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Forum Replies

  1. Hi Gabriel,

    Your example is correct, good job! Also thanks for reporting those errors, just fixed them.

    Rene

  2. Sure, which exact subnet mask do you refer to?

  3. Hi Lynkaran,

    I'll explain how to do the first example, see if you can solve them with my technique. Let's start with 112.10.78.40/22:

    First we need to figure out what the subnet mask since /22 doesn't tell us much. You need to write this down in binary and convert it to decimal:

    first 8 bits = 11111111 (255 in decimal)
    next 8 bits = 11111111 (255 in decimal)
    next 8 bits = 11111100 (252 in decimal)
    next 8 bits = 00000000 (0 in decimal)

    So now we know the subnet mask is 255.255.252.0

    How many hosts do we have per subnet? There are 2 + 8 host bits so 10 host bits in total. You can use this formula for this:

    210 - 2 = 1024.

    You need to remove 2 since one address is the network address and the other one is the broadcast address.

    What subnets do we have? There's a quick way to calculate this. Take number 256 - subnet mask = size of subnet:

    256 - 252 = 4

    Now we can write down the subnets, I'll start with 0:

    1 112.10.0.0/22

    2 112.10.4.0/22

    3 112.10.8.0/22

    4 112.10.12.0/22

    5 112.10.16.0/22

    6 112.10.20.0/22

    7 112.10.24.0/22

    ...and so on

    You can see it increases with 4 every time, you now have your network addresses. What about the broadcast address? That's the last address in each subnet.

    For subnet #1 that will be 112.10.3.255 and for subnet #5 it's 112.10.19.255.

    See if you can do the other examples on your own with these steps. All you need is practice, practice, practice.

    Hope this helps.

    Rene

  4. Hello Apurva.

    That is exactly correct! Keep in mind as well that the first and last addresses in the range are the network and broadcast addresses respectively so you will have 126 addresses available for hosts.

    I hope this has been helpful!

    Laz

  5. Hi Rene,

    I was trying to figure out a way of "knowing" the amount of bits needed for huge hosts without the need of a calculator, like:

    • 2^X-2= hosts (too complicated).

    This way sorts the problem!

    "
    255.255.255.0 /24 – 256 addresses
    255.255.255.128 /25 – 128 addresses
    255.255.255.192 /26 – 64 addresses
    255.255.255.224 /27 – 32 addresses

    You can see that whenever the subnet becomes smaller, we can only use half of the addresses. I can apply the same logic if I move into the other direction:

    255.255.240.0 /21 – 2048 addresses
    255.255.248.0 /22 – 1024 addresses
    255.255.254.0 /23 – 512 addresses
    255.255.255.0 /24 – 256 addresses"

    Thanks!

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