I have a table, Sheet1$ that contains 616 records. I have another table, Rates$ that contains 47880 records. Rates contains a response rate for a given record in the sheet for 90 days from a mailing date. Within all 90 days of a records Rates relation the total response is ALWAYS 1 (100%)
Example:
Sheet1$: Record 1, 1000 QTY, 5% Response, Mail 1/1/2009
Rates$: Record 1, Day 1, 2% Response
Record 1, Day 2, 3% Response
Record 1, Day 90, 1% Response
Record N, Day N, N Response
So in that, I've written a view that takes these tables and joins them to the right on the rates to expand the data so I can perform some math to get a return per day for any given record.
SELECT s.[Mail Date] + r.Day as Mail_Date, s.Quantity * s.[Expected Response Rate] * r.Response as Pieces, s.[Bounce Back Card], s.Customer, s.[Point of Entry]
FROM Sheet1$ as s
RIGHT OUTER JOIN Rates$ as r
ON s.[Appeal Code] = r.Appeal
WHERE s.[Mail Date] IS NOT NULL
AND s.Quantity <> 0
AND s.[Expected Response Rate] <> 0
AND s.Quantity IS NOT NULL
AND s.[Expected Response Rate] IS NOT NULL);
So I save this as a view called Test_Results. Using SQL Server Management Studio I run this query and get a result of 211,140 records. Elapsed time was 4.121 seconds, Est. Subtree Cost was 0.751.
Now I run a query against this view to aggregate a piece count on each day.
SELECT Mail_Date, SUM(Pieces) AS Piececount
FROM Test_Results
GROUP BY Mail_Date
That returns 773 rows and it only took 0.452 seconds to execute! 1.458 Est. Subtree Cost.
My question is, with a higher estimate how did this execute SO much faster than the original view itself?! I would assume a piece might be that its returning rows to management studio. If that is the case, how would I go about viewing the true cost of this query without having to account for the return feedback?