wildcard mask range number question in chapter exam ?

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  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    pinkiaiii wrote: »
    so following your summary it comes out like 128,64,[32],16,8,4,2,1 getting 192 followed up to here.

    ohh.... okay.
    So you are Adding, right?
    128 and 64 gives you the 192?

    That's your mistake.
    What i tried to say in my Summary... there is NO Adding.
    pinkiaiii wrote: »
    then if id was to apply by some mistake another bit (32) it would be too far thus that third number indicates that all calculations should be done in 3 octet-thats the tricky part i dont understand since in my head looking at that table,i would be hard pressed to know at what point it would turn over to 4th octet,

    Nope.
    All calculations Begin in the 3rd octet.
    All calculations End in the 3rd octet.
    There is NO JUMPING to another octet.

    Read what i said again:
    Take a CLOSE look at the SUbnet MAsk.
    255.255.192.0

    From Left to Right... LOOK for the OCTET that is NOT "255".

    The 1st octet is "255".
    The 2nd octet is also "255".
    But, the 3rd octet is NOT "255". (it's 192).

    So THAT is why the Subnet Ranges start/stop in the 3rd octet:
    172.18.0.0
    172.18.64.0
    172.18.128.0
    172.18.192.0

    there is NO adding being done.
    You just LOOK at the subnet mask.
    Then you find the octect that is NOT 255 :]

    For example,
    if the subnet Mask was:
    255.128.0.0

    ~then the Octet you work in is: 2nd octet.

    If the subnet Mask was:
    255.255.252.0

    ~then the Octet you work in is: 3rd octet.

    If the subnet Mask was:
    255.255.255.192

    ~then the Octet you work in is: 4th octet.

    Just use your EYEBALLS. there is NO math being done here :]
  • pinkiaiiipinkiaiii Posts: 216Member
    finally have to say i understood it LOL hard sweat,even thou it looks really obvious now.

    just following as it was any other type of class subneting and treating bits in same manner

    1=2
    2=4
    3=8
    4=16
    5=32
    6=65
    7=128
    8=256
    ________________________
    9=512

    and that combined with magic number gives blocks to use. 1=128 2=64 3=32 4=16 5=8,then converting into masks 128,192,224,240,248

    to bad cant give you some sort of gold metal logo icon_wink.giffor help provided.
  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    hmmmm,

    Okay, so it sounds like you think you've got it :]

    Let's do one last Q just to make sure!

    Scenario:
    You have been asked to subnet your customer's network.
    Their network is: jj.kk.ll.mm
    Their subnet-mask is /27

    A) Identify the Magic#
    B) Identify the Octet you will work in.
    C) Identify all the Subnet Ranges (just like you've done before)

    Good Luck!!
  • pinkiaiiipinkiaiii Posts: 216Member
    /27

    using 3 bits

    mask 224

    255.255.255.224 class c forth octet, or could be class b as well since no ip given.
    magic number 32

    8 subnets

    0 valid ranges 1-31 and another minus one for usable ones 1-30
    32 | 33-63| 33-62
    64 |65-97 |65-96
    98 |99-127 | 99-126
    128 | 129-161| 129-160
    162 |163-191 |163-190
    192 |193-223 |193-222
    224 |224-255 |224-254

    for some reason it gets squeezed up thus the | once posted.ok didnt look at any charts all from head,but looking at my post noticed mistake made at 98,should been 96 thus messed up :/

    thus again,0,32,64,96,128,162,194,224

    ranges 1-31,33-63,65-95,95-127,129-161,163-193,194-223,225-255
    valid ones up to 30,62,94,126,150,192,222,254
  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    You GOT it, pinkiaiii!!

    Your ADDING is a little off.... but you Nailed the concept :]
    pinkiaiii wrote: »
    /27
    using 3 bits
    mask 224

    255.255.255.224 class c forth octet, or could be class b as well since no ip given.
    magic number 32

    Do NOT worry about Identifying if its CLASS B or Class C, etc.
    It does NOT matter.
    The only thing that matters is recognizing the correct octet (4th, in this case)
    pinkiaiii wrote: »
    8 subnets

    0 valid ranges 1-31 and another minus one for usable ones 1-30
    32 | 33-63| 33-62
    64 |65-97 |65-96
    98 |99-127 | 99-126
    128 | 129-161| 129-160
    162 |163-191 |163-190
    192 |193-223 |193-222
    224 |224-255 |224-254

    for some reason it gets squeezed up thus the | once posted.ok didnt look at any charts all from head,but looking at my post noticed mistake made at 98,should been 96 thus messed up :/

    thus again,0,32,64,96,128,162,194,224

    ranges 1-31,33-63,65-95,95-127,129-161,163-193,194-223,225-255
    valid ones up to 30,62,94,126,150,192,222,254

    You got it.
    The actual syntax i was looking for was:
    jj.kk.ll.0
    jj.kk.ll.32
    jj.kk.ll.64
    jj.kk.ll.96
    ... etc
    but you got it.

