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  1. Member
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    #1

    Default Hardest 4 questions from CCNA 1-2. Need Good explainer!!

    Hey guys, I thought I finally figured out subnetting, ive read all the examples in this form too.

    I managed to answer lots of subnetting questions, but these 4 questions really stumped me.

    Question 1


    Question 2


    Question 3


    Question 4


    IF ANYONE IS WILLING TO GO OUT THEIR WAY TO LOOK AT THEM, AND ASSIST IN EXPLAINING HOW THE ANSWER IS WHAT IT IS, I'D BE EXTREMLY GRATEFUL!!

    In case the images didnt come up.
    question 1 http://i162.photobucket.com/albums/t...ninja84/Q4.jpg
    question 2 http://i162.photobucket.com/albums/t...ninja84/Q3.jpg
    question 3 http://i162.photobucket.com/albums/t...ninja84/Q2.jpg
    question 4 http://i162.photobucket.com/albums/t...ninja84/Q1.jpg


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    #2
    The first one i believe is 151.100.156.1 that is not n the same subnet as the rest.


    and the next three are the same problem?!?!?!

    regardless i believe it's 192.168.100.63
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  4. Senior Member deneb829's Avatar
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    #3
    for the first one:
    You get the network address by ANDing the IP and the subnet mask. In binary, ANDing means that 1+1=1; 1+0=0; 0+0=0
    So where the ip address and subnet mask both have a 1, you put a 1 in the bottom row. If the IP address or the subnet mask bit is a 0, you put a 0.

    To get the Broadcast address you invert the subnet mask (make the 1's 0's). and OR the results. OR is a 1+1 = 1; 1+0=1; 0+0=0

    10010111.01100100.11100001.00000001 (151.100.225.1) IP Address
    11111111.11111111.11000000.00000000 (255.255.192.0) Subnet Mask
    10010111.01100100.11000000.00000000 (151.100.192.0) Network Address

    10010111.01100100.11100001.00000001 (151.100.225.1) IP Address
    00000000.00000000.00111111.11111111 Inverted Subnet Mask
    10010111.01100100.11111111.11111111 (151.100.255.255) Broadcast Address
    -----------------------------------------------------------------------------------------------
    10010111.01100100.11000001.01100100 (151.100.193.100) IP Address
    11111111.11111111.11000000.00000000 (255.255.192.0) Subnet Mask
    10010111.01100100.11000000.00000000 (151.100.192.0) Network Address

    10010111.01100100.11100001.00000001 (151.100.225.1) IP Address
    00000000.00000000.00111111.11111111 Inverted Subnet Mask
    10010111.01100100.11111111.11111111 (151.100.255.255) Broadcast Address
    -----------------------------------------------------------------------------------------------
    10010111.01100100.10011100.00000001 (151.100.156.1) IP Address
    11111111.11111111.11000000.00000000 (255.255.192.0) Subnet Mask
    10010111.01100100.10000000.00000000 (151.100.128.0) Network Address

    10010111.01100100.11100001.00000001 (151.100.225.1) IP Address
    00000000.00000000.00111111.11111111 Inverted Subnet Mask
    10010111.01100100.10111111.11111111 (151.100.191.255) Broadcast Address
    -----------------------------------------------------------------------------------------------
    10010111.01100100.11111111.11010010 (151.100.255.210) IP Address
    11111111.11111111.11000000.00000000 (255.255.192.0) Subnet Mask
    10010111.01100100.11000000.00000000 (151.100.192.0) Network Address

    10010111.01100100.11100001.00000001 (151.100.225.1) IP Address
    00000000.00000000.00111111.11111111 Inverted Subnet Mask
    10010111.01100100.11111111.11111111 (151.100.255.255) Broadcast Address

    You can see that the network address for 151.100.156.1 is 151.100.128.0 and the other three have a network address of 151.100.192.0

    For the 2nd set

    11000000.10101000.01100100.00100000 (192.168.100.32)
    11111111.11111111.11111111.11100000 (255.255.255.224)

    Using the inverted subnet process listed above
    11000000.10101000.01100100.00100000 (192.168.100.32)
    00000000.00000000.00000000.00011111 Inverted Subnet Mask
    11000000.10101000.01100100.00111111 (192.168.100.63)
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    #4
    Youre a legend, thanks!!
    That illustrates it really well.

    Ive just spent few hours on other broadcasts and ive got it right.

