# Statistics (Part 8) – Nulls

This is a continuation of my blog series about statistics. The previous post is here.

So the question is what happens to your cardinality estimates when the column is nullable and contains many nulls (or is it more accurate to say doesn’t contain…regardless I think you know what I mean)?

Under this condition the face value of the density in the density vector is not used but adjusted to account for the presence of nulls in the column.

In part 2 of this series I created a table called AWStatTest which contains a subset of the columns from the Person.Address table in the Adventureworks database, AddressLine1, AddressLine2, City, and Postal Code.

The column AddressLine2 is nullable and, given the nature of addresses the second address line doesn’t always contain data. This is the column I will be using for my test.

The first step is to create a statistic object on this column which I will do so with this statement. I got into the habit of using full scans in my examples because it is often easier to understand how cardinality estimations work when the numbers predicted by the stats match the actual numbers in the table. This isn’t a blanket recommendation for full scans though.

And to see the stats it produced. Recall from part 2 of this series that density is calculated by the formula 1/(number of unique values). The number of unique values in the column can found with this statement.

SELECT COUNT(*) FROM
(
) A

In our case the value returned is 39. So, 1/39 = .02564103 which is the density shown in the density vector above.

In order to test cardinality estimation under these conditions I will need to use a predicate which causes the optimizer to use the density value for estimation instead of the histogram. Comparing against a variable instead of a static value will force this.

Based on the knowledge gained in previous posts I would expect the cardinality estimate to be calculated from the density vector using the formula (Number or rows in the table) * (Density) or in our case (4040008 * .02564103.) which would be 103590 rows.

But the estimate produced by the optimizer is quite different: 1275.79 My next guess would be that the optimizer is using a magic density. But that formula (4040008^.75) produces an estimate of 90112.84, still way off from the estimate the optimizer produced.

If the database option ANSI_NULLS is ON then a null comparison via an equality statement will never resolve to true. Even a comparison of NULL = NULL does not resolve to true. Given this, the only way our predicate AddressLine2 = @AddressLine2 can be true is if the value of @AddressLine2 is not null and the value of this field in the row is also not null. In other words any row which has AddressLine2 of null would never be returned by this statement even if the value of @AddressLine2 is null.

So, what if we recalculated density and cardinality but disregarded nulls. There would be only 38 unique values in the columns instead of 39 so the density would be 1/38 or .02631578. The histogram shows there are 3991528 null values in AddressLine2. The total number of rows in the table minus the number of null entries should give us the number of non-null values in the column 4040008 – 391528 = 48480.

Using the new density and table row count to calculate cardinality would give us,
48480 * .02631578 = 1275.79, the same cardinality estimate produced by the optimizer.

This is with ANSI_NULLS turned on. What if ANSI_NULLS is disabled? With ANSI_NULLS a comparison of NULL = NULL would resolve true and all of the nulls in the column could be returned when using the equality predicate construct.

SET ANSI_NULLS OFF

Running the same query again with ANSI_NULLS off produces a cardinality estimate of 103590 which is the estimate we came up with using the original density and full table row count. Neat, huh?

# Statistics (Part 7) – Magic Density

Sometimes the optimizer has absolutely no idea how to estimate cardinality. There are either no statistics and auto-create statistics has been disabled or the predicate isn’t of a form that SQL Server can use any of the existing statistics. There was an example of the latter problem in my last blog post about stats on computed columns. On these occasions the optimizer must be like Spock.

SPOCK
In order to return us to the exact moment at which we left the 23rd Century, I have used our journey back through time as a referent, calculating the coefficient of elapsed time in relation to the acceleration curve.

BONES
Naturally.

SPOCK
Acceleration is no longer a constant.

BONES
Well, you’re gonna have to take your best shot.

SPOCK
Best shot?

BONES

SPOCK
Guessing is not in my nature.

BONES
Well nobody’s perfect.

