1. CSE123A discussion session 2007/03/08 Ryo Sugihara

2. Topics Review –Network Layer (2,3): Route computation Distance vector Link state

3. Where are we now?

4. Route computation Router needs to know “which port to forward” –Done by “routing table” But how to make routing table? –By exchanging routing information –What kind of information? Router A B C D PrefixLink 001*A 0*B 01*C 11*C 110*D defaultB Routing table

5. Distance vector & Link state Two ways to exchange routing info –Distance vector Exchange “distance to a destination” Each router knows the next hop for each dest –One hop of the stortest path to the dest –Link state Exchange “my neighbors (and costs)” Each router knows whole network topology –compute the shortest paths by Dijkstra’s algorithm

6. Example R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE (132.239.10.0/24) 1

7. General strategy R1 (somehow) knows CSE is distance one from itself R1 tells the neighbors (=R2, R3) about it –DV: “CSE is distance one from me” –LS: “I have link with CSE (d=1), R2 (d=5), and R3 (d=1)” R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE 1

8. Distance vector R1 tells “CSE is distance one from me!” –to R2 and R3 R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE “CSE = 1” 1

9. What happens at R2? 5 2 3 p1 p2 p3 “CSE = 1” R2 Netdist... CSE6... Netdist... CSEinf... Netdist... CSEinf... Netdist... CSE6... iface... p1... Routing table Table for each interface One routing table

10. Distance vector Router forwards distance info to others when its routing table is updated R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE “CSE = 6” “CSE = 2” “CSE = 6” 1

11. What happens at R2? 5 2 3 p1 p2 p3 R2 Netdist... CSE6... Netdist... CSE4... Netdist... CSEinf... Netdist... CSE6... iface... p1... “CSE = 2” 4p2 Routing table

12. In case of link failure, Switch to backup route Problem: “Count-to-infinity problem” (next) 5 2 3 p1 p2 p3 R2 Netdist... CSE6... Netdist... CSE4... Netdist... CSEinf... Netdist... CSE4... iface... p2... Routing table inf 6 p1

13. Count-to-infinity problem R1 2 1 3 CSE p1 p2 p3 p4 Netdist CSE2 Netdist CSE6 Netdist CSE2 iface p3 Routing table Netdist CSE3 Netdist CSE5 Netdist CSE3 iface p1 Routing table inf 5 p26 p4 “CSE = 5” 8 8 p4 “CSE = 8” 11 11 p2 R2 R3 1

14. Link state R1 broadcasts “I have link with CSE (d=1), R2 (d=5), and R3 (d=1)” –It will be flooded to the entire network R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE LSP R1 R2 R3 CSE 1 5 1 LSP = 1

15. Link state Each router has the entire graph –In DV, it only knew next hop for each dest. Each router computes next hop for each dest. –By Dijkstra’s algorithm R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE 1

16. Intelligent flooding Without “intelligence”, LSP may loop Solution: add “sequence number” to LSP –If a router received a “newer” LSP, it forwards –Otherwise, it discards Other issues: –Subtle problems: “source jumping”, “aging”  lecture notes –How Dijkstra’s algorithm works  http://en.wikipedia.org/wiki/Dijkstra's_algorithm http://en.wikipedia.org/wiki/Dijkstra's_algorithm R1 R2 R3 R4 R5 5 1 2 3 8 10 4 CSE LSP 1

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