vector and histogram are the most interesting stuff when it comes to understanding how cardinality estimates are done.

In order to understand the density vector you have to understand what density is. Density is a measurement of how selective an index is. It can be calculated by the formula: 1/(Number of Unique Values in the set). In our case the set is the column (or columns) upon which the statistics are built.

If ‘A’ is in the column 10 times, ‘B’ is in the column 20 times and ‘C’ is in the column 1000 times the number of unique values is 3, ‘A’, ‘B’ and ‘C’.

To get the number of unique values you can use this query:

FROM (SELECT DISTINCT addressline1 FROM dbo.AWStatTest) AA

In our case there are 2815 unique values for AddressLine1. 1/2815 = 0.0003552398. Now take a gander at the All Density value for the column AddressLine1 in the 2nd section of the statistics example. It is .0003552398. This is how the density value in the second section is calculated.

Take the density and multiply it by the number or rows in the table .0003552398 * 4040008 = 1435.1715. This means that if there were an even distribution of each of the unique values there would be roughly 1435 rows each. We can check this by multipying 1435.1715 by 2815 (the number of unique values) which comes out to 4040007.7725.

It can also be calculated by the formula (Average Number of Duplicates/Number of Rows).

In this example the average number of duplicates is going to be the (number of rows)/(number of distinct values). The average number of duplicates is 4040008/2815 = 1435.1715 (look familiar?). Divide this by the number of rows and you get the same density value as the 1st formula. We can actually derive the first formula from the second.

Knowing that the two formulas reduce to the same thing I will use 1/2815 for simplicity.

This is for a single column. How about the density on multiple columns? When we look at the combination of columns the density is based on the uniqueness of those two columns.

SELECT DISTINCT AddressLine1,AddressLine2 FROM dbo.AWStatTest
) AA

This query tells us there are 2833 unique values for the combination of AddressLine1 and AddressLine2. 1/2833 = 0.0003529827, which matches the All density value in the second section for AddressLine1 and AddressLine2.

So what can these density values do for us? They can help us estimate how many times a specific value (or combination of values) exists in a set of data by multiplying the density by the number of rows in the set (table).

This example shows the optimizer using the Density value from the density vector. I’m specifically using the variable here in the predicate to cause SQL Server to use the density vector. I will explain in a future post what makes this happen.

DECLARE @address nvarchar(50)
SET @address = 'test'
SELECT * FROM dbo.AWStatTest WHERE AddressLine1 = @address

A snapshot of the estimated number of rows to be returned.

1435.17. Does that number look familiar? It is the density of AddressLine1 .0003552398, multiplied by the number of rows in the table, 4040008.

If we selected on AddressLine1 and AddressLine2 the density used would have been 0.0003529827.

When SQL Server creates statistics on multiple columns it creates combinations of density vectors in the order the columns are specified in the CREATE STATISTICS command (or in our case the CREATE INDEX command because statistics are created as a by-product of index creation). For the statistics associated with the IX_AWStatTest index the columns are AddressLine1, AddressLine2, City and PostalCode. The four density values computed are
AddressLine1, AddressLine2
AddressLine1, AddressLine2, City
AddressLine1, AddressLine2, City, PostalCode.

Recall that density is a factor of the number of unique values and if the density is computed on multiple columns it is the number of unique values of that combination of values.

In these statistics the number of unique values of AddressLine1 are computed, the number of unique values of the combination of AddressLine1 and AddressLine2, the number of unique values of the combination of AddressLine1, AddressLine2, and City, and finally the number of unique values of the combination of AddressLine1, AddressLine2, City, and Postal Code.

The important thing to note is that just because two columns are in the same statistic doesn’t mean the density can be used. In this example the columns AddressLine1 and PostalCode are both represented in forth density value but this line is not the combination of AddressLine1 and PostalCode only. That combination doesn’t exist in these statistics. If you need that combination you would have to create another statistics object.

Density if a great tool for calculating cardinality estimates and is used often by the query optimizer but there is a problem. This is a good graphical representation of what the problem is.