    Ive got couple of questions for you, this might sound dumb, but on the revision material I used, it wasnt there. what does the slash represent after the ip.

    E.G What valid host range is the IP address 172.28.229.7/28 a part of?
    And what is your second part related too?

    Mucho Gracius for your help!!
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    #5
    Quote Originally Posted by monkey_ninja3
    Youre a legend, thanks!!
    That illustrates it really well.

    Ive just spent few hours on other broadcasts and ive got it right.

    Ive got couple of questions for you, this might sound dumb, but on the revision material I used, it wasnt there. what does the slash represent after the ip.

    E.G What valid host range is the IP address 172.28.229.7/28 a part of?
    And what is your second part related too?

    Mucho Gracius for your help!!
    You should really go back and read the material you have on subnetting again. Also, go to the FAQ in this forum and read every post on subnetting. It has its own section. If you're asking what the /28 stands for then you may not be as comfortable with this as you think. Re-read your books chapter on addressing and subnets.
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    #6
    Quote Originally Posted by monkey_ninja3
    E.G What valid host range is the IP address 172.28.229.7/28 a part of?

    And what is your second part related too?
    /28 is the same as 255.255.255.240

    Could you give us an idea of what study material your using? On most texts, turn the page after subnetting classfull and it explains the "/" notation. Classfull, classless, cidr, vlsm, etc. These are all part of the same topic.

    Let us know. You're hooked on the drug that is learning this topic and we can give you some pointers as to where to go next.
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    #7
    Quote Originally Posted by monkey_ninja3
    Ive got couple of questions for you, this might sound dumb, but on the revision material I used, it wasnt there. what does the slash represent after the ip.

    E.G What valid host range is the IP address 172.28.229.7/28 a part of?
    And what is your second part related too?

    Mucho Gracius for your help!!
    Hey monkey ninja

    Just count the 1's in a binary subnet mask (not inverted) and you've got your slash. For example:

    11000000.10101000.01100100.00100000 (192.168.100.32)
    11111111.11111111.11111111.00000000 (255.255.255.0)
    That subnet has 24 1's so it's a /24

    11000000.10101000.01100100.00100000 (192.168.100.32)
    11111111.11111111.11111111.11100000 (255.255.255.224)
    This one from the earlier example would be a /27

    10010111.01100100.11000001.01100100 (151.100.193.100)
    11111111.11111111.11000000.00000000 (255.255.192.0)
    Also from the earlier example would be a /18

    172.28.229.7/28
    would be:
    10101100.00011100.11100101.00000111 (172.28.229.7)
    11111111.11111111.11111111.11110000 (255.255.255.240) (Remember to AND)
    10101100.00011100.11100101.00000000 (172.28.229.0) Network Address

    10101100.00011100.11100101.00000111 (172.28.229.7)
    00000000.00000000.00000000.00001111 Inverted Subnet Mask (Remember to OR)
    10101100.00011100.11100101.00001111 (172.28.229.15) Broadcast Address

    Since we can't use the network address and the broadcast address for hosts, then the valid IP addresses in this subnet are 172.28.229.1 - 172.28.229.14

    Hope this makes sense.
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    #8
    Hey guys, thanks for your advice - you dudes rock!!!
    I read the FAQ again, and other sites, I also re-read the cisco book.

    your've helped me a lot, I think I get it finally!!

    I understand that (correct me if i'm wrong) :-

    - to find the network IP address for a host IP, you logical ANDing its subnet mask

    - to find the broadcast IP address for a host IP, you locigical Anding its 'inverse' subnet mask.

    The slash after a network ip address represents the ammount of bits borrowed.for the subnets

    2 * bits borrowed (-2) = useable subnets
    2 * bits remaining (-2) = useable hosts

    Class A can borrow up to 22 bits for subnetting
    Class B can borrow up to 14 bits for subnetting
    Class C can borrow up to 6 bits for subnetting

    But the final bit thats confusing me, is how do you work out what specific hosts Ip Address's are usable on subnets.
    Im asking - for example, you have class B address of 172.24.142.78/21
    So you have 5 bits borrowed for subnetting, and 11 bits for hosting.
    Max available subnets = 30 subnets, and 2046 hosts on each subnet.
    How do you work out what IP addrees's you can use on each of the 30 subnets?

    If you can guide me somewhere where it answers it that i've missed, or if you can kindly illustrate it, would be awesome! I have the ccna exam in 2 weeks, im spending 10 hours a day learning this - and it's taking a lot longer than I thought.

