We're Sorry, Full Content Access is for Members Only...

If you like to keep on reading, Become a Member Now! Here is Why:

  • Learn any CCNA, CCNP and CCIE R&S Topic. Explained As Simple As Possible.
  • Try for Just $1. The Best Dollar You've Ever Spent on Your Cisco Career!
  • Full Access to our 588 Lessons. More Lessons Added Every Week!
  • Content created by Rene Molenaar (CCIE #41726)

 

312 New Members signed up the last 30 days!

satisfaction-guaranteed

100% Satisfaction Guaranteed!
You may cancel your monthly membership at any time.
No Questions Asked!


Forum Replies

  1. Rene,

    A good way to explain this subject that is a little confuse.
    About challenge, I tried to solve it...

    A network 10.0.0.0

    One subnet for 600 hosts -> It's need a block 1024
    One subnet for 250 hosts. -> It's need a block 256
    One subnet for 120 hosts. -> It's need a block 128
    One subnet for 30 hosts. -> It's need a block 32
    One subnet for 2 hosts. -> It's need a block 4

    A block of 1024 is like 4x256, which is need 10 hosts bits.

    So...

    Subnet 1: (size 1024)

    network address: 10.0.0.0/22
    netmask: 255.255.252.0
    first host: 10.0.0.1
    last host: 10.0.3.254
    broadcast address: 10.0.3.255

    Subnet 2: (size 250)

    network address: 10.0.4.0/24
    netmask: 255.255.255.0
    first host: 10.0.4.1
    last host: 10.0.4.254
    broadcast address: 10.0.4.255

    Subnet 3: (size 120)

    network address: 10.0.5.0/25
    netmask: 255.255.255.128
    first host: 10.0.5.1
    last host: 10.0.5.126
    broadcast address: 10.0.5.127

    Subnet 4: (size 30)

    network address: 10.0.5.128/27
    netmask: 255.255.255.224
    first host: 10.0.5.129
    last host: 10.0.5.158
    broadcast address: 10.0.5.159

    Subnet 5: (size 2)

    network address: 10.0.5.160/30
    netmask: 255.255.255.252
    first host: 10.0.5.161
    last host: 10.0.5.162
    broadcast address: 10.0.5.163

    I hope I'm right...

    When you have a free time, please, fix the details below.
    "Subnet 2: 256 – 32 = <>" so the subnet mask is 255.255.255.224"
    and
    on subnet2 the last host IP should be 172.16.2.254

    Hug and more a time, thanks for excellent work!!!

  2. Hi Rene,

    Can you pls elaborate on how you got this subnet mask? I didn't get it. thanks
    <>

  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!

16 more replies! Ask a question or join the discussion by visiting our Community Forum