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Thread: Errors in LAB 5
02-21-2012, 12:03 PM #1Registered User
- Join Date
- Feb 2012
Errors in LAB 5
I believe that most of the answers to this lab are wrong.
220.127.116.11 to 18.104.22.168
15 = 00001111
so the first 20 bits are unchanged i.e. 22.214.171.124 /20
The answer given is 126.96.36.199/16 which would include for example 188.8.131.52
But the question is unclear because the CIDR notation is missing. without that this would be a Class A network and the IP addresses would both be on 184.108.40.206/8
192.168.1.0 - 192.168.120.0 again CIDR is missing so say we assume class C
the first 17 bits are unchanged 120<128
So a good summary would be 192.168.0.0/17
The answer given 192.168.0.0/25 which is only 192.168.0.0 - 192.168.0.127
Q2 the question gives two /13 networks. the answer /16 how can that be a summary.
220.127.116.11/12 I think works.
I could go on