Hi everyone. I need some help. i'm studying for the CISCO CCNA exam and am having a lot of problems. one of the qestions i have is:- What is the last valid host address on the subnet that the host 192.168.216.65/29 belongs to. A. 192.168.216.71 B. 192.168.216.70 C. 192.168.216.65 D. 192.168.216.64 i have been talking to my tuter about this via email but am not able to work out the answer with what she is explaning to me. please could you explain how to work out the answer. Thanks Matt

hint: bottom and top ips of the subnet [network and broadcast] are not valid, correct? What does that leave you with?

I use a Subnetting table. This one does not include CIDR Notation but you can just write /24 to the left and fill it to the right. http://www.mcmcse.com/articles/subn... So if you use the Chart you will see they used 7 bits for Networks leaving 2 for the Hosts. Being that you need one address for the network and the other for the broadcast there is no hosts on this network only the broadcast which is the last address in the network which the answer. D. 192.168.216.64 This is a useless network by the way.

192.168.216.65 /29 note /29 not /24 which results in 6 hosts per subnet which makes it a valid network range. at /24 you have 254 hosts per subnet I believe if neckchopper remembers that network and broadcast ips are detracted from the total number of available hosts he will have his answer.

Ya, I just realized I screwed up. Your right it was 6 bits.

Hi guys sorry for taking so long to get back to you guys. Thanks for all your help. it helped out alot. Cheers Matt (Neckchopper)