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simple BGP problem

cisc0.ameer
Level 1
Level 1

hello

very simple topology i create 4 routers as a i put scenario's photo below

No special configuration very basic eBGP peering with physical interface

but output is weird

5.5.5.5 is also on R1

i've checked several times No special BGP config i did , but each time i face different outputs from command " show ip bgp "

R1 config :

router bgp 100
bgp log-neighbor-changes
network 1.1.1.1 mask 255.255.255.255
network 5.5.5.5 mask 255.255.255.255
network 11.11.11.1 mask 255.255.255.255
network 111.111.111.1 mask 255.255.255.255
neighbor 192.168.12.2 remote-as 200
neighbor 192.168.13.3 remote-as 300

 

and similarly on R2 ...& others

BGP1.png

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

but in R2 when i take output from show ip bgp

BGP1.png

From R3 Output of show ip bgp

BGP1.png

From R4 output as below

BGP1.png

 

 !!!!why R3 shows just 1 Best valid path while R2 shows both paths

i use "debug ip bgp neighbor 192.168.13.3 update" and i received below

logically yes it gets from originator and other paths it cant install

but why it should show like that !!!!

BGP1.png

5 Replies 5

omz
VIP Alumni
VIP Alumni

Hi

The AS numbers on the AS_PATH list are used for loop detection; a router receiving a BGP route with its own AS number in the AS_PATH assumes a loop and discards the route.

http://www.ciscopress.com/articles/article.asp?p=2738462&seqNum=2

 

 

thanks

and with as-override on R1 or allowas-in on R3 we are not enable to install both path same time

correct?

Martin L
VIP
VIP

BGP uses 2 was to preventing loops: AS Path attribute for e BGP and sort of "BGP split horizon" rule for iBGP
AS path is mentioned above; the other rule iBGP will not advertise iBGP routes to others.

thanks

what you mean by rule for iBGP AS path?in topology there aren't any iBGP peering ....

Hello

Its loop avoidance initiating it isn't a problem- Both R2 -R3 will advertise that prefix originating from R1 to R4 and R4 sees both paths from those rtts to that prefix as equal so it will possibly choose based on the oldest route or lowest bgp rid.

R4 will choose R2 as its best path so it will advertise the prefix it receives from R2 towards R3, As such R3 will also have two paths to that prefix and rtr2 will only have the one unless it loses connection towards R1 then it will obtain the prefix via rtr 4 -rtr3 path.


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Kind Regards
Paul