split horizon

Question

Assume that network shown in below diagram uses distance vector routing with forwarding tables at each node. If each node periodically announces their vectors to the neighbour using the split-horizon strategy, what is the distance vector routing advertised from A to B after stabilization?

Comments

So is this right ??

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yep..

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@gatesjt: What you're saying is true for split horizon with poison reverse. But the question here asks only split horizon.

Answer 1

Directly quoting from Peterson & Davie (even the example in definitions) :

Split Horizon: 

The idea is that when a node sends a routing update to its neighbors, it does not send those routes it learned from each neighbor back to that neighbor. For example, if B has the route (E, 2, A) in its table, then it knows it must have learned this route from A, and so whenever B sends a routing update to A, it does not include the route (E, 2) in that update.
 

Split Horizon with poison reverse:

A stronger variation of split horizon. B actually sends that route back to A, but it puts negative information in the route to ensure that A will not eventually use B to get to E. For example, B sends the route (E, ∞) to A.
 

So in this question, after stabilization forwarding table for A will be [triplets in form(Destination, cost, next hop) ] :

1. (A, 0, -)

2. (B, 5, B)

3. (C, 4, C)

4. (D, 7, B)

Now entry no. 2 and 4 indicate that A learned these routes from B itself, so when following split horizon, A doesn't forward these entries to B. Hence the distance vector advertised to B after stabilization will consist of only entry no. 1 and 3

Comments

From forouzan :

Using the split horizon strategy has one draw- back. Normally, the Distance Vector Protocol uses a timer, and if there is no news about a route, the node deletes the route from its table. When node $B$ in the previous scenario eliminates the route to $X$ from its advertisement to $A$, node $A$ cannot guess that this is due to the split horizon strategy (the source of information was $A$) or because $B$ has not received any news about $X$ recently. The split horizon strategy can be combined with the poison reverse strategy. Node $B$ can still advertise the value for $X$, but if the source of information is $A$, it can replace the distance with infinity as a warning: “Do not use this value; what I know about this route comes from you.”

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Peterson & Davie: in the same paragraph as mentioned in the pratyush madhukar 's answer:

In a stronger variation of split horizon, called split horizon with poison reverse, $B$ actually sends that (2 and 4) route back to $A$, but it puts negative information in the route to ensure that $A$ will not eventually use $B$ to get to $E$. For example, $B$ sends the route ($E$, $\infty$) to $A$. The problem with both of these techniques is that they only work for routing loops that involve two nodes

Answer 2

According to me this is the answer can anyone verify