Trouble with VLSM questions...

mappsy91mappsy91 Posts: 3Registered Users ■□□□□□□□□□
You are issued address 123.123.0.0/16. Note this is the correct public address, correctly allocated, not a misprint.

a)
Using VLSM Provide a table listing:

i)
Each additional sub-network address only for the subnets identified in figure 1
ii)The range of hosts for each subnet identified
iii)The broadcast address for each subnet identified

The topology is here:
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I don't really want the answers as I'd like to know what the hell I'm doing, just feel horribly stuck if someone could help me get going in the right direction that would be greatly appreciated!

Thanks,

Comments

  • MooseboostMooseboost Senior Member Posts: 767Member ■■■■□□□□□□
    What part of the question are you stuck on? Tell us what you have figured out so far, and we can help point you in the right direction.
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  • mappsy91mappsy91 Posts: 3Registered Users ■□□□□□□□□□
    So to get at least that many subnets I'd need to borrow 9 host bits to make it a /25 which would allow up to 512... Do I then just calculate the individual addresses in the figure? To work that out I just use the block size, which should be 128?
    For the range of hosts for each subnet, it should be the same amount of hosts in each subnet. So I use the subnet masks host bits and 2x-2 to get the range? Then the broadcast address is the last address of each subnet?
    Thanks for any help. I know these seem like easy questions, I just want to be sure I'm on the right path.
  • OctalDumpOctalDump Posts: 1,722Member
    Maybe I don't understand the question but I see 9 subnets, 3 of which have "sub-subnets". So you'd need to break up your /16 space to accommodate at least 9 subnets. The nearest power of 2 is 16, 2^4, so you'd take 4 more bits and have /20 networks with (2^12)-2 hosts. You could further break up one of those supermassive subnets to give you sufficient addresses for your routers etc.
    2017 Goals - Something Cisco, Something Linux, Agile PM
  • mappsy91mappsy91 Posts: 3Registered Users ■□□□□□□□□□
    Sorry, here's the question in full:

    you must devise a scheme allowing for more than 402 subnets (0 to 401) with 401 being the last subnet allocated for users (see figure 1). You are issued address 123.123.0.0/16. Note this is the correct public address, correctly allocated, not a misprint.

    a)Using VLSM Provide a table listing:

    i)Each additional sub-network address only for the subnets identified in figure 1
    ii)The range of hosts for each subnet identified
    iii)The broadcast address for each subnet identified

    Given the point of devising a scheme allowing for more than 402 subnets, does what I've said above make sense?
  • MenyusMenyus Posts: 1Registered Users ■□□□□□□□□□
    HEY

    did you manage to work this out?
  • OctalDumpOctalDump Posts: 1,722Member
    Ah, ok. Yeah, 9bits (2^9 = 512) for your subnets, leaving 7bits for hosts (2^7 -2 =126). Not sure if "including subnet xxx" means that it is a sub-subnet. Also it's really odd to have multiple subnets on one VLAN. Not sure what they mean by LAN, either - if they mean a switched LAN, or a maybe a campus network with Layer 3 routing. But then it's really odd to give all the devices on a LAN public IPs.

    Anyway your subnet mask is 255.255.255.128, so you've got subnets incrementing every 128 in the right most octet. You'd increment the second most octet by one every 2 subnets. So to get the subnet address for any particular subnet, you'd take the subnet number (0-401) and divide by 2. If you end up with a half (1.5 or 200.5 or whatever) then you put 128 in the last octet. That's probably really unclear, but for 401 (the 402nd subnet, since we are counting from 0), you'd have an address like 123.123.200.128.
    2017 Goals - Something Cisco, Something Linux, Agile PM
  • GS270GS270 Posts: 1Registered Users ■□□□□□□□□□
    Hello, can you expound further on this? Thanks. I don't understand the divide by two concept.
  • OctalDumpOctalDump Posts: 1,722Member
    Each octet is 256 values, so if the subnets are on a 128 boundary, there will be 2 per octet. So you count 10.0.0.0, 10.0.0.128, 10.0.1.0 10.0.1.128, 10.0.2.128 and so on. So for the third octet goes up by 1 for every 2 subnets. so if you have 4 subnets, the third octet will increase 2 times (10.0.0.0, 10.0.1.0), and for 40 subnets, then third octet increases 20 etc.

    Dotted decimal is like a really odd numbering system.
    2017 Goals - Something Cisco, Something Linux, Agile PM
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