How many possible host addresses are available for an IPv4 address where the first and second octets are reserved for network addresses?
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is this combinatorics? – Nosrati Jul 31 '18 at 08:12
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$2^8\times 2^8 = 2^{16}$ if you can only change the last two octets – Sonal_sqrt Jul 31 '18 at 08:21
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SO: What is the total amount of public IPv4 addresses? – Martin Sleziak Sep 01 '18 at 13:12