    E.G. What is the first valid host on the subnetwork that the node 172.24.142.78/21 belongs to?

    NinjaMonkey
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    #9
    Hi im sorry to say you still have a bit more work in subnetting to go (don’t want to offend you)
    Let’s look at the first example
    We are dealing with a class b address so as default the first 16 bits of the mask are (255.255.0.0) 0r a /16 (as there are 16 bits)
    Now we can see the mask they give is 255.255.192.0 (or can be said as /18 as there is 18 bits) this means they have borrowed 2 bits.
    128 = 1bit borrowed
    192 = 2 bits borrowed
    224 = 3 bits borrowed and so on
    Now if we have 192 in the third octet and we know we are working with a class b address this means we are subnetting in this octet -192 from 256 leaves us with a block size of 64. so are subnets are .
    151.100.0.0
    151.100.64.0
    151.100.128.0
    151.100.192.0
    Now we look at the example they give us and we can see 151.100.156.1 is in a different subnet, all the other addresses are in the 151.100.192.0 network


    Now if we wanted to know how many networks and how many hosts would be in these networks that is a bit more work
    Again we know it is a class b address so the first 16 bits are the default and there mask is 255.255.192.0 this equals 18 bits (but we know 16 are the default) so we have two bits to subnet with
    For the networks 2 to the power of 2 = 4 networks, and as I have shown above these networks are 151.100.0.0
    151.100.64.0
    151.100.128.0
    151.100.192.0
    For the hosts 2 to the power of what bits are left (as there is only 32 bits to start with and 18 bits are being used in the mask) so we have 14 bits left
    So 2 to the power of 14 = 16384 hosts take away your first address (the actual network) and the last address (broadcast address) leaves us with 4 networks and 16382 hosts on each of these 4 networks
    Now if we are now using ip classless you would have to take away 2 from the networks.

    It is easier talking about this than trying to explain it like this as I don’t know how much you know and some of what im trying to say does not go down on text as well as I could say it. If this is a bit much say so and I will break it down a bit more
    The best examples of subnetting are in Todd lammle book by sybex (ccna Cisco certified network associate) worth it for the subnetting chapter alone.

    to show you were the hosts work in lets look at the first network
    151.100.0.0 the network
    151.100.0.1 the first valid address
    151.100.63.254 the last valid host as 151.100.64.0 is the next network
    151.100.63.255 the broadcast address

    hope this helps

    stephen
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  11. Senior Member deneb829's Avatar
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    #10
    Quote Originally Posted by monkey_ninja3
    - to find the broadcast IP address for a host IP, you locigical Anding its 'inverse' subnet mask.

    NinjaMonkey
    The Broadcast is going to be a logical OR, not an AND - otherwise, you'll get some whacked out numbers.
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    #11
    For the 172.24.142.78/21 question the mask is 255.255.248.0 and if we take away 248 from 256 leaves us with a block size of 8 so are networks are
    172.24.0.0
    172.24.8.0
    172.24.16.0
    172.24.24.0 As you can see we are going up in blocks of 8 all the way up to the 172.24.248.0 network.
    Now your pc is on the 172.24.136.0 network
    172.24.136.1 The first valid address
    172.24.143.254 the last valid address
    172.24.143.255 the broadcast address
    The reason 172.24.143.255 is the broadcast address is 172.24.144.0 is the next network up.

    For how many networks 2 to the power of 5 = 32 networks

    For the hosts 2 to the power of 11 = 2048 – 2 = 2046 hosts on each network

    Please remember we are using ip classless here

    stephen
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    #12
    hey dude, thanks for the answer.
    Im still a bit confuzzled.
    I wasnt reffering to the cisco questions anymore, 'cause that geeza earlier illustrated it perfect for me!
    I was stating the 172.24.142.78/21 example.
    Your way is an SWEET way of showing the cisco questions in more depth ,Thanks!

    Ive tried copying your explainations to work out the 30 subnets Ip addreeses on 172.24.142.78/21
    So you have 5 bits borrowed for subnetting, and 11 bits for hosting.
    2 the the power of 5 = 32. -2 = 30 useable subnets?
    So the subnet mask will be 255.255.248.0
    So 256 - 248 = .8
    So subnets go up in increments of 8?

