1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.

subnetting

Discussion in 'Networks' started by Danmurph, Apr 8, 2010.

  1. Danmurph

    Danmurph Byte Poster

    127
    1
    27
    Hi Guys,

    I have a quick question, I was just flicking through the network plus book by Mike Myers validating everything I've done over the past month in this new job when I got stopped on the subnetting chapter,
    Mike is saying you are not allowed in any circumstance to use only 1 borrowed bit from the host portion of an IP address and I just wanted to make sure I wasn't goin mad in disagreeing with this?
    Surely I could have two subnets like this:

    192.168.0.1/25 :-

    network address - 192.168.0.0
    hosts - 192.168.0.1 - 192.168.0.126
    broadcast - 192.168.0.127

    Network address - 192.168.0.128
    hosts - 192.168.0.129 - 192.168.0.254
    Broadcast -192.168.0.255

    Im sure this is valid (even double checked in packet tracer), it also appears in the working example in chapter 7 his subnet masks are wrong?

    Can someone please tell me if I am right or am I missing something

    Thanks :D
     
    Certifications: MCDST, MCP, A+
    WIP: Everything!!
  2. SimonD

    SimonD Terabyte Poster Moderator

    3,463
    397
    199
    You certainly can, it's required for VLSM (we were doing the math the other day where we had /25 as an allowed subnet).

    All a /25 subnet would do would be to give you two networks of 124 hosts (for a class C network obviously) or from there breaking the subnet down into even smaller chunks for further networks.
     
    Certifications: CNA | CNE | CCNA | MCP | MCP+I | MCSE NT4 | MCSA 2003 | Security+ | MCSA:S 2003 | MCSE:S 2003 | MCTS:SCCM 2007 | MCTS:Win 7 | MCITP:EDA7 | MCITP:SA | MCITP:EA | MCTS:Hyper-V | VCP 4 | ITIL v3 Foundation | VCP 5 DCV | VCP 5 Cloud | VCP6 NV | VCP6 DCV | VCAP 5.5 DCA
    WIP: VCP6-CMA, VCAP-DCD and Linux + (and possibly VCIX-NV).
  3. Danmurph

    Danmurph Byte Poster

    127
    1
    27
    I can see why they say this now, this seems to be just for exam purposes (not using all 1's and all 0's).
    I am however stuck on another part and am so close to cracking this I can feel that I am over thinking it! :x
    I understand how to figure out how many hosts and how many subnets for any given subnet mask, what I am now stuck on is this, I understand how to seperate the subnets of addresses in a class C environment e.g.
    192.168.1.0/27
    192.168.1.33 - 192.168.1.62
    192.168.1.65 - 192.168.1.94
    etc
    etc
    But how do i do this for a class a or b say 172.21.0.0
    with a mask of 255.255.255.192?
    I know that the hosts are in blocks of 62 and that there are 1024 subnets but how do they start and how are they divided into blocks of 62, if I could just see a pattern of the first maybe 5 - 6 blocks of hosts this would be a huge help for me,

    If I could get to the bottom of this today this would be great, I fell like I've wasted my day bashing my head with this :oops:
     
    Certifications: MCDST, MCP, A+
    WIP: Everything!!
  4. SimonD

    SimonD Terabyte Poster Moderator

    3,463
    397
    199
    A class B address usually wouldn't be broken up that small, you can still do VLSM with class A and B subnets but instead of using the 4th octet you would be using the 2nd and 3rd octets to break down the size of the subnets\hosts, again the math is still fairly straight forward.

    A good idea is to write out a table like this

    2048 - 1024 - 512 - 256 - 128 - 64 - 32 - 16 - 8 - 4 - 2
    0012 - 0011 - 010 - 008 - 007 - 06 - 05 - 04 - 3 - 2 - 1
    (the 0's are to pad it out to make it a tad easier to understand, you can go further up to 4096 and beyond but I am doing this to show you how to work it out).

    It's important to know that with VLSM you have to list the hosts in size order (largest to smallest), that allows us to create the subnets correctly (failing to do that means you will have serious problems later on).

    Looking at the top line above we have to know how many hosts are required for each subnet, for instance if we wanted 1000 hosts we would have to use the 1024 point which is 11 bits from the host part of the mask, with that in mind we then know that we need to use 11 bits, what we do then is take 11 away from 32 (leaving us 23), that then means that we have a /26 subnet mask for the first network so our first subnet would actually be 172.21.0.0/23 the first host would be 172.21.0.1, the last usable host would be 172.21.3.254. Why? because to use all 1022 addresses would mean that we had to use all of the last octet 4 times, therefore increasing the 3rd octet from 0 - 3).

    ok?
     
    Certifications: CNA | CNE | CCNA | MCP | MCP+I | MCSE NT4 | MCSA 2003 | Security+ | MCSA:S 2003 | MCSE:S 2003 | MCTS:SCCM 2007 | MCTS:Win 7 | MCITP:EDA7 | MCITP:SA | MCITP:EA | MCTS:Hyper-V | VCP 4 | ITIL v3 Foundation | VCP 5 DCV | VCP 5 Cloud | VCP6 NV | VCP6 DCV | VCAP 5.5 DCA
    WIP: VCP6-CMA, VCAP-DCD and Linux + (and possibly VCIX-NV).
  5. Danmurph

    Danmurph Byte Poster

    127
    1
    27
    Thanks Simon,

    This is a little clearer but I am still unclear as to where I start the subnets, I think my original understanding of this topic is all wrong,
    If you take a look at this example I have been following
    Example 2: 255.255.240.0 (class b)

