# question about subneting

Aug 5th, 2007

i read this in some book:

i have ip addrees 172.31.128.255/18 and it says that its unicast address!

how can that be? it says also that its broadcast address is 172.31.131.255.

from my calculations the broadcast is 172.31.191.255 and in the address

172.31.128.255 the 3rd octet is network number of sort so how can that be a unicast?!

Overall Rating: 3 (1 ratings)

## Replies

mohammedmahmoud Sun, 08/05/2007 - 05:59

Hi,

172.31.128.255/18 is a unicast indeed, please check its subnet mask:

255.255.192.0

The r3d octet is 1 1 0 0 0 0 0 0 (the first two bits must match, which is 128 meaning that they must be "1 0"

and the 4th octet is 0 0 0 0 0 0 0 0 (which means that the 4th octet can have any values from 0 to 255)

and thus having all ones (255) in the 4th octet is fine to be unicast as the broadcast of this subnet is 172.31.191.255 (all ones in the rest of the bits in the 3rd octets and all ones in the 4th octet).

And accordingly you are right in that the broadcast is 172.31.191.255.

HTH,

Mohammed Mahmoud.

Pavel Bykov Sun, 08/05/2007 - 12:07

For better understanding, write the address and mask in binary.

For an address to be a broadcast address, all HOST bits need to be 1's. If at leas one host bit is 0, than it's not broadcast.

Network address on the other hand is when all host bits are 0's. If at least one host bit is not 0, than it's not broadcast.

Your example in binary:

Address: 1010110.100011111.10 | 000000.11111111

Mask: 11111111.11111111.11 | 000000.00000000

I put "|" mark where network and host portions are divided. Mask specifies that (where string of 1's end)

As you can see, some of host bits in the address are not 1's, that means it's not broadcast address.

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