05-17-2016 12:29 PM - edited 03-08-2019 05:47 AM
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.
05-17-2016 01:05 PM
Subnet ID would be 10.100.0.0/17
Broadcast IP would be 10.100.127.255
05-17-2016 01:13 PM
The binary math in Joseph's post is better than what I had posted. +5 for Joseph.
HTH
Rick
05-17-2016 01:14 PM
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.
05-17-2016 01:22 PM
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
05-18-2016 02:43 AM
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.
05-18-2016 02:37 AM
Thanks Rick, but truth be told, I used an on-line subnet calculator. ;)
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