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Subnetting Help

Discussion in 'Windows Server 2003 / 2008 / 2012 Exams' started by masterchi, Feb 20, 2012.

  1. masterchi

    masterchi New Member

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    Hopefully someone can help me with subnetting. I found some good articles and i have the majority of it down on how to do it but i'm still rusty in some places.

    I did learn subnetting about 5 years ago to complete my CCNA but haven't used it since then so now that i'm trying to finish up my MCSA i have to relearn subnetting for test 70-291. Any help would be appreciated as i'm going through subnettingquestions.com right now and a few trick me up.


    - How many subnets and hosts per subnet from 10.0.0.0 /20?
    Answer = 4096 subs and 4094 hosts.

    ----how did they get that answer. I thought i was doing good because i figured out /20 was a 255.255.240.0 mask which means each subnet increments by 16 so i counted subnets like "10.0.0.0 (1), 10.0.16.0 (2), 10.0.32.0 (3), 10.0.48.0 (4)..." and so on by increments of 16 which gave me 17 subs and of course 254 hosts for each subnet. I have no idea how they gathered 4096 subs and 4094 hosts.



    -What is the valid host range from 172.31.1.134 /28?
    Answer = 172.31.1.129 - 172.31.1.142
    **I worked it out to be 172.31.1.128-172.31.1.144. I see my mistake as from number to number should be 14 hosts but remember this question as it will affect my reasoning for the next one pertaining to it being a Class B address yet the subnet 255.255.240.0 is only causing me to make adjustments to the last octet**

    -What is the broadcast of 10.190.208.0 255.255.240.0?
    Answer = 10.190.223.255
    **I answered incorrectly of 10.190.208.254. What the hell? This is the exact same subnet (255.255.240.0) as the first question and the only difference is that this is a Class A address. Since the Class B address above had me only changing the last octet (in the d class range) so i assumed it was the same for this question as its the same subnet. Now it seems like if i have a Class A address i should always change the 3rd octet while Class B address's have me change the 4th octets. Makes no sense to me that its skipping an octet. I would assume by logic that if i have a Class A address such as this on (10.190.208.0) then i have free range to change numbers after the first octet (the 190.208.0) and if i had a Class B address then i could change the third octet over. In these two questions it turned out that a Class A can only change the third octet and on and Class B changes only 4th octet.

    Please tell me how stupid i am and where i am going wrong? Am i overthinking it?
     
  2. Nyx

    Nyx Byte Poster

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    was actually thinking of these questions before falling asleep, I've learned subnetting at least 4 times in my life but never needed it at work and it fades away...
    all the answers are correct.

    re q1.
    you have maximum of 4096 subnets, each having two addresses, network and broadcast. you'd never see that in real life but what do I know. 4094 is the maximum number of hosts if you don't subnet it(minus network and broadcast). I don't understand your reasoning here, don't really know where to start explaining.

    re q2.
    network is 172.131.1.128, broadcast 172.31.1.143, between them you have hosts range.

    re q3.
    with mask of 20 you have last 12 bits of network address to change, change them all to 1's and you get 10.190.223.255

    have a google for subnetting examples, there's quite a few posts about subnetting on this forum as well. Once it 'clicks' you'll be fine.

    obviously if I'm wrong at any point please correct me:)
     
  3. soundian

    soundian Gigabyte Poster

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    1.
    there are 12 bits that are used in the subnet: the 8 in the second octet and 4 in the third. 2^12=4096
    there are also 12 bits on the host range so 4094=4096-2 for network and broadcast addresses
    2. You got that one.
    3. I only take into account the class of the address as far as making sure the subnet mask is viable (for example, a class B address must start 255.255.x.x, it cannot start 255.0.0.0 for example). Which octect the subnet mask is not equal to 0 or 255 is the "major" octet, as I like to think of it. If the major octet is the fourth octet your reasoning is correct, if it falls on the 2nd or 3rd octets the broadcast address will always end in 255.
    A quick sanity check:
    The 'major' octet will have broadcast address=network address+block size-1
    all other octets to the right of the this will have 0 for network or 255 for broadcast
    any to the left will be unchanged.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  4. masterchi

    masterchi New Member

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    Thank you both for that help, i'm going through some more questions online but hopefully this helped me. So Nyx, to find broadcast address's you would take the left over bits and then change them into ones after converting the address into binary? I did that and it did give me the right answer.

    I guess my biggest confusion (and maybe others) is that there is about 3-4 different equations depending on the scenario and i just get confused on which one to do when. Is there a catchall equation that can do it all or maybe could you explain "use the Anding method for finding subnet masks, use powers of 2 to figure out hosts and subnets, etc"

    Soundian - "1.
    there are 12 bits that are used in the subnet: the 8 in the second octet and 4 in the third. 2^12=4096
    there are also 12 bits on the host range so 4094=4096-2 for network and broadcast addresses"

    I'm looking over question 1 and i'm confused how you got an 8 and a 4 from 10.0.0.0 /20.
     
    Last edited: Feb 22, 2012
  5. masterchi

    masterchi New Member

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    Got it now. Since 10.0.0.0 was a class A address and /20 in binary is 11111111.11111111.11110000.00000000 that means the zero's are hosts and the 1's are network. Since its a class A we ignore the first octet of 1's and whats left is 12 one's by combining the second and third octet. As for the host, simply count the remaining zero's and then just "power of 2" for each of those numbers and don't forget to minus 2 digits from the host side. I tried this out on about 4 different problems with success. Thanks so much all of you.
     
  6. RichGK

    RichGK Bit Poster

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    As I'm just about to seriously work on my MCITP this topic has really helped me get to grips with subnets but I'm confused over one aspect of the answer for question 1, which is why is the first octet ignored when using a subnet /20 ?
     
    Certifications: Dip Comp (Open)
    WIP: MCITP
  7. soundian

    soundian Gigabyte Poster

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    You don't borrow network bits with subnetting, only host bits. That means that this class A address only has the last 3 octets available to borrow from.
    Similarly class B and C addresses only have 2 and 1 available respectively.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  8. RichGK

    RichGK Bit Poster

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    So say for example I've been allocated 10.0.0.0 and subnet /20 then with my network I have got 4096 subnets and 4094 host addresses to allocate.

    eg this would give the following situation where IP 10.0.32.1 (mask 255.255.240.0) is on one subnet and 10.0.16.1 (mask 255.255.240.0) is on another subnet?
     
    Certifications: Dip Comp (Open)
    WIP: MCITP

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