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## UNIT V INTER-NETWORKING

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**OVERVIEW**• Principles – Repeaters – Bridges Routers – Gateways – other devices • Routing algorithms – Distance Vector routing – Link state routing • TCP/IP - Protocol structure -Internet Protocol - Transmission control Protocol - Applications**Repeaters**• Repeater • Extends distance limitation on networks (both voice and data) • Filters noise • Regenerates signals • For twisted pair wire, repeaters are placed every 100 meters**Repeaters**• Repeaters are very simple devices. They allow a cabling system to extend beyond its maximum allowed length by amplifying the network voltages so they travel farther. • Repeaters are nothing more than amplifiers and, as such, are very inexpensive. • Repeaters can only be used to regenerate signals between similar network segments. • For example, we can extend an Ethernet 10Base2 network to 400 meters with a repeater. But can’t connect an Ethernet and Token Ring network together with one. • The main disadvantage to repeaters is that they just amplify signals. These signals not only include the network signals, but any noise on the wire as well. • Eventually, if you use enough repeaters, you could possibly drown out the signal with the amplified noise. For this reason, repeaters are used only as a temporary fix.**Bridges,Routers**• Bridge • Connects two LANs using same protocol • Single path between LANs • Minimal sophistication • Router • Connects multiple LANs using same protocol • Choice of paths between LANs • Mainstay of internetworking**Bridges**• They join similar topologies and are used to divide network segments. • For example, with 200 people on one Ethernet segment, the performance will be mediocre, because of the design of Ethernet and the number of workstations that are fighting to transmit. If you divide the segment into two segments of 100 workstations each, the traffic will be much lower on either side and performance will increase. • If it is aware of the destination address, it is able to forward packets; otherwise a bridge will forward the packets to all segments. They are more intelligent than repeaters but are unable to move data across multiple networks simultaneously. • Unlike repeaters, bridges can filter out noise. • The main disadvantage to bridges is that they can’t connect dissimilar network types or perform intelligent path selection. For that function, you would need a router.**Routers**• Routers are highly intelligent devices that connect multiple network types and determine the best path for sending data. • The advantage of using a router over a bridge is that routers can determine the best path that data can take to get to its destination. • Like bridges, they can segment large networks and can filter out noise. • However, they are slower than bridges because they are more intelligent devices; as such, they analyze every packet, causing packet-forwarding delays. Because of this intelligence, they are also more expensive. • Routers are normally used to connect one LAN to another. • Typically, when a WAN is set up, there will be at least two routers used.**Switch**• Connects multiple LAN segments using the same protocol • Connections may use twisted pair, coaxial cable, or fiber optics wiring • Faster than bridges • Enables simultaneous communication between multiple network segments**Switch**• A network switch is a computer networking device that connects network segments. • Low-end network switches appear nearly identical to network hubs, but a switch contains more "intelligence" (and a slightly higher price tag) than a network hub. • Network switches are capable of inspecting data packets as they are received, determining the source and destination device of that packet, and forwarding it appropriately. • By delivering each message only to the connected device it was intended for, a network switch conserves network bandwidth and offers generally better performance than a hub. • A vital difference between a hub and a switch is that all the nodes connected to a hub share the bandwidth among themselves, while a device connected to a switch port has the full bandwidth all to itself. • For example, if 10 nodes are communicating using a hub on a 10-Mbps network, then each node may only get a portion of the 10 Mbps if other nodes on the hub want to communicate as well. . • But with a switch, each node could possibly communicate at the full 10 Mbps.**Gateway**• Gateway • Connects multiple LANs using any protocol • Very sophisticated • Supports today’s internet by providing access points to several networks • Hub • Connects nodes to a network • Sometimes acts as repeater**destination router**5 first-hop routerdefault router 3 B D 2 5 2 3 1 F A 1 1 2 C E least-cost path source router Routing Algorithms • Determining the path(route) destination host source host**Routing Algorithm**• The algorithm used for the routing protocols defines the following processes: • Mechanism for sending and receiving routing information. • Mechanism for calculating the best paths and installing routes in the routing table. • Mechanism for detecting and reacting to topology changes.