TOPIC: IP Addressing Scheme - Confusing...

Hello People i dont know if i should be asking this. I have been studying Linux Network Administrators Guide. (arrrh 800+ pages) i i came across this which seemed different from what i saw in Craig hunts TCP/IP for Unix Admin. And guys please just clear me and make me understand may be the difference i see is me (lol)

This is it:

The book said : the size of the host part depends on the size of the network? mmmhmmmm.

Please explain this:
Class A 1.0.0.0 through 127.0.0.0. the network number is contained in the first octet. this class provides for a 24 - bit host part allowing roughly 1.6million host per network

Class B 128.0.0.0 through 191.255.0.0. the network number is in the first 2octets. this class allows for 16, 320 nets with 65, 024 hosts each

Class C 192.0.0.0 through 233.255.255.0. the network number is contained in the first 3 octet. this class allows for nearly 2 million networks with up to 254 hosts

I think i need a clearer understanding regarding the Host and net calculation on class A and B


Cheers People

Subnetting is always fun, it's what my students get hung up on the most...

For Class A networks the default subnet mask is this: 255.0.0.0
Represented as Network (N) bits and Host (H) bits:
NNNNNNNN.HHHHHHHH.HHHHHHHH.HHHHHHHH

For Class B networks the default subnet mask is this: 255.255.0.0
Represented as Network (N) bits and Host (H) bits:
NNNNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH

For Class C networks the default subnet mask is this: 255.255.255.0
Represented as Network (N) bits and Host (H) bits:
NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH

The equation for the amount of networks you can have in your new subnet after borrowing Host bits to make more Network bits is this: 2 ^ (# of borrowed H bits)

The equation for the amount of hosts you can have in your new subnet after borrowing is this: [2 ^ (# of remaining H bits)] - 2
(The "-2" at the end is to account for the Network ID and Broadcast ID addresses)

The equations remain the same regardless of what IP class you're dealing with. As the number of network (N) bits increase, the number of host (H) bits decrease - more networks, but less hosts in each network. As the number of network (N) bits decrease, the number of host (H) bits increases - less networks, huge blocks of hosts.

I hope this helped - if not lemme know and I'll explain it another way.

it shd b 223 instd of 233 !

@ Katzebnt ur explanation came in handy but the calculations u did i mean the 2^... does it match with the calculation i stated in my initial post?

for example how did u compute this or prove the computation :-
Class B 128.0.0.0 through 191.255.0.0. the network number is in the first 2octets. this class allows for 16, 320 nets with 65, 024 hosts each


@Hades
Man believe me thats how i saw it in the book (233) infact many stuff i saw is not matching with others i have seen else where. the whole thing is......?

That book is really confusing you! It has some typing errors.

For class A network:

Network bits are in the first octet.......8 network bits.
The first leftmost bit in class A network is the network identifier. It is always set to a '1XXXXXXX' to show you its 'Class A'. Keeping that in mind, the usable network bits are only 7 (count the Xs) from the octet. Therefore 2^7 = 128 networks.

Host bits occupy the next 3 octets. That gives you 24 host bits. Translating to 2^24 = 16,777,216 hosts (roughly 16 million hosts/network not 1.6 million). Also keep in mind that the 1st address (network I.D) and last address (broadcast address) cannot be assigned to a host. Thus you subtract from the total addresses (16,777,216). Also the whole of 127.0.0.0 is reserved for loopback testing. Thus subtract from the total number of addresses to get total number of USABLE address.

For Class B network:

Network bits are in the 1st and 2nd octet.......16 bits.
The two leftmost bits again in Class B network are the network identifier. They are always set as '10XXXXXX XXXXXXXX' to show you its 'Class B'. Keeping that in mind, the usable network bits are only 14 (count the Xs) from the two octets comprising network bits. Therefore total number of networks is 2^14.

Host bits occupy the remaining 2 octets. That gives you 16 host bits. Translating to 2^16 = 65,536 hosts. Also remember that the 1st address (network I.D) and the last address (broadcast address) of each network cannot be assigned to any host, so the must be subtracted from total number of hosts to get total number of USABLE hosts.

Nice Analysis Losh. I have a better catch of what i saw and i am very sure the problem was not coming from me but the book right?

Their calculations of the Octet for example in class B 2^14 ~ 16384.
The book gave 16320. Even if it had to be approximated their figure was far apart. Surprisingly it's an O'reily Compilation.

Anyways thanks Losh u presented a nice analysis. Thanks to everyone who contributed. I will make pen corrections in the book before i place it in my library and some other g33k finds the mistake. Lol.

Gracias.


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