binary and IP addresses

You are forgetiing to add the .0 address on the next subnet.

It's easier to use the whole octet (ie 255 not 256) and add 2 (1 per octet). 255 128 64 32 16 8 4 2 1 + 2 = 512

That does NOT give you 512 hosts anyways, it gives you 510 hosts, 1 network and 1 broadcast.

You do know about 2^N formula correct?

2^N = number of addresses including network and broadcast. 2^N-2= number of hosts 2^9-2=510

I found writing out the table a couple dozen times commits it to memory quite easily.

Hi

I was wondering if someone could clear something up for me:

when converting decimal to binary then 255 = 11111111 or 8 places: 128 64

32 16 8 2 4 1

When dealing with subnet masks and hosts, to get 512 hosts you have to use 9 places ie

256 128 64 32 16 8 2 4 1 but all ones add up to 511? what am I doing wrong here?

Thank you all! guess I need to revisit subnetting :)

Regards

Mark

Hi

255 is the last number from 0 ( zero )

form INCLUDING zero to 255 is 256

00000000 = 0 11111111 = 255

-----------------------------

dont make the mistake with hosts A subnet can handle x hoste -2 ( network and host address !!

192.168.0.0 / 24 also 255.255.255.0

192.168.0.0 Network address

192.168.0.1 Fist host 192.168.0.254 Last host 192.168.0.255 Broadcast address

a subnet of 8 free bits can be seen as 256 -2 == 254 host !

Rijk van Harn Ex CCNA ( planning exam next month )

"Mark Scott" schreef in bericht news:yTFdf.24283$ snipped-for-privacy@fe2.news.blueyonder.co.uk...