markpm

Trying to get my head around IP subnetting, again!
Ive tried several times, and it gets a bit easier every time, but something still gets my head hurting...

Found the following on a Cisco webpage. I can follow the whole bit no problem, apart from the end where it gives the broadcast address for that particular subnet. How the hell do they get the .191???
Just cant work it out at all!

"Example 3: Class B
Let's say you have a need for more subnets than 254. (Remember this is the maximum number of subnets in a single octet.) Sticking with our class B address, let's configure an 11 bit subnet. This means we will use all 8 bits from our 3rd octet and the first three bits from the 4th octet. The subnet mask is now 255.255.255.224 (128+64+32=224). Now you need to find out what subnet the following address is in: 172.16.10.170 255.255.255.224. First, denote the address in binary representation (just octets 3 and 4 for a class B address) like this:

00001010 10101010 (address representation 10.170)
11111111 11100000 (subnet mask representation 255.224-first 11 bits subnet)
-----------------
00001010 10100000 (results of logical "AND") 10 160
So, the address here is in subnet 172.16.10.160. The valid addresses for this subnet are 172.16.10.161 through 172.16.10.190 (.191 is the broadcast address). As soon as you hit 10.192, the bits in the subnet change and you move into subnet 10.192. "

Any subnetting wizards that can explain how the broadcast address is .191?

Cheers
Mark...

dsmith

Broadcast Address is the address with all bits of the Host portion set 1.

so in your example

00001010 10101010 (address representation 10.170)
11111111 11100000 (subnet mask representation 255.224-first 11 bits subnet)

Broadcast = Subnet + Host as all "1"
e.g
00001010 101/11111

gives 10.191

Deano

GavinP

I find it easiest to get my head around using small subnets:

Using the example of a 192.168.1.x network and a mask of 255.255.255.252, the valid addresses are:

192.168.1.0 - Network address
192.168.1.1 - Valid IP address
192.168.1.2 - Valid IP address
192.168.1.3 - Broadcast address

If you write 252 out as binary, it is shown as 11111100 (i.e. only the last two digits are available - everything else is masked).

If you can only use the last two digits, you can only make four binary numbers:

00000000 - Zero
00000001 - One
00000010 - Two
00000011 - Three

Having said all that, I cheat and use this :

http://www.cotse.com/networkcalculator.html

Thanks

Gavin

BazH

Find it easier in decimal, eg for above use 256-252 =4 so subnets are 192.168.1.0,4,8 and so on taking away 1 from the next network address to get the broadcast address. (explained much better by Todd Lammle - sybex)

dsmith

Decimal. pppffffft.

There 10 types of people, those that understand binary and those that don't

Deano

stevencotton

I love that joke I also love the non-null terminated string joke, but I'll spare ya.

Steve.

BazH

......... and there are those that are lazy barstewards

dsmith

so there's 11 types then ?

Deano

markpm

Thanks all,

I read something else today that sorted it out for me. Basically the last bit position in the subnet mask is the same as how the subnets will increment up.

ie. as in my example, the subnet x.x.11111111.11100000, the last bit position there (in the last octet) is at 32(decimal)
So as we know the subnet ID is 172.16.10.160 then your next subnet will start at 32 up from there. ie. 172.16.10.192 and so on
And as long as you remember to take out the first and the last addresses youve got your 'useable' host addresses.

I know what Im on about!

And Deano, would that be B types?