1. Hugo Kornelis
Now where does THAT estimate come from? The nuts and bolts of cardinality estimation

2. WHAT estimate????

3. WHAT estimate????

4. BIG Thanks to SQL Sat Denmark sponsors
GOLD
SILVER
BRONZE

5. Raffle and goodbye Beer
Remember to visit the sponsors, stay for the raffle and goodbye beers 
Join our sponsors for a lunch break session in :
cust 0.01 and cust 1.06
We hope you’ll all have a great Saturday.
Regis, Kenneth

6. Hugo Kornelis
I make SQL Server Fast

7. Hugo Kornelis I make SQLServerFast.com
Blog:
Execution Plan Reference:
Resources
Deck and demo for this session are there
Articles (planned)

8. Hugo Kornelis I make SQLServerFast.com I do other community things
Technical editor for various books
SQL Server Execution Plans, 3rd edition (“Real Soon Now”™)
11 years MVP (2006 – 2016, SQL Server/Data Platform)
Lots of other things

9. Hugo Kornelis I make SQLServerFast.com I do other community things
I work
Independent database consultant
Will do (almost) anything for money

10. Hugo Kornelis I make SQLServerFast.com I do other community things
I work
Contact details

11. Which version? “Old” (legacy) cardinality estimator
Introduced in SQL Server 7.0
Unchanged until SQL Server 2012
“New” cardinality estimator
Introduced in SQL Server 2014
Small changes in later versions
Better?
Database compatibility level
Trace flags 2312 (force new) / 9481 (force old)
Usually used with OPTION (QUERYTRACEON nnnn) query hint
SQL 2016+: ALTER DATBASE SCOPED CONFIGURATION SET LEGACY_CARDINALITY_ESTIMATION = [ ON | OFF ];

12. Overview Cardinality estimation How: What? Why? How?
The usual suspects
Statistics
Single-table queries (simple filters, complex filters)
Multiple tables (single equijoin condition, single non-equijoin condition, multiple join conditions)

13. Cardinality estimation: What is it?
Prediction of the number of rows that an operator will return

14. Cardinality estimation: Why is it important?
Determines plan choice
Bad join strategy
Non-optimal index choice
Serial or parallel plan

15. Cardinality estimation: Why is it important?
Determines plan choice
Determines memory grant
Operators that store data in memory: Sort, Hash Match (Join), Hash Match (Aggregate)
Required total memory computed based on cardinality estimates
Query execution waits until memory available
Insufficient memory: Spill to tempdb

16. The usual suspects Table variables No statistics
Estimated table cardinality
Always 1 row
Except …
Fix: temporary tables

17. The usual suspects Table variables
Multi-statement table-valued functions
Implemented using table variable
No statistics
Estimated cardinality: always 1 row
Changed to 100 rows in SQL Server 2014
Fixes:
Inline table-valued function
Copy results to temporary table
SQL Server 2017: Interleaved Execution
Rest of plan recompiled after table-valued function is evaluated
Restrictions apply

18. DEMO The usual suspects Table variables
Multi-statement table-valued functions

19. The usual suspects Table variables
Multi-statement table-valued functions
Stale and outdated statistics
AUTO_UPDATE_STATISTICS
AUTO_UPDATE_STATISTICS_ASYNC
Trace flag 2371
SQL Server 2008R2 and up
On by default as on SQL (requires compatibility level 130)

20. The usual suspects Table variables
Multi-statement table-valued functions
Stale and outdated statistics
Unrepresentative statistics
Use higher sampling rate, or use FULLSCAN
Drawbacks: slower, more resources used
Cannot be configured for automatic statistic updates
Not guaranteed to work

21. The usual suspects Table variables
Multi-statement table-valued functions
Stale and outdated statistics
Unrepresentative statistics
Parameter sniffing
Estimates based on value of first execution
Plan reused for later executions
Even when variables change

22. The usual suspects Table variables
Multi-statement table-valued functions
Stale and outdated statistics
Unrepresentative statistics
Parameter sniffing
OPTIMIZE FOR hint
Estimates based on hard-coded value
Even when actual value is different

