A confederated AS is comprised of three members. One member is comprised of 4 routers, while the other two members are comprised of 3 routers each. Assuming member ASs are fully meshed and use full…

Question

Options

• A12
• B13
• C14
• D15

• A
  14% (5)
• B
  9% (3)
• C
  3% (1)
• D
  74% (26)

Explanation

D (15) is correct because you must count both the full-mesh iBGP sessions inside each member AS and the confederation eBGP sessions between member ASs. Each member's internal sessions use the formula n(n-1)/2: the 4-router member needs 6 sessions, and each 3-router member needs 3 sessions - totaling 12 internal sessions. The three member ASs are also fully meshed with each other, adding 3(3-1)/2 = 3 inter-member sessions, giving 12 + 3 = 15. Choice A (12) is the most common trap - it correctly calculates the internal iBGP sessions but ignores the confederation eBGP links between members entirely; B (13) and C (14) suggest counting only 1 or 2 inter-member links instead of the full 3. Memory tip: Think of a confederation as "two layers of mesh" - always apply n(n-1)/2 twice: once inside each member, and once across the members themselves, then sum everything.

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