Subnetting Class B & C

If I subnet the Class B address 172.16.0.0 255.255.192.0

Then

Subnets = 2
Host = 16,382

I understand that in subnetting class B addresses you use the same subnet numbers in the 3rd octect as you do in the 4th octect with class C addresses.

The first subnet is 64.0 while the first host is 64.1 : the last host is 127.254
Thats all the info that i got from my CCNA book.
It didnt tell me the valid host in between 64.1 -- 127.254.

So I did a little math....... I figured that 64.1 - 64.254 then it repeats to 65.1 - 65.254
254 x 63 = 16,002 Im off 380....... so

My confusion comes from the 4th octect. What numbers does the fourth octect generate while the third octect uses the "regular" subnet numbers? What are the valid host, lets say the first 65, that I could use?

Later

I forgot the class C........

If I have the class C address 192.168.10.32 my subnet mask should be 255.255.255.0

Then how would I subnet the main address if my subnet mask was 255.255.255.240?
My confusion stems from not knowing how to subnet a Class C address with a number already in the 4th octect.

Sorry for the double post.

Are you studying for the ccna?

If so what books did you get or are using?
I have the most current from cisco they are the intro/cnda(or whatever)
and in the intro book in the 5th chapter it covers in great detail the ip adressing and subnet masking, classes a/b/c,and much more. Its funny that you posted this cuz I just got done reading this area...

/me shudders at the memories of the CCNA studies.


When I figure out how to get my CCNA study guide out from under my desk without it collapsing, I'll post ya an answer.

(Don't feel bad, I never got this advanced math like stuff either.)

Well I got Sybex Fourth Edition
Also have CCNA 1 and 2 Companion Guide 3rd Edition

:)
If you have a Class B address your last two octets are defined as host bits. That's why the subnet mask is 255.255.0.0. When subnetting,
you are "stealing" host bits beginning from the left side, the most significant bits. In your CCNA books there should be a formula:

n to the power of "number of bits you stole" - 2. So if you are stealing two bits, that's
2 to the power of 2 minus 2, or 4 minus 2= 2. That's two subnets or a mask of 255.255.192.0
Consider how many of the original 16 host bits are left. 14. You apply the same formula,
2 to the power of 14 minus 2, you'll end up with 16,382 hosts.

How did the number 192 in the 3rd octet come about?

11111111.11111111.00000000.00000000

This is the mask in binary, or 255.255.0.0 in decimal. The left most bit in the 3rd octet

represents the value of 128 in decimal, or 10000000. Add the next most significant bit

like 11000000, and the decimal representation would be...192, because the second bit

holds the value of 64, and 128 + 64 is 192. Likewise, by stealing 8 bits you'd end up

with...11111111 in the third octet, or 255 decimal. The 3rd octet is now full. Want to

steal more bits? OK, you continue to ravage you "Class B address by stealing 10 bits.

Remember, we now are involving the fourth octet as well:

11111111.11000000 , so now the decimal representation of the fourth octet is...192.

Lets say you need a **** load of subnets, like more than 255. Steal 10 bits. Using the formula you'll end up with 1022 subnets and having now only 6 host bits left, that'll give you 62 hosts per subnet. Your subnet mask will look like this then: 255.255.255.192.

With a Class C address and a subnet mask of 255.255.255.0, only the last octet defines your host bits, so in order to subnet, you can only steal bits from that octet. The max number of subnets (useful) would be 64. Steal more bits and you'll have to read RFC 3021.

In classless inter domain routing, CIDR, and supernetting with VLSM the use of IP subnet zero is allowed, but a routers' software must support it as well.

Old fashioned IP classes are defined by the starting bits in the first octet. There was a time when subnet masks weren't used, but it was a wasteful and inefficient way to assign IP addresses, especially in light of the growth of the internet. Hence subnet masks. You'll have to get a bit into boolean algebra and learn the concept of "ANDing". You are dealing with binary numbers ( 8 bits per octet ) that are written in decimal to make it easier for humans to read, but the underlying concept of an IP address "ANDed" with a mask creates a mathematically constrained result, which defines your subnet, range of hosts and broadcast address. By your choice of number of stolen bits, or sliding the boundary, you have more control over the needs of a networks IP address assignment.

If you want to get into networking, there is simply no way around the need to know this subject. But you can also d/l a free subnet calculator from "Solarwinds" and play with it.
It will give you a bit of a visual reference as to what is happening. In decimal at least.

Anyway, I hardly believe that I did the subject much justice, it takes more than a few paragraphs, but I tried.

OK, now on to IPv6... just joking

OK, I have to finish what I started...

A Class C address has been assigned to you. 192.168.10.64 with a mask of 255.255.255.192. Your ISP has effectively constrained the number of host bits that are available for you to use.

A look at the subnet mask : 11111111.11111111.11111111.11000000

By virtue of the mask, you only have 6 host bits. That is a total of 62 hosts.

But look at the subnet mask again. From the ISP's perspective, they have subnetted

192.168.10.0 into 4 subnets .0 to .63, .64 to .127, .128 to .191 and .192 to .255.

They assigned the 192.168.10.64 to 192.168.10.127 block to you. ( Never mind that

this is a private address). By assigning a 255.255.255.240 mask !within your network!,

you can actually steal some of those host bits again, and create a few of your own

subnets with a few hosts of their own. I'm going to leave the rest for you to figure

out...

I couldnt ask for more......

Sincerely prop