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  1. Member
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    #1

    Default Wildcards - Range and Even/Odds

    Subnet Mask | Wildcard | Range (My guess?)
    255 - 1111 1111 | 0 0000 0000 | 1 Address 256 times (really 255 - 256 means you move across the dot 0.0->.0.0)
    254 - 1111 1110 | 1 0000 0001 | 1-2 128x subnets = 256
    252 - 1111 1100 | 3 0000 0011 | 1-4 64x '' =256
    248 - 1111 1000 | 7 0000 0111 | 1-8 32x '' =256
    240 - 1111 0000 | 15 0000 1111 | 1-16 16x '' = 256
    224 - 1110 0000 | 31 0001 1111 | 1-32 8x '' =256
    192 - 1100 0000 | 63 0011 1111 | 1-64 4x '' =256
    128 - 1000 0000 | 127 0111 1111 | 1-128 2x '' = 256

    Am I right on this if not please correct my mistakes
    here is a pdf that kinda helped me
    https://nsrc.org/workshops/2009/summ...subnetting.pdf

    Also I remember reading and seeing video that even or odd numbers represent a different range
    https://www.youtube.com/watch?v=Sg7UY5a9RkE
    Help me understand the even and odd wild card mask idea.

    Q178 - (Referencing a question that if need be I can go back and find it)

    Does the range all begin with 1-2, 1-4, 1-8, 1-16 ...or do you, if given an address that is your starting point
    Last edited by AndreL; 07-10-2017 at 04:45 PM.
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  3. Senior Member CryptoQue's Avatar
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    #2
    When you state range, are you trying to find the subnet range of an IP address? For example, if you have the mask of 255.255.255.252 or 0.0.0.3 then you have a total range of 4 IP addresses. Only 2 of those addresses are usable since the first and last are for the network address and broadcast. Therefor, if you have a class C address of 192.168.0.2 it would fall in the range of the 192.168.0.0 subnet. There can be a maximum of 64 total /30 subnets used in a class C range.

    192.168.0.0 - network
    192.168.0.1 - host IP
    192.168.0.2 - host IP
    192.168.0.3 - broadcast

    Mask is /30 or 255.255.255.252 or 1111.1100 or 0.0.0.3 or 0000.0011
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  4. Member
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    #3
    Range
    Which was one of my questions. When range is said in the context of wildcard, ranges are they pre determined blocks like 1-2, 1-4, 1-8, 1-16 when given a address like 192.168.10.4 the range is 192.168.10.4 - .7 (where .4 and .7 are not used (cause they are broadcast and network ID) so when 1-3 or is it 0-3 come into play when considering the range at the begins. And.4 is the beginning of the range in question.

    Any Ideas on the even or odd wildcards

    Also 255 and 254 cannot be used inthe C class or /25 range cause they represent /31 and /32.

    Found a mistake I think ???
    255 - 1111 1111 | 0 0000 0000 | 1 Address 256 times (really 255 - 256 means you move across the dot 0.0->.0.0)
    254 - 1111 1110 | 1 0000 0001 | 0-1 128x subnets = 256
    252 - 1111 1100 | 3 0000 0011 | 0-3 64x '' =256
    248 - 1111 1000 | 7 0000 0111 | 0-7 32x '' =256
    240 - 1111 0000 | 15 0000 1111 | 0-15 16x '' = 256
    224 - 1110 0000 | 31 0001 1111 | 0-31 8x '' =256
    192 - 1100 0000 | 63 0011 1111 | 0-63 4x '' =256
    128 - 1000 0000 | 127 0111 1111 | 0-127 2x '' = 256

    Just Brain storming here

    .0 give 0 - no bits - 1 addresses
    .1 give 0, 1 - 1 bit - 2 addresses
    .3 give 0, 1, 2, 3 - 2 bits - 4 addresses
    .7 give 0, 1, 2, 3, 4, 5, 6, 7 - 3 bits - 8 addresses
    .15 give 0-15 4 bits - 16 addresses
    .31 give 0-31 5 bits - 32 addresses
    .63 give 0-63 6 bits - 64 addresses
    127 give 0-127 7 bits - 127 addresses
    255.255
    /8 0000 0000.1111 1111.1111 1111
    /9 0000 0001
    /10 0000 0011
    /11 0000 0111
    /12 0000 1111
    /13 0001 1111
    /14 0011 1111
    /15 255 255 255
    /16 1111 1111.1111 1111 1111 1111

    I was think you only have to worry about IP address in the 4th octet taking away your host for Broadcast and Network ID cause in the 1st 2nd and 3rd octet the broad cast is in the last octet. Not sure how it works with the network ID think its the first address.
    Last edited by AndreL; 07-11-2017 at 12:02 AM.
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  5. Senior Member CryptoQue's Avatar
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    #4
    Yes, ranges are predetermined based on the specific mask and/or IP provided. The links below should help you better understand binary even and odd wildcards.

    Binary Math - Part I
    Binary Math - Part I Answers
    Binary Math, Part II
    Binary Math - Part II Answers
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    #5
    Am I right in my range

    255 - 1111 1111 | 0 0000 0000 | 1 Address 256 times (really 255 - 256 means you move across the dot 0.0->.0.0)
    254 - 1111 1110 | 1 0000 0001 | 0-1 128x subnets = 256
    252 - 1111 1100 | 3 0000 0011 | 0-2 64x '' =256
    248 - 1111 1000 | 7 0000 0111 | 0-6 32x '' =256
    240 - 1111 0000 | 15 0000 1111 | 0-15 16x '' = 256
    224 - 1110 0000 | 31 0001 1111 | 0-31 8x '' =256
    192 - 1100 0000 | 63 0011 1111 | 0-63 4x '' =256
    128 - 1000 0000 | 127 0111 1111 | 0-127 2x '' = 256

    I did change it cause I found a another mistake
    Last edited by AndreL; 07-13-2017 at 12:45 PM.
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  7. Senior Member CryptoQue's Avatar
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    #6
    /30 = .252 should reflect 0-3 64x
    /29 = .248 should reflect 0-7 32x

    The rest looks correct. Your had these accurate in previous post. What made you change your mind about it?
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    #7
    mis counted
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    #8
    No one replayed to my even and odd wild cards can someone help me with that.

    Also how are wildcards applied, to IP addresses directly or are they applied on a per block basis etc.

    When applying .3 or a .31 to 192.168.1.150
    Do I use ... directly
    .48, .47, .46 for .3 or

    .50, .49, .48

    and for .31

    .18-.49 or
    .0-.31 block size
    Last edited by AndreL; 08-07-2017 at 07:12 PM.
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  10. Member
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    #9
    This page helped me understand wildcard more for people who find this page and need more help
    https://people.richland.edu/dkirby/143aclwildcards.htm
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