IP Subnet - Class C address?

The network mask, when represented in binary, is a number of Ones then a number of Zeros.

The ones part represent the Network half of the address, the Zeros the Host part.

(That's why net masks can also be written as eg. address/24 or address/28, that's the number of network bits out of the 32 Bit address).

Dotted decimal notation represents each eight-bit byte of the 32 bit address or mask as it's decimal equivalent.

Each 255 is 'all ones' in binary, so with just those it would be a /24 mask

192 = 1100 0000 binary; the Host part is six bits.

Those extra two bits mean it's a /26; 32-26 = 6 as another way of getting the number of host bits.

Six bits have a numeric range of 64; from 0 to 63 inclusive.

The Zero host is the 'network address' itself, the 'all ones' host is the broadcast address for that subnet.

That leaves 62 addresses usable for hosts.

Hi with a decimal dot address system of four octets it is the last decimal dot which determines this the 192 to 255 is the allocated number except the last one is a broadcast address.

this means you have 62 user available addresses.

192 = sshhhhhh

32+16+8+4+2+1 = 63

1 address should be allocated for broadcast

So, the number of hosts per subnet is 62

This easiest way to work this out using a class C network is this

255 - the last octet in your subnet mask....192

255 - 192 = 63, take off one for broadcast and that's your answer.

to work out class B do the same but with an earlier octet.

255.255.192.0 would give you the following

255 - 192 = 63

with 255 hosts per single number in the subnet you get

(63 * 255) - 1 for your broadcast, this gives you the hosts per class B!

In this case it is 16064 hosts per subnet.

The same for class A!

Easy innit?

You use a Yoda mask then you put it inside the switch gate. You turn the nozzle. You should then have C and B assigned by then.