I hope someone can enlighten me on the following:

Say for example I have a server that has an allocation of 20 IPs configured for the server.

Now each time when I created a private nameservers for a reseller and use a combination of 2 different IPs from the pool of 20 IPs.

How many private nameservers do you think that can be created ???

If all the name servers point to the same server where the DNS is located then all you need is two IP addresses, and just setup each different name server usign the two IP addresses. If setting up Reverse DNS, then you can set this to your main name server address.

I don't see how hosting 10 nameservers can justify the use of 20 IP addresses.

Alan

I am sorry my question was not clear enough. What I meant was:

Say the IPs 1-20 so each time when I create a nameservers I would do the following:

1st nameservers using IPs 1 & 20
2nd nameservers using IPs 2 & 20
-
-
-
19th nameservers using IPs 19 & 20

The next level of namesevers:
1st nameservers using IPs 1 & 19
2nd nameservers using IPs 2 & 19
---
---
18th nameservers using IPs 18 & 19


So for each level, there is less nameservers that can be created.


All the nameservers resolves to only one server having the 20 IPs allocated for the server.


What I want to know is... will it work if I use the above system to create as many nameserver as possible instead of 20 IPs where I can only create a maximum of 10 nameservers if I use the standard system.

But if they are all pointing to the same DNS records on the same server, then why would you even need more than 2 IPs for all the nameservers?

Alan

If you used the Combination scheme you mentioned above there would be 210 unique combinations.

However, I don't understand why you don't just use the same 2 IP's for all nameservers.

I would suggest to use only 2 IPs, it'll eb cheaper and easier overall. Just create the new nameservers using the different domains, but with the 2 same IPs that point to the DNS server(s).

Alan

Thanks very much for the suggestions and advice.

Regards.