Subnets and Adresses

banderas1978

What is the right way of doing this task, please?
I have a class C network, with an id of 198.255.8.0 with 4 subnets (3 of which are in use). What are the starting addresses of the available subnets?
Thank you.

dynamik

198.255.8.0
198.255.8.64
198.255.8.128
198.255.8.192

Those are the network numbers. Just increment each by one for the first assignable address. That's an easy one to do in your head because you can just split the last octet into fourths.

banderas1978

Thanks a lot!

websponge

wouldnt this be dependent on how many subnets you have in the first place?

if its 4 subnets you have, you borrow 3 bits from the host portion (giving 8 subnets, but 2 are unusable, unless you are using VLSM)

so you have 6 subnets

address are:

192.255.8.32
192.255.8.64
192.255.8.96
192.255.8.128
192.255.8.160
192.255.8.192

192.255.8.0 and 192.255.8.224 cannot be used as its part of the overall network address and broadcast address.

thats what I think anyway, but if Im wrong please correct me
but, your address` work on borrowing 2 bits only from the host allowing VLSM,

dynamik

You're able to use those two subnets with any modern equipment/protocol.

http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml

Here's the highlight if you don't feel like reading all of it:

Cisco.com wrote:

On the issue of using subnet zero and the all-ones subnet, RFC 1878 states, "This practice (of excluding all-zeros and all-ones subnets) is obsolete. Modern software will be able to utilize all definable networks." Today, the use of subnet zero and the all-ones subnet is generally accepted and most vendors support their use. However, on certain networks, particularly the ones using legacy software, the use of subnet zero and the all-ones subnet can lead to problems.

royal

To just go into the detail a little more:

You need 2 bits because 2^2 = 4 and you can use both 00 and 11 so you don't need to do 2^2-2 as you do with client bits.

So we have 2 bits to work with so we have:
00
01
10
11

That means that our last octet will look like:
00000000
01000000
10000000
11000000

00 = 0
01 = 64
10 = 128
11 = 192

That means that we can have:
198.255.8.00000000 = 192.255.8.0
192.255.8.01000000 = 192.255.8.64
192.255.8.10000000 = 192.255.8.128
192.255.8.11000000 = 192.255.8.192

shednik

Just to add to the great responses thus far everyone has a different way of looking and calculating it. For me if ip subnet zero is enabled, I usually take the network 198.255.8.0/24 and you said you have four subnets. So my mind immediately thinks ok so 256/4 =64 since your network is a /24 which makes the block size 64 and would make each subnet have a /26(255.255.255.192). Then from there you just interval going up by 64.

198.255.8.0
198.255.8.64
198.255.8.128
198.255.8.192

There is my quick explanation of how to subnet the /24 network.