what is the mathematical formula that the DUAL algorithm uses in Cisco EIGRP to calculate the shortest path?
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http://www.cisco.com/univercd/cc/td/doc/cisintwk/ito_doc/en_igrp.htm
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Input Description: A graph G = (V,E) with weighted edges.\n\nProblem: The subset of E of G of minimum weight which forms a tree on V.\n\nExcerpt from The Algorithm Design Manual: The minimum spanning tree (MST) of a graph defines the cheapest subset of edges that keeps the graph in one connected component. Telephone companies are particularly interested in minimum spanning trees, because the minimum spanning tree of a set of sites defines the wiring scheme that connects the sites using as little wire as possible. It is the mother of all network design problems. Minimum spanning trees prove important for several reasons:\n\n    * They can be computed quickly and easily, and they create a sparse subgraph that reflects a lot about the original graph.\n    * They provide a way to identify clusters in sets of points. Deleting the long edges from a minimum spanning tree leaves connected components that define natural clusters in the data set, as shown in the output figure above.\n    * They can be used to give approximate solutions to hard problems such as Steiner tree and traveling salesman.\n    * As an educational tool, minimum spanning tree algorithms provide graphic evidence that greedy algorithms can give provably optimal solutions.
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I am not sure if I got the correct meaning of your question, but I guess the answer, if you are asking the number of possible paths in such a network, is 2*n*(n-1) in the case of one-way paths and n*(n-1) when talking of two-way paths.\n\nThe reason is that, in the one-way case, each of the n nodes may be linked to any of the resting (n-1) nodes in one of the two possible ways, making 2*n*(n-1). In the case of two way paths, only one path linking each of the nodes exists thus the n*(n-1) number.\n\nYou may be asking the number of possible networks, which would be a more interesting problem. If this is so, please ask another question.
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u spelt it wrong :-(\nit should have been Dijikstra\n\nSTDMETHODIMP CDijkstra::ShortestPath(long node1, long node2)\n{\n    ReleaseGraph();\n    // init d and pi\n    InitializeSource(g, g.m_nodes[node1-1]);\n    // Set S empty\n    VTYPE_NODE S;\n    // Put nodes in Q\n    VTYPE_NODE Q;\n    VTYPE_NODE::iterator kl;\n    for(kl=g.m_nodes.begin(); kl<g.m_nodes.end(); kl++)\n    {\n        CNode node = (*kl).Copy();\n        Q.push_back(node);\n    }\n    // Algorithm\n    while(Q.size())\n    {\n        CNode nod = ExtractMin(Q); // the minim value \n                // for the shortest path up to this step\n        S.push_back(nod);\n        // each vertex v which is a neighbour of u\n        VTYPE_NODE::iterator kl;\n        for(kl=g.m_nodes.begin(); kl<g.m_nodes.end(); kl++)\n        {\n            if(ExistEdge(nod, (*kl)))\n            {\n                bool gasit = false;\n                VTYPE_NODE::iterator kll;\n                for(kll=Q.begin(); kll<Q.end(); kll++)\n                {\n                    if((*kll).m_NodeNr == (*kl).m_NodeNr)\n                    gasit = true;\n                }\n                if(gasit)\n                    Relax(nod, (*kl), GetEdgeVal(nod, (*kl)));\n            }\n        }\n    }\n    RefreshDone(node1, node2);\n    return S_OK;\n}
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You first define an array NodesList with index 0 as origin (C standard) containing all your nodes. The variable NodesNumber is the number of nodes. The key element is to define the operators S (for successor) and P (for predecessor) that will make the array behave as a circular list.\nS(NodesList[ i ]) = NodesList[ i + 1] if i < NodesNumber - 2\nand for the last node\nS(NodesList[ NodesNumber - 1 ]) = NodesList[ 0 ] \nSimilarly for P\nP(NodesList[ i ]) = NodesList[ i - 1] if i > 0\nand for the first node (origin)\nP(NodesList[ 0 ]) = NodesList[  NodesNumber - 1 ] \nThe algorithm has 2 parts\n// forward\nStartNode := NodesList[ NodesNumber - N  ]\nfor I := 1 to N\n  begin\n     StartNode := S(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := S(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\n// backward\nStartNode := NodesList[ N  ]\nfor I := 1 to N\n  begin\n     StartNode := P(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := P(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\nIn the second part the starting point is changed and S() are changed by P()
928288
It would depend on the context, however what is most likely meant by "Dual Router" is a pair (two) routers for which one is a backup.\n\nParticularly where potentially unstable WAN links are concerned, companies often build a backup router that would take over the wide area connection in the event that the primary (active) router or WAN link fails. \n\nTo support the above Active/Standby configuration, a function known as HSRP (Hot Standby Router Protocol), which is proporietry to Cisco routers may be used. \n\nThe Hot Standby Router Protocol (HSRP) provides network redundancy for IP networks, ensuring that user traffic immediately and transparently recovers from first hop failures in network edge devices or access circuits. \n\nAn alternative to HSRP is a function called VRRP (Virtual Redundant Router Protocol). VRRP is similar to HSRP but non-cisco proprietry (can run on multiple vendors equipment).\n\nTo sum up, deploying dual routers is all about resilience and fault tolerance when building high-availability networks.
