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Thread: Errors in LAB 5

  1. #1
    Join Date
    Feb 2012
    Posts
    12

    Unhappy Errors in LAB 5

    Hello
    I believe that most of the answers to this lab are wrong.

    e.g. Q5
    66.66.0.0 to 66.66.15.0
    15 = 00001111
    so the first 20 bits are unchanged i.e. 66.66.0.0 /20
    The answer given is 66.66.0.0/16 which would include for example 66.66.100.0
    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 66.0.0.0/8

    Q6
    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.
    172.144.0.0/12 I think works.

    I could go on

  2. #2
    Join Date
    Mar 2013
    Posts
    2

    Question

    Quote Originally Posted by sgall View Post
    Hello
    I believe that most of the answers to this lab are wrong.

    e.g. Q5
    66.66.0.0 to 66.66.15.0
    15 = 00001111
    so the first 20 bits are unchanged i.e. 66.66.0.0 /20
    The answer given is 66.66.0.0/16 which would include for example 66.66.100.0
    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 66.0.0.0/8

    Q6
    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.
    172.144.0.0/12 I think works.

    I could go on
    Hello,

    I totally agree with Q2 and Q6, but I cannot follow you on Q5, as a possible answer for that would be 66.66.0.0 /20, because you have to summarize 66.66.0.0 through 66.66.15.0, that is block of 16, so 8+8+4=20 to get to 255.255.240 that gives you a block of 16. But again this is probably wrong but I have no idea why???

    Also, I cannot understand the Q4 Q7 and Q8!!!

    4. 203.168.6.0/24 and 203.168.60.0/24
    Interesting octet is the 3rd and to summarize this range I suppose that you should use block of 64, that is 8+8+2=16 to get 255.255.192.0 or /20.

    7. 172.16.1.0 through 172.16.7.0
    Why the summarize address is not the 172.16.0.0? Instead the answer is 172.16.1.0!!!?? ? We have a block size of 8, so it should be the 0-7 block of addresses!

    8. 192.168.128.0 through 192.168.190.0
    Again, interesting octet is the 3rd and to summarize this range I suppose that you should use block of 64, that is 8+8+2=16 to get 255.255.192.0 or /20.


    I am pretty sure that I haven't understand something, but I don't no what. I am so confused!!!

    Please someone help!!!

    Thanks in advance!

  3. #3
    Join Date
    Dec 2006
    Posts
    2,186

    Default

    There were errors and they are all documented here:
    http://www.lammle.com/discussion/showthread.php?t=4521

    Todd

  4. #4
    Join Date
    Mar 2013
    Posts
    2

    Smile

    Quote Originally Posted by lammle View Post
    There were errors and they are all documented here:
    http://www.lammle.com/discussion/showthread.php?t=4521

    Todd
    Thank you very much Mr. Lammle!!!
    I am sorry I didn't see that thread!!!
    Now I feel very pleased that I actually get summarization at first place.
    Thanks again!

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