[Internet-e-architetture-di-rete] riepilogo 20160321 e esercizi

[Antonio Prado]

> On 05 Apr 2016, at 23:30, daniele magiera <daniele.magiera at gmail.com> wrote:
> 
> Q: 128.127.128.127 /1
> A: 128.0.0.0 - 255.255.255.255
Ok

> Q: 128.127.128.127 /31
> A: 128.127.128.126 - 128.127.128.127
Ok

> Q: 128.127.128.127 /17
> A: 128.127.128.0 - 128.127.255.255
Ok

> Q: 128.127.128.127 /15
> A: 128.126.0.0 - 128.127.255.255
Ok

> Q: 128.127.128.127 / 255.254.0.0
> A: 128.126.0.0 - 128.127.255.255
Ok

> Q: 128.127.128.127 / 224.0.0.0
> A: 128.0.0.0 - 159.255.255.255
Ok

> Q: 128.127.128.127 / 255.255.224.0
> A: 128.127.128.0 - 128.127.159.255
Ok

> Q: 128.127.128.127 / 255.255.255.64
> A:  La net mask non è composta da bit contigui ( 255.255.255.64 = 11111111.11111111.11111111.01000000 ), si ottiene l’equivalente di una netmask 255.255.255.0
> Con indirizzi alternati

direi di no, spiega meglio
grazie
--
antonio