What is the last valid host on the subnetwork 172.25.190.0/24?

gouki2005

Not sure with this one./24 is last one is 254. but is a class B adress.help me

networker050184

Why would it being a class B have anything to do with it? Break it down to binary and see what you get.

azaghul

You're correct, .254 is the last usable address. Class A, B & C can be considered legacy terms, while 172.25.x.x is in the class B address space, the /24 is what is important.

MrBrian

Like the others have said, always look at the mask that is given to you for the IP space. It will tell you the bit boundary for the network and host bits. If no mask was given, then you'd assume /16 in this case, since 172 in the first octet is a class B by default (Class B = 128-191 in the first octet).

Since it uses a /24 bit mask, that means those first 24 bits are "frozen" or "locked" in. They don't change now. In addition, the /24 means that the remaining bits, 8 in this case, are for hosts. This makes it kind of easy because you can then simply disregard the first 24 bits and zoom your focus into those last 8 bits. Then just remember that the first and last possible bit counts in this space are reserved for: the actual network, and the broadcast address of the network.

So in this case, 172.25.190.0 and 172.25.190.255 can't be delegated to any hosts.

Meaning, the usable host address range is from 172.25.190.1 - 172.25.190.254... You're done!

TehToG

Class A, B, C addressing is just a convention used by IANA (Internet Assigned Numbers Authority) to assign public IP address spaces (and some private spaces too). It was introduced to stretch the IPv4 addresses further. The slash notation is an actual descriptor for the network.