23. Statistics Number of rows sampled Total number of rows in table
Last update of this statistics

24. Statistics 1.0 / COUNT(DISTINCT LastName)
1.0 / COUNT(DISTINCT LastName, FirstName, MiddleName)

25. Statistics 7 rows have LastName = ‘Brown27’
146 rows have LastName > ‘Brook6’ and LastName < ‘Brown27’
23 distinct LastName values in those rows
So on average ~ 6.35 (146 / 23) rows for each of those values

26. Statistics
DEMO
DBCC SHOW_STATISTICS

27. Statistics Assumptions made when using statistics: Independence
Predicates are not correlated
Uniformity
Values are evenly spread
Containment / Inclusion
Values searched for will exist

28. Single-table queries
Comparison with a variable
Equality
E-05

29. Single-table queries Comparison with a variable Equality
* 29750
sys.dm_db_partition_stats

30. Single-table queries Comparison with a variable Equality
* 29750 =
DBCC SHOW_STATISTICS

31. Single-table queries Comparison with a variable Equality Inequality
Fixed selectivity assumption: 30%
0.3
* 29750
= 8925
sys.dm_db_partition_stats

32. DEMO Single-table queries Comparison with a variable Equality
Inequality

33. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality 7
* / = 7
29750
29750
sys.dm_db_partition_stats
DBCC SHOW_STATISTICS

34. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
* / 29750
29750
29750

35. DEMO Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality

36. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality
= 37
* / = 37

37. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality
= 37 36
* / = 37 36

38. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality
+ (~67% * 34)
= ~24.78 * / = ~24.78
(actual estimate: )

39. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality
+ (~67% * 34) = ~24.78 * / = ~24.78
(actual estimate: )

40. DEMO Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality

41. Single-table queries
Comparison with a constant, sniffed parameter, or sniffed variable
Equality
Inequality
Changes on SQL Server 2014:
Different interpolation
Difference between < and <= is observed
> ‘Zwilling4’ 
>= ‘Zwilling4’ 

42. DEMO Ascending Key problem Not just for keys!
(And not just ascending either)

43. Ascending Key problem Workaround since SQL Server 2005 SP1:
Trace flag 2389 (“known ascending columns”)
Trace flag 2390 (“all other columns”)
Value out of range and column indexed?
find MIN/MAX from table
If within actual range, use average density
Example:
Statistics: rowcount 10,000; highest value 250; density 0.004
Actual: rowcount 11,500; highest value 300
WHERE value = 300
Density * Rowcount = * 10,000 = 40
WHERE value = 310
Out of actual range, so estimate is still 1
WHERE value = 290
In new range. Estimate should be 40, but is 1

44. Ascending Key problem Workaround since SQL Server 2005 SP1:
Trace flag 2389 (“known ascending columns”)
Trace flag 2390 (“all columns”)
Value out of range and column indexed?
find MIN/MAX from table
If within actual range, use average density
Example:
Statistics: rowcount 10,000; highest value 250; density 0.01
Actual: rowcount 11,500; highest value 300
WHERE value > 290
Interpolation based on 1,500 rows in range: 332
WHERE value > 310
Out of actual range, so estimate is still 1

45. Ascending Key problem Change in SQL Server 2014: WHERE value = 300
Value out of range? (Regardless of “known ascending” and index!)
Estimate as if variable was used, based on number of rows added
Example:
Statistics: rowcount 10,000; highest value 250; density 0.01
Actual: rowcount 11,500; highest value 300
WHERE value = 300
Density * Rowcount = * 1,500 10,000 = 40
WHERE value = 310
Density * Rowcount = * 10,000 = 40
WHERE value = 290
Density * Rowcount = * 10,000 = 40

46. Ascending Key problem Change in SQL Server 2014: WHERE value > 290
Value out of range? (Regardless of “known ascending” and index!)
Estimate as if variable was used, based on number of rows added
Example:
Statistics: rowcount 10,000; highest value 250; density 0.01
Actual: rowcount 11,500; highest value 300
WHERE value > 290
Inequality assumption * New rows = 0.3 * 1,500 = 450
WHERE value > 310
Inequality assumption * New rows = 0.3 * 1,500 = 450

