OSPF Election on Ethernet(broadcast)

Hi guys,

I'm am currently simulating labs on how OSPF and elections work. Currently, I have to routers which are connected via crossover cable on fast Ethernet ports.

Both interfaces have the default priority 1 and the IP address are:
R1: 192.168.10.49/29 fa0/0
R1: 10.255.255.9/30 s0/0
R1: 10.255.255.82/30 s0/1

R2: 192.168.10.51/29 fa0/0

From what I read in the books and Internet, if the priority are the same on both routers, the router with the highest RID will win. However, when I run "show ip ospf neighors" and "show ip ospf int fa0/0" I see that R1 is currently the DR.

Can anyone help me with this? I am running this on the latest version of packet tracer.

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Comments

pham0329

The higher RID wins, but there's no preemption. This means if R1 boots up first, it will be the DR no matter what its RID or priority is.

nycbp

pham - yes. I realized this right after I typed all of the stuff. Either way, thank you for your quick response!

method115

pham0329 wrote: »

The higher RID wins, but there's no preemption. This means if R1 boots up first, it will be the DR no matter what its RID or priority is.

So really in a real world scenario (lets say you have 4 routers connected) which ever router boots up first will be the DR? Unless you somehow manage to have every router boot up at the same time. I'm assuming in a real world scenario most people manually configure the DR/BDR? From what I remember when studying for CCNA manually configuring settings was almost always suggested/preferred way of doing it over automation.

pham0329

Yes, if you want to ensure that certain routers do not get elected, you would set their priority to 0.

thedrama

Thats true. Routers with priority 0 is disqualified from the DR/BDR election process. One more thing, you need to care whether there are
loopback interfaces on that segment.