# The dreaded topic of subnetting...

• 25th April 2010, 02:04 AM
Resolver
The dreaded topic of subnetting...
Hi all,

I'm preparing for 070-291, and am going over some test prep. I thought I had a pretty firm grasp on subnets, but I'm caught up on an idea, and need some clarification.

Lets say my network is 10.0.0.0/20

That makes my mask

11111111 . 11111111 . 11110000 . 00000000 ( 255.255.240.0 )

So my networks can range from

00001010 . 00000000 . 00000000 . 00000000 ( 10.0.0.0 )
to
00001010 . 11111111 . 11110000 . 00000000 ( 10.255.240.0 )

My question involves the 3rd octet.

If the network in use is

00001010 . 00000001 . 10110000 . 00000000 ( 10.1.176.0 )

and the host portion in binary is

xxxxxxxx . xxxxxxxx . xxxx0110. 00000010 ( x.x.6.2 )

Is the resultant IP address for the host

00001010 . 000000001 . 10110110 . 00000010 ( 10.1.182.2 )

I'm mentally caught up on combining the network and host portion of the binary digits into a final decimal. Just want to make sure I got this straight in my head.
• 25th April 2010, 06:16 AM
SYNACK
Yes you are correct, it does combine that way adding to 182.
• 25th April 2010, 08:05 AM
mac_shinobi
Are there any good books that go through cidr / vlsm / subnetting / super netting etc ??
• 25th April 2010, 02:10 PM
RabbieBurns
Quote:

Originally Posted by mac_shinobi
Are there any good books that go through cidr / vlsm / subnetting / super netting etc ??

the cisco ccna book covers all of that
• 25th April 2010, 03:11 PM
Resolver
Thanks Synack.
• 28th April 2010, 02:35 AM
Resolver
The study guide I'm using shows a diagram with the following information...

Clients
IPs 10.10.10.66-10.10.10.70
SM 255.255.255.192
GW 10.10.10.65

Then another portion of the network with
10.10.10.94
255.255.255.192
10.10.10.93

The question the diagram is asking is irrelevant to my question. As I understand it, the only IPs available for a 255.255.255.192 subnet would be x.x.x.0-63

Subnet 11111111 . 11111111 . 11111111 . 11000000
allows for hosts xxxxxxxx . xxxxxxxx . xxxxxxxx . xx000000 in the ending zeros. Those zeros wouldn't allow a host number high enough to access a gw at 10.10.10.65, or 93, or be an IPs at 66-70 or 94.

Is this yet another poorly written question in my guide, or am I at a misunderstanding?
• 28th April 2010, 07:28 AM
bio
Quote:

Originally Posted by Resolver
As I understand it, the only IPs available for a 255.255.255.192 subnet would be x.x.x.0-63

Nope this is not true.

lets break down the last octect 192 to binary -> 11000000
now lets break it down to the possible number of network id's (the first 2 digits are important here)

00000000 = invalid
01000000
10000000
11000000 = invalid

You broke it down to 2 possible network ids. 01 = 64 and 10 = 128.
Thus your 2 network ID's are 10.10.10.64 and 10.10.10.128
Now the host ip adress will be 10.10.10..65 (GW ip) to 10.10.10.127 (broadcast ip) for network id 10.10.10.64
Now the host ip adress will be 10.10.10..129 (GW ip) to 10.10.10.191 (broadcast ip) for network id 10.10.10.128

Learn To Subnet: is a great resource to learn :)

bio..
• 28th April 2010, 01:07 PM
Resolver
If the subnet is 11111111 . 11111111 . 11111111 . 11000000, that means the network portion of the address is where the 1s are. It cannot be all 0's or all 1's, as you mention, but having two 0's in the last octet doesn't violate that. The network could be 10.0.10/26, in which case the two digits in the 4th octet would be zeros, but the network address wouldn't be ALL 0s, and therefore ok ....correct?

Assuming the last two octets are 00 (Yes, I should have said 1 to 62, but for sake of math, I'm including the entire range) the range is 0-63, right?

What you said makes sense to me, the question in the book doesn't specify if the network portion includes any digits in the last octet. I guess it's up to me to figure that out based on the given subnet and host addresses.
• 28th April 2010, 01:11 PM
ChrisH
Quote:

Originally Posted by mac_shinobi
Are there any good books that go through cidr / vlsm / subnetting / super netting etc ??

Most of the books are as good as each other, you (or I did at least) just have to keep reading them over and over until it sinks in :P
• 28th April 2010, 03:01 PM
bio
Quote:

Originally Posted by Resolver
If the subnet is 11111111 . 11111111 . 11111111 . 11000000, that means the network portion of the address is where the 1s are. It cannot be all 0's or all 1's, as you mention, but having two 0's in the last octet doesn't violate that. The network could be 10.0.10/26, in which case the two digits in the 4th octet would be zeros, but the network address wouldn't be ALL 0s, and therefore ok ....correct?

Assuming the last two octets are 00 (Yes, I should have said 1 to 62, but for sake of math, I'm including the entire range) the range is 0-63, right?

What you said makes sense to me, the question in the book doesn't specify if the network portion includes any digits in the last octet. I guess it's up to me to figure that out based on the given subnet and host addresses.

In practice you could use the 00 and 11 networks ID's but your routers should be able to cope with RFC1812. I also remember from my cisco exams that the cisco books considers 00 and 11 to be invalid (for exam taking ). So your way of thinking is good :)

bio..
• 28th April 2010, 04:53 PM
Resolver
Quote:

Originally Posted by ChrisH
Most of the books are as good as each other, you (or I did at least) just have to keep reading them over and over until it sinks in :P

I read about subnetting in at least 3 different books. The thing that confused me the most were the different classes (A, B, C). Most books I've seen, even new books, still treat the internet as if we're all using public addresses, and that no one subnets.

I didn't really grasp subnetting until I took some MCSA night classes, and even now I still need to check my answers with you guys.
• 29th April 2010, 01:12 PM
MYK-IT
I found Advanced Subnet Calculator from Solarwinds very useful in understanding this topic further.