on
Let's assume for the simplicity sake that we have a flat layer 2 network - no routers - 500 PC's and one big subnet.

Let's say you had one big private Class B network 10.120.x.x... which would give you like 65k IP's...  and by using 255.255.0.0 as the subnet mask.. each device understands 10.120 is the network and everything from 10.120.0.0 to 10.120.255.255 are hosts.

That makes sense to me.

A simple Class C - 256 IP addresses makes sense as well.

But what about subnetting to 255.255.254.0 which combines 2 Class C's to 512 or so hosts...

My question is...  where is the RANGE defined?  Let's say Host A is 10.120.50.50 with a subnet of 255.255.254.0 and Host B is 10.120.60.50....  they can talk right?  But could 10.120.60.50 talk to 10.120.40.50?  Shouldn't right?  But could 10.120.50.50 talk to 10.120.40.50?

I know this is a simple answer... to a question that probably doesn't make sense.

Thanks!
Comment
Watch Question

Do more with

EXPERT OFFICE® is a registered trademark of EXPERTS EXCHANGE®
Commented:
Break it down into binary and you'll get your answer; The network portion gets 23 bits of the address, and the host portion gets 9.  To get the range for the hosts, you'd take the number of bits available and multiply it to the power of 2 minus 2 (as all 1s and all 0s are not allowed).  Thus you'd have a total of 512 minus 2 hosts, which is 510.  You'd be able to to assign 510 hosts in a single range of each of the networks.  You'd do the same thing to get the number of networks you'd have available except that you wouldnt minus 2.  Thus with a /23 mask (255.255.254.0), you'd have available 8,388,608 networks each with 512 hosts.

Commented:
>Let's say Host A is 10.120.50.50 with a subnet of 255.255.254.0 and Host B is 10.120.60.50....  they can talk right?

Is my basic understanding of Subnetting flawed?  Those two IP's are on different networks if the subnet is as shown.

1111 1111.1111 1111.1111 1110.0000 0000

The bits that are 1 need to be identical on two addresses for them to be on the same network.  With the network shown there are 512 addresses (less the reserved ones).

If you had said 10.120.50.50 and 10.120.51.50 then I would agree they could talk.

Commented:
I was typing when Henry was posting, apologies for an resulting duplication.

Commented:
Theres also a trick for figuring out what the network is and what the host is for individual addresses using the net mask.  It's called ANDING.  Its a mathmatical process that converts the IP address and the subnet mask into binary, places them both on top and compares the two.  The comparison creates a third number which will be the network address.  This is accomplished by a 1 and a 1 equal a 1.  A 1 and a 0 equal a 0.  A 0 and a 0 equal a 0.  Line the numbers of exactly, top to bottom and do this AND process and you'll be left with the network address.  So in your first example, you had the number 10.120.60.50.  Here's what it'd look like;

11111111-11111111-11111110-00000000 = Subnet Mask of 255.255.254.0
00001010-01111000-00111100-00000000 = AND functions results, the network for the address.

So converted back to dotted decimal, that address is 10.120.60.0.  That would be the network that 10.120.60.50 belongs to and any other address that works out to that same address would be in the same subnet.

Commented:
There are all kinds of programs online that not only help you to understand subnetting better, but will allow you to practice your subnetting, which if you want to get into the networking field, is an important skill to maintain.  CCIE level subnetting is something you have to practice on a regular basis.

Commented:
For the range it is the 512 adjacent addresses in the mask starting from X.X.0.0 - X.X.1.255 minus the network of 0.0 and broadcast of 1.255.  This would keep going through the addresses.  With address x.x.2.0 a network address for a new network and x.x.3.255 the broadcast for the next network and so on, where the 0.0 network and the 2.0 network are 2 separate networks.

and so on below just small example

4.0 - 5.255
6.0 - 7.255
8.0 - 9.255
10.0 - 11.255
12.0 - 13.255
14.0 - 15.255
16.0 - 17.255
18.0 - 18.255

in these networks the even numbers start the network with the top including the broadcast are the odd numbers in the second octet.

the x.x.50.0 network is different the the x.x.60.x
and the x.x.50.x and x.x.40.x are  both different so neither of your example networks are able to communicate with out a router.

x.x.50.50 could talk to the x.x.51.50 network no problem hope this clears it up a little.

Commented:
I think you mean:-

4.0 - 7.255
8.0 - 15.255
16.0 - 31.255

where:-

4 to 7 are all the possible combinations using 3 bits
8 to 15 are all the possible combinations using 4 bits
16 to 31 are all the possible combinations using 5 bits

(less the special cases of course)

Commented:
I was pretty sure we were discussing the 255.255.254.0 subnet, which is only 4.0 through 5.255 a total of 510 usable addresses and 512 total minus the network and broadcast?

We only freed up a single bit in the second octet, and all 8 in the last, the subnet that allows the numbering you state in your response moorhouselondo would be

255.255.252.0 not 255.255.254, unless I am really missing something?

Commented:
Cheever, sorry I thought you were outlining available address ranges if the number of bits in the subnet were expanded, which on re-reading I see your explanation is clear.  Yes please ignore my last comment.

Do more with