OSPF Version 2

RFC 2328 §10.10 Figure 14 shows an example of an adjacency forming. Routers RT1 and RT2 are both connected to a broadcast network. It is assumed that RT2 is the Designated Router for the network, and that RT2 has a higher Router ID than Router RT1.

1. 1. RT1 → RT2: Hello(DR=0,seen=0)

   RT1's interface comes up and it sends Hellos without knowing any Designated Router or neighbor. RT2 hears this Hello and moves RT1 to Init.

   RT1: — · RT2: Init

   RFC 2328 §10.10 Hello(DR=0,seen=0) ------------------------------> Hello (DR=RT2,seen=RT1,...) Init

2. 2. RT2 → RT1: Hello (DR=RT2,seen=RT1,...)

   RT2's next Hello says it is the Designated Router and that it has heard RT1. RT1 now starts to bring up the adjacency and goes to ExStart.

   RT1: ExStart · RT2: —

   RFC 2328 §10.10 Hello (DR=RT2,seen=RT1,...) Init <------------------------------ ExStart D-D (Seq=x,I,M,Master)

3. 3. RT1 → RT2: D-D (Seq=x,I,M,Master)

   RT1 sends an empty Database Description (D-D) packet claiming to be master, with the I, M and MS bits set.

   RT1: — · RT2: ExStart

   RFC 2328 §10.10 ExStart D-D (Seq=x,I,M,Master) ------------------------------> D-D (Seq=y,I,M,Master) ExStart

4. 4. RT2 → RT1: D-D (Seq=y,I,M,Master)

   RT2 also claims master. RT2 has the higher Router ID, so RT1 becomes slave, adopts RT2's DD sequence number y and moves to Exchange.

   RT1: Exchange · RT2: —

   RFC 2328 §10.10 D-D (Seq=y,I,M,Master) ExStart <------------------------------ Exchange D-D (Seq=y,M,Slave)

5. 5. RT1 → RT2: D-D (Seq=y,M,Slave)

   RT1 answers as slave using sequence number y. RT2 moves to Exchange.

   RT1: — · RT2: Exchange

   RFC 2328 §10.10 Exchange D-D (Seq=y,M,Slave) ------------------------------> D-D (Seq=y+1,M,Master) Exchange

6. 6. RT2 → RT1: D-D (Seq=y+1,M,Master)

   The master (RT2) polls with the next sequence number, describing more of its database.

   RFC 2328 §10.10 D-D (Seq=y+1,M,Master) Exchange <------------------------------

7. 7. RT1 → RT2: D-D (Seq=y+1,M,Slave)

   The slave (RT1) responds with the same sequence number. Polls and responses continue like this.

   RFC 2328 §10.10 D-D (Seq=y+1,M,Slave) ------------------------------>

8. 8. RT2 → RT1: D-D (Seq=y+n, Master)

   The last poll from the master has the M (more) bit off.

   RFC 2328 §10.10 D-D (Seq=y+n, Master) <------------------------------

9. 9. RT1 → RT2: D-D (Seq=y+n, Slave)

   The exchange ends when both the poll and its response have the M bit off. RT2's database is already up to date, so it goes straight to Full. RT1 still needs some LSAs, so it goes to Loading.

   RT1: Loading · RT2: Full

   RFC 2328 §10.10 D-D (Seq=y+n, Slave) Loading ------------------------------> LS Request Full

10. 10. RT1 → RT2: LS Request

    RT1 asks RT2 for the LSAs it is missing with a Link State Request packet.

    RFC 2328 §10.10 LS Request Full ------------------------------> LS Update

11. 11. RT2 → RT1: LS Update

    RT2 answers with a Link State Update packet carrying those LSAs.

    RFC 2328 §10.10 LS Update <------------------------------ LS Request

12. 12. RT1 → RT2: LS Request

    RT1 requests the rest of what it still needs.

    RFC 2328 §10.10 LS Request ------------------------------> LS Update

13. 13. RT2 → RT1: LS Update

    Once RT1 has updated the necessary parts of its database, it is Full too. The two routers are fully adjacent.

    RT1: Full · RT2: —

    RFC 2328 §10.10 LS Update <------------------------------ Full