And just to let you know, I'm hoping to re-record this presentation so the slides appear more clear. When it's uploaded I'll link to it in the comments, and we'll update the APNIC Training website to point to the new video.

Hello, Sir. If possible you can also share with me that notes. My Eamil id is- [email protected]. Thanks.

Hi, you have some really interesting questions. * No, that is a recursive server. The client is the resolver. It sends query to the the local dns (a recursive server). * Negative caching allows the negative responses to be cached for a maximum time specified in the SOA record. (see RFC2308) * When a resolver sends a recursive query, the recursive dns server acts in its behalf and queries other servers until it finds the answer. When a resolver sends an iterative query, the DNS server (who’s not authoritative for that zone) only provides a referral to the next DNS server to query. * Yes, you can configure the dns server to listen only to specific interfaces. By default, it listens to all. For BIND, there is a listen-on{} and a listen-on-v6{} option in named.conf

Thanks for the reply, and some further questions. Use Windows for example, the DNS Client Service is the resolver in this case right? So it is the resolver decide which type of query(recursive/iterative) to send? How does it decide?

Hi Ramprasad, all of Trainer Philip Smith's slides can be viewed at www.bgp4all.com.au/dokuwiki/workshops/start thanks APNIC Training

Hi Yogesh, you may wish to join the mailing list of the Network Operators Group (NOG) in your country.

you can make it by using Static IP not DHCP

you right SIR, me neither could not get it

You may need to get your mindset right to learn

A stub resolver is a resolver with minimum functionality. Most operating systems (*nix and Windows) act as stub resolvers. It works alongside a recursive/caching dns server that can perform recursive queries.

thanks APNIC

+Ridwan Omar Hi Ridwan, you can email [email protected] with any questions or comments you may have. thanks

+fbifido fbifido your answer for both Ex1.1 and 1.2 are correct. Good work!

Hi, good work there. But I've got a problem with the incremental value, how does one determine that? /36 was going on a leap of 1 but /35 is 2, confused a lil bit. Care to explain?

+hn031 The calculation of /36 and /35 are as follow, first converted to binary, then back to hex: 2 4 0 6 : 6 4 0 0 : 0 0 0 0 :: / 3 6 0000 | 0000 0000 0000 ---> 2406:6400:0000::/36 0001 | 0000 0000 0000 --->2406:6400:1000::/36 0010 | 0000 0000 0000 ---> 2406:6400:2000::/36 0011 | 0000 0000 0000 ---> 2406:6400:3000::/36 2 4 0 6 : 6 4 0 0 : 0 0 0 0 :: / 3 5 000 | 0 0000 0000 0000 ---> 2406:6400:0000::/35 001 | 0 0000 0000 0000 ---> 2406:6400:2000::/35 010 | 0 0000 0000 0000 ---> 2406:6400:4000::/35 011 | 0 0000 0000 0000 ---> 2406:6400:6000::/35

+hn031 2406:6400:x000::/36 x = 0 to f = 16 so x is increment by 12406:6400:x000::/35 x=0,2,4,6,8,A,C,E(8) if 0 to f = 8, which is wrong cause 0 to f is 16 so for 0 to f to equal 8, we do 16/8 and we get 2. so if we count from 0 to f with increments of 2 we get 8. so x is an increment of 2.hope that make sense.

Thanks , just saw your explanation by I'm busy now. Will give you feedback when I'm done going over it. Thanks for your time.

Hi, thank you for your comment. We have tested the recording and the volume seems ok? You may need to turn the volume up on your computer? thanks APNIC Training

+Michael Taylor Hi, yes you are correct for the first item. Did you also try the next exercise?