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For each customer, the running total of her first order is equal to the order total. The running total of the second order is then equal to the order total plus the first running total, and so on. This is the key to a solution that uses set-based iteration to determine the running total for the first orders of all customers, then calculate all second running totals, and so forth. This algorithm, for which the code is shown in listing 4, needs as many iterations as the highest number of orders for a single customer 28 in this case. Each individual iteration will probably be slower than a single iteration of the iterative solution, but because the number of iterations is reduced from more than 30,000 to 28, the total execution time is faster.

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Attach Script To Folder Click this item to display the snappily named Select Compiled Script File(s) Containing Folder Actions dialog box (see Figure 11-10) Select the script and click the OK button; then choose the folder in the Choose A Folder dialog box, and click the Choose button

USE AdventureWorks; SET NOCOUNT ON; DECLARE @Results TABLE (CustomerID int NOT NULL, OrderDate datetime NOT NULL, SalesOrderID int NOT NULL, TotalDue money NOT NULL, RunningTotal money NULL, Rnk int NOT NULL, PRIMARY KEY (Rnk, CustomerID)); INSERT INTO @Results (CustomerID, OrderDate, SalesOrderID, TotalDue, RunningTotal, Rnk)

SELECT CustomerID, OrderDate, SalesOrderID, TotalDue, TotalDue, RANK() OVER (PARTITION BY CustomerID ORDER BY OrderDate, SalesOrderID) FROM Sales.SalesOrderHeader; DECLARE @Rank int, @RowCount int; SET @Rank = 1; SET @RowCount = 1; WHILE @RowCount > 0 BEGIN; SET @Rank = @Rank + 1; UPDATE SET = + FROM @Results AS INNER JOIN @Results AS ON prv.CustomerID = AND prv.Rnk = WHERE nxt.Rnk = SET @RowCount = @@ROWCOUNT; END; nxt RunningTotal prv.RunningTotal nxt.TotalDue nxt prv nxt.CustomerID @Rank- 1 @Rank;

Figure 11-10

Loss: The Less, The Better! The overriding concern in any power transmission system is minimizing the loss Power wastage occurs almost entirely as heat in the transmission line conductors and dielectric, and in objects near the line Some loss can take the form of unwanted electromagnetic radiation from the line Loss also occurs in transformers Power loss in an electrical system is analogous to the loss of usable work produced by friction in a mechanical system The less of it, the better! In an ideal power transmission system, all of the power is VA power; that is, it is in the form of ac in the conductors and an alternating voltage between them It is undesirable to have power in a transmission line or transformer exist in the form of true power, because that translates into either heat loss, or radiation loss, or both The place for true power dissipation or radiation is in the load, such as electrical appliances or radio antennas Power Measurement in a Transmission Line In an ac transmission line, power is measured by placing an ac voltmeter between the conductors, and an ac ammeter in series with one of the conductors (Fig 17-9) Then the power P (in watts) is equal to the product of the rms voltage E (in volts) and the rms current I (in amperes) This technique can be used in any transmission line But this is not necessarily an indication of the true power dissipated by the load at the end of the line Recall that any transmission line has a characteristic impedance This value, Zo, depends on the diameters of the line conductors, the spacing between the conductors, and the type of dielectric material that separates the conductors If the load is a pure resistance R containing no reactance, and if R = Zo, then the power indicated by the voltmeter/ammeter scheme will be the same as the true power dissipated by the load provided that the voltmeter and ammeter are placed at the load end of the transmission line.

SELECT CustomerID, OrderDate, SalesOrderID, TotalDue, RunningTotal FROM @Results ORDER BY CustomerID, OrderDate, SalesOrderID;

The Attach Script To Folder command on the Script menu displays this dialog box for picking a script to attach to a folder