This might be a very trivial question.

What I understand is that different networks maintained by different ISP's (or the same) are assigned CIDR IP addresses which they can use to create their own subnets.

If say ISP's 'A' and 'B' are assigned and respectively, then how is the issue resolved from the fact that every host IP address in ISP 'B' is a valid host IP address in ISP 'A'?

What am I not understanding?

  • RIRs do not assign overlapping networks to their ISPs. – Ron Maupin Dec 8 '17 at 21:06
  • In the context of route aggregation how is this resolved then? Or is aggregation only done in the local scale? – deftextra Dec 8 '17 at 21:11
  • Aggregation, if possible, is done by an ISP at its edge. The Internet is, in general, not able to be aggregated. An ISP assigned will subnet that into a bunch of smaller networks for its own use, but it will advertise the aggregate to its neighbor ASes. – Ron Maupin Dec 8 '17 at 21:14
  • BTW i just used those numbers as an example. I was not bothered to look up real IP. – deftextra Dec 8 '17 at 21:14

In your example you are using rfc1918 addresses, which are not announced to internet. If you have a private wan with two ISP, it will be up to routing configuration where the traffic will go to. An CIDR with /24 netmask length is preferred over /16 since it's more specific.

| improve this answer | |

Stian is correct about the IP addresses being RFC1918 private-use addresses. The missing part though, is when using private-use addresses to access the internet, there must be a router involved that uses RFC1631, network address translation. This involves changing the private addresses that cannot be routed over the internet to public addresses that can. ISPs do this to reduce their costs by sharing fewer public ip addresses among many more clients.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.