Count to infinity problem The Bellman–Ford algorithm does not prevent routing loops from happening and suffers from the count to infinity problem. The core of the count-to-infinity problem is that if A tells B that it has a path somewhere, there is no way for B to know if the path has B as a part of it. To see the problem, imagine a subnet connected like A–B–C–D–E–F, and let the metric between the routers be "number of jumps". Now suppose that A is taken offline. In the vector-update-process B notices that the route to A, which was distance 1, is down – B does not receive the vector update from A. The problem is, B also gets an update from C, and C is still not aware of the fact that A is down – so it tells B that A is only two jumps from C (C to B to A). Since B doesn't know that the path from C to A is through itself (B), it updates its table with the new value "B to A = 2 + 1". Later on, B forwards the update to C and due to the fact that A is reachable through B (From C's point of view), C decides to update its table to "C to A = 3 + 1". This slowly propagates through the network until it becomes infinity (in which case the algorithm corrects itself, due to the relaxation property of Bellman-Ford).
Count to infinity problem The Bellman–Ford algorithm does not prevent routing loops from happening and suffers from the count to infinity problem. The core of the count-to-infinity problem is that if A tells B that it has a path somewhere, there is no way for B to know if the path has B as a part of it. To see the problem, imagine a subnet connected like A–B–C–D–E–F, and let the metric between the routers be "number of jumps". Now suppose that A is taken offline. In the vector-update-process B notices that the route to A, which was distance 1, is down – B does not receive the vector update from A. The problem is, B also gets an update from C, and C is still not aware of the fact that A is down – so it tells B that A is only two jumps from C (C to B to A). Since B doesn't know that the path from C to A is through itself (B), it updates its table with the new value "B to A = 2 + 1". Later on, B forwards the update to C and due to the fact that A is reachable through B (From C's point of view), C decides to update its table to "C to A = 3 + 1". This slowly propagates through the network until it becomes infinity (in which case the algorithm corrects itself, due to the relaxation property of Bellman-Ford).