Each of these columns has the same density even though the different values (red and green) are only distributed evenly (i.e. the same number of each not the fact they are ordered red to green) in one of the columns. Estimates caluculated based on density be grossly inaccurate if the data distribution is skewed. In this case there skew isn’t significant because there are only ten rows. How about the case where there are 100,000 rows and only one is read. The estimate would be 50,000, wildly over for one and wildly under for the other.

To get much, much better estimates SQL Server uses histograms which will be the topic of the next post.

Statistics (Part 1) – What would you say you do here?

This is the first post in a series of posts I will be doing about SQL Server statistics.

As I got more serious about learning SQL Server one of the things I had a hard time wrapping my mind around was statistics. In the abstract I could understand what they were for but beyond that I didn’t understand how they worked. Now that I’ve learned a lot about them I figure others could use some simple instructions on what they are. Hopefully this is simple enough.

Stats are the unsung hero of query optimization. With good stats you can get a well optimized query. With bad stats…not so good. I’ve seen 100x performance gains from implementing good statistics.

Statistics help the query optimizer determine the best execution plan to use to fulfill a query by predicting the number of rows that will be returned from operations. Knowing the number of rows that will be returned by different operations the optimizer can choose the combinations of operators that will best (lowest cost – quickest, less I/O, less CPU) fulfill the query.

The best way to explain this is with an example. I’m going to create a table with three columns, an IDENTITY column, a char(1) column and a varchar(10) column. There will be 10,000 rows. In 9999 of them the character column will be the value “A” while 1 of them is the value “B”. The varchar(10) column is some throw away data. I will put an index on the character column, Col1. The index will automatically create statistics. Then I can run a couple of queries against Col1 to see what types of query plans are generated.

(ID int IDENTITY(1,1) NOT NULL,Col1 char(1),Col2 varchar(10))


CREATE INDEX IX_StatTest ON dbo.StatTest(Col1)

INSERT INTO dbo.StatTest (Col1,Col2) VALUES ('A','ABCDEFGHIJ')
GO 9999

INSERT INTO dbo.StatTest (Col1,Col2) VALUES ('B','ABCDEFGHIJ')

The first test will be this query with the predicate searching for records where Col1 matches “A”.

SELECT * FROM dbo.StatTest WHERE Col1 = 'A'

This is the query plan produced. It is a clustered index scan.

Looking at the properties of the clustered index scan operator it shows and estimate of 9999 rows to be returned.

However with this execution plan of this query using a different predicate results in a different query plan.

SELECT * FROM dbo.StatTest WHERE Col1 = 'B'

This plan uses an index seek/bookmark lookup combination. The optimizer estimated that 1 row would be returned for this value. In the previous query an index scan was the optimal way to retrieve all of the rows where Col1 = A. In the second query and index seek and book mark lookup are the optimal.

It was the statistics that enabled the optimizer to craft a query plan that was optimal for each data condition. Without the statistics the optimizer wouldn’t have had any idea how many rows would be returned or what the best method of accessing the data was.

The consequences of using the wrong plan are fairly noticeable. Imagine if the query optimizer had no way of knowing there was only one value of ‘B’ in Col1 and chose to do a table scan instead of a bookmark lookup.

By using SET STATISTICS IO ON we can see how may logical reads were required to fulfill this query when using a bookmark lookup.

Table ‘StatTest’. Scan count 1, logical reads 4, physical reads 3, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0.

Now forcing it to use a table scan instead.

Table ‘StatTest’. Scan count 1, logical reads 40, physical reads 1, read-ahead reads 40, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0.

The table scan required 41 physical reads (physical reads and read-ahead reads) while the bookmark lookup required only 3. The number of physical reads to perform the scan is indefinite and related to the size of the table. The number of physical reads to perform the bookmark lookup may increase slightly as the size of the index grows (and the B-tree grows deeper) but nothing like the number a table scan will need. Why use a table scan to find a single record when a key lookup is faster? Likewise why use key lookups to lookup up multiple records when a table scan is faster? It is statistics that helps the optmizer to determine the best method to access the data.

In the next post I look at the statistics object and start to examine how the optimizer uses it to get cardinality estimates.