47. Single-table queries Multiple predicates
SQL 7 – SQL 2012: Assume independency
Estimate selectivity of each predicate
(Estimate / Table cardinality)
Find combined selectivity
(Selectivity 1 * Selectivity 2 * Selectivity 3 * …)
Find rowcount
(Table cardinality * Combined selectivity)

48. Single-table queries Multiple predicates
Example: Table has 100,000 rows
WHERE Col1 = ‘A’  4,500 rows
Selectivity = 4,500 / 100,000 = 0.045
WHERE Col2  Density = 0.025
Selectivity = 0.025
WHERE Col3  Fixed selectivity
Selectivity = 0.3

49. Single-table queries Multiple predicates
Example: Table has 100,000 rows
WHERE Col1 = ‘A’ AND Col2 AND Col3
Selectivity = * * 0.3 =
Estimate = 100,000 * = 33.75

50. Single-table queries Multiple predicates
SQL 2014: Assumed partially dependent
Estimate selectivity of each predicate
Sort in order of ascending selectivity
Most selective selectivity comes first
Find combined selectivity (“exponential backoff”)
(Selectivity 1× Selectivity 2 × Selectivity 3 ×…)
Find rowcount

51. Single-table queries Multiple predicates
Example: Table has 100,000 rows
Selectivity values: 0.045, 0.025, 0.3
Ascending order: 0.025, 0.045, 0.3
Selectivity = 0.025× × ≈
Estimate = 100,000 * =

52. Single-table queries Multiple predicates Fix (for SQL 7 – SQL 2012)
Create multi-column statistics
(These are never created automatically)
Will use average density for column combination
Not used on SQL 2014 RTM
Was fixed in SQL 2016

53. Single-table queries
DEMO
Multiple predicates

54. Multiple tables
DEMO
Simple joins
One equality predicate

55. One equality predicate
Aligning histograms
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)

56. One equality predicate
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
2 * 2 = 4
2 * 2 = 4
4 + 4

57. One equality predicate
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(98/49) * 2 = 4
1 * (61/55) ≈ 1.1
(149/149) * 1 = 1

58. One equality predicate
~20% * 48 = 9.6 distinct values
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
9.6 * (98/49) * (84/42) = 38.4

59. One equality predicate
~80% * 48 = 38.4 distinct values
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
~80% * 54 = 43.2 distinct values
38.4 * (98/49) * (61/55) ≈ 85.2

60. One equality predicate
~10% * 148 = 14.8
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
~20% * 54 = 10.8
10.8 * (149/149) * (61/55) ≈ 12.0

61. One equality predicate
~90% * 148 = 133.2
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
100 * (149/149) * (100/100) = 100

62. One equality predicate
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(actual estimate: )
= 249.7
* / = 249.7

63. One equality predicate
SQL Server 2014: Much simpler!
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)

64. One equality predicate
SQL Server 2014: Much simpler!
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
2 * 2 = 4
2 * 2 = 4
4 + 4

65. One equality predicate
SQL Server 2014: Much simpler!
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
199 * (248/199) * (248/199) ≈ 309.1
= * / = 317.1
(actual estimate: 316.5)

66. One equality predicate
SQL Server 2014: Actual
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
200 * (250/200) * (250/200) = 312.5
2 * 2 = 4
= * / = 316.5
(actual estimate: 316.5)

67. Multiple tables Simple joins One inequality predicate
Nothing documented
So … let’s speculate!
SQL Server 7.0 – 2012:
Variation on equality algorithm

68. One inequality predicate2
* 250 = 500
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
= 250
500

69. One inequality predicate
~20% * 48 = 9.6 distinct values
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2) 84
84 * 0.5
= 166
9.6 * (98/49)
* (166 + (84 * 0.5))
* 166 =
500

70. One inequality predicate
98/49 * 164 = 328
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
= 164

71. One inequality predicate
38.4 * 98/49 * ( ((43.2 * 61/55) * 0.5) =
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(43.2 * 61/55) * 0.5
(11.8 * 61/55) =

