In a block of addresses, we know the ip address of one host is 182.44.82.16/26.what are the first address (network address) and the last address in this block
Answers
Following this answer https://serverfault.com/a/327782/288201 I assume that:
/23 means 23 bits for the subnet. At 8 bits per byte we get 16 bits for the first two bytes, and then seven (7) for the third => 1111 1111. 1111 1111. 1111 1110. 0000 0000, right?
It means our last two bytes will be of the form (nnnn nnnH. HHHH HHHH), where n is a subnet bit and H a host bit.
Thus the network address is (nnnn nnn|0 0000 0000) -> .0, the broadcast is (nnnn nnn|1 1111 1111) -> 2^9 is broadcast? Impossible. Here I miss something, something simple. So what would be the first and last IP? This is what I don't get following the answer mentioned in the link above (by b0fh).
First 192.168.176.1, last 192.168.177.254?
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