    1. 2-2=14 subnets
    2. 2-2=4094 hosts per subnet
    3. 256-240=16.0, 32.0, 48.0, 64.0, etc.
    4. Broadcast for the 16.0 subnet is 31.255. Broadcast for the 32.0 subnet is 47.255, etc.
    5. The valid hosts are:


    Subnet 16.0 32.0 48.0 64.0
    first host 16.1 32.1 48.1 64.1
    last host 31.254 47.254 63.254 79.254
    broadcast 31.255 47.255 63.255 79.255

    What I don't understand is why the first host is 16.1 and why does it go up in 16's? I'm probably over thinking this now as I have been trying to get this all day, do you understand what I mean?
    When you are working in class C as I understand you always divide the 255 in the last octet into equal groups in the subnets calculated or is this wrong? damn I think I'm confusing myself now :oops:
     
    Certifications: MCDST, MCP, A+
    WIP: Everything!!
  6. SimonD

    SimonD Terabyte Poster Moderator

    3,463
    397
    199
    It goes up in increments based on where you're taking the last bit from, so in your example you're taking the last bit at 128+64+32+16 (which gives you your 240), as the 16 is the last bit used that's the increment that you need to use for your hosts.
     
    Certifications: CNA | CNE | CCNA | MCP | MCP+I | MCSE NT4 | MCSA 2003 | Security+ | MCSA:S 2003 | MCSE:S 2003 | MCTS:SCCM 2007 | MCTS:Win 7 | MCITP:EDA7 | MCITP:SA | MCITP:EA | MCTS:Hyper-V | VCP 4 | ITIL v3 Foundation | VCP 5 DCV | VCP 5 Cloud | VCP6 NV | VCP6 DCV | VCAP 5.5 DCA
    WIP: VCP6-CMA, VCAP-DCD and Linux + (and possibly VCIX-NV).
  7. Danmurph

    Danmurph Byte Poster

    127
    1
    27
    Simon, I have read text upon text and not one place has mentioned this simple but now obvious fact!
    Thanks to you I can sleep easy now, this has made everything seem so much clearer, so THANK YOU DUDE!!
    :D
     
    Certifications: MCDST, MCP, A+
    WIP: Everything!!
  8. alebleicker

    alebleicker Bit Poster

    13
    0
    21
    Hi Simon,

    I can't understand why taking 11 bits from 32 results in 23 and not 21, I believe there is something so stupid I'm doing so far that I can't notice what it is now. it is the same with the example in the other topic, 255.255.255.0 take 3 bits and ends with 255.255.224. I just can't see what is wrong :/ I'm quite frustrated with this, sorry
     
    Certifications: A+,N+,70-270,70-290
    WIP: MCSA
  9. DC Pr0Mo

    DC Pr0Mo Kilobyte Poster

    265
    6
    41
    I think its cause there is a typo in simons table, it goes from right to left 008 - 010 (no 009). looking like you need 11 bits for 1000 address when you only need 10 bits, meaning a /22 network.
     
    Certifications: MCDST | BSc Network Computing
    WIP: 70-291 | 70-293 | 70-294 | 70-297
  10. SimonD

    SimonD Terabyte Poster Moderator

    3,463
    397
    199
    You're correct, there is a typo, well spotted :)
     
    Certifications: CNA | CNE | CCNA | MCP | MCP+I | MCSE NT4 | MCSA 2003 | Security+ | MCSA:S 2003 | MCSE:S 2003 | MCTS:SCCM 2007 | MCTS:Win 7 | MCITP:EDA7 | MCITP:SA | MCITP:EA | MCTS:Hyper-V | VCP 4 | ITIL v3 Foundation | VCP 5 DCV | VCP 5 Cloud | VCP6 NV | VCP6 DCV | VCAP 5.5 DCA
    WIP: VCP6-CMA, VCAP-DCD and Linux + (and possibly VCIX-NV).
  11. bazzawood30

    bazzawood30 Byte Poster

    176
    4
    17
    Just going back to the start for a min

    QUOTE]Mike is saying you are not allowed in any circumstance to use only 1 borrowed bit from the host portion of an IP address and I just wanted to make sure I wasn't goin mad in disagreeing with this?[/QUOTE]

    Did the book perhaps mean that if you had a /31 subnet you would only have a network address and a broadcast address and no hosts so it not usable? Making a /30 the first usable subnet for point to point links.
     
    Certifications: ECDL,A+,N+,CCENT,CCNA,MCP,MCDST
  12. bazzawood30

    bazzawood30 Byte Poster

    176
    4
    17
    there is a simple was to calculate the number of subnets you are creating using your binary chart.

    7 6 5 4 3 2 1 0
    128 64 32 16 8 4 2 1

    overlay the chart with 7 to 0 as shown above, us it in the following way.

    example /27

    3 extra network bits = 128 + 64 + 32 = /224

    last network bit 32 = sunbet jumps every 32 usable hosts 32-2=30

    number of subnets = count your extra network bits backwards on the binary chart counting zero
    as you can see on the chart 3 is over the 8 so you have 8 networks
    you can confim this by 256/32=8

    Draw this chart out for you exams and you can work out your subnets hosts in no time. I extend the chart out to 1024 just to be on the safe side.
     
    Certifications: ECDL,A+,N+,CCENT,CCNA,MCP,MCDST
  13. Johnd76

    Johnd76 Megabyte Poster

    704
    16
    64
    Certifications: MCP, MCDST
    WIP: Not a thing

Share This Page

Loading...