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XXX.XXX.XXX.XXX 189.136.123.130 Subnet C XXX Subnet A 192.168.0.128 255.255.255.192 Fa0/3 ?.?.?.? Subnet B Fa0/1 192.168.0.0 255.255.255.128 192.168.0.177 Fa0/2 192.168.0.98 192.168.0.0/25 74 Hosts 192.168.1.128/25 66
XXX.XXX.XXX.XXX 189.136.123.130 Subnet C XXX Subnet A 192.168.0.128 255.255.255.192 Fa0/3 ?.?.?.? Subnet B Fa0/1 192.168.0.0 255.255.255.128 192.168.0.177 Fa0/2 192.168.0.98 192.168.0.0/25 74 Hosts 192.168.1.128/25 66 Hosts 192.168.2.0/30 2 Hosts 192.168.0.128/25 XX Hosts 192.168.1.0/25 68 Hosts Subnet C XXX Subnet A 192.168.169.0 255.255.255.128 Fa0/3 Fa0/1 Subnet B 192.168.169.128 255.255.255.128 Fa0/2 Subnet C XXX Subnet A 192.168.0.64 255.255.255.224 Fa0/3 ?.?.?.? Fa0/1 192.168.0.80 Fa0/2 Subnet B 192.168.0.96 255.255.255.224 192.168.0.112
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Computer Networks
Authors: Andrew S. Tanenbaum, David J. Wetherall
5th edition
132126958, 978-0132126953
