ipv4 question

[IZZYJOV76]

i have an ipv4 10.100.2.255 / 17

What is the subnet ID and the broadcast?

Not able to understand the broadcast id, i believe it is 10.100.2.255, but for the cisco lab is 10.100.3.255.

[Joseph W. Doherty]

Subnet ID would be 10.100.0.0/17

Broadcast IP would be 10.100.127.255

[Richard Burts]

The binary math in Joseph's post is better than what I had posted. +5 for Joseph.

HTH

Rick

[IZZYJOV76]

Consider whether it is a valid unicast IPv4 address and therefore will not be rejected when configured on a router interface.

10.100.2.255 255.255.254.0, it is /23..i believed i made a mistake putting /17..sorry

Answer :The number 10.100.2.255, with mask 255.255.254.0, is a valid IPv4 address. It is in a subnet whose subnet ID is 10.100.2.0 with subnet broadcast address 10.100.3.255.

[Richard Burts]

Thank you for posting more information from the lab. It does clarify what they lab was asking, and does clarify that the mask they were using was /23 rather than /17.

Joseph's binary math was correct for a /17 mask and the lab is correct that for a /23 mask that 10.100.2.255 is a valid unicast host address. 10.100.2.255 would be the broadcast address if the mask were /24 but not for /23.

Part of the point of this lab is that we have an instinctive reaction that 10.100.2.255 looks like a broadcast address (we are much more used to thinking in /24) but we always need to be aware of the impact of the mask.

HTH

Rick

[Joseph W. Doherty]

Consider whether it is a valid unicast IPv4 address and therefore will not be rejected when configured on a router interface.

10.100.2.255 / 17 is a valid unicast IPv4 address, although with an unusually large network.

Yes, for 10.100.2.255/23, network is 10.100.2.0/23 and broadcast is 10.100.3.253.

BTW, if it were 10.100.2.255/22, network would be 10.100.0.0/22, but broadcast would still be 10.100.3.255.

[Joseph W. Doherty]

Thanks Rick, but truth be told, I used an on-line subnet calculator.  ;)