Subnetting - 3rd octet

J_86J_86 Posts: 262Member ■■□□□□□□□□
Ok, so I seem to be really struggling when subnetting in the 3rd octet for some reason.

I have been watching the CBT nuggets videos, if any of you are familiar with that, but all of those example have been in the 4th octet.

Take for example this question I got from subnettingquestions.com:

How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?
Answer: 4096 subnets and 4094 hosts
So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer? icon_confused.gif:

Comments

  • gosh1976gosh1976 Posts: 441Member
    It can be difficult applying what you have learned with class C addresses to Class A & B.

    I know that it takes 4 bits to get 240: 128+64+32+16= 240 and there are already 8 bits being borrowed from the second octet. So you have 2^12 = 4096 subnets. That leaves 4 host bits in the third octet and 8 in the last octet so 2^12-2= 4094.

    Not sure that helps but that was my thought process when I saw the question.
  • SaeedonWebSaeedonWeb Posts: 18Member ■□□□□□□□□□
    Well, you need to remember the IP address classes. This address belongs to the Class A with the default subnet mask of 255.0.0.0. The subnet mask they give in the question is already showing that the bits borrowed are from the 2nd AND 3rd octate, hence 12 bits as mentioned by the poster above.

    I hope it further clarifies your concept.

    Saeed
  • JockVSJockJockVSJock Posts: 1,118Member
    J_86 wrote: »
    Ok, so I seem to be really struggling when subnetting in the 3rd octet for some reason.

    I have been watching the CBT nuggets videos, if any of you are familiar with that, but all of those example have been in the 4th octet.

    Take for example this question I got from subnettingquestions.com:

    How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?
    Answer: 4096 subnets and 4094 hosts
    So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer? icon_confused.gif:


    Actually, it should be no problem to use binary to solve this problem.

    Find the # of subnets/hosts for the following:
    10.0.0.0
    255.255.240.0

    First convert everything down to binary
    00001010.00000000.00000000.00000000
    11111111.11111111.11110000.00000000
    

    Looking at the ip address, you know that the 2nd Octet is dedicated to the subnet and along with the bites that are turned on in the 3rd Octet which gets us this:
    
    2x12=4096
    
    
    4096 Subnets
    
    

    Everything else that is 0 in the 3rd and 4th Octet goes for the hosts. Remember to subtract 2 for the Network and Broadcast Address:
    
    2*12 = 4096 
    
    4096 - 2 = 4094 hosts
    
    
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  • J_86J_86 Posts: 262Member ■■□□□□□□□□
    JockVSJock wrote: »
    Actually, it should be no problem to use binary to solve this problem.

    Find the # of subnets/hosts for the following:
    10.0.0.0
    255.255.240.0

    First convert everything down to binary
    00001010.00000000.00000000.00000000
    11111111.11111111.11110000.00000000
    

    Looking at the ip address, you know that the 2nd Octet is dedicated to the subnet and along with the bites that are turned on in the 3rd Octet which gets us this:
    
    2x12=4096
    
    
    4096 Subnets
    
    

    Everything else that is 0 in the 3rd and 4th Octet goes for the hosts. Remember to subtract 2 for the Network and Broadcast Address:
    
    2*12 = 4096 
    
    4096 - 2 = 4094 hosts
    
    


    Ah, now I get it! Thanks guys, I think I was just over thinking it.
  • mella060mella060 Posts: 196Member
    J_86 wrote: »
    So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer? icon_confused.gif:

    Well you should know straight off the bat that an address starting with 10 is a class A address.

    The default subnet mask for a class A address is:

    255.0.0.0...in binary that is 11111111.00000000.00000000.00000000

    255.255.240.0 in binary is...11111111.11111111.11110000.00000000

    how many extra 1's are there ? 12

    what is 2^12 ? 4096

    How many 0's are there ? 12

    what is 2^12 - 2....4096-2 = 4094

    You need to know what the subnet formulas are where n is the number of bits borrowed from the host portion of the address. In your example you are borrowing 12 bits.

    2^n for subnets

    2^n - 2 for hosts

    Also you really need to know your powers of 2
  • thedramathedrama Posts: 291Member
    J_86 wrote: »
    Ok, so I seem to be really struggling when subnetting in the 3rd octet for some reason.

    I have been watching the CBT nuggets videos, if any of you are familiar with that, but all of those example have been in the 4th octet.

    Take for example this question I got from subnettingquestions.com:

    How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?
    Answer: 4096 subnets and 4094 hosts
    So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer? icon_confused.gif:

    A different point of view :

    As you know, by default 10.0.0.0 belongs to Class A. And by default, subnet mask of Class A is 255.0.0.0
    According to this question : the resulted subnet mask is 255.255.240.0

    That means, 8 bits from the second octet and the 11110000(number of 1s) 4 bits from the third octet taken.

    Besides, the number of subnets is calculated as 8 plus 4 = 12 indicating 2"12 subnets = 4096 subnets

    How many zeroes remained? 4 bits on the third octet and 8 bits on the last octet indicating 2"12 -2 = 4094 hosts per subnet

    Where does it lead us to in case of block size? as you have seen, third octet shows us the 240. now, 256-240 = 16.

    this means block size = 16 is equal to incremental value = 16

    So,

    if the first subnet is 10.0.0.0
    the second will be 10.0.16.0
    .
    .
    .
    .
    until the third octet becomes the 256(i mean which passes to second octet)
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