    1 ) 172.24.8.0
    2 ) 172.24.16.0
    3 ) 172.24.24.0
    4 ) 172.24.32.0
    5 ) 172.24.40.0
    6 ) 172.24.48.0
    7 ) 172.24.56.0
    8 ) 172.24.64.0
    9 ) 172.24.72.0
    10) 172.24.80.0
    11) 172.24.88.0
    12) 172.24.96.0
    13) 172.24.104.0
    14) 172.24.112.0
    15) 172.24.120.0
    16) 172.24.128.0
    17) 172.24.136.0
    1 172.24.144.0
    19) 172.24.152.0
    20) 172.24.160.0
    21) 172.24.168.0
    22) 172.24.176.0
    23) 172.24.184.0
    24) 172.24.192.0
    25) 172.24.200.0
    26) 172.24.208.0
    27) 172.24.216.0
    2 172.24.224.0
    29) 172.24.232.0
    30) 172.24.240.0

    I know there are 32 subnets , but only 30 subnets that can be used, have I made a mistake, as 172.24.248.0 would be 31, and 172.24.255.0 would be reserved for broadcast?
    So there are the 30 subnets for the Network Ip 172.24.142.78/21, with subnet mask
    255.255.248.0 ... IS that right?? (stupid of me to not pick out a shorta example, but didnt wanna leave anything to assumtion)

    So if thats correct and those are the 30 useable subnets? then what Ip address's can I use for the hosts on each one ? as I have a bodocious ammount of 2046 hosts on each one ??

    So you have 5 bits borrowed for subnetting, and 11 bits for hosting.
    Max available subnets = 30 subnets, and 2046 hosts on each subnet

    Any clarirication would be great, dont worry about offending me, this has been the bane of my existence for past 3 days.
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    #13
    Quote Originally Posted by monkey_ninja3
    I wasnt reffering to the cisco questions anymore, 'cause that geeza earlier illustrated it perfect for me!
    geeza? I'm not that old!

    Unless you're from the U.K. then I guess it means something different
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    #14
    haha, you truely are a diomand geeza mate, young man doesnt have the same ring to it.
    I see, bradcast is logical OR. I just got the term wrong - thanks for clearing that 1 up.

    I tried answering that question b4 I saw your 2nd post stephen.
    I get it, so there are 32 subnets, as you dont minus 2.

    So 172.24.0.0 would be another 1
    and finally 172.24.148.0 would be 32.

    So I understand that, but how do I apply the host ip's to it.
    2046 to each subnet, right?

    I understand now its CIDR, classless. Thanks.

    NinjaMonkey
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    #15
    I just checked it on an internet subnet calculator.

    172.24.142.78/21


    Address: 172.24.142.78 10101100.00011000.10001 110.01001110
    Netmask: 255.255.248.0 = 21 11111111.11111111.11111 000.00000000
    Wildcard: 0.0.7.255 00000000.00000000.00000 111.11111111
    => Network: 172.24.136.0/21 10101100.00011000.10001 000.00000000 (Class B)
    Broadcast: 172.24.143.255 10101100.00011000.10001 111.11111111
    HostMin: 172.24.136.1 10101100.00011000.10001 000.00000001
    HostMax: 172.24.143.254 10101100.00011000.10001 111.11111110
    Hosts/Net: 2046 (Private Internet)


    It's Saying that 172.24.136.1 - 172.24.143.254 is the Ip range!!

    Stephen, can you please explain hwo this relates to what you told me. Thanks buddy.
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    #16
    Hi again well the 172.24.248.0 network can be used (depends if we are using ip subnet zero its default on Cisco routers since version 12 I think) this means you don’t have to take away two from the networks.
    So for our last network =
    172.24.248.0 The network
    172.24.248.1 The first address
    172.24.255.254 the last valid host
    172.24.255.255 the broadcast address

    As you can see from the example above the network runes from 172.24.248.0 to 172.24.255.255 the broadcast for a total of 2048 addresses (remember to -2) and this is only one of your networks and there are 32 (2 to the power of 5 using ip classless you don’t need to takeaway 2)

    For another of your subnets take the 172.24.8.0 network
    172.24.8.1 The first add
    172.24.15.254 the last valid add
    172.24.15.255 the broadcast address
    As 172.24.16.0 is the next network up
    I’ve typed quite a bit here and as its late here, I may have made a few mistakes if so I will read over what I have typed tomorrow and check it but feel free anyone to point them out

