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 629 Lessons. More Lessons Added Every Week!
  • Content created by Rene Molenaar (CCIE #41726)

 

420 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 add

    ... Continue reading in our forum

  2. 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

    ... Continue reading in our forum

  3. 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

  4. Hi Rene,

    Shouldn’t be for 600 hosts, 3 blocks would suffice? 256*3
    Instead of taking 4 blocks in above solution?

    Rahul

  5. Hello Rahul

    If you were to use three separate subnets to accommodate 600 hosts then you could create them, but they would still be separate subnets. For example, you could use

    192.168.0.0/24
    192.168.1.0/24
    192.168.2.0/24

    That would give you 256*3 = 768 IP addresses.

    However, you would still have three SEPARATE subnets each requiring a network address, a broadcast address and a default gateway. You would also require routing to communicate between the subnets. For example, a host at 192.168.0.26 needs to go through a router to reach 192.168.2.26.

    If you want

    ... Continue reading in our forum

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