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In a more general framework, articulated structures can be obtained by de ning a skeleton model, given by a tree of single-body transforms, specifying the intermediate frame locations between connected bodies (Fig. 2.11). These relationships rst de ne the location of each body with respect to its parent (where the parent of the root body is W) at zero pose, through a set of reference transforms between links l1 and l2 , which we denote by Tl1,l2. Subsequently, we can add the desired degrees of freedom to this structure by means of the local transforms Tl1,l2 , de ned by single-body homographies. For example, in Fig. 2.11 we have a structure with two-dimensional translations between palm and ngers, plus local transforms de ned by pure rotations around the z axes. The interlink matrices are shown in Fig. 2.12, and the transformation between two adjacent links ( 0, 1) is the product of three terms: 1. A reference T0,1, de ning the structural transform between links 0 and 1, at zero pose p = 0 2. A local transform T0,1 ( p) , the function of a subset of the overall pose parameter p 3. An incremental transform T0,1 ( p ) , with respect to the local T , that will be estimated during tracking and used to update the local T through one of the rules above Therefore, the overall world T of frame 1 is computed from the parent node 0 as

Service request = S, PKS {Tauth, Kr, Auth} with Auth = C, certificate, [Kr, S, PKC, Tauth]SKc where Tauth is the initial time for authentication, Kr is a one-time random number that the server will use as a symmetric key to encrypt its answer, {. . .} represents encryption with the server public key, PKS, while [. . .] represents the seal computed with the client s private key, SKC. This architecture improves speed and security. The operations of public key Kerberos are described in IETF RFC 1510 [83]. The official Web page for Kerberos is located at: http://web.mit.edu/kerberos/www/index.html. A frequently asked questions (FAQ) file on Kerberos can be consulted at the following address: ftp://athena-dist.mit.edu/pub/kerberos/KERBEROS.FAQ. Tung [84] contains a good compendium of information on Kerberos. The Swedish Institute of Computer Science is distributing a free version of Kerberos, called Heidmal. This version was written by Johan Danielsson and Assar Westerlund, and includes improvements in security protocols, such as the support of Triple DES. A commercial version is TrustBroker available from CyberSafe at http://www.cybersafe.com.

The network design assumes 100 PEs and two interfaces in each VRF The first question is how many VRFs can be expected on each PE When doing this computation, one could fall into one of two pitfalls: 1 Assume all PEs have sites from all VPNs (similar to the Paris example we saw earlier) and require support of 10000VRFs per PE 2 Compute the number of VRFs naively by dividing the number of VPNs by the number of PEs, and require 10000/100 = 100 VRFs per PE In both cases, the problem is that the number of customer sites in each VPN is completely left out from the computation, so the results do not reflect reality The number of customer interfaces affects the total number of VRFs in the network In the example network, each customer has 50 interfaces and the design decision assumes two interfaces per VRF.

The architectural process view shows the processes involved in the applications that run within that architecture. This section covers this topic in a general sense for Web applications built on an enterprise platform. Figure 2.9 is a diagram showing how various components might interact in a typical Webbased enterprise application.

= (0.4762) = 0.8942

Therefore, the customer s 50 interfaces translate to 25 separate VRFs, each on a different PE In this case, a VRF translates to a single customer site (VRF = customer_interfaces/interfaces_in_VRF) The number of VPNs in the network must be multiplied by the number of sites in each VPN to obtain the total number of VRFs, yielding 10000 25=250000 VRFs in the network (VRF_total=VPN_ total customer_VRFs) Assuming an average distribution of VRFs over PEs, each PE must service 250000/100 = 2500 VRFs (PE_VRFs = VRF_total/ PE_number) Compare this number to the two numbers computed.

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