    Ask anymore and if I can help I will
    All the best

    Stephen
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    #17
    to put another way take the 172.24.0.0 network
    it runes from 172.24.0.1 the first add 172.24.0.2 the second 172.24.0.3 the third add and so on / cut a few hundred an go to 172.24.5.24 to the next 172.24.5.25 / cut a few hundred more to 172.24.6.1 / and a few hundred more to the 172.24.7.1 to 172.24.7.2 to 172.24.7.3 cut another few hundred to 172.24.7.254 the last valid address and finally 172.24.7.255 the broadcast address ( remember the range in each octet is from 0 to 255 giving us a total of 256) so when we reach the 172.24.7.255 ( the broadcast address) the next address up is the 172.24.8.0 network. and there is all your 2048 addresses in the 172.24.0.0 network ( remember 172.24.0.0 the network and 172.24.7.255 the broadcast)
    now i really am off to bed


    when you reach 255 in an octet the next one up will move the octet on the left up one and reset the one you were working in to 0
    eg
    172.24.0.255 the next address will be
    172.24.1.0 and this will run to 172.24.1.255 and again the next one up will be
    172.24.2.0 and this will run from 176.24.2.0 to 176.24.2.255 and when we reach 172.24.2.255 in this octet the next host will be 172.24.3.0 to 172.24.3.1 to 172.24.3.2 to 172.24.3.3 and so o.
    untill you reach 172.24.7.255 (the broadcast address) because the next one up will be the 172.24.8.0 network and it will work the same eg
    172.24.8.0 network
    172.24.8.1 first add
    172.24.15.254 the last host
    172.15.15.255 the broadcast add

    and as you know know the next network is the 182.24.24.0 network


    stephen
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    #18
    Thanks man, I think ive finally painted whole picture in my head.

    I understand the Ip Network address is 172.24.142.78

    This Ip is then seperated into 32 Sub-networks.

    On EACH of these subnetworks, there can be up to 2046 hosts.

    Now I think i've finally got it!

    172.24.0.0 is the first subnet , which has hosts 172.24.0.1 - 172.24.7.254
    Then the next subnet 172.24.8.0 - which has hosts 172.24.8.1 - 172.24.8.254
    and so on ...

    And this is called classless - I saw it on subnetting questions dot com.

    Ok so if this was classfull, I would have to take away 2 from what? The subnets?
    So if this was classfull, it'd only have 30 subnetworks, what woudl they be?
    What instance woudl it be classful as cisco has it as default.

    Damn - Youre a legend mate, thanks for your help so much!


    MonkeyNinja.
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    #19
    when thinking classful the networks are the default
    eg
    take the 10.0.0.0 network its default mask is 255.0.0.0
    this means all its hosts are on the same network and since it is classful you can not subnet it to do something like this 10.1.1.1/16 (we are subnetting in the second octet and since the mask is /16 now my networks can be
    10.0.0.0
    10.1.0.0
    10.2.0.0
    10.3.0.0
    all the way up to 10.254.0.0 which all classless as they are not in ther class boundaries
    or subnet it at all eg 10.45.56.2/24 if we are using classful our 10.0.0.0 net work will always be the 10.0.0.0 network with its dfault mask 255.0.0.0

    or the 172.16.0.0 network with its classful mask 255.255.0.0 since it is classful we can not subnet it at all. each classful network stays with its default subnet mask

    ill explain this better tomorrow if you want me to as i really tired now .
    just always think classful use its default mask only
    classless now you can use different masks ie subnet



    stephen
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    #20
    Thanks dude - I'll check out this forum tommorow afternoon for your explanations and spend rest of day revising it.

    I think I understand

    if you subnet class A, B or C address whatsoever, than any of those subnet address's are classless?

    or ..

    If you borrow more than an octet of host bits, it becomes classess?
    E.G
    a)10.0.0.0/17 or more
    b)172.16.0.0/25 or more

