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Classless (CIDR) IPs

Discussion in 'Network+' started by Japanese Elvis, Jan 16, 2006.

  1. Japanese Elvis

    Japanese Elvis Nibble Poster

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    I'm having a bit of a mental block with this :blink

    The following question popped up on my exam simulator, and (despite reading the explanation) it just isn't sinking in.

    Does anyone fancy trying to (simply) explain the process?

    You manage a computer running Windows 2000 Server and the Dynamic Host Configuration Protocol (DHCP) server service. You need to add addresses for a subnet that has been installed for a new department.

    What is the highest Internet Protocol (IP) address that you can assign to a host for the subnet ID 172.16.32.0/21?


    Thanks
     
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  2. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Let's see if I can do this. :biggrin

    21 bits is 2 bytes and 5 bits (2 bytes = 16 bits)

    So the mask is FF FF F8 00 (in hex - I find it easier to convert the /number format to hex first), which is 255. 255. 248. 0.

    The broadcast address is defined as having an unit address of all 1s, so will be 07 FF (hex) or 7. 255 in decimal.

    So the full broadcast address is 172. 16. 32 + 7. 255 which is 172. 16. 39. 255.

    So the last useable address is one less - 172.16.39.254

    Harry (now tries to find what he got wrong there!)
     
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  3. Clyde

    Clyde Megabyte Poster

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    cidr notation simply mentions the number of bits in the mask set to 1 so /21 means
    11111111.11111111.11111000.00000000
    or 255.255.248.0
    once you know that yer in business!
     
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  4. Japanese Elvis

    Japanese Elvis Nibble Poster

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    Thanks Harry, that is indeed the correct answer.

    I like the way you've done that, i.e. converting to hex first.

    I think that may be easier for me to work with.

    The explanation given was;

    You can use the range of addresses 172.16.32.1 to 172.16.39.254 for hosts on this subnet. The network prefix /21 indicates that the first two octets and five bits in the third octet are reserved for the subnet ID. The lowest order bit of the five bits in the third octet represents an increment of 8 for the subnet IDs. The valid subnet IDs include 8, 16, 24, 32, 40, and 48 and so forth. The range of addresses from 172.16.32.1 to 172.16.39.254 are valid host IDs for the subnet 172.16.32.0/21.

    I reckon your explanation is a bit clearer.

    Cheers!

    :biggrin
     
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  5. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Very many thanks for that vote of confidence!

    Harry.
     
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  6. Boycie
    Honorary Member

    Boycie Senior Beer Tester

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    Brilliiant explanation Harry :thumbleft
     
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  7. Bluerinse
    Honorary Member

    Bluerinse Exabyte Poster

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    I must be missing something here then. How does converting it to Hex make it simpler :blink Now we are juggling between three number formats, binary, decimal and hex?

    Again the hex part is not clear. What am I missing :rolleyes:

    I know you are correct Harry but a bit more of an insight into why you prefer to convert to hex would be appreciated.
     
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  8. ffreeloader

    ffreeloader Terabyte Poster

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    It was my understanding from reading Harry's post that he was just more comfortable working with hex than with binary. Remember, he's a developer. :twisted: :biggrin
     
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  9. Bluerinse
    Honorary Member

    Bluerinse Exabyte Poster

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    Yes but what has processing photographs got to do with it :blink
     
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  10. Japanese Elvis

    Japanese Elvis Nibble Poster

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    I think what I liked about that method was that by stating the mask in hex it is easier to take in visually, i.e. FF is less visually confusing than 11111111.
    Also, converting the octets to hex is quite a simple process (split into two and convert each half).
    For me it seemed to make sense and was easier than getting dizzy looking at a big line of 1s.

    I do take the general point that it may appear to be making things more complex by using every number system within grasp, but for me it was just one of these occasions when - even reading the answers and explanations - I just wasn't getting it. I hoped someone on here would offer a different perspective, they did, and I'm now full tilt towards my next mental block :blink
     
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  11. Clyde

    Clyde Megabyte Poster

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    the only think about doing it via hex is, as was earlier said, you need to do more calculations, and also, the method will be non standard and therefore, all printed materials will be less familiar in the long run...
     
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  12. ffreeloader

    ffreeloader Terabyte Poster

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    Because that's how Harry "pictures" it..... :biggrin
     
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  13. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Well - first because I've been working in binary/octal/decimal/hex for over 20 years. So I'm comfortable in switching about and using the most appropriate format - appropriate to me that is.

    We were talking about a string of one bits - represented by /21. Now I find binary unwieldy in long strings, and the decimal representations 'non-obvious'. OTOH hex is easy for me, you are only dealing with 4 bits at a time for the digit, and the weights of each bit, 8, 4, 2, and 1 are hard wired in my brain by now! So to contruct the subnet mask in hex is quicker than any other method *for me*.

    To get the network broadcast address is a logical NOT of the network address. Again, for me I can do this in my head easily if it is in hex. If in decimal I'd have to use a calculator.

    I hope this goes a little way to explain why I did it in this way!

    Harry.
     
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  14. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    :biggrin Wonderful!

    Of course a real old-timer would be using octal.....

    Little known fact - the train numbers on the front of Northern Line trains are in octal.

    Harry. (who at one time had an Underground fanatic for a boss)
     
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  15. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    That is precisely why I do it that way.

    Harry.
     
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  16. Bluerinse
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    Bluerinse Exabyte Poster

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    Yes thanks Harry, everything is much clearer to me now :D
     
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