subnetting on 291?

nelnel Member Posts: 2,859 ■□□□□□□□□□
just a quick q?
do you have to know this like the back of your hand? also does anything like vlsm etc come into the exam or not?

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  • doom969doom969 Member Posts: 304
    I got a few questions that needed subnetting on 291. Like, calculating the correct subnet or mask was a part icon_rolleyes.gif in finding the good answer. I also got 2 sims were subnetting was involved, so you had to be pretty sure, or you'd miss the sim. I wouldnt say that you have to know it like the back of your hand, but you definitly have to be able to do it. As for Variable length subnets, I did not get it on the test, but its in the Ms-Press Book so i gues you could get it on the exam. Maybe I was just lucky. icon_eek.gif

    GLuck to you

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  • Scooty70Scooty70 Member Posts: 2 ■□□□□□□□□□
    For me I had one question that was related to VLSM
    If you intend on doing CCNA then yes know it like the back of your hand and be able to do it in your head. I saw a previous post asking why you would use it and it's something that you will come across in networking, especially in the ISP world and is very useful for consverving addres space. We have for example a /28 assigned by our ISP for things like and etc
    To work this out simply we know that a /24 (Class C) gives you one network and 254 hosts (with the .255 being the broadcast address and .0 being the network address)
    In the above example we have a /28 which means we have borrowed 4 bits of the network address range to subnet
    We add these 4 bits up
    128 + 64 + 32 + 16 = 240
    Remembering that if we add up all the bits in an octet it would equal 255 (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1), but we can't forget we actually start at 0, so this gives us a number of 256 to work with (0 to 255 equals 256 numbers)
    If we subtract 240 from 256 we end up with 16
    Remembering we must have a broadcast and a network address this leaves us with 14 hosts
    Using say as an eample this means that is the network address - are the available hosts is the broadcast address is the next network address - are the availale hosts is the boradcast address

    and so on

    We can work out with all certainty that .0, .16, .32, .48, .64 etc are the networks (just keep adding 16)
    Subtract 1 will always be the broadcast address and plus one will be the first available host

    A question states that you need x hosts or x number of networks, what is the closest power of 2
    Example is we need 20 hosts, closest power of 2 is 5, 2 times 2 five times equals 32, the same to the power of 4 is only 16. We need to then steal 5 bits from the network, add the last 5 bits in the octet up and subtract from 256 and this is the / VLSM you need, in this case a /27 or (256 less 32 equals 224)

    I hope that makes some sort of sense, just remember it's all to the power of 2

  • nelnel Member Posts: 2,859 ■□□□□□□□□□
    cheers for the replies.

    i know how to subnet and how to use vlsm but i cant do it in my head lol!
    so i was just wondering how much MS push it.

    cheers again
    Xbox Live: Bring It On

    Bsc (hons) Network Computing - 1st Class
    WIP: Msc advanced networking
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