    I was just trying to Show you that the Subnet's ADDRESS doesn't actually matter.
    The ONLY thing that matters is the Subnet MASK.


    Way to go, Grasshopper :]
  • pinkiaiiipinkiaiii Posts: 216Member
    Thanks again :))
    one tree down dozen more to go,any chance your good at ipv6 ? :]
  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    pinkiaiii wrote: »
    Thanks again :))
    one tree down dozen more to go,any chance your good at ipv6 ? :]

    Not with the same Authority that i can explain subnets.

    However,
    if you understand the basic of ipv4, then you can understand some of the basics of ipv6.

    Danscourses, Videos 68 & 69 are good to watch:
    https://www.youtube.com/playlist?list=PL33E07ECCA73C0755


    and Professor Messer, Section 1.8, is also a winner (in case your teacher stinks):
    Professor Messer's CompTIA N10-006 Network+ Training Course | Professor Messer IT Certification Training Courses

    Give them a watch; it's only 35-40 minutes total :]


    With that being said,
    i think you should work on ROUTE SUMMARIZATION instead.
    Logically, it's the next part of subnetting...
  • pinkiaiiipinkiaiii Posts: 216Member
    Cool thx for video numbers to watch and think already seen few of dans ones.Sadly with being tough ipv6 i think there was mention of it once that ipv6 is 128bits maybe a line or two that windows can get their ipv6 from router,but rest what i know is just by reading stuff myself and couple videos watched,thus i call it remember the small bits i was thought and seek help online icon_wink.gif

    Since you did great job on explaining whole class b subneting and it took about a week or so back and forth with replies and different scenarios to play,but you seen where i made mistakes,thus eventually coming out where i could point in the right direction and spot mistakes while doing questions you asked,just by refocusing and trying again,to understand reasoning behind each question you gave, and pointing my mistakes where my thinking was wrong.seems easy now,but it took time,not a lot in terms of doing it,but just to focus how to understand it right when laying it out from scratch.

    this whole topic started as acl but turned into something of a lesson that i think someone having issues and looking trough topics would easily get hang of it.

    yeah route summarization sounds like next option to go actually once you mentioned since its easier then subneting,but depending on how many networks it takes, one bit wrong will get whole mask wrong for given addresses.
  • pinkiaiiipinkiaiii Posts: 216Member
    Also to keep this topic of ACLs and wildcard masks active there was quite interesting question today in the lab book,that went back to basics if anyone wants a go at it,since i did needed explanation on it icon_study.gif

    Basically you have written acl to deny telnet on vty line for most ip except few.

    heres show access-lists output looks like

    permit 10 192.168.1.3 (2 matches)

    permit 20 192.168.1.4 wildcard bits 0.0.0.3 (2 matches)

    30 deny any (3 matches)


    Why failed telnet produced more matches then sucesfull connection :) ?
  • pinkiaiiipinkiaiii Posts: 216Member


    A company has several networks with the following IP address requirements:
    IP phones - 50
    PCs - 70
    IP cameras - 10
    wireless access points - 10
    network printers - 10
    network scanners - 2

    Which block of addresses would be the minimum to accommodate all of these devices if each type of device was on its own network?





    Correct
    Response

    Your
    Response



    tick.gif

    172.16.0.0/24



    172.16.0.0/25



    172.16.0.0/23



    172.16.0.0/22


    I feel like im beating a dead cat here,but how come /24 is better then /25 given that its class b address and its asking for minimum that would be /25 providing there would be 126 hosts per subnet,rather having 254 hosts and subnets-which is way too much when asking for minimum.Is it ccna logic or am i missing some quirk hidden in the question icon_confused.gif



  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    Agreed.

    If they were All on the same Subnet.... then you'd need 152 addresses; /24

    But this is saying "if each type of device was on its own network".
    Which means... look for the LARGEST address block... "PCs - 70".
    So, the block needs to be... /25

    Unless someone can explain this better..... i say the book's answer is WRONG.
    lol
  • pinkiaiiipinkiaiii Posts: 216Member
    thanks for your input wolfkhat as always icon_wink.gif its question from netacad chptr exam.

    posted another question but took me couple tries just to see that i messed up with binary-thus had to delete it for my own sanity doing route summarization-back to doing some drawing on paper.
  • volfkhatvolfkhat Posts: 947Member ■■■■■■■□□□
    Route Summarization is kind of EASY; if you approach it correctly.

    Feel free to post a Q, and i can probably teach you a couple of tricks.
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