And if you are allowing IP subnet-zero, then you have subnets=(2^18)-1 or 262143 subnets.
As far as network and broadcast addresses are concerned: For your example of 10.100.0.0/26, the network address is 10.100.0.0, and the broadcast address is 10.100.0.63 Don't even get me started on the new feature available in 12.2(T) where you can use /31 networks. ""Hunt Lee"" wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > so = /18 = 2^18 - 2 > > Number of subnet is 262142. > > > Hunt Lee wrote: > > > It would be great if someone can give me a hand on this: I know how to > > calculate the number of subents and number of hosts per subent, but I'm > > very confused about the Network address and the Broadcast address: > > > > Say I have a network: 100.10.0.0 255.255.255.192: > > > > 1) To work out the subnet: > > > > 100.10.0.0 is a Class A, so = /8 > > > > 255.255.255.192 = /26 > > > > Therefore, /26 - /8 = /14, > > > > The number of subnets = 2^14-2= 16382 > > > > 2) To work out the number of host: > > > > /32 - /26 = /6 > > > > The number of hosts = 2^6-2 = 62 hosts per subnets > > > > Thanks so much for your help in advance. > > > > Best Regards, > > Hunt Lee Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=23643&t=23632 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
