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Quick subnet help

  • 19-12-2007 9:59pm
    #1
    Closed Accounts Posts: 5,240 ✭✭✭


    I have this address 172.16.1.0/24
    Is the wild card for this not 0.0.0.255? When i try to enter it into packet tracer it says "Bad mask 0xFF for address 172.16.1.0"

    Ill be the first to admit that i suck at the networking side of things, im just trying to get this assignment done.What am i doing wrong :o.


Comments

  • Registered Users, Registered Users 2 Posts: 9,957 ✭✭✭trout


    http://en.wikipedia.org/wiki/Subnetwork

    The answer you seek is here ... but you'll have to read it.
    Subnetting is a bit of a pain to get your head around at first ... persevere. :)


  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    Well yes I would have guessed that that was the Wildcard for it but then again, a bit 1 could mean ignore or check. So it could just as easily be 255.255.255.0.

    Have you tried this?


  • Closed Accounts Posts: 5,240 ✭✭✭Endurance Man


    Someone please just tell me what these wild cards are :(, iv been up half the nite. I got hold of a wildcard calculator but packet tracer still tells me the masks are wrong.

    172.16.1.0/24
    172.16.10.0/24
    172.16.20.0/24

    The first one was 255.255.0.0, and i get 0.0.0.3 for the second? Help please!
    Edit, the second one must be 0.0.0.255, why wont it work ><.


  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    Ok I don't know much about Packet Tracer, but i take it you are using a wildcard mask to filter out only the addresses you are interested in?

    Is this what you want?


  • Registered Users, Registered Users 2 Posts: 640 ✭✭✭CraggyIslander


    Someone please just tell me what these wild cards are :(, iv been up half the nite. I got hold of a wildcard calculator but packet tracer still tells me the masks are wrong.

    172.16.1.0/24
    172.16.10.0/24
    172.16.20.0/24

    The first one was 255.255.0.0, and i get 0.0.0.3 for the second? Help please!
    Edit, the second one must be 0.0.0.255, why wont it work ><.

    The /24 means the subnet mask is the first 24 bits i.e. 255.255.255.0 for all 3 of those

    the address ranges are
    172.16.1.1 - 172.16.1.254
    172.16.10.1 - 172.16.10.254
    172.16.20.1 - 172.16.20.254

    Edit: not sure what you mean by wildcard address... but if its the broadcast address (i.e. broadcasts to the whole range then its the following three:

    172.16.1.255
    172.16.10.255
    172.16.20.255


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  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    The /24 means the subnet mask is the first 24 bits i.e. 255.255.255.0 for all 3 of those

    the address ranges are
    172.16.1.1 - 172.16.1.254
    172.16.10.1 - 172.16.10.254
    172.16.20.1 - 172.16.20.254
    Subnet masks and wildcard masks seem similar but are a different thing


  • Registered Users, Registered Users 2 Posts: 640 ✭✭✭CraggyIslander


    http://www.subnet-calculator.com/wildcard.php

    Quck glance tells me it uses no of hosts as the basis for calculation.

    Think i'll stick to the traditional way of calculating netmasks for subnets, supernets, etc.

    the /24 still indicates 255 hosts so would mean a wildcard mask of 0.0.0.255


  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    Yes, 172.16.1.0/24 means a range of subnets between: 172.16.1.1 - 172.16.1.254 excluding broadcast/network address.

    Now in wildcard masks normally, a 0 bit means 'Check' and a binary 1 bit means 'Ignore'

    So to match addresses from 172.16.1.1 - 172.16.1.254 is a wildcard mask of 0.0.0.255 because:

    172.16.1.0 = 172.16.1.00000000
    172.16.1.255 = 172.16.1.11111111

    You can see that the only thing we need to check is the first 3 octets as no matter what the end, it will always be between 0 - 255 , or 1 - 254 for valid addreses. So for this reason we place a 1 for each bit of last octet - to ignore these.

    Take for example if we want to match addresses between 172.16.1.0 and 172.16.1.64 then:

    172.16.1.16 = 172.16.1.00000000
    172.16.1.64 = 172.16.1.01000000

    You can see here that its the 7th bit that changes, so we really only have to ensure that bit 7 above doesn't change to 1 and also that bit 8 doesn't change to a 1 to form 127, so to check this we place a 0 in these places and ignore the rest as no matter what the remaining 6 bits are they will always lie between 0 and 64.
    So to ignore the remaining 6 bits we have 1's in there places.

    So we get 0.0.0.00111111 = 0.0.0.63

    Hope this makes sense


  • Registered Users, Registered Users 2 Posts: 640 ✭✭✭CraggyIslander


    so what is the use of these wildcard masks? to make calculations 'easier'?

    ps for 64 hosts the subnet mask would be 255.255.255.192 :D no calculator needed for that


  • Closed Accounts Posts: 583 ✭✭✭monkey tennis


    so what is the use of these wildcard masks? to make calculations 'easier'?

    Wildcard masks are more versatile than subnet masks.
    ps for 64 hosts the subnet mask would be 255.255.255.192 :D no calculator needed for that

    Actually, that's for 62 hosts. Only 62 out of the range of 64 IPs are host-assignable. And the topic under discussion is wildcard masks; subnet masks are child's play :p


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  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    so what is the use of these wildcard masks? to make calculations 'easier'?

    ps for 64 hosts the subnet mask would be 255.255.255.192 :D no calculator needed for that
    Take it you havn't dealt with Access Control Lists then :)

    Work out how to only let hosts that end with an even address to get through. That might take a bit more thinking


  • Closed Accounts Posts: 5,240 ✭✭✭Endurance Man


    Thanks for the responses guys, i figured it out, handed in and now the holidays start :D.


  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    Thanks for the responses guys, i figured it out, handed in and now the holidays start :D.
    Lucky you, I'm stuck in a java assignment due for tomorrow and i've a year project to have completed for january thats supposed to be a 3 person group but unfort one dropped out and other couldn't give a crap!

    Enjoy the christmas :)


  • Closed Accounts Posts: 5,240 ✭✭✭Endurance Man


    Unlucky ^^, i had a similar experience with group memebers buggering off during our business assingment. Would not have been so bad if there where more than two of us i the group!
    Give me java over cisco any day!
    Merry Christmas


  • Registered Users, Registered Users 2 Posts: 640 ✭✭✭CraggyIslander


    Webmonkey wrote: »
    Take it you havn't dealt with Access Control Lists then :)

    Work out how to only let hosts that end with an even address to get through. That might take a bit more thinking

    For the ACL work i've done subnet, supernet and CIDR were always sufficient. Mostly comes down to how the network was designed in the first place tho. :)

    only allowing hosts with even addresses? makes no sense, redesign and use two smaller networks :p


  • Closed Accounts Posts: 583 ✭✭✭monkey tennis


    only allowing hosts with even addresses? makes no sense, redesign and use two smaller networks :p

    Wouldn't it be nice if things were always that simple...


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