776599
http://www.purplemath.com/modules/fcns2.htm
234352
Try this link. Tons of that kind of info.\nhttp://www.dslreports.com/faq/578
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Split horizon is a method of preventing a routing loop in a network. The basic principle is simple: Information about the routing for a particular packet is never sent back in the direction from which it was received.\n\nSplit horizon can be achieved by means of a technique called poison reverse. This is the equivalent of route poisoning all possible reverse paths - that is, informing all routers that the path back to the originating node for a particular packet has an infinite metric. Split horizon with poison reverse is more effective than simple split horizon in networks with multiple routing paths, although it affords no improvement over simple split horizon in networks with only one routing path. \n\nRIP for IP, like most distance vector routing protocols, announces its routes in an unsynchronized and unacknowledged manner. This can lead to convergence problems. However, you can enable modifications to the announcement algorithms to reduce convergence time in most situations.\n\nCount-to-Infinity Problem\nThe classic distance vector convergence problem is known as the count-to-infinity problem and is a direct result of the asynchronous announcement scheme. When RIP for IP routers add routes to their routing table, based on routes advertised by other routers, they keep only the best route in the routing table and they update a lower cost route with a higher cost route only if is being announced by the same source as the current lower cost route. In certain situations, as illustrated in Figures 3.1 through 3.5, this causes the count-to-infinity problem.\n\nAssume that the internetwork in Figure 3.1 has converged. For simplicity, assume that the announcements sent by Router 1 on Network 1 and Router 2 on Network 3 are not included.\n\n \nFigure 3.1 Converged Internetwork\n\nNow assume that the link from Router 2 to Network 3 fails and is sensed by Router 2. As shown in Figure 3.2, Router 2 changes the hop count for the route to Network 3 to indicate that it is unreachable, an infinite distance away. For RIP for IP, infinity is 16.\n\n \nFigure 3.2 Link to Network 3 Fails\n\nHowever, before Router 2 can advertise the new hop count to Network 3 in a scheduled announcement, it receives an announcement from Router 1. The Router 1 announcement contains a route to Network 3 which is two hops away. Because two hops away is a better route than 16 hops, Router 2 updates its routing table entry for Network 3, changing it from 16 hops to three hops, as shown in Figure 3.3.\n\n \nFigure 3.3 Router 2 After Receiving Announcement From Router 1\n\nWhen Router 2 announces its new routes, Router 1 notes that Network 3 is available three hops away through Router 2. Because the route to Network 3 on Router 1 was originally learned from Router 2, Router 1 updates its route to Network 3 to four hops. (See Figure 3.4.)\n\n \nFigure 3.4 Router 1 After Receiving Announcement From Router 2\n\nWhen Router 1 announces its new routes, Router 2 notes that Network 3 is available four hops away through Router 1. Because the route to Network 3 on Router 2 was originally learned from Router 1, Router 2 updates its route to Network 3 to five hops. (See Figure 3.5.)\n\n \nFigure 3.5 Router 2 After Receiving Another Announcement from Router 1\n\nThe two routers continue to announce routes to Network 3 with higher and higher hop counts until infinity (16) is reached. Then, Network 3 is considered unreachable and the route to Network 3 is eventually timed out of the routing table. This is known as the count-to-infinity problem.\n\nThe count-to-infinity problem is one of the reasons why the maximum hop count of RIP for IP internetworks is set to 15 (16 for unreachable). Higher maximum hop count values would make the convergence time longer when count-to-infinity occurs. Also note that during the count-to-infinity in the previous example, the route from Router 1 to Network 3 is through Router 2. The route from Router 2 to Network 3 is through Router 1. A routing l
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You really need to clarify your question. \n\nI could give you the min and max path weight in a undirected graph between two vertices, but in a weighted, undirected graph, an edge's weight is an edge's weight. There's no formula needed, unless I really, really misunderstood your question.
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Well, physical connection happens through satellites and trans-continental cabling.  However, most connection is done virtually through routing from one internet location to another.\n\nIf you go to a DOS command prompt, and type in "tracert xxx.xxx.xxx.xxx" (the x's represent the ip address), then you can see how many "hops" across the internet that are being taken to get to the destination.\n\nNOTE: You can do a tracert (Trace Route) to a domain name as well.
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U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
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Imagine that a travelling salesman has to hit, say, 100 cities in the USA, and he knows road distances between them, what's the minimum distance he has to travel to do all of them and come back home?  There are numerous variations on this problem, but this kind of a problem usually defies mathematical analysis.  It's called a "NP-hard' problem, which is a measure of the time it takes a computer to determine the optimum path.
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It's a computer term:\n\nhttp://www.cisco.com/en/US/products/hw/optical/ps2011/products_implementation_design_guide_chapter09186a00802212a7.html
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OSBF and EIGRP r routing protocols \nOSPF : open shrted path first .... it is a hirarcial routing protocols and it is used to run za intenet ... \nit is partioning za world to OSPF areas ... and every OSBF area seperate the routing traffic to increase za effichioncy...itis a Link state roting protocol\n\nEIGRP : it is a routing protocol devolped by CISCO and is za best to run on za cisco devices ... it calculates za metric using a metric equation its tech is not a link state or a distance vector .. it is a new tech
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Below you will find the specifications for the AES (Rijndael) and Three-DES algorithms.  \n\nAs for the writing of the programs themselves, I'm almost done with my CSE degree--sounds like your pain is just beginning :-D
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http://www.networkphysics.com/products/literature/case_studies/mpls.shtml