Unlike Mr. Spock guessing is definitely in the nature of the optimizer and when it has no other way to estimate cardinality it will simply guess. Well, it is always guessing but most of the time the guess is educated, informed by stats. Without information to go by the guesses are so unbelievably bad you’d think you’d stumbled into a screening of Highlander 2: The Quickening. There are some rules it has to go by but ultimately it is taking completely uneducated guesses. However I’ve found that knowing how it is going to guess has been helpful in discovering that the optimizer is guessing. To my knowledge there is no flag or other notification that it has taken a complete WAG. And knowing that it is guessing helps track down why we have crappy query plans and points us in a direction where we might be able to fix them.

For years I’ve read that if the optimizer doesn’t have statistics to go on it will estimate 10% of the rows in the table for straight equality searches.
Consider this query.

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

On a table with 12,000 rows this would be estimated at 1,200 rows returned. In preparation for this series of blog posts I tested everything I was asserting before putting it on the page. The only time I could come up with a 10% of row count cardinality estimate was with 10,000 rows on the table. I also could not figure out what percentage was being used. If I changed the table size the percentage changed. At that point I suspected that the cardinality estimate was logarithmic in nature and changed based on the size of the base table.

So I created many test tables with each with a different number of rows, 1000, 2000, 3000, 4000, 5000, 10000, 11000, etc. I plotting the numbers of rows estimated on a graph and, with the help of someone who knows what he is doing, was able to figure out the formula that fit the curve. It turns out that the instead of 10% the estimate is ROWS ^ .75. This explained why my 10000 row table gave me a 10% estimate (10000^.75 = 1000 = 10000 * .10.) It was pure coincidence.

I then tested a query whose predicate queried against two columns.

SELECT * FROM dbo.StatTest WHERE Col1 = 'A' AND Col2 = 'B'

I expected to use the same formula but it didn’t work. Plotting the numbers on a graph showed that the formula is ROWS ^ .6875. I was about to give up when I ran across this blog post from Ian Jose which showed the factor constant used for cardinality estimates with a different number of equality matches in the predicate. I’m repeating the values here in case that page ever goes away.

1 column – Rows ^ (3/4) = Rows ^ .75
2 column – Rows ^ (11/16) = Rows ^ .6875
3 column – Rows ^ (43/64)
4 column – Rows ^ (171/256)
5 column – Rows ^ (170/256)

175 column – Rows ^ (0/256)

I tested this out to four column equality comparisons and decided that was good enough and I would believe Ian’s numbers.
For other situations the magic densities I’ve read about all were correct. BETWEEN uses 9% of row count and <, >, <=, =>, use 30% of row count.

Ultimately if the optimizer is using magic densities it is a good idea to figure that out and create the stats or fix the code that allow it to make educated guesses. Your instance will love you for it.

# Statistics (Part 6) – Computed Columns

This topic is unbelievable dry. Dryer than ancient Egyptian pharaoh bone dust dry. Dryer than the surface of the sun dry. The layman might say “you’re talking about SQL Server and you expect it to be interesting?” Fair point. Maybe the topic of SQL Server is dry to the laymen but almost every topic is interesting to someone. Some people collect thimbles or bugs. Some people juggle geese. And as interested in SQL Server as I am this particular aspect of stats has got to be the driest I’ve ever written about. Death Valley dry. Dryer than Aunt Edna’s thanksgiving turkey dry.

However I feel it is necessary to include this information as part of the series. Please pardon me if I try to spice it up a bit. I already shorted out one keyboard from the boredom induced drool. This builds on my previous and hopefully less boring series on stats.

So…stats on computed columns. You can do this. Even if the column isn’t persisted. There is a caveat though. The formula for the column must be deterministic.

What does it mean to be deterministic? Wikipedia defines determinism as “a metaphysical philosophy stating that for everything that happens there are conditions such that, given those conditions, nothing else could happen.” In computer science this means that given the same input you will always have the same output. In a deterministic universe if we went back 14 billion years and reran the big bang I’d still be typing this boring topic and you would have still stopped reading this two paragraphs ago (see what I did there?)

