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Note that if you have a good estimation of page_density, you don t need to oor this value because the fact that a row cannot span pages (with the aforementioned exceptions) is already accounted for in the page_density value In such a case, you want to use the result number as is even if it s not an integer On the other hand, if you just estimate that page_density will be close to 100 percent, as it is in our case, omit the page_density operand from the calculation and oor the result In our case, rows_per_leaf_page amount to oor((8192 - 96) / 200) = 40 The number of pages maintained in the leaf (call it num_leaf_pages) This is a simple formula: num_rows / rows_per_leaf_page In our case, it amounts to 1,000,000 / 40 = 25,000.

Again, you can see a nicely shaped parabola in the set-based solution s graph, and now you know how to explain it: remember if traffic increases by a factor of f, the number of leaf-level rows inspected by the set-based query grows by a factor of f 2 . In other words, the algorithmic complexity of the set-based solution is O(n2).

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The average gross nonleaf row size (call it non_leaf_row_size) A nonleaf row contains the key columns of the index (in our case, only orderdate, which is 8 bytes); the 4-byte uniqui er (which exists only in a clustered index that is not unique); the page pointer, which is 6 bytes; a few additional bytes of internal overhead, which total 5 bytes in our case; and the row offset pointer at the end of the page, which is 2 bytes In our case, the gross nonleaf row size is 25 bytes The number of rows that can t in a nonleaf page (call it rows_per_non_ leaf_page) The formula to calculate this value is similar to calculating rows_per_leaf_ page.

Notice the usage of parentheses here for the input expression to the TOP option. Because SQL Server 2005 supports any self-contained expression as input, the expression must reside within parentheses. For backwardcompatibility reasons, SQL Server 2005 still supports SELECT TOP queries that use a constant without parentheses. However, it's good practice to put TOP constants in parentheses to conform to the new requirements.

For the sake of simplicity, I ll ignore the nonleaf page density factor and calculate the value as oor((page_size - header_size) / non_leaf_row_size), which in our case amounts to oor((8192 - 96) / 25) = 323 The number of levels above the leaf (call it L-1) This value is calculated with the following formula: ceiling(logrows_per_non_leaf_page(num_leaf_pages)) In our case, L-1 amounts to ceiling(log323(25000)) = 2 Obviously, you simply need to add 1 to get L, which in our case is 3..

set-based code have to do with problems where cursor code has a complexity of O(n) and set-based code has a complexity of O(n ), where n is the number of rows in the table . These are just convenient problems to demonstrate performance differences . However, you might face problems for which the solutions have different complexities . The important point is to be able to estimate complexity and performance . If you want to learn more about algorithmic complexity, see 5 in the book T-SQL Querying.

This exercise leads me to a very important point that I will rely on in my performance discussions. You can play with the formula and see that with up to about several thousand

As an example of the PERCENT option, the following query returns the most recent one percent of orders, generating the output shown in Table 7-2: SELECT TOP(1) PERCENT OrderID, CustomerID, OrderDate FROM dbo.Orders ORDER BY OrderDate DESC, OrderID DESC;

Interestingly, this is yet another type of problem where a more complete implementation of the OVER clause would have allowed for a set-based solution to perform substantially faster than the cursor one . Here s what the set-based solution would have looked like if SQL Server supported ORDER BY and ROWS in the OVER clause for aggregations:

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SELECT app, MAX(concurrent) AS mx FROM (SELECT app, SUM(event_type) OVER(PARTITION BY app ORDER BY ts, event_type ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS concurrent FROM (SELECT app, starttime AS ts, 1 AS event_type FROM dbo.Sessions UNION ALL SELECT app, endtime, -1 FROM dbo.Sessions) AS D1) AS D2 GROUP BY app;

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