CSE123A discussion session

2007/03/08 Ryo Sugihara

Topics

• Review– Network Layer (2,3): Route computation

• Distance vector• Link state

Where are we now?

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?

RouterA

BC

D

Prefix Link001* A0* B01* C11* C110* Ddefault B

Routing table

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

Example

R1

R2

R3

R4

R5

5

1

2

3

8

10

4

CSE(132.239.10.0/24)

1

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

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”

“CSE = 1”1

What happens at R2?

5

2

3

p1

p2

p3

“CSE = 1”

R2

Net dist... ...

CSE 6... ...

Net dist... ...

CSE inf... ...

Net dist... ...

CSE inf... ...

Net dist... ...

CSE 6... ...

iface...p1...

Routing table

• Table for each interface• One routing table

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 = 2”

“CSE = 2”

“CSE = 6”

1

What happens at R2?

5

2

3

p1

p2

p3

R2

Net dist... ...

CSE 6... ...

Net dist... ...

CSE 4... ...

Net dist... ...

CSE inf... ...

Net dist... ...

CSE 6... ...

iface...p1...

“CSE = 2”

4 p2

Routing table

In case of link failure,

• Switch to backup route• Problem: “Count-to-infinity problem” (next)

5

2

3p1

p2

p3

R2

Net dist... ...

CSE 6... ...

Net dist... ...

CSE 4... ...

Net dist... ...

CSE inf... ...

Net dist... ...

CSE 4... ...

iface...p2...

Routing table

inf

6 p1

Count-to-infinity problem

R1

2

1

3

CSE

p1

p2

p3

p4

Net distCSE 2

Net distCSE 6

Net distCSE 2

ifacep3

Routing table

Net distCSE 3

Net distCSE 5

Net distCSE 3

ifacep1

Routing table

inf

inf

5 p2

6 p4

“CSE = 5”

8

8 p4

“CSE = 8”

11

11 p2

R2

R31

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

LSP

R1

R2

R3CSE 1

5

1

LSP =

1

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

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

R1

R2

R3

R4

R5

5

1

2

3

8

10

4

CSE

LSP

LSP1

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