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Suppose we have three routers R1 R2 and R3 with full mesh connectivity in OSPF broadcast network (LAN). R1 becomes DR. R2 BDR and R3 is forming adjacency with both of them. Now let router r4 be attached but it can only hear and send to R3 only, (because of say network malfunctioning) which sends hello saying his DR is R1 and BDR are r2 but it himself does not assume any of the duties. The questions is will r4 elect r3 as his dr (because according to rfc 9.4 nothing makes him non eligible for dr election algorithm. Assume r3 priority is not zero. Can someone shed a light what would happen according to standard on such situations.

 R1(DR)--R2(BDR)  
     \      /
      \    /
       \  / 
        \/  
        R3 HELLO=(DR=R1, BDR=R2)
         |
         | 
        R4 ELECTED DR - ???
  • If r4 can only hear, it will never become adjacent – Ron Trunk Jul 26 '16 at 18:59
  • Sorry r4 is full duplex with r3 but only with it – Boris Jul 26 '16 at 19:06
  • I'm curious, how do you think this situation could happen on a broadcast medium? Every router can hear every other router on a broadcast medium. It's an all-or-none situation. If R4 can communicate with R3, it can communicated with R1 and R2 if R3 can. – Ron Maupin Jul 26 '16 at 19:16
  • Also, an OSPF router on a broadcast medium will not nominate a different router for DR or BDR, so R4 cannot nominate R3 for the DR; R4 will assume R4 is the DR until R3 tells it otherwise. Elections are not preemptive; a router with a higher priority that any other will not trigger an election when it comes up. R3 will not participate in an election as long as its DR and BDR are up. – Ron Maupin Jul 26 '16 at 19:21
  • @RonMaupin, you've not been in networking long enough; I've seen this happen too many times... vlan misconfiguration (shouldn't be possible, but is), etherchannel fault, plain o' Broken Switch(tm), etc. – Ricky Beam Jul 26 '16 at 19:28
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REVISED:

I'm not sure the RFC covers this odd situation, so it would be dependent on the implementation.

After further reflection (and little prodding by @ronmaupin), it seems clear that R4 will never get beyond the 2way state with R3.

R3 will send hellos to R4 with the RID of the DR and the BDR. R4 will accept these values and attempt to contact R1 and R2. Since it can't, it will never load its database, so there will be no OSPF routes in the routing table.

The question is, since R4 can't contact the DR or BDR, will it start a new election? If it decides that it or R3 is the BDR, it "should" start sending hellos on the AllDRRouters address (224.0.0.6). But R3 isn't listening, because it already knows who the DR and BDR are.

So it should stay in the 2way state, thinking it's the DR, but no one else will pay attention to it.

  • I don't think R4 will form a full adjacency with R3. That requires R3 to form a full adjacency with R4, and R3 should not do that. R3 will only form a full adjacency with the DR and BDR on the broadcast medium. I think R3 will go to 2-Way with R4, and R4 will be stuck in EXSTART. – Ron Maupin Jul 26 '16 at 20:01
  • I tend to agree RFC assumes full mesh. – Boris Jul 26 '16 at 20:17
  • @RonTrunk BTW Will R3 accept full adjacency request in this case? You mean, that being a BG (not a DR or BDR) in broadcast network does not prevent the router from accepting the DB synchronization requests? – Boris Jul 26 '16 at 20:46
  • @Boris, I don't think R3 will form a FULL adjacency with any routers other than R1 (DR) and R2 (BDR) on that one interface. R3 should get to a 2-Way adjacency with R4 and stop. That would leave R4 stuck in the EXSTART state with R3, possibly jumping back and forth between 2-WAY and EXSTART, depending on the OSPF implementation. A FULL adjacency means that both routers have exchanged LSAs, and their databases are synchronized, but R3 will only do that with a DR and BDR on a broadcast medium, and it already has that with R1 and R2. – Ron Maupin Jul 27 '16 at 0:04
  • I do believe the answer is "RFC does not give an answer" to this weird situations and actual behavior are left to implementors as RFC assumes full connectivity e.g broadcast is real broadcast reaching all nodes. and deals only with broadcast networks that can actually converge with one DR. As pointed out correctly the R3 does not listen to AllDRRouters (and should discard all packets to according to RFC as it is not in correct state) "... so full adjacency cannot be formed in any case.Marking it as answer. – Boris Jul 27 '16 at 5:42
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In your topology R3 & R4 is a OSPF point to point connection. So there are no DR or BDR election. The neighbourship between r3 and r4 in a FULL state always. But r3 and r2 only from adjacency no neighbourship is fromed between them because r1 elected as a DR among them. So r3 and r1 , r2 and r1 will only from neighbourship

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