72. One inequality predicate
1 * =
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(10.8 * 61/55) =

73. One inequality predicate
14.8 * 149/149 * ( ((10.8 * 61/55) * 0.5) =
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(10.8 * 61/55) * 0.5 =

74. One inequality predicate
149/149 * 102 = 102
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
= 102

75. One inequality predicate
133.2 * 149/149 * (2 + (100 * 0.5)) =
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
100 * 0.5 2

76. One inequality predicate
Abercrombie0 (2)
Abercrombie0 (2)
Abolrous0 (1)
Ackerman0 (2)
(98/49) (149/149)
(84/42) (61/55) (100/100)
Abercrombie0 (2)
Abercrombie48 (2)
Abolrous9 (1)
Ackerman0 (2)
(actual estimate: ) =

77. Multiple tables Simple joins One inequality predicate
Nothing documented
So … let’s speculate!
SQL Server 7.0 – 2012:
Variation on equality algorithm
SQL Server 2014:
Same with the simplified histogram

78. One inequality predicate2
* 250 = 500
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
= 250
500

79. One inequality predicate
250
* (250 * 0.5) = 31250
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
250 * 0.5
500

80. One inequality predicate
Abercrombie0 (2)
Ackerman0 (2)
248/199
248/199
Abercrombie0 (2)
Ackerman0 (2)
(actual estimate: 31750)
500
= 31750

81. Multiple tables Complex joins (actual estimate: 13.1007)
Two equality predicates
SQL Server 7.0 – 2012
Compute selectivity for each predicate
Multiply (assume independence)
Cartesian product (estimate): (2975 * 1975)
Join on LastName only (estimate):
Selectivity = / = 4.242E-04
Join on FirstName only (estimate):
Selectivity = / = 5.256E-03
Estimate = 2.230E-06 * =
Combined selectivity = 4.242E-04 * 5.256E-03 = 2.230E-06
(actual estimate: )

82. Multiple tables Complex joins (actual estimate: 21327.1)
Two equality predicates
SQL Server 7.0 – 2012
Compute selectivity for each predicate
Multiply (assume independence)
Or use density for column combination
Multi-column statistics
Statistics for multi-column index
DISTINCT p.FirstName, p.LastName (estimate): 27550
DISTINCT p2.FirstName, p2.LastName (estimate): 19200
DISTINCT matching combination: 19200
Rows per combination in p: rows * density = 1.08
Rows per combination in p2: rows * density = 1.03
Estimate: * 1.08 * 1.03 =
(actual estimate: )

83. Multiple tables Complex joins Two equality predicates SQL Server 2014
Estimate #distinct combinations on each side
Multiply smallest with estimated densities
 Same as SQL Server with multi-column statistics
(even when there are no multi-column statistics)

84. Multiple tables Complex joins Equality and inequality
SQL Server 7.0 – 2012
Compute selectivity for each predicate
Multiply (assume independence)
Same as for two equality predicates
Multi-column statistics never used

85. Multiple tables Complex joins Equality and inequality SQL Server 2014
Estimate cardinality of each input
Assume each row from large input matches one row from smaller input

86. Multiple tables Complex joins Join with extra filter predicates
on other columns
SQL Server 7.0 – 2012
Assume filters are correlated
Determine filter selectivity
Scale down histograms
Align scaled-down histograms

87. Multiple tables Complex joins Join with extra filter predicates
on other columns
SQL Server 2014
Assumes no correlation between filters
Align original histograms (simplified)
Determine filter selectivity
Reduce estimate after histogram alignment

88. Twitter: @Hugo_Kornelis
T H E E N D
Questions?

89. Raffle and goodbye Beer
Remember to visit the sponsors, stay for the raffle and goodbye beers  Join our sponsors for a lunch break session in : cust 0.01 and cust 1.06 We hope you’ll all have a great Saturday. Regis, Kenneth

90. BIG Thanks to SQL Sat Denmark sponsors
GOLD
SILVER
BRONZE

91. Twitter: @Hugo_Kornelis
T H E E N D
Questions?