It would be impossible to generate statistics on a column that is created from a non-deterministic expression. Imagine a column that was based on the result of RAND(). Every select would produce a new value on each row. Stats would be wrong the minute they are created. So SQL Server won’t let you create stats on non-deterministic computed columns.

However given the predictability of determinism (this is actually the third time you’ve read this in the last 42 billion years you just don’t remember the first two times) SQL Server is easily able to create statistics on a computed column, even if the computed column isn’t persisted. If the computed column C is the sum of column A and column B then the computed column will always be A + B even if C is never saved on disk.

According to the documentation the optimizer can make estimates even if the predicate contains the formula the computed column is calculated from instead of the column name. I will test that too. Fun.
Continue reading Statistics (Part 6) – Computed Columns

# Statistics (Part 5) – String Index

I’ve been talking about different aspects of stats and cardinality estimates. My last post is hiding here.

The String Index portion of a statistics object is sparsely documented. The existence of the string index in a stats object can be discovered by looking at the header output of DBCC SHOW_STATISTICS. The documentation from Microsoft often refers to this as the String Summary.

The string index assists in cardinality estimates for substrings in predicates containing the LIKE clause with wildcard characters. It only does this for the leading column of a stats object and only for character data types, i.e. the string index is constructed when the leading column of a stats object is a character column.

The index contains substrings from the first and last 40 characters of a string. If the string is 80 characters long then the string index is created for all characters. If the string is 120 characters long the middle 40 characters would not be included in the string index.

To show this I created a test table with column that is 120 characters long. This column can logically be divided into three sections, the first 40 characters, the middle 40 characters, and the last 40 characters. In each of these sections I create a substring that is unique to that section and does not exist in the other sections. By querying the table with LIKE and wildcards I can see if the cardinality estimates for the middle section are different from the estimates for the beginning and ending sections. In theory the cardinality estimate for the middle section should be the same for predicate whose substring is not part of the string index. I also include several thousand rows that have no noteable substrings at all just to give the table soem mass.

CREATE TABLE StringTest (ID int IDENTITY(1,1) PRIMARY KEY NOT NULL, Col1 varchar(120))
INSERT INTO StringTest (Col1) SELECT REPLICATE('A',120)
GO 1000
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',10) +
REPLICATE('B',10) +
REPLICATE('A',100)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',10) +
REPLICATE('C',10) +
REPLICATE('A',100)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',120)
GO 1000
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',50) +
REPLICATE('D',10) +
REPLICATE('A',60)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',50) +
REPLICATE('E',10) +
REPLICATE('A',60)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',120)
GO 1000
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',90) +
REPLICATE('F',10) +
REPLICATE('A',20)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',90) +
REPLICATE('G',10) +
REPLICATE('A',20)
GO 30
INSERT INTO StringTest (Col1)
SELECT REPLICATE('A',120)
GO 1000

CREATE STATISTICS StringTestStat ON dbo.StringTest(Col1) WITH FULLSCAN
DBCC SHOW_STATISTICS('StringTest','StringTestStat') In the first section there are 30 rows with the substrings ‘BBBBBBBBBB’ and ‘CCCCCCCCCC’, in the second section there are 30 rows with the substrings ‘DDDDDDDDDD’ and ‘EEEEEEEEEE’, and in the last section there are 30 rows with the substring ‘FFFFFFFFFF’ and ‘GGGGGGGGGG’. What we should see are cardinality estimates for the B string and C string similar to the cardinality estimates for the F string and the G string (giggle). But the cardinality estimates for the D and E strings should be similar to cardinality estimates for a substring that doesn’t exist.