**2 main classes:**Centralized all routers have complete topology, link cost info “link state” algorithms Distributed: Each router knows link costs to neighbor routers only “distance vector” algorithms Routing Algorithm classification**Dijkstra’s algorithm**Link costs known to all nodes computes least cost paths from one node (‘source”) to all other nodes gives forwarding table for that node iterative: after k iterations, know least cost path to k dest.’s A Link-State Routing Algorithm**5**3 5 2 2 1 3 1 2 1 x z y u w v Notation: • c(x,y): link cost from node x to y; = ∞ if not direct neighbors • D(v): current value of cost of path from source to dest. v Dijkstra’s Algorithm 1 Initialization: 2 N' = {u} 3 for all nodes v 4 if v adjacent to u 5 then D(v) = c(u,v) 6 else D(v) = ∞ 7 8 Loop 9 find w not in N's.t. D(w) is a minimum 10 add w to N' 11 update D(v) for all v adjacent to w and not in N' : 12 D(v) = min( D(v), D(w) + c(w,v) ) 13 /* new cost to v is either old cost to v or known 14 shortest path cost to w plus cost from w to v */ 15 until all nodes in N'**x**z w u y v destination link (u,v) v (u,x) x y (u,x) (u,x) w z (u,x) Dijkstra’s algorithm: example (2) Resulting shortest-path tree from u: Resulting forwarding table in u:**Distributed: Distance Vector**• To find D, node S asks each neighbor X • How far X is from D • X asks its neighbors … comes back and says C(X,D) • Node S deduces C(S,D) = C(S,X) + C(X,D) • S chooses neighbor Xi that provides min C(S,D) • Later, Xj may find better route to D • Xj advertizes C(Xj,D) • All nodes update their cost to D if new min found**Distance Vector Algorithm**Bellman-Ford Equation (dynamic programming) Define dx(y) := cost of least-cost path from x to y Then dx(y) = min {c(x,v) + dv(y) } where min is taken over all neighbors v of x v1 y x v2 v**5**3 5 2 2 1 3 1 2 1 x z w u y v Bellman-Ford example Clearly, dv(z) = 5, dx(z) = 3, dw(z) = 3 B-F equation says: du(z) = min { c(u,v) + dv(z), c(u,x) + dx(z), c(u,w) + dw(z) } = min {2 + 5, 1 + 3, 5 + 3} = 4 Node that achieves minimum is next hop in shortest path ➜ forwarding table**1**4 1 50 x z y Distance Vector: link cost changes Link cost changes: • if DV changes, notify neighbors At time t0, y detects the link-cost change, updates its DV, and informs its neighbors. At time t1, z receives the update from y and updates its table. It computes a new least cost to x and sends its neighbors its DV. At time t2, y receives z’s update and updates its distance table. y’s least costs do not change and hence y does not send any message to z. When can it get complicated ?**60**4 1 50 x z y Distance Vector: link cost changes Link cost changes: • Y thinks Z’s best cost is 5 • Thus C(y,x) = 5 + 1 = 6 • Announces this cost • Z thinks C(z,x) = 6 + 1 … Poissoned reverse: • If Z routes through Y to get to X : • Z tells Y its (Z’s) distance to X is infinite (so Y won’t route to X via Z) • will this completely solve count to infinity problem?**Message complexity**LS: with n nodes, E links, O(nE) msgs sent DV: exchange between neighbors only convergence time varies Speed of Convergence LS: O(n2) algorithm requires O(nE) msgs may have oscillations DV: convergence time varies may be routing loops count-to-infinity problem Robustness: what happens if router malfunctions? LS: node can advertise incorrect link cost each node computes only its own table DV: DV node can advertise incorrect path cost each node’s table used by others error propagate thru network Comparison of LS and DV algorithms**Bellman–Ford algorithm**• The Bellman–Ford algorithm computes single-source shortest paths in a weighted digraph.Forgraphs with only non-negative edge weights, the faster Dijkstra'salgorithm also solves the problem. • Thus, Bellman–Ford is used primarily for graphs with negative edge weights. The algorithm is named after its developers, Richard Bellman and Lester Ford, Jr**Algorithm**• Bellman–Ford is based on dynamic programming approach. In its basic structure it is similar to Dijkstra's Algorithm, but instead of greedily selecting the minimum-weight node not yet processed to relax, it simply relaxes all the edges, and does this |V | − 1 times, where |V | is the number of vertices in the graph. • The repetitions allow minimum distances to propagate accurately throughout the graph, since, in the absence of negative cycles, the shortest path can visit each node at most only once. Unlike the greedy approach, which depends on certain structural assumptions derived from positive weights, this straightforward approach extends to the general case.**Finding negative cycles**• When the algorithm is used to find shortest paths, the existence of negative cycles is a problem, preventing the algorithm from finding a correct answer. However, since it terminates upon finding a negative cycle, the Bellman-Ford algorithm can be used for applications in which this is the target to be sought - for example in cycle-cancelling techniques in network flow analysis.**Disadvantages**• It does not scale well. • Changes in network topology are not reflected quickly since updates are spread node-by-node. • Count to infinity (if link or node failures render a node unreachable from some set of other nodes, those nodes may spend forever gradually increasing their estimates of the distance to it, and in the meantime there may be routing loops).