Hello gurus! hoping for a BGP expert to chime in here. Im studying for my CCIE, and there is something in Jeff Doyle's Routing TCP/IP vol2 book that I just cant seem to figure out and its really stalling my understanding of the BGP path selection algorithm.
Its on pg 195, example 3-57, attached as an image in this post (Ive also attached the network diagram that this output refers to). Basically its an output of "show ip bgp" and whats stumping me is simply: for the aggregate route 192.168.192.0/21, why has this router selected as best (>) the one via next hop 192.168.1.254?? I would have thought based on the presence of the LocalPref = 100 on the 192.168.1.237 route that would have been selected. But apparently not! Heres a walk through of the path selection logic as i understand it:
1/WEIGHT: both 0, so skipped.
2/LOCAL_PREF: this is my problem, .237 should win, but ignoring for now...
3/ORIGINATED LOCALLY: neither are they are learnt from BGP peers, so skipping.
4/AS_PATH: both identical, AS100 only, so skipping
5/ORIGIN CODE: both are 'i' (IGP), both were created from "aggregate-address" statements on their originating routers downstream in AS100
6/MED: both empty, so skipping
7/PREFER [eBGP] over [confedBGP] over iBGP: so the .254 route apparently wins on this condition... which in isolation, i agree with (clearly the eBGP .254 route is better than the .237 iBGP candidate).
.... however what about step 2/LOCAL_PREF!?
looking forward to some expert guidance here to help me squash this one :)
thank in advance,
Keiran
