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GATE CS 2022 Q47 — Routing Protocols: Shortest Path, Flooding, Distance Vector, Link State

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Consider a network with three routers P, Q, R\text{P, Q, R} shown in the figure below. All the links have cost of unity.

The routers exchange distance vector routing information and have converged on the routing tables, after which the link Q-R\text{Q-R} fails. Assume that P\text{P} and Q\text{Q} send out routing updates at random times, each at the same average rate. The probability of a routing loop formation (rounded off to one decimal place) between P\text{P} and Q,\text{Q}, leading to count-to-infinity problem, is _______________.

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Answer: 0.5

Initially, the network is connected as P−Q−RP - Q - R with unit link costs. The routing paths to destination RR before the link failure are:

  • QQ reaches RR directly with a cost of 11 (next hop: RR).
  • PP reaches RR via QQ with a cost of 22 (next hop: QQ).

When the link Q−RQ-R fails, QQ immediately detects the failure and updates its distance to RR to ∞\infty.

Since PP and QQ send routing updates independently at random times with the same average rate, the next update event is equally likely to be initiated by either router (each with probability 0.50.5):

  • Case 1: PP sends its update to QQ first (Probability = 0.50.5)
    Before QQ can inform PP about the failure, PP sends its periodic update advertising a path to RR with cost 22. QQ receives this and, having no other path, updates its route to RR via PP with cost 2+1=32 + 1 = 3. This creates a routing loop (P→Q→PP \to Q \to P), leading to the count-to-infinity problem.
  • Case 2: QQ sends its update to PP first (Probability = 0.50.5)
    QQ informs PP that RR is unreachable (cost ∞\infty). PP updates its routing table accordingly, and no loop is formed.

Thus, the probability of a routing loop formation is 0.5\mathbf{0.5}.

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