A regional ISP owns three head-end blocks for route advertisement to its upstream peer: 192.168.8.0/24, 192.168.9.0/24, and 192.168.11.0/24. The block in between, 192.168.10.0/24, belongs to a different provider and is not part of this ISP's allocation. Which route advertisement correctly summarizes what the ISP can safely send upstream?
Select an answer to reveal the explanation.
Short Explanation
Summarization only works when the boundary math lines up with what you actually own. The first two blocks share a bit pattern that collapses cleanly into one /23, but folding in the third block would silently scoop up a neighbor's /24 sitting in the middle — so that one has to travel on its own.
Full Explanation
A summary route is only valid if the block it describes contains nothing the advertiser doesn't own. 192.168.8.0/24 and 192.168.9.0/24 differ only in their last host-relevant bit and share a common /23 boundary, so 192.168.8.0/23 legitimately covers both with zero foreign addresses included. 192.168.11.0/24 cannot join that summary because the next mask up that would encompass it, a /22, spans 192.168.8.0 through 192.168.11.255 and therefore includes 192.168.10.0/24, which belongs to another provider — advertising that block would claim reachability for address space the ISP does not control. The single /22 option makes exactly this mistake: it picks the smallest containing block without checking who owns every address inside it. The all-/24s option overcorrects, assuming that one non-contiguous block forfeits summarization everywhere, when the two genuinely adjacent blocks can still be merged. The /23-for-all-three option is a mask-size error: a /23 only ever spans two consecutive /24s, so it structurally cannot represent three blocks regardless of how the count divides. An operational check is to write out the binary of each block's third octet and confirm the candidate summary's fixed prefix bits match every owned block and no unowned one before it's redistributed upstream.