Netmask Ordering

JdotQ · August 2007

So I was going over the review questions in the MS Press 2nd Ed. book, and came across this question where the answer confused me...

(let me know if I'm not allowed to post a screen cap of the question - it has a diagram, so kinda hard to describe the question - Question is on page 5-55 [#4])

Now, Netmask Ordering is a simple concept, if a FQDN has multiple resource records, reorder the DNS answer so the local IP that is the closest (to the local subnet/network) is at the top of the list, then follow with the remote IP's (out of the local subnet). So we're looking for where the client would fit in, so the FQDN in the local subnet would be the first choice...

So, with that diagram, /20 would mean a subnet mask of 255.255.240.0. With that subnet mask, the ranges would be (255-240 = 15);

192.168.0.1 - 192.168.14.254
192.168.15.1 - 192.168.29.254
192.168.30.1 - 192.168.44.254
192.168.45.1 - 192.168.59.254
192.168.60.1 - 192.168.74.254
etc

So, in the question, the client that is querying the DNS server has an IP of 192.168.33.5 (which would put it in the 192.168.30.0 - 192.168.44.0 range) -- but there are two servers in that range (IP's 192.168.30.2 & 192.168.42.40); so now we're down to two possible answers.

How would go about figuring out the order from this point forward? Unless my ranges are incorrect, I'm not seeing a clear answer?

(( The answer the MS Press book gives is "192.168.42.40" ))

georgemc · August 2007

JdotQ wrote:

So, with that diagram, /20 would mean a subnet mask of 255.255.240.0. With that subnet mask, the ranges would be (255-240 = 15);

192.168.0.1 - 192.168.14.254
192.168.15.1 - 192.168.29.254
192.168.30.1 - 192.168.44.254
192.168.45.1 - 192.168.59.254
192.168.60.1 - 192.168.74.254
etc

So, in the question, the client that is querying the DNS server has an IP of 192.168.33.5 (which would put it in the 192.168.30.0 - 192.168.44.0 range) -- but there are two servers in that range (IP's 192.168.30.2 & 192.168.42.40); so now we're down to two possible answers.

How would go about figuring out the order from this point forward? Unless my ranges are incorrect, I'm not seeing a clear answer?

(( The answer the MS Press book gives is "192.168.42.40" ))

With a /20 mask your subnet increment would be 256-140=16 (not 15).

192.168.0.1 - 192.168.15.254
192.168.16.1 - 192.168.31.254
192.168.32.1 - 192.168.47.254
192.168.48.1 - 192.168.63.254
192.168.64.1 - 192.168.79.254

Try it like that and re-read the section in your book on subnetting.

JdotQ · August 2007

George,

Thanks for the reply. Subtracting 240 from 256 would yield the supplied answer (rather than subtracting from 255). And it all makes sense now.

Thanks for pinpointing my error...not sure how I ended up thinking 255 instead of 256

Smallguy · August 2007

JdotQ wrote:

George,

Thanks for the reply. Subtracting 240 from 256 would yield the supplied answer (rather than subtracting from 255). And it all makes sense now.

Thanks for pinpointing my error...not sure how I ended up thinking 255 instead of 256

i did the3 same thing forever couldn't figure out why I knew how ot subnet but got wrong answers

JdotQ · August 2007

Smallguy,

Glad to hear I'm not alone!

I actually went back to some notes I took from CBT Nuggets about figuring out the range of IP's...

Convert the subnet into binary;

(255.255.240.0)
11111111.11111111.11110000.00000000

Look for the right most subnet bit;

11111111.11111111.11110000.00000000

And notice that bit is in the binary position of "16";

128  64  32  16  8  4  2  1
1    1   1   1   0  0  0  0

This is the increment that the range of IP's should take!

At least this is something I won't forget it

Netmask Ordering

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