Preparation

KMAN24KMAN24 Posts: 44Member ■■□□□□□□□□
As far as VLSM, CIDR, SUPERNETTING and Route Summarization goes do you think one justs needs to know concepts well or should one expect to see questions on the exam such as the following :

You are given a Class C network 192.168.1.0/24. You need three networks with 60 hosts and two networks with 30 hosts. What are the subnet mask values you could use ? (Choose all Correct Answers)

a. 255.255.255.128 and 255.255.255.192
b. 255.255.255.224 and 255.255.255.240
c. 255.255.255.192 and 255.255.255.224
d. none

Answer to this is A,C

Another Example

You are given a class c network and you have four Lan segments with the following number of devices 120,60 and two with 30. What subnet mask values would you use to accommodate these segments?

a. /24 /25 /26
b /25 /26 /27
c /26 /27 /28
d none

answer none

So I am debating whether I need to be an expert in figuring these types of questions out quickly before attempting the exam. Any feedback and/or help understanding these would be much appreciated.

Comments

  • fuseboxfusebox Posts: 87Member ■■□□□□□□□□
    General consensus is that if you are able to memorize as much as possible with regards to subnetting etc, the better. I am not trying to memorize too much as I like to work things out in front of me on pad and paper when necessary, hopefully I will be allowed to do this in the exam.

    But having looked at your question, I am alittle confused with the answer to the second question.
    Im a newbie.... please be easy on me.
  • JerzJerz Posts: 86Member ■■□□□□□□□□
    fusebox wrote:
    But having looked at your question, I am alittle confused with the answer to the second question.

    I guess I'm confused on both. I would think "A" would be the answer to the first one and "B" would be the answer to the second one. Of course I haven't started studying for CCNA yet so that probably adds to my confusion.
  • tunerXtunerX Posts: 447Member ■■■□□□□□□□
    I would think answers C and B, respectively.

    The first 1 is 5 networks.

    255.255.255.192=/26 192.168.1.0
    255.255.255.192=/26 192.168.1.64
    255.255.255.192=/26 192.168.1.128
    255.255.255.224=/27 192.168.1.192
    255.255.255.224=/27 192.168.1.224

    The second question is 120, 60, 30

    255.255.255.128=/25 126 hosts
    255.255.255.192=/26 60 hosts
    255.255.255.224=/27 30 hosts

    192.168.1.0/25
    192.168.1.128/26
    192.168.1.192/27
    192.168.1.224/27
  • KMAN24KMAN24 Posts: 44Member ■■□□□□□□□□
    Yes you are right on the second one, the answer I should have posted is B per the book however on the first one the book claims A is also a correct answer because : it creates one 126 host segment and two 62 host segments. Note: I don't agree with this either
  • JerzJerz Posts: 86Member ■■□□□□□□□□
    OK, my reasoning:

    First Question:
    3 subnets-1=2bin =10 = 2 bit positions for subnets leaving 6 for hosts
    255.255.255.192 or /26
    256 - 192 = 64
    192.168.1.0
    192.168.1.64
    192.168.1.128
    192.168.1.192

    Second Question:
    2 subnets - 1 = 1bin = 1bit position for subnets leaving 7 for hosts
    256 - 128 = 128
    192.168.1.0
    192.168.1.128

    And now I think I see where I messed up…
    First mask: 255.255.255.11000000
    Second mask: 255.255.255.11100000 (since two bits were already being used up for 2 subnets in the first question I *guess* you have to take the next bit position for the other two subnets).

    SO:
    First mask: 255.255.255.192
    Second mask: 255.255.255.224 instead of the .128 that I previously thought.

    But I guess you could probably flip that around too (second part of the second question):
    2 subnets - 1 = 1bin = 1bit position for subnets leaving 7 for hosts
    256 - 128 = 128
    192.168.1.0
    192.168.1.128
    First Mask: 255.255.255.128

    3 subnets-1=2bin =10 = 2 bit positions for subnets leaving 6 for hosts
    255.255.255.11000000
    But since the first subnet bit position is already being used you would do:
    255.255.255.11100000 making this subnet mask 255.255.255.224
    Second mask: 255.255.255.224

    192.168.1.0
    192.168.1.32
    192.168.1.64
    192.168.1.96
    192.168.1.128
    192.168.1.160
    192.168.1.192
    192.168.1.224

    hmmm.... maybe I'm confusing myself.
  • tunerXtunerX Posts: 447Member ■■■□□□□□□□
    The first question says 3 subnets with 60 hosts and 2 subnets with 30 hosts. 3+2=5. With answer A you can only get a maximum of three networks. The question specifically asks for two things.

    3 networks with 60 hosts.
    2 networks with 30 hosts.

    Did you post the question incorrectly? Were you trying to ask for three networks total?

    1 network with 60 hosts.
    2 networks with 30 hosts.
  • xetrevxetrev Posts: 59Member ■■□□□□□□□□
    this is how i remember subnetting

    128 192 224 240 248 252 254 255

    if you have a /27 that means 27 zero's so you have 8 zeros in the first octet 16 in the next and 24 in the 3rd so 27-24 = 3 3 bits borrowed

    count 3 over from the left and you get 224

    then you just do 2^2-2 for hosts and subnets


    i hope this makes sense, also just practice, once you start to do a couple it sorta just makes sense after a while
  • sooners8507sooners8507 Posts: 6Member ■□□□□□□□□□
    xetrev wrote:
    this is how i remember subnetting

    128 192 224 240 248 252 254 255

    if you have a /27 that means 27 zero's so you have 8 zeros in the first octet 16 in the next and 24 in the 3rd so 27-24 = 3 3 bits borrowed

    count 3 over from the left and you get 224

    then you just do 2^2-2 for hosts and subnets


    i hope this makes sense, also just practice, once you start to do a couple it sorta just makes sense after a while

    Wow that would have helped me a lot on my test today. I was trying to do it all the complicated way. Thanks for the insight.
  • keenonkeenon Posts: 1,921Member ■■■■□□□□□□
    thats a great way i'm adding it to my brain
    Become the stainless steel sharp knife in a drawer full of rusty spoons
  • sunny_evandersunny_evander Posts: 126Member
    xetrev wrote:
    this is how i remember subnetting

    128 192 224 240 248 252 254 255


    This is a very interesting method and i learned it from a Cisco press book(Cisco Intro Exam) . Before learning it i was always confused and it took heaps of time to solve a question but this formula *we can call it* helped me a lot and i can do a question in seconds :D
    :santa:
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