I'm reviewing the redistribution of routing protocols on the above link, and I've got a mental block. I can't even visualize the underlined portion below.
Can someone try to dumb this down a bit more for me?
"Note: The mechanics of route redistribution is proprietary on Cisco routers. The rules for redistribution on a Cisco router dictate that the redistributed route be present in the routing table. It is not sufficient that the route be present in the routing topology or database. Routes with a lower Administrative Distance (AD) are always installed in the routing table. For example, if a static route is redistributed into IGRP on R5, and then IGRP subsequently redistributed into RIP on the same router (R5), the static route is not redistributed into RIP because it never got entered into the IGRP routing table. This is due to the fact that static routes have an AD of 1 and IGRP routes have an AD of 100 and the static route is installed in the routing table. In order to redistribute the static route into IGRP on R5, you need to use the redistribute static command under the router rip command."
Are they trying to say that the static route is being learned by R5 through IGRP or are they trying to say that the static route is being redistributed by R5 into IGRP?
this is good question for routing redistribution
the description mentioned above sate that if you redistribute static into IGRP in R5( static has lower AD ) then in R5 will still see this route via static
then if you redistribute IGRP into RIP in R5 now then do not expect the redistributed static into IGRP to be redistributed to RIP
redistribute static into IGRP>>> redistribute IGRP into RIP ( exclude static )>>>>
in fact this is the case with any three routing protocols get redistributed in the same router
if you redistribute any route between two of them then you redistributed one of then in the other the first one will not get redistributed
in your question the solution is by redistributing the static directly into RIP you will get the static route apeared in RIP domain
if helpful Rate
We need to analyze the source of the route from the router's perspective. Let's replace the route from static learned to EIGRP learned and we are going to redistribute this route into BGP.
BGP has a lower Administrative Distance (AD) so you would think the route will be seen as a BGP route on this router, but it won't. The route will be shown as learned via EIGRP because that's its original source. How this could change into BGP? If another router feeds back this route via eBGP. Now, BGP has loop prevention mechanism which will prevent that situation but it can happen if those mechanisms are disabled.
To summarize, don't focus on the AD of the route but the source. On the example above, the router learned the route via EIGRP and it passed this route into BGP so other BGP speaking routers could have reachability to it.
In your case, the route was learned via static and if this router wants to pass this information into other routing protocols, you must redistribute static into each of the routing processes.
Hope this helps.
Good question. Look at it like this:
If a static route is redistributed into IGRP, it will be advertized to other routers running that protocol.
That is essentially what redistribution does.
However, because the static route has an AD of 1 it will also be in the local routing table, NOT the IGRP redistributed route. (distance 100)
Because the route is not in the local routing table as an IGRP route, it is not redistributed into RIP.
Hope this clarifies your problem.
it is not only because the AD value as Edison stated above the routing source has effect as well because the router still reading the source is other protocol
example was you have RIP, eBGP and EIGRP
if you redistribute EIGRP into BGP ( eBGP AD is 20 less than EIGRP ) then you redistribute BGP into RIP then the EIGRP will not get redistributed because the router still see the route source as EIGRP even through eBGP AD is less
unless it get fed through another router via eBGP and this will cause a routing loop
The only semi-unclear difference about paying attention to the source of the route is with connected routes. If a protocol runs over that interface via a network statement (as with OSPF, EIGRP, AND RIP) then those protocols set a bit on the connected route declaring partial ownership of the route. For example, if an interface is 10.1.1.1/24 and OSPF has a network statement of 10.0.0.0, it sets a bit on the connected route. If EIGRP then does "redistribute OSPF 1", the connected route "owned" by OSPF is redistributed in, even though the route in the table is connected, not OSPF. Clear?
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well you remained with a tricky point of redistributing connected, if you redistribute connected into EIGRP , and then you do redistribute EIGRP into OSPF, do NOT expect to see any connected route in OSPF unless you also redistribute connected into OSPF. In other words, redistribution doesn't change the source of the router, so double redistribution does NOT work
Absolutely agree. When you cover an interface with a network statement, the routing protocol sets a bit declaring partial ownership. When you redistribute connected, it doesn't set that bit so it isn't considered an owner.
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