03-19-2015 01:28 AM - edited 03-07-2019 11:09 PM
As the Routing Protocol RIP its max hop count is 15
Every interface of the Router receives and forwards the updates then why this limits to 15
consider Router 1 sends its routing table to R2 then R3... R4 to R5 and ... R15 then of course R15 will also forward the routing table updates.
Every router involves sharing its routing table to others.. Router16 must have updates of Router 15 because as they re neighbors. then why says the limit is 15 ?
I don't say that R1 should have sent its updates to R16, but its neighbor should. isn't it ?
Please help to solve this problem.....
03-19-2015 05:33 AM
Have a look at rfc 1058. It explains how this limitation can be remedy:
https://www.ietf.org/rfc/rfc1058.txt
shouldn't be any bigger than required. Thus the choice of infinity is a tradeoff between network size and speed of convergence in case counting to infinity happens. The designers of RIP believed that the protocol was unlikely to be practical for networks with a diameter larger than 15. There are several things that can be done to prevent problems like this. The ones used by RIP are called "split horizon with poisoned reverse", and "triggered updates". 2.2.1. Split horizon Note that some of the problem above is caused by the fact that A and C are engaged in a pattern of mutual deception. Each claims to be able to get to D via the other. This can be prevented by being a bit more careful about where information is sent. In particular, it is never useful to claim reachability for a destination network to the neighbor(s) from which the route was learned. "Split horizon" is a scheme for avoiding problems caused by including routes in updates sent to the gateway from which they were learned. The "simple split horizon" scheme omits routes learned from one neighbor in updates sent to that neighbor. "Split horizon with poisoned reverse" includes such routes in updates, but sets their metrics to infinity. If A thinks it can get to D via C, its messages to C should indicate that D is unreachable. If the route through C is real, then C either has a direct connection to D, or a connection through some other gateway. C's route can't possibly go back to A, since that forms a loop. By telling C that D is unreachable, A simply guards against the possibility that C might get confused and believe that there is a route through A. This is obvious for a point to point line. But consider the possibility that A and C are connected by a broadcast network such as an Ethernet, and there are other gateways on that network. If A has a route through C, it should indicate that D is unreachable when talking to any other gateway on that network. The other gateways on the network can get to C themselves. They would never need to get to C via A. If A's best route is really through C, no other gateway on that network needs to know that A can reach D. This is fortunate, because it means that the same update message that is used for C can be used for all other gateways on the same network. Thus, update messages can be sent by broadcast. In general, split horizon with poisoned reverse is safer than simple split horizon. If two gateways have routes pointing at each other, advertising reverse routes with a metric of 16 will break the loop immediately. If the reverse routes are simply not advertised, the erroneous routes will have to be eliminated by waiting for a timeout. However, poisoned reverse does have a disadvantage: it increases the
HTH
03-19-2015 09:29 AM
Thanks for reply..
I'm new to the lesson and I don't understand that much deep...
Can you please explain it simply?
Thank you again....
03-19-2015 09:56 AM
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I should double check, but it's also possible only 4 bits were set aside for hop count. Yes, today, it can be difficult to appreciate the importance of saving just a few bits, but likely it was back when RIP was being developed. Plus, after all, who would ever have their own internal network that needs more than 15 hops (or more than 32 bits for a global network, or more than 640 KB RAM in a PC, etc.)? ;)
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