Q & A on RIP Heuristics for Routing Loops Q: One of my students asked me the difference between the two solutions of the count-to-infinity problem, i.e., the poisoned reverse and the split horizon hack. I answered that the poisoned reverse converges faster than the other one, but was not able to come up with any example so support my answer. Kindly give me the solution of this problem along with some good examples. A: SIMILARITIES: Both techniques are designed to handle the same problem: routing loops that arise in a distance-vector algorithm when routes ``backwash'' from a receiver to a sender. Neither is needed until routes change. Finally, they are designed to be used together -- one does not usually choose between them. DIFFERENCES: Split horizon is implemented by a receiver to prevent the receiver from advertising a route back over the interface on which it arrived. Poison reverse is implemented by a sender which transmits negative information after a previously-advertised route has been lost. Most distance-vector protocols use timeout to accommodate packet loss -- each route persists until the timer expires. Usually, the timeout, T, is a multiple of the update interval (e.g., T = 3*I). Thus, poison reverse has the advantage of knocking routes out of the reeceiver's table in the next update interval rather than waiting for the timeout interval. One more note: poison reverse is often coupled with ``triggered updates'', a heursitic that causes routers to propagate new information immediately rather than waiting for the next update interval. Thus, poison reverse with triggered updates causes almost immediate convergence.