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Possible subnets?

Discussion in 'General Cisco Certifications' started by datarunner, Nov 6, 2007.

  1. datarunner

    datarunner Byte Poster

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    hi all

    need some help here. how can i work out the number of possible subnets if say 4 or 5 or 6 etc bits are borrowed from the host. we were taught this in class but i missed this part.
     
    Certifications: A+, N+, MCP 210, 270, HNC Networking
    WIP: MCSA
  2. onoski

    onoski Terabyte Poster

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    Hi datarunner, do a check on google as you'd get a lot of info on your quest. I did this whilst studying for the CCNA and found a lot of useful info on subnetting etc. You also want to check www.cisco.com/learning

    Cheerio:biggrin
     
    Certifications: MCSE: 2003, MCSA: 2003 Messaging, MCP, HNC BIT, ITIL Fdn V3, SDI Fdn, VCP 4 & VCP 5
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  3. r.h.lee

    r.h.lee Gigabyte Poster

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    datarunner,

    Let's look at the situation where a single bit is borrowed...
    Code:
    Subnet Mask .1|0000000.
    
    Possible subnets
    1) .0|0000000.
    2) .1|0000000.
    
    Total number of possible subnets = 2 subnets
    
    Let's look at the situation where two bits are borrowed...
    Code:
    Subnet Mask .11|000000.
    
    Possible subnets
    1) .00|0000000.
    2) .01|0000000.
    3) .10|0000000.
    4) .11|0000000.
    
    Total number of possible subnets = 4 subnets
    
    Let's look at the situation where three bits are borrowed...
    Code:
    Subnet Mask .111|00000.
    
    Possible subnets
    1) .000|000000.
    2) .001|000000.
    3) .010|000000.
    4) .011|000000.
    5) .100|000000.
    6) .101|000000.
    7) .110|000000.
    8) .111|000000.
    
    Total number of possible subnets = 8 subnets
    
    So as a summary, here's a chart of number of bits borrowed and total number of possible subnets...
    Code:
    Bits Borrowed   Possible Subnets
    -------------   ----------------
    1               2
    2               4
    3               8
    
    Now, the mathematical relationship between "bits borrowed" and "possible subnets" is...
    Code:
    # of possible subnets = 2 ^ bits borrowed
    
    So for an exercise, using the above equation, find and post the # of possible subnets for:
    1. 4 bits borrowed
    2. 5 bits borrowed
    3. 6 bits borrowed

    Note, since you won't have a calculator in the Cisco exam, do the above exercises without a calculator.
     
    Certifications: MCSE, MCP+I, MCP, CCNA, A+
    WIP: CCDA
  4. datarunner

    datarunner Byte Poster

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    hi there

    thanks very much 4 that. i was totally stuck with this.
     
    Certifications: A+, N+, MCP 210, 270, HNC Networking
    WIP: MCSA
  5. datarunner

    datarunner Byte Poster

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    # of possible subnets = 2 ^ bits borrowed

    can u explain this or show me the sum/formula as i dont really understand what the 2 ^ means.sorry 4 being thick.
     
    Certifications: A+, N+, MCP 210, 270, HNC Networking
    WIP: MCSA
  6. datarunner

    datarunner Byte Poster

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    heres a sample question, now im not looking for the answer, just how to work it out

    On a class c network with 3 bits borrowed for subnets to which subnet would the ip subnet and host range 01100001 belong?

    a. second subnet
    b. third subnet
    c. fourth subnet
    d. fifth subnet

    im guessing the forth but im totally stumped here
     
    Certifications: A+, N+, MCP 210, 270, HNC Networking
    WIP: MCSA
  7. Sparky
    Highly Decorated Member Award

    Sparky Zettabyte Poster Moderator

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  8. r.h.lee

    r.h.lee Gigabyte Poster

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    datarunner,

    The "2 ^" refers to "2 raised to the power of." For example, "2 ^ 2" translates to "2 raised to the power of 2." In other words, "2 squared." The answer to "2 ^ 2" is '4'. Another way of thinking of powers as...
    Code:
    2 ^ 1 = 2 = 2
    2 ^ 2 = 2 x 2 = 4 
    2 ^ 3 = 2 x 2 x 2 = 8
    2 ^ 4 = 2 x 2 x 2 x 2 = 16
    
    Does this help?
     
    Certifications: MCSE, MCP+I, MCP, CCNA, A+
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  9. r.h.lee

    r.h.lee Gigabyte Poster

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    datarunner,

    For a "class c network" what is the default mask?
     
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  10. greenbrucelee
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    greenbrucelee Zettabyte Poster

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    I have been reading alot about subnets on here latley and am slightly curios, am I right in saying 4 bits is 16, 5 is 32 6 bits is 64, 7 bits 128 or is that way off and too obvious.
     
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  11. datarunner

    datarunner Byte Poster

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    so using 1 bit would be 2 ^ 1
    2 bits 2 ^ 2
    3 bits 2 ^3
    4 bits 2 ^ 4
    5 bits 2 ^ 5
    6 bits 2 ^ 6
    7 bits 2 ^ 7

    is that correct?
     
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  12. r.h.lee

    r.h.lee Gigabyte Poster

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    datarunner,

    Yes.
     
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  13. onoski

    onoski Terabyte Poster

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  14. Sparky
    Highly Decorated Member Award

    Sparky Zettabyte Poster Moderator

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    Yup, handy reference book to have though. :study
     
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  15. datarunner

    datarunner Byte Poster

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    thanks man now to working out the hosts per network if a certain amount of bits are used. is there an easy way to do that? ie if 4 bits are used on a class c then how many hosts per network. sorry 2 b a pain.
     
    Certifications: A+, N+, MCP 210, 270, HNC Networking
    WIP: MCSA
  16. r.h.lee

    r.h.lee Gigabyte Poster

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    datarunner,

    The number of hosts really depends on how many bits were borrowed for the subnets.

    Code:
    1 bit borrowed for subnetting = 8 bits minus 1 bit borrowed = 7 bits left. 62 hosts possible.
    2 bits borrowed for subnetting = 8 bits minus 2 bits borrowed = 6 bits left. 30 hosts possible.
    3 bits borrowed for subnetting = 8 bits minus 3 bits borrowed = 5 bits left. 14 hosts possible.
    
    Now, you might be wondering, 2^7 = 64 so why is there only 62 hosts possible? Well, that's because the "all zeros hosts" is reserved for the subnet ID/address and the "all ones hosts" is reserved for the subnet broadcast address so you subtract two from the total hosts possible to come up with a new hosts possible number.
     
    Certifications: MCSE, MCP+I, MCP, CCNA, A+
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  17. BosonMichael
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    BosonMichael Yottabyte Poster

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    You're missing a power of 2 in each. 2^7 = 128, so 2^7-2 = 126 possible hosts.

    Same with the others... 2^6 = 64, 2^5 = 32, etc.
     
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