Substrings in the first 40 characters:

SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%BBBBBBBBBB%' SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%CCCCCCCCCC%' Substring in the middle 40 characters:

SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%DDDDDDDDDD%' SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%EEEEEEEEEE%' Substrings in the last 40 characters:

SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%FFFFFFFFFF%' SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%GGGGGGGGGG%' Substring that doesn’t exist:

SELECT * FROM dbo.StringTest WHERE Col1 LIKE '%SSSSSSSSSS%' As predicted the estimates for the substrings in the first and last sections, the B and C substrings are similar to the estimates for the substrings in the last section, the F and G substrings. However the estimates for the middle section, the D and E substrings are similar to the estimates for substrings that do not exist in the column.

Given that there is so little documentation on this topic this is the best I can do to show how the string index can influence cardinality estimates.

My next post will be about statistics on computed columns.

# Statistics (Part 4) – Multi-column cardinality

In the previous posts the predicates for the example queries have all had only one element, i.e. only one constraint against the table in the WHERE clause. While troubleshooting cardinality estimate problems it is helpful to understand what happens when there are two constraints against two different columns in the same table.

Update 2/22/2014:
Before reading this you might want to check out Paul White’s article on this topic where he gives a much better explanation than I did. He also includes the case of OR selectivity which didn’t even occur to me to investigate.

This is the sample query that will be used to show how cardinality estimates happen when there are two constraints.

SELECT * FROM dbo.AWStatTest where AddressLine1 = N'100, rue des Rosiers' and PostalCode = N'2010'

In Post 2 of this series I created a table called AWStatTest and filled it with data for demo purposes. So you don’t have to refer back to that post here is the DDL for that table and the index I created.

CREATE TABLE AWStatTest
(
ID int identity(1,1),
City nvarchar(30),
PostalCode nvarchar(15)
)

The index creation also includes a creation of statistics. This is a picture of the statistics for IX_AWStatTest and will be referenced in this post. The predicate in the sample query is constraining the search by AddressLine1 and PostalCode. As shown in my post on density, the statistics IX_AWStatTest alone cannot be used for cardinality estimates for queries constrained by AddressLine1 and PostalCode.

There are at least two ways this can be estimated, 1) by using multiple stats or 2) by using a multi-column statistic

Using Multiple Stats
One of the features of SQL Server is the ability to create single column statistics on the fly in order to help out with optimization (if the database is set to auto create them anyway…my database is set up to do so). When I run the query above SQL Server created statistics for the column PostalCode. The name of those statistics is machine generated, _WA_Sys_00000005_5AEE82B9 (this will be named differently on your machine). You can see these stats with DBCC SHOW_STATISTICS.

DBCC SHOW_STATISTICS ('dbo.AWStatTest','_WA_Sys_00000005_5AEE82B9') This statistics object has a density vector and histogram as well.

SQL Server can use the histogram for AddressLine1 from the statistic IX_AWStatTest and the histogram for PostalCode from the statistic _WA_Sys_00000005_5AEE82B9 to make a cardinality guess. The histogram for PostalCode shows that our constraint value ‘2010’ is a RANGE_HI_KEY and therefore EQ_ROWS will be used to provide cardinality estimates for this value. The number or rows in the table estimated to match postal code 2010 is 13714.8. Looking at the histogram for AddressLine1 shows that ‘100, rue des Rosiers’ is a RANGE_HI_KEY and EQ_ROWS would give us an estimate of 2020 rows if the predicate included a search for just this value.

So how are these used together? A quick example problem that should be easy for anyone familiar with playing cards to understand which helps us understand using two different histograms to make a cardinality estimate. There are thirteen cards in a suit and four suits per deck of standard playing cards. What is the probablity of drawing a face card that is a heart? In this case we are looking for cards that have two specific characteristics, a face card and a heart. There is a 13/52 chance of drawing a heart. There are 12 face cards (Jack, Queen, King) in a deck, so there is a 12/52 chance of drawing a face card.

The probability of drawing a face card that is also a heart is:
13/52 * 12/52 = .0576923.
If we were to treat this like a density and estimate the number of cards in the deck that would match our double criteria we would multiply .0576923 by the number or cards (rows) in the deck, .0576923 * 52 = 3. This is the exact number or cards matching the criteria we actually have.

