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INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 2, NULL, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 3, NULL, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 4, NULL, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 5, NULL, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 6, 1, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 7, 1, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(10, 1, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(14, 1, 'mL', 230.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 6, 2, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 7, 2, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(10, 2, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(14, 2, 'mL', 205.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(11, 2, 'mL', 25.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 6, 3, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 7, 3, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(11, 3, 'mL', 225.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(12, 3, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 9, 4, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(12, 4, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES( 9, 5, 'EA', 1.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(12, 5, 'EA', 2.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(13, 10, 'g' , 5.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(14, 10, 'mL', 20.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(14, 12, 'mL', 20.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(16, 12, 'g' , 15.00); INSERT INTO dbo.BOM(partid, assemblyid, VALUES(17, 16, 'g' , 15.00);

Simply use the last day of the month calculation as the reference date in the last weekday calculation . For example, the following expression returns the last occurrence of a Monday in the reference month:

This query generates the following output:

SELECT DATEADD(day, DATEDIFF(day, '19000101', -- last day of month DATEADD(month, DATEDIFF(month, '18991231', SYSDATETIME()), '18991231') ) /7*7, '19000101');

Finally, you create a derived table out of the previous step s query, and you keep only pairs where nxt cur is greater than one. You add one to cur to get the actual start of the gap and subtract one from nxt to get the actual end of the gap:

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unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty) unit, qty)

To address a different weekday, substitute the anchor date in the last weekday calculation to the applicable one . For example, the following expression returns the last occurrence of a Tuesday in the reference month:

SELECT cur + 1 AS start_range, nxt - 1 AS end_range FROM (SELECT seqval AS cur, (SELECT MIN(B.seqval) FROM dbo.NumSeq AS B WHERE B.seqval > A.seqval) AS nxt FROM dbo.NumSeq AS A) AS D WHERE nxt - cur > 1;

Note that this solution got rid of the point before in nity with no special treatment because the nxt value for it was NULL. Run this solution against BigNumSeq to test its performance:

SELECT DATEADD(day, DATEDIFF(day, '19000102', -- last day of month DATEADD(month, DATEDIFF(month, '18991231', SYSDATETIME()), '18991231') ) /7*7, '19000102');

BOM represents a directed acyclic graph (DAG). It holds the parent and child node IDs in the assemblyid and partid attributes, respectively. BOM also represents a weighted graph, where a

SELECT cur + 1 AS start_range, nxt - 1 AS end_range FROM (SELECT seqval AS cur, (SELECT MIN(B.seqval) FROM dbo.BigNumSeq AS B WHERE B.seqval > A.seqval) AS nxt FROM dbo.BigNumSeq AS A) AS D WHERE nxt - cur > 1;

In a manner very similar to calculating the first and last occurrence of a weekday in the reference month, you can calculate the first and last occurrence of a weekday in the

The plan for this query is shown in Figure 6-12.

weight/number is associated with each edge. In our case, that weight is the qty attribute that holds the quantity of the part within the assembly (assembly of sub-parts). The unit attribute holds the unit of the qty (EA for each, g for gram, mL for milliliter, and so on).

reference year . Simply substitute the first or last month day calculation with the first or last year day calculation . Following are a few examples . First occurrence of a Monday in the reference year:

6

This solution is signi cantly slower than the previous one. It ran on my system for 48 seconds and incurred 31,875,478 logical reads. The reason for the large number of reads becomes apparent when you examine the plan. The plan shows a full scan of the index to retrieve all sequence values (close to 10,000,000 of them), and per each row, an index seek operation is used to return the next value. With a cost of 3 reads per seek (for the 3 levels of the index), you get about 30,000,000 reads for all seeks. To apply the solution to a temporal sequence, use the DATEADD function to add or subtract an interval, and the DATEDIFF function to calculate the difference between cur and nxt:

SELECT DATEADD(day, DATEDIFF(day, '19000101', -- first day of year DATEADD(year, DATEDIFF(year, '19000101', SYSDATETIME()), '19000101') -1) /7*7 + 7, '19000101');

The Road System that I will use is that of several major cities in the United States, and it is depicted graphically in Figure 9-4.

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