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Is a scenario possible, where split horizon with poisoned reverse can’t prevent routing circles? E.g., if router A does announce all routes that route over B to B as infinity, B will never route over A to nodes that A routes to over B.

But is there a scenario, where a failing link causes a routing circle, despite all nodes having using split horizon with poisoned reverse? Maybe with multiple routers involved? Or does this always prevent routing circles? Maybe this is trivial, but I couldn’t think of an example right now. Thanks in advance!

1 Answer 1

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Consider this:

     B
     | \
   1 |  \1
     |   \
     A -- C 
     |  1
     +
     |
     D

At time t, link between A and D is broken.

A tells B & C that D is unreachable by DA(D) = inf

B computes new route to D through C. DB(D)=CB(C)+DC(D)=1+2=3. B will tell C that D is unreachable by DB(D)=inf (poisoned reverse). B still tells A it has a path to D of cost 3 (split horizon doesn't apply).

A computes new route through B. A tells C that D is now reachable

Etc…

2
  • What does DB(D), C(B,C), DC(D) and DB(D) indicate? What is the convention you are using?
    – Deepak
    Feb 9 at 17:26
  • DB(D) means from D to B begin at D. C(B, C) should be CB(C) (it has been corrected). Feb 10 at 14:26

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