Let’s translate this example to our cardinality problem. We are looking for rows of data that have two specific characteristics, AddressLine1 and PostalCode equal to specific values. In this situation the probability of a match is for AddressLine1 and Postal Code are 2020/4040008 and 13714.8/4040008 respectively.

Multiply these together by the number of rows in the table and the cardinality estimation is for the query is: (2020/4040008) * (13714.8/4040008) * 4040008 = 6.85739. SELECT * FROM dbo.AWStatTest where AddressLine1 = N'11, quai de l´ Iton' and PostalCode = N'94010' and City = N'Bellingham'

The cardinality estimates for each of these values individual are, 9090, 42062.04, and 58405.71 respectively.

(9090/4040008) * (42062.04/4040008) * (58405.71/4040008) * 4040008 = 1.36819 The examples above all used values that were RANGE_HI_KEYS from the respective histograms. In my tests I have seen that estimates using values from the AVG_RANGE_ROWS column work exactly the same way.

Using multiple statistics objects isn’t the only way SQL Server can estimate cardinality on a table with multiple predicates.

Using a multi-column density
In Post 2 I talked about density vectors and how they can be used. They can also be used for multi-column estimates. To force SQL Server to use a multi-column density for cardinality estimation I am going to create a new table with the same data as the old one, turn off auto creation of statistics and create a statistics object for just PostalCode and AddressLine1. The primary reason I’m creating a new table is because I’ve done so much testing on my other one that I fear being unable to get pure estimates.

SELECT * INTO dbo.AWAL1PCTest from dbo.AWStatTest
ALTER DATABASE TestDB SET AUTO_CREATE_STATISTICS OFF

I should end up with a statistics object for PostalCode and AddressLine1 that looks like this: The density value for the combination of AddressLine1 and PostalCode is .000249066
The estimate should be .000249066 * 4040008 = 1006.23

Running this query:

SELECT * FROM dbo.AWStatTest where AddressLine1 = N'100, rue des Rosiers' and PostalCode = N'2010'

In this post we have seen that cardinality estimates for predicates that contain multiple matching conditions for two different columns on the same table can be estimated in at least two ways.
1. By using the histogram values from multiple statistics objects
2. By using a multi-column density value

The point of the previous three posts is not to outline all of the ways the density vector and histogram parts of statistics objects are used make cardinality estimates. That information is not publically available (that I can find anyway). Most of what I talk about in the previous three posts comes from performing a lot of tests and comparing results to calculations. The purpose is to highlight how the various components can be involved so that one can have more success in troubleshooting cardinality estimate issues. I’ve often been surprised at some of the estimates that come up and they don’t always align themselves to the “rules” I’ve laid out above, though they usually do. That being said, tracing cardinality estimates back to specific statistics used has been helpful in identifying skewed data distributions and possible places for optimizations.

The next topic is an area that is very opaque and, I suspect, is one of the causes of some of the odd cardinality estimates I’ve seen: the String Index.

# 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.

DBCC SHOW_STATISTICS('AWSTatTest','IX_AWStatTest')
WITH HISTOGRAM 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.

SELECT COUNT(*) FROM dbo.AWStatTest
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
(
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.

CREATE TABLE AWStatTest
(
ID int identity(1,1),
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.

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.

GO
DBCC SHOW_STATISTICS('dbo.AWStatTest','IX_AWStatTest')
GO

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:

SELECT COUNT(*)
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 COUNT(*) [ALL DENSITY]
FROM
(
) 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.

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

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.

CREATE TABLE dbo.StatTest
(ID int IDENTITY(1,1) NOT NULL,Col1 char(1),Col2 varchar(10))

CREATE CLUSTERED INDEX IC_StatTest ON dbo.StatTest(ID)

CREATE INDEX IX_StatTest ON dbo.StatTest(Col1)
GO

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

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

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.