Stuck on Subnets

horizonuser

Hi,
I am having some trouble with getting my head around subnetting. I have been reading up for the exam, and at this stage, I think I would pass. But I am a little concerned that I know HOW to subnet but I don't nessacerally understand it. Heres my pickle:

Suppose you are given 192.168.10.0/26 and asked to get mask and the usuall hosts, subnets, broadcast etc. Here what I would do:

This is a class C address which means three network nodes and 1 host, therefor mask must be 255.255.255.x. A 26 bit address means 2 bits on the 4th octet meaning 11000000. This means the mask is 128+64+0+0+0+0+0+0 = 192. The Subnet mask is 255.255.255.192.
So far so good. I then use the following formula to get the subnets and host bit. 2^-2 = 2. There are 2 subnets. Then 2^6-2=64 hosts.
I then get useable subnets by using 256-192=64 as base. 1st subnet is 64, second is 64+64=128. 192 is not useable.
Right here is my problem. What does 2 subnets actually mean in reality? It follows from this that I can use the following hosts - 65-126 and 129-190. Is this correct? Does this mean that the addresses 192.168.10.65 up to 192.168.10.126 and then 192.168.10.129 up to 192.168.10.190?

Does the subnet mask change from the .126 to the .129 range? how can this be used to phsyically split a network into two subnets?

Subnetting is tough!!!

Kaminsky

Your very close to getting this down solid. Just a matter of seeing where it all fits in.

Firstly, you don't necessarily discount the first and last subnets. If the question says that subnet zero is allowed, you're going to have 4 subnets. The first starting at 0 - first host 1.

Apart from all the binary and the numbers and ranges and stuff, what is actually happening is you are creating seperate networks that have no idea about each other without routing in place from one to the other. Broadcasts on one will not be seen by the other at all. Collisions may still be a factor as they are both physically using the same wire.

What the host ranges mean is that on one subnet, that you spread around say personell, you can assign those individual host ip addresses wherever you like within that department and then use the other subnet for the hosts in say the marketting department. The "logical" (layer 3) address broadcasts on one subnet will not cause problems for the host of the other department. Don't just think of hosts being PCs. A host address could easily be assigned to a server located in that department.

You will still use the same subnet mask for both subnets as the mask denotes the structure of those subnets. IE you won't have a different subnet mask for each subnet. The routers etc will see a host address/mask combination and know what subnet it lies within.

Another confusing thing that you might come across is when you see a host address ending in 255. Now your brain screams at you that this is a broadcast but not necessarily so. Depending on how the subnet is broken up, it may be that the subnet crosses the 3rd octet boundary and .255 will be legitimate. This confused me a lot initially. What you understand about Class C can confuse you when you start working with As and Bs.

In server terms, hosts in one department will not be able to get to (see) hosts based in the other department unless there is routing in place to allow that to happen. Even with routing in place to allow that, restrictions could still be put in place via access lists on the routers so that only certain hosts can see the other subnet.

If you are still a bit unsure, may I suggest www.subnettingquestions.com for lots and lots of practice before your exam. If you get questions wrong, don't just move onto the next question, work out why you were wrong and that will really drill subnetting into your head for good and you should never have to relearn ever again.. well.. in theory.


I have yet to sit my CCNA but from all I read it is really important that you are really comfortable with these as they are usually just a part of a bigger question and doing them with ease can save you lots and lots of valuable exam time for use elsewhere.

Good Luck

horizonuser

Thanks for your reply. It does make a little more sense.

So, it would be possible to have two hosts, say 2 PC's, sitting on the same wire. One with an IP address of 192.168.10.66 and the other with address 192.168.10.128, both with subnet mask of 255.255.255.192 but they wouldn't be able to communicate without a router? Is this correct?

Kaminsky

Yep!

With subnet 0 (just in case)


1) Class C

2) 192 = 2 bits = 4 seperate subnets
(If this were Class B it would be 255.192 = 10 bits = 1,022 subnets)
(If this were Class A it would be 255.255.192 = 18 bits = 262,142 subnets)

3) 256-192 = 64 -2 = 62 hosts + Subnet + BCast
(If this were class B the SN range would be 0.64)
(If this were class A the SN range would be 0.0.64)


SN Ranges
1st) 192.168.10.0 hosts 192.168.10.1 - 192.168.10.62 BC 192.168.10.63
2nd) 192.168.10.64 hosts 192.168.10.65 - 192.168.10.126 BC 192.168.10.127
3rd) 192.168.10.128 hosts 192.168.10.129 - 192.168.10.190 BC 192.168.10.191
4th) 192.168.10.192 hosts 192.168.10.193 - 192.168.10.254 BC 192.168.10.255

192.168.10.66 is 2nd possible host ID on 2nd Subnet BUT your 192.168.10.128 is a subnet address. 192.168.10.129 would be fine in the 3rd subnet.

As long as router has a route between SN 2 and SN 3 they can communicate. If you wanted only one particular host in SN 3 to access SN 2 then that could be accomplished with access list blocking the rest of SN 3.


See how easy subnets are (Once the light bulb goes on that is.... )