1.) Suppose there are three nodes 1,2,3 performing AODV routing protocol.
2.) Node1 generates the traffic to Node3 and the route is via node2.
3.) the rate of generation of the packet by node1 is kept less, so route_table_timeout occurs. Suppose this erases the route in node2 table..
4.) Now node1 sends the packet to node2.
5.) node2 generates a route Request.
6.) node1 replies to node2 since node1 has the route to node3.
7.) so node node2 will forward the packet to node1.
So a loop gets set up...node1 - node2 - node1.... How can this be avoided?
Note that AODV packets have only Next hop information whereas DSR has the entire route.
Very interesting question! In my opinion, the answer is as follows. Intermediate node answering a path request includes the amount of time the associated route will still be valid. Hence, node 2 will include the amount of time the path it is offering for node1 to reach node 3, is valid. Thus, the path will be timed out at node 1 at the same point in time (or preferably a little bit earlier to compensate for message transmission delay) as node 2, thus node 1 cannot answer node 2.
In fact, the question is interesting. As far as I know, apart from what Anne-Lena comments, the RREQ is broadcast and thus Node3 receives the RREQ so that Node2 establishes the correct route.
On the other hand, we have also to consider the precursors list since it would not have any sense that one of the precusors of an entry is the gateway. If Node1 has an entry for Node3 with gateway Node2 and receives a RREQ from Node2 towards Node3, it shouldn't add Node2 into the precursor list of Node3 since Node2 is the gateway. By this way Node1 locally solves the loop.
It is true that the loops avoidance carried out by AODV has been greatly discussed but more focusssed on the seq. numbers handling.
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