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As part of its ongoing efforts to help secure its PC operating systems, Microsoft created and released a free utility called Windows Defender in February 2006 The stated purpose of Windows Defender is to protect your computer from spyware and other unwanted software (http://wwwmicrosoftcom/athome/security/spyware/software/ defaultmspx) Windows Defender is standard with all versions of the Vista operating system and is available via free download for Windows XP Service Pack 2 or later in both 32- and 64-bit versions It has the following capabilities: Spyware detection and removal Windows Defender is designed to find and remove spyware and other unwanted programs that display pop-ups, modify browser or Internet settings, or steal personal information from your PC Scheduled scanning You can schedule when you want your system to be scanned or you can run scans on demand

However, configuring CQ is not as simple as setting a 192,000-byte limit on Queue 1, a 77,000-byte limit on Queue 2, and a 499,000-byte limit on Queue 3 If you did that, the latency would be a bit high because each queue is processed in order, and CQ will not switch from one queue to another until either all bytes up to the byte limit have been transferred or the queue is empty For instance, if Queue 2 and Queue 3 were full, Queue 1 would have to wait at least 750 milliseconds (075 seconds) to be serviced an eternity in computer terms To put that time in perspective, right now, over a modem connection, a ping to yahoocom takes me approximately 150 ms That means, through 12 routers, it took less than 75 ms for my packet to reach the server and less than 75 ms for it to return Obviously, a 750-ms wait (just on the sending side; once you count the return packet, it becomes a 15-second wait) is unacceptable Tip The following example contains my own calculations for figuring out how to set byte limits in CQ Although I think this way works better for understanding what you are doing, Cisco actually has a quicker method to perform the same task, which is detailed at the following web address: http://wwwciscocom/ univercd/cc/td/doc/ product/software/ios121/ 121cgcr/qos_c/qcprt2/ qcdconmghtm#xtocid840033 A better way to set byte limits is to chop the total bandwidth into smaller time slices and use those to determine the byte limits For instance, if you chopped the bandwidth up into 1/40th of a second, Queue 1 would get a byte limit of 4800 (384,000 divided by 40 equals 4800), Queue 2 would get 1925 (77,000 divided by 40), and Queue 3 would get 12,475 (499,000 divided by 40) This arrangement would give Queue 1 a wait time of about 19 ms if both Queues 2 and 3 were full This scenario is much better, but in reality, it should probably be even lower The problem with setting the queue significantly lower than this is that, even though you set the limit in bytes, the router can send only full packets So, if your packet sizes were always 1500 bytes (the maximum size for Ethernet), when Queue 2 is processed, it will actually send 3000 bytes instead of 1925 Applying the same logic to Queue 1 and Queue 3, you end up with 6000 bytes for Queue 1 and 13,500 bytes for Queue 3, giving you a bandwidth split of 266 percent for Queue 1, 133 percent for Queue 2, and 60 percent for Queue 3, which isn't exactly desirable So, to correctly set the byte limits, you need to take a look at what the average packet size is for each queue Say, for instance, that the application used for Queue 1 has an average packet size of 1000 bytes, Queue 2 has an average packet size of 1200 bytes, and Queue 3 has an average packet size of 800 bytes Well, to figure the correct ratio out, you need to figure out what the ratio of packets is for each queue In other words, how many packets do you want sent out of Queue 1, as compared to packets from Queue 2 and Queue 3 Well, based on the bandwidth breakdown, you want 192,000 bytes allocated for Queue 1, which is 192 packets You want 77,000 bytes from Queue 2, which is about 64 packets (641, rounded down so that the result is even), and 499,000 bytes from Queue 3, which is about 624 packets (6237, rounded up so that the result is even) Tip For this calculation to work, the packet numbers must be even Now you need to figure out what the byte size is if you divide the second interval into the smallest possible pieces, and the easiest strategy is to find the highest common denominator (HCD) for all of the packet results (Break out the calculator!) In this case, the HCD for these.

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The heart of the Complexity Matrix is populated by a figure called PCE Destruction (PCED)

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