Computer Networks: Split horizon with Poisson Reverse Frouzan

Question

As soon as I have studied I got that count to infinity problem will not occur if A advertise its window first to B before B can. As a result B gets information that X is un reachable now, and system will be in stable state.

Screen shot of above theory:-

Now I also know that how to solve this using split horizon i.e. B sends its routing table to A but it will not sends that row in the advertising table which contain entry which shows that "B knows the path to X which passes through A". and thus it will not send this infor to A.

But I am getting difficulty in understanding what is split horizon.

In the line saying: "Node B can still advertise the  value for X......." till last line.

Comments

Split horizon says:

When router sends Distance vector to its neighbor it must send ∞ for the entry for which it is depending on it. It is saying "Don't depend on me for which I depend on you".

It means B will send its routing table to A but It will contain ∞ in the table entry whose path is via A.

Answer 1

---

The Last Screenshot is an Split Horizon With Reverse Poision

Here B will advertise X in its routing table but the value is set as infinity stating that X is unrechable and giving a hint that A should not use this value since it was computed from A's Routing Table....

The first Screenshot is Simple Split Horizon where the entry of X in B's Routing Table itself is not Shared with A.

Hence the only diff is that in Reverse Poison Split Horizon the entry for X is shown as infi and in Simple Split Horizon the entry for X itself is ommited from the table while sharing the Routing Table info..

Comments

Thanks for Conforming.!!!