Subnetting scenario

jenna28jjjenna28jj Posts: 4Registered Users ■□□□□□□□□□
But I need some help. Just starting off in a CCNA discovery class, 2nd semester. We are involving ourselves in subnetting, VLSM etc. Can't really go to the instructor for help since it is an online satellite class and takes him 16 years to respond to any emails, so I thought I would come to you guys....
My question...I have to design an IP addressing scheme which I can do..then the question asks...determine the number of hosts needed for each subnet, which I am assuming i need 6 subnets. Even if that is wrong and I dont need 6, my real question is how the heck to I figure out the number of hosts I need per subnet? Am I making this out to be harder than it is? I can figure out subnet masks, I can do this little chart I was taught, but I just cant wrap my head around how to figure out the number of hosts per subnet?? Help?

Design an IP addressing scheme. Done
Determine the number of hosts needed for each subnet. I am thinking 6 because i need 2 subnets for each serial link and 3 for each ethernet link so that equals 5 but 6 would be better.
Determine an appropriate addressing scheme using VLSM. I will get to that eventually.
The scenario uses a Class C private IP addressing scheme of 192.168.25.0

Comments

  • SharkDiverSharkDiver Posts: 844Member
    You are correct in needing 6 subnets, but they are asking for the number of hosts per subnet.

    First off, you have 4 separate LANs, so that is 4 subnets that you need.
    Secondly, each of the WAN links need their own subnet, so will need 2 subnets there.

    You MUST have 6 subnets, you can't do this with 5.

    Now, what they are really trying to get you to figure out is how many hosts you need in each subnet.
    Subnets have either 126, 62, 30, 14, 6 or 2 hosts in them depending on your subnet mask.
    The subnets actually have either 128, 64, 32, 16, 8 and 4 IP addresses in them, but you can't use the first or last for hosts.

    What they want you to see is that you need to use a subnet of 128 (255.255.255.128 mask) for LAN B to fit all 92 hosts in it. Then you need two subnets of 32 (255.255.255.224 mask) for LAN A and LAN D. Then, you will need a subnet of 8 (255.255.255.248 mask) to fit the 3 hosts in LAN C. Lastly, you will need a subnet of 4 (255.255.255.252) for each of the two WAN links.

    I hope this helps.
  • jenna28jjjenna28jj Posts: 4Registered Users ■□□□□□□□□□
    yes. how do I figure that out?
  • martell1000martell1000 Posts: 389Member
    by learning how to subnet.

    a little hint: like beer subnets come in various package size. your job is to figure out the size of the subnets you need to fit in the given number of hosts.

    imagine like 92 friends are coming over - how many sixpacks will you need to get em all drunk. same with subnets.
    And then, I started a blog ...
  • WebmasterWebmaster Posts: 10,292Admin Admin
    Determine the number of hosts needed for each subnet. I am thinking 6 because i need 2 subnets for each serial link and 3 for each ethernet link so that equals 5 but 6 would be better.
    That won't work. You cannot determine the number of hosts simply by counting the required amount of subnets (which are 6 in your diagram, not 5, notice R3 has two Ethernet segments attached). The more subnets, the less hosts per subnet.
    The scenario uses a Class C private IP addressing scheme of 192.168.25.0
    So you basically have 254 addresses (all the host addresses in the Class C address you have been given) to divide over the subnets by wasting as little address space as possible, which is the VLSM part.

    IPv4 Subnetting TechNotes
  • jenna28jjjenna28jj Posts: 4Registered Users ■□□□□□□□□□
    Ok, I have been working all day. ( and thank you martell1000 I understand the basic's of subnetting) and here is what I came up with...just wondering if you guys think I have worked it out alright??
    The complex network consists of:
    · Class C private address 192.168.25.0
    · Three Routers (R1 – 3)
    · Four LAN’s (LAN A – D)
    · Two WAN Links (WAN 1 – 2)

    The network connections are as follows:
    · LAN A is connected to Rl
    · LAN B is connected to R3
    · LAN’s C and D are connected to R2
    · WL 1 connects R1 to R2
    · WL 2 connects R1 to R3

    The number of hosts needed for each subnet are:
    · LAN A – 19
    · LAN B – 92
    · LAN C – 3
    · LAN D – 19

    Determined that basic sub-netting will not work with this network, because the number of hosts in LAN B will only allow two subnets and the requirement is six. VLSM will work using the following IP addressing scheme:
    VLSM IP Addressing Scheme for
    Network 192.168.25.0
    Name / Required Subnet Address Address Range Broadcast Network Prefix
    Addresses Address___________________
    LAN B / 92 192.168.25.0 .1 - .126 .127 192.168.25.0 /25
    LAN A / 19 192.168.25.128 .129 - .158 .159 192.168.25.128 /27
    LAN D / 19 192.168.25.160 .161 - .190 .191 192.168.25.160 /27
    LAN C / 10 192.168.25.192 .193 - .198 .199 192.168.25.192 /29
    WAN - 1 / 2 192.168.25.200 .201 - .202 .203 192.168.25.200 /30
    WAN - 2 / 2 192.168.25.204 .205 - .206 .207 192.168.25.204 /30
  • SharkDiverSharkDiver Posts: 844Member
    Looking it over quickly, it looks like you nailed it.

    I would say JOB WELL DONE!!!
  • jenna28jjjenna28jj Posts: 4Registered Users ■□□□□□□□□□
    really? thanks!!
  • WebmasterWebmaster Posts: 10,292Admin Admin
    Just a minor mismatch between the table and the diagram, LAN C in your table should have 3 instead of 10 addresses. Doesn't effect the rest, which is indeed correct.

    In a class C network subnets can have the following amount of hosts:
    256-2=254
    128-2=126
    64-2=62
    32-2=30
    16-2=14
    8-2=6
    4-2=2

    The 2 is obviously subtracted for the subnet's network and broadcast address. Determining the minimum required amount of hosts based on your diagram is a matter of rounding up the number of hosts to a number in the list above. Which you did, so good job indeed. You can verify the results by using our subnet calculator.

    As you can see it still leaves some useable/unassigned addresses, which in a real-world situation might have been used to allow some growth in LAN C. For the same reason it would be wise to use the last 8 addresses of the Class C subnet for the WAN link subnets, allowing LAN C to expand without affecting (having to re-address) the WAN link subnets.
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