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ibps-so 2016 · Question 181 of 394

Three routers of Chandrabhaga Urban Bank — X at Pandharpur, Y at Sangli and Z at Kolhapur — are joined in a line, X to Y and Y to Z, and run a distance-vector routing protocol that advertises its whole table to its neighbours every thirty seconds. A network attached directly to X now fails. Before X can tell Y that the network is unreachable, Y advertises to X the route to that network which Y had itself learned from X, at a cost one greater. X believes it, and advertises back at a cost greater still, and the two go on raising each other's metric by one at every exchange while packets for the dead network shuttle between them. The condition so described, and the two standard measures that between them contain it, are respectively:

  1. AThe count-to-in nity problem; de ning a xed maximum metric that counts as in nity (16 hops in RIP), and applying split horizon with poisoned reverse, so that a route is never advertised back — or is advertised back only as unreachable — over the interface upon which it was learnedCorrect
  2. BA routing loop caused by unequal-cost load sharing; disabling load sharing, and increasing the update interval from thirty seconds to two minutes
  3. CCongestion collapse of the leased circuits; applying tra c shaping at X, and increasing the bu er at Y
  4. DA broadcast storm upon the link between X and Y; enabling the spanning tree protocol, and placing the redundant port in the blocking state
  5. EFlapping of the interface at X; damping the interface, and raising the administrative distance of the protocol above that of the static route

Answer: A. The count-to-in nity problem; de ning a xed maximum metric that counts as in nity (16 hops in RIP), and applying split horizon with poisoned reverse, so that a route is never advertised back — or is…

Official answer key verified. Detailed explanation coming soon.

Source: IBPS SO 2016 IT Officer (held 28-29 Jan 2017, single-stage CWE) - Prepp · memory-based

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