Router A reaches network X with cost 3 via Router B. B's direct link to X then fails. Before B can propagate the failure, A sends its periodic update saying 'I reach X with cost 3.' B concludes it can now reach X via A with cost 4, and begins advertising cost 4. A then updates to cost 5, B to 6, and so on. This scenario illustrates:
AThe Bellman-Ford algorithm failing to converge due to incorrect initialization
BThe count-to-infinity problem, where stale routing information creates a self-reinforcing feedback loop of increasing costs
CSplit horizon successfully preventing the loop from forming
DPoison reverse correctly marking the route as unreachable
This is the textbook count-to-infinity scenario. After the failure, B no longer has a valid route to X, but before it can propagate that information, it receives A's stale advertisement and mistakenly believes it can reach X through A. A in turn updates based on B's updated cost. The two routers bounce the route back and forth with increasing costs until the metric hits the maximum (16 in RIP). Options C and D describe mitigations that would have prevented this, not the problem itself.
Question 2 Multiple Choice
RIP uses a maximum hop count of 15 and treats 16 as 'infinity' (unreachable). The primary reason for this hard cap is:
ATo ensure RIP converges faster than OSPF on small networks
BTo bound the count-to-infinity problem, preventing counting from running indefinitely when a destination becomes unreachable
CTo limit the size of routing tables that routers must maintain
DTo enforce a maximum network diameter for administrative reasons
Without a maximum metric, the count-to-infinity problem would cause routers to increment a failed route's cost forever. The max-15 cap means the loop terminates after at most 15 increment cycles, at which point both routers agree the destination is unreachable. The cap makes the protocol eventually correct, though it still converges slowly. Options A and C are not the primary motivation; option D is a side effect, not the design purpose.
Question 3 True / False
Split horizon substantially solves the count-to-infinity problem by preventing routers from ever advertising stale routes back to the neighbor from which they learned them.
TTrue
FFalse
Answer: False
Split horizon reduces the problem but does not fully solve it. It prevents a two-router loop (A learned from B, so A won't advertise back to B), but multi-router loops involving three or more routers can still cause count-to-infinity. For example, if A and C both have routes through B, and B's link fails, A might hear C advertising a stale route and update, then C hears A's update, etc. This is why larger networks prefer link-state protocols with a complete topology view rather than relying on split horizon.
Question 4 True / False
In distance-vector routing, each router periodically sends its entire routing table — its distances to all known destinations — to its directly connected neighbors.
TTrue
FFalse
Answer: True
This is the defining characteristic of distance-vector protocols and the source of both their simplicity and their problems. Each router shares its complete distance vector (not just directly connected links), which allows knowledge to propagate through the network in successive rounds. It also means that stale information can spread — if a router's routing table contains an outdated entry, it will share that bad information with neighbors, who may then rely on it. This is contrast to link-state protocols, where routers share only the state of their directly connected links.
Question 5 Short Answer
Explain why distance-vector protocols are susceptible to the count-to-infinity problem and describe one mitigation technique along with its limitation.
Think about your answer, then reveal below.
Model answer: Distance-vector routers only know their distances to destinations, not the full topology. When a link fails, neighbors may have already cached the failed router's distance advertisement and believe they have an alternate route through it — creating a routing loop. Each router increments the cost based on what its neighbor advertises, which was itself based on the original router's stale distance. Split horizon partially fixes this by refusing to advertise a route back to the neighbor from which it was learned, breaking two-node loops. But split horizon fails for loops involving three or more routers, which can still count up to the maximum metric before convergence.
The fundamental problem is that distance vectors carry no path information — a router can't tell whether the route it's hearing about loops back through itself. Link-state protocols solve this by flooding the complete topology to every router, so loops are immediately detectable. The price of distance-vector simplicity is slow convergence and vulnerability to these feedback loops.