172.31.128.255/18 Is Unicast Address ???

Unanswered Question
NateBattle Thu, 10/25/2007 - 06:23

.255 is the broadcast on that network. The network is 172.31.128.0, the first and last usable is .1 and .254. The last octet would be

11111111 which equals 255.

As far as the quickest way to do it, I couldn't tell you. I just do it the long way.

srue Thu, 10/25/2007 - 06:36

172.31.128.255/18 is indeed a unicast address.

the subnet mask is: 255.255.192.0

in binary it is: 11111111.11111111.11000000.0

the ip address s: 172 . 31 .10000000.11111111

there are still 6 bits unused for host addresses.

the broadcast address would be: 172.16.191.255

xs_echoss Thu, 10/25/2007 - 08:43

Yes, it is.

the subnet number is 172.31.128.0, and first usable address is 172.31.128.1, the broadcast address is 172.31.159.255(cuz next subnet number is 172.31.160.0). So all others between 172.31.128.1~172.31.159.255 will be consider a usable host address, a unicast address.

GoBucks82 Thu, 10/25/2007 - 18:27

172.31.160.0 would not be the next subnet. The next subnet would be 172.31.192.0

scottmac Thu, 10/25/2007 - 10:26

The thing to remember is that to be a broadcast address, all host bits must be Ones, no matter how many bits there are in the host portion of the address, the must all be ones to be a broadcast.

If any of the host-portion bits are a zero, then it is not a conventional broadcast adress.

To be a "network address" all of the host-portion bits must be a zero. If any host-portion bits are not zero, then it is not a "network address" (aka "network nuber").

So, in your example, even though the entire last octet is ones, there are still six more positions that must also be ones in the third octet (mask = 255.255.192.0 = 11111111.11111111.11000000.000000 ... all of the zeros in the third octet would need to be ones for this to be a broadcast address).

IF you ever get confused, just map it on paper, all host ones = broadcast, all host zeros = network address/network number.

Good Luck

Scott

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