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the most common form of nontransparent load balancing. This means that you are balancing the load between multiple hubs, but there is some knowledge of the data or work going into the balancing decisions. You can add hubs transparently when you have the special situation where all hubs are equal. This might be the case on a Web server for example, where all the servers might have the same Web site. In this case, you may be able to add hubs and load balance across the set. When hubs are equivalent, you can do transparent load balancing, which means that you can just add hubs all day long to fix the problem. When all of your hubs are equal, you throw hardware at the problem, which is perhaps the easiest solution. It may or may not cost more to use this technique than the other, but in a lot of situations, it can be done with little rearchitecting. Where redesigning costs time, hardware costs money. If the redesigning is quick, the time is less costly than the hardware. If the redesign takes a long time, it may exceed the cost of new hardware. Neither throwing hardware at a problem nor adding hubs has to be purely reactionary. These fixes can be part of the original plans. For example, the development machine might be a single CPU PC with the express plan to make it a multiple process PC at deployment. The refactorings to apply to this AntiPattern are Partition Data and Work, Plan for Scaling, and Throw Hardware at the Problem, all found in this chapter.

One-sided t. Two-sided t. b= 1.1 1.2 1.3 1.4 1.5 0.5 9 8 8 7 7 6 6 6 6 5 8 8 8 7 7 6 5 5 5 5 7 7 6 6 6 6 6 6 6 6 6 5 7 7 6 6 6 6 6 5 12 11 10 10 9 10 9 8 8 7 9 8 7 7 7 7 7 7 6 6 5 10 9 8 8 7 8 7 7 6 6 7 6 6 6 5 6 5 8 8 7 7 6 6 6 5 6 6 6 5 6 5 0.01 21 18 16 14 13 0.05 16 14 13 11 10 0.5 8 7 6 6 6 0.01 18 15 14 12 11 0.5 6 5 0.01 15 13 11 10 9 0.05 11 10 8 8 7 8 8 8 7 7 7 6 6 6 6 5

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N OT E Depending on the machine settings and the path settings, the message may differ from that displayed in Figure 12.4.

using the Cauchy Schwarz inequality and the continuity of f ( ) since is in ( L1 . Summing up, (10.28) and (10.29) imply that lim 0 g ) (0) = f ( ) (0), and we are done. Theorem 10.9. Let X be a Gaussian process satisfying the hypotheses of Proposition 10.8 and F a function in L2 ( (x)dx) with Hermite rank 1. Then, as t + , 1 t Tt = t

In a similar way, if j4 < j1 = j2 = j3 (and similarly, if any one of the indices is strictly smaller than the others and these are all equal), the corresponding term vanishes since in this case

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