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Week11 lec2

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Week11 lec2

1. 1. Chapter 4 Network Layer Computer Networking: A Top Down Approach 4th edition. Jim Kurose, Keith Ross Addison-Wesley, July 2007.
2. 2. Classless Addressing •To overcome address depletion •No classes, but the address are still granted in blocks. •The size of the block( the number of addresses) varies based on the nature and size of the entity. • Household: 2 addresses •Large organization: thousands of addresses • ISP: thousands or hundreds of thousands based on the number of customers it may serve. •The number of addresses in a block must be power of two (2,4,8…..). •Classless Inter Domain Routing (CIDR)
3. 3. Mask:Slash Notation • Each mask is made of some ones from the left and followed by some 0s. • Instead of 255.255.255.224 – Mask has 27 1s – Attach this number to a classless address –x.y.z.t/n • n defines the number of bits that are same in every block.
4. 4. Subnetting Home Assignment An organization is granted a block of addresses with the beginning address 14.24.74.0/24. There are 232−24= 256 addresses in this block. The organization needs to have 11 subnets as shown below: a. two subnets, each with 64 addresses. b. two subnets, each with 32 addresses. c. three subnets, each with 16 addresses. d. four subnets, each with 4 addresses. Design the subnets.
5. 5. Internet and Routing Basic function of the Internet  To allow any two hosts to talk to each other using IP packets Routing enables data packets to find the way through the Internet Depending on the locations of the two hosts, the delivery can be  Direct, or  Indirect
6. 6. Direct delivery
7. 7. Indirect delivery
8. 8. How do hosts make routing decisions? When a host X receives a packet to be delivered to Y Host X checks whether Y is within the same subnet If yes, directly deliver the packet to host Y If no, deliver the packet to the appropriate router How can host X tell whether Y is in the same network?
9. 9. Exercise Host X with IP address 130.130.10.10 and network mask 255.255.255.128 receives the following two packets:  Packet A destined for 130.130.10.56  Packet B destined for 130.130.10.156 Determine whether they will be delivered directly or indirectly?.
10. 10. Solution Host IP address is 130.130.10.10 Subnet mask is 255.255.255.128 Network address is 130.130.10.0 Address range 130.130.10.0 to 130.130.10.127 Packet A will be delivered directly Packet B will be delivered indirectly
11. 11. CIDR and Routing Information Company X : ISP X owns: National ISP 206.0.68.0/22 206.0.64.0/18 204.188.0.0/15 209.88.232.0/21 ISP y : 209.88.237.0/24 Organization z1 : Organization z2 : 209.88.237.192/26 209.88.237.0/26
12. 12. CIDR and Routing Information Routers in National ISP do not know anything about Company X, ISP Y, or Organizations z1, z2. Company X : ISP X does not know about Organizations z1, z2. National ISP 206.0.68.0/22 ISP X owns: ISP y sends everything which matches the prefix: 206.0.64.0/18 209.88.237.192/26 to Organization z1 204.188.0.0/15 209.88.237.0/26 to Organization z2 209.88.232.0/21 ISP X sends everything which matches the prefix: 206.0.68.0/22 to Company X, 209.88.237.0/24 to ISP y National ISP sends everything which matches the prefixes 206.0.64.0/18, 204.188.0.0/15, 209.88.232.0/21 to ISP X. ISP y : 209.88.237.0/24 Organization z1 : Organization z2 : 209.88.237.192/26 209.88.237.0/26
13. 13. Routing Protocols • Define how routers exchange network information – What type of information – The format of information exchange – When to exchange – Which router to exchange information with • Examples – Routing Information Protocol (RIP) – Enhanced Interior Gateway Routing Protocol (EIGRP) – CISCO Proprietary – Open Shortest Path First (OSPF) – Border Gateway Protocol (BGP)
14. 14. Routing Algorithms Given a set of routers a routing algorithm finds a “Good” path from source router to destination router Least cost path routing algorithm local forwarding table header value output link 0100 0101 0111 1001 A graph is used to formulate routing problems A Graph G=(N,E) is a Set of N nodes and a collection E of edges Nodes in the graph represent Routers Edges represent physical links 3 2 2 1 packet’s header 0111 1 3 2
15. 15. Graph Abstraction 5 2 u v 2 1 x 3 w 3 1 5 z 1 y 2 Graph: G = (N,E) N = set of routers = { u, v, w, x, y, z } E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w,z), (y,z) }
16. 16. Graph Abstraction: costs 5 2 u v 2 1 x • c(x,x‟) = cost of link (x,x‟) 3 w 3 1 5 z 1 y 2 - e.g., c(w,z) = 5 • Cost can be •Physical length of the link •Delay etc. Cost of path (x1, x2, x3,…, xp) = c(x1,x2) + c(x2,x3) + … + c(xp-1,xp) Question: What‟s the least-cost path between u and z ? Routing algorithm: Algorithm that finds least-cost path
17. 17. Routing Algorithm Classification Global Routing Algorithm Decentralized Routing Algorithm  Computes least cost path  No node has complete     using complete global knowledge about the network. Takes connectivity between all nodes and all link costs as input. All routers have complete topology, link cost information Also called “Link State” Algorithms Used by Open Shortest Path First Protocol (OSPF)     information about the cost of all links. In the beginning knowledge of its own directly attached links. Computes least cost path by an iterative process of calculation and exchange of information. Also called Distance Vector (DV) Algorithm Used by Routing Information Protocol (RIP)
18. 18. Link-State Routing Algorithm  Network topology and link costs are known to all nodes  Each node broadcast link state packets to all other nodes in the network  Each link state packet contains the identities and cost of its attached links Dijkstra‟s Algorithm  Computes least cost paths from one node („source”) to all other nodes  Iterative: After k iterations, least cost paths to k destinations are known Notation:  D(v): Current value of least cost path from source to destination (v).  p(v): Predecessor node along path from source to v  N': Subset of nodes whose least cost path is definitively known
19. 19. Dijkstra‟s Algorithm: Example Step 0 1 2 3 4 5 N' u ux uxy uxyv uxyvw uxyvwz D(v),p(v) D(w),p(w) 2,u 5,u 2,u 4,x 2,u 3,y 3,y 5 2 u v 2 1 x D(x),p(x) 1,u D(y),p(y) ∞ 2,x D(z),p(z) ∞ ∞ 4,y 4,y 4,y Resulting forwarding table in u: 3 w 3 1 5 z 1 y destination 2 link x (u,x) y (u,x) v w z (u,v) (u,x) (u,x)
20. 20. Dijkstra‟s Algorithm 5 1 Initialization: 3 v w 2 N' = {u} 2 3 for all nodes v u 2 1 4 if v is a neighbor of u 3 1 5 then D(v) = c(u,v) x y 1 6 else D(v) = ∞ 7 8 Loop 9 find w not in N' such that D(w) is a minimum 10 add w to N' 11 update D(v) for each neighbor v of 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' 5 z 2
21. 21. Dijkstra‟s Algorithm-Example Find the shortest path from S to all nodes using Dijkstra‟s Algorithm?
22. 22. Solution Step 0 N’ s D(x), p(x) D(t),p(t) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z) ∞ 1,s 4,s ∞ ∞ ∞ ∞ 1 st ∞ 3,t 5,t ∞ 8,t 6,t 2 stu ∞ 5,t 6,u 8,t 6,t 3 stuv 8,v 6,u 6,v 6,t 4 stuvy 8,v 6,u 5 stuvyz 8,v 6,u 6 stuvyzw 8,v 7 stuvyzwx 6,t
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