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93 NETWORK BOUNDARY TRAFFIC CLASSIFICATION AND CONDITIONING Traffic conditioners are various QoS functions needed on a network boundary The boundary routers classify or mark traffic by setting the DSCP field and monitor incoming network traffic for profile compliance The DSCP field indicates what treatment the packet should receive in a Diffserv domain The QoS functions can be those of packet classifier, DSCP marker, or traffic metering function, with either shaper or dropper action Figure 93 presents the logical structure of traffic classification and conditioning functions The classification of packets can be done in one of two ways, depending on the connectivity of the boundary router Some boundary routers are connected to customer networks, and some boundary routers are connected to other network operators ISPs..

kinter is a package used for building a graphical user interface (GUI) in Python. It runs on many operating systems, including UNIX, Windows, and Macintosh. Tkinter is the de-facto standard GUI library for Python, and is often bundled with it. Tkinter is very easy to use; it is built on top of the high-level scripting language Tcl.

A substance consisting of molecules characterized by the repetition (neglecting ends, branch junctions, and other minor irregularities) of one or more types of monomeric units Number of electrons needed to ll out the outermost shell of an atom Example: a carbon atom has 6 electrons, with an electron shell con guration of 1s22s22p2 Hence, carbon has a valence of 4, since 4 electrons can be accepted to ll the 2p orbital Some exceptions exist; therefore, the more general de nition of valence is the number of electrons with which a given atom generally bonds or number of bonds an atom forms (eg, iron may have a valence of 2 or a valence of 3)..

A boundary router that is connected to a customer network uses six fields in an incoming IP packet to determine the PHB that the packet should receive in the core network These six fields are the IP source address, IP destination address, transport protocol, Diffserv field in the incoming packet header, source port in the transport headers, and destination port in the transport header A rule that maps a packet to a PHB does not need to specify all six fields We refer to such a rule as a classification rule When a classification rule does not specify any value for a field, that field s value is not used for the purpose of classification Boundary routers could use just one field in the incoming IP packet to determine the PHB for their network This field could be the Diffserv field contained in the incoming packet.

A boundary router would simply change the Diffserv field to some other value that corresponds to a specific PHB at the core routers This type of classification would be the one expected at the exchange points of other ISPs The neighboring ISP domain might have been using a different set of PHBs or might use different Diffserv field values to represent the same PHB In addition to the classification process, the access routers should change the Diffserv field value so that it corresponds to the correct PHB in the core network It can also exert rate control or shaping on all the packets that have.

Since it was believed by many, including most of the participants in the test, that the diagnosis of AIH was an all or none question, it was also desirable to present the results in some conventional indicators that were more familiar to clinicians and were based on a yes/no type of response The scores were placed into two categories of 1 5 and 6 10 that dichotomized the results into absence/presence of AIH The sensitivity/speci city pair and positive/negative predictive values were calculated accordingly In addition the frequency by which the use of CAD resulted in a change of diagnosis during the experiment, as opposed to mere change in con dence of one particular diagnosis or another, was examined The diagnosis of absence/presence of AIH for individual case was determined based on the aforementioned method of dichotomizing the score ratings.

The eigenvalues of A are 0 = 1 , and 1 = 3 . These eigenvalues are both 2 2 negative, and so the solution to (10.108) happens to be stable. In fact, the exact solution can be shown to be x(t) = 3 e t/2 + 7 e 3t/2 , 4 4 y(t) = 9 e t/2 + 7 e 3t/2 2 2 (10.109a) (10.109b)

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