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Inside Microsoft SQL Server 2008: T-SQL Querying
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Pointer to first IAM 1:1791 Pointer to first 1:47120 Pointer to root 1:47186
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Single Page Allocation @0x3642C08E Slot 0 = (0:0) Slot 1 = (0:0) - - Extent Alloc Status Slot 1 80x3642C0C2 (1:0) (1:22064) = NOT ALLOCATED (1:22072) (1:24256) = NOT ALLOCATED (1:24264) = NOT ALLOCATED (1:24272) (1:32344) = ALLOCATED (1:32352) = NOT ALLOCATED ---
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20061231 20061231 20061231
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999716 999717 999718 999757 999758 999759
C0000015545 C0000011129 C0000011129 C0000010921 C0000012275 C0000003426
... ... ... ... ... ...
20061231 20061231 20061231 1:47186
NULL 20030118 20030206 --20061116 20061205 20061225
unq page
NULL 1:47184 258 1:47585 210 1:47187 603 1:36836 540 1:36837 367 1.36838
1:47184
NULL 20021208 20021210 --20030118 20030118 20030118
1:47185
page# od
20030118 20030118 20030118 --20030206 20030206 20030206
1:36838
20061225 20061225 20061225 --20051231 20051231 20051231
unq page#
258 298 338 90 130 170 1:47508 1:47509 1:47510 1:47829 1:47830 1:47831
unq page#
367 1:41380 407 1:41381 447 1:41382 496 1:41477 536 1:41478 576 1:41479
NULL 1:47120 1 1:47121 1 1:47122 138 178 218 1:47505 1:47506 1:47507
1:47120
od unq morecols
------------20021202 NULL 20021203 NULL 20021203 1 --20021207 1 20021207 5 20021208 NULL
1:47121
20021208 20021208 20021208 --20021210 20021210 20021210
1:47122
20021210 20021210 20021210 --20021210 20021210 20021210
1:41478
20061231 20061231 20061231 --20061231 20061231 20061231
1:41479
20061231 20061231 20061231 --20061231 20061231 20061231
unq morecols
1 5 3 1 5 6 -------------
unq morecols
7 8 9 5 6 7 -------------
unq morecols
536 - - 537 - - 538 - - 573 - - 574 - - 575 - - -
unq morecols
576 - - 577 - - 578 - - 613 - - 614 - - 615 - - -
FIGURE 4-16 Clustered table/index
The rest of the discussion in this section is relevant to both clustered and nonclustered indexes unless explicitly stated otherwise. When SQL Server needs to perform ordered scan (or ordered partial scan) operations in the leaf level of the index, it does so by following the linked list. Note that in addition to the linked list, SQL Server also maintains an IAM page (or pages) to map the data stored in the index by le order. SQL Server may use the IAM pages when it needs to perform unordered scans of the index s leaf level. This type of scan based on IAM pages is known as an allocation order scan. A scan that is done in index order is known as an index order scan. The performance difference between the two types of scans depends on the level of fragmentation in the index. Remember that the most expensive part of an I/O operation is the movement of the disk arm (that s at least the case with traditional disk drives that have moving parts, as opposed to solid-state disks). An index order scan in an index with no fragmentation at all performs similarly to an allocation ordered scan, while an index order scan will be substantially slower in an index with a high level of fragmentation. Fragmentation (known as logical scan fragmentation) evolves mainly because of splits of pages at the leaf level of the index. A split of a leaf page occurs when a row needs to be inserted into the page (because of the insert of a new row or an update of an existing row) and the target page does not have room to accommodate the row. Remember that an index maintains the data in an ordered fashion based on index key order. A row must enter a certain page based on its key value. If the target page is full, SQL Server will split the page. That is, it will allocate a new page, then move half the rows from the original page to the new
4
Query Tuning
one, then insert the new row either to the original or to the new page based on its key value, and then adjust the linked list to re ect the right logical order of the pages. The new page is not guaranteed to come right after the one that split it could be somewhere later in the le, and it could also be somewhere earlier in the le. Logical scan fragmentation is measured as the percentage of the out-of-order pages in the leaf level of the index with respect to the total number of pages. An out-of-order page is a page that appears logically after a certain page according to the linked list but before it in the le. Note one exception to the rule that an insert to a full index leaf page will cause a split: When the inserted row has a higher key than the highest key in the index, the rightmost index leaf page is not split. Instead, a new empty page is allocated, and the new row is inserted into that page. This architecture is designed to avoid costly splits and empty space that will not be reclaimed in ever-increasing indexes. On top of the leaf level of the index, the index maintains additional levels, each summarizing the level below it. Each row in a nonleaf index page points to a whole page in the level below it. The row contains two elements: the key column value of the rst row in the pointed index page and a 6-byte pointer to that page. The pointer holds the le number in the database and the page number in the le. When SQL Server builds an index, it starts from the leaf level and adds levels on top. It stops as soon as a level contains a single page, also known as the root page. SQL Server always starts with the root page when it needs to navigate to a particular key at the leaf, using an access method called an index seek, which I ll elaborate on later in the chapter. The seek operation will jump from the root to the relevant page in the next level, and it will continue jumping from one level to the next until it reaches the page containing the sought key at the leaf. Remember that all leaf pages are the same distance from the root, meaning that a seek operation will cost as many page reads as the number of levels in the index. The I/O pattern of these reads is random I/O, as opposed to sequential I/O, because naturally the pages read by a seek operation will seldom reside next to each other. In terms of our performance estimations, it is important to know the number of levels in an index because that number will be the cost of a seek operation in terms of page reads, and some execution plans invoke multiple seek operations repeatedly (for example, a Nested Loops join operator). For an existing index, you can get this number by invoking the INDEXPROPERTY function with the IndexDepth property. But for an index that you haven t created yet, you need to be familiar with the calculations that allow you to estimate the number of levels that the index will contain. The operands and steps required for calculating the number of levels in an index (call it L) are as follows (remember that these calculations apply to clustered and nonclustered indexes unless explicitly stated otherwise):
The number of rows in the table (call it num_rows)
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