    MonkeyNinja.
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    #21
    Morning ok lets talk about how the router sees the ip address and the mask, im going to use a class c address to try make this a bit easies
    Ip add 192.168.128.0 the network
    255.255.255.0 With it class-full mask or /24
    Now if I need to subnet this to get 4 networks (my mask is going to change to show how many bits I have to borrow and my new mask will be a class-less mask as it no longer falls within the boundaries of a default mask for a class c)

    when you are talking about -2 from the networks it depends on if the router is using the ip subnet zero command or not
    192.168.128.0 The network
    255.255.255.0 The mask
    This is how the router sees this
    11000000.10101000.10000000.00000000 the network
    11111111.11111111.11111111.00000000 the default mask
    Now as I said above we need 4 networks so let’s start borrowing bits in the example below I have borrowed 2 bits
    11000000.10101000.10000000.00000000 the network (stays the same)
    11111111.11111111.11111111.11000000 the new mask with 2 bits on
    Now let’s just concentrate on the fourth octet in the mask for a second I know if I borrow 2 bits I will get 4 networks (2 to the power of 2 =4) lets see the 4 networks in binary
    Just the fourth octet
    000000000 = 0 or subnet 192.168.128.0
    010000000 = 64 or subnet 192.168.128.64
    100000000 =128 or subnet 192.168.128.128
    110000000 =192 or subnet 192.168.128.192
    As you can see this is how the router sees it but there is a problem here. without the ip subnet zero command in the router we are not allowed to use the first and the last networks because you cannot have a network of all 0s or all 1s so we are left with only the two middle networks and to get our 4 networks we need to borrow more bits lets try 3 bits
    Again only the fourth octet the new mask is 255.255.255.224 (borrowed 3 bits)
    192.168.128.0 00000000 network of 0
    00100000 network of 32
    01000000 = 64
    01100000 = 96
    10000000 = 128
    10100000 = 160
    11000000 = 192
    11100000 = 224
    Now again without the ip subnet zero command we must -2 networks the first and the last as you cannot have a network of all 0s or all 1s ( for want of a better explanation the router cant understand it ) so to get our 4 networks we need to borrow 3 bits (2 to the power of 3 -2 =6 networks)
    Now we will use the ip subnet zero command on a router (on by default since version 12 I think)
    Again lets only look at what is happening in the fourth octet
    192.168.128.0 Network
    255.255.255.0 default mask now changed to 255.255.255.192( 2 bits for subnetting)
    so are networks will be
    1982.168.128.0
    192.168.128.64
    192.168.128.128
    192.168.128.192
    And let’s just look at the fourth octet in binary
    00000000 = 0
    01000000 =64
    10000000 =128
    11000000 =192
    But this time the router can have a network of all 0s or all 1s and it can understand it because we are using the ip subnet zero command
    Now as you can see this has nothing at all to do with class-less and class-full networks or cidr which is classless
    It is just a way to let the router use all the bits in the mask (have a network of all 0s or all 1s)
    As far as I know if you are asked a question on subnetting assume it is talking about using ip subnet zero command unless otherwise stated ( ie you don’t need to -2 for the networks )
    If someone can explain this better please feel free
    if you need any more help just ask

    sorry had to edit this thanks to carveone for pointing my mistake out


    Stephen
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  23. Junior Member
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    #22
    I really don't wish to be nasty. Honest. But from my own experience you need to answer all 4 questions in less than 4 minutes. Seeing as they are easy, try to do it in 2 minutes. The CCNA is a cruel mistress

    (Yes, I know the thread has gone off that topic but I thought I'd go back to the first questions there).

    Eg: 1st question, the first thing you should think when looking at it is "increment is 64". That should be right on your fingers. If it isn't and you can't remember it then hell, write it down. You'll have plenty of time before the exam starts to write all that stuff down. I did cause I can't get the numbers for say, /27 in 5 seconds. And you really really need to. And I've better things to fill my pained brain with than subnet masks.

    Anyway. Increment is 64. Then 192 to 255 is the range (0-63, 64-127, 128-191, 192-255) and C is the answer. You should be able to get that in 30 seconds. Really!

    Ditto with the second. Increment 32, 63 is broadcast. Bam! C
    And the third. Increment 16. Broadbast 64+16-1 = 79. Last usable 78.
    Fourth. Increment 32. 128 is the next multiple above 100. 127 is broadcast.

    Oh yes. And as topstar says, classful means the default. Ie: If you didn't have the subnet mask there, what would you assume it to be. Classful routing protocols like RIP (v1) send just the classful networks. (1-126).0.0.0 implies 255.0.0.0. I think topstar has wandered off into supernetting, vlsm and all that.
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  24. Junior Member
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    #23
    I learnt that you subtract 2 from hosts and 2 from subnets unless specifically told that "ip subnet-zero" is enabled and then you have one more subnet. I admit it can be quite confusing when the question "how many networks" is posed. Subnets or Networks

    "ip classless" enables classless routing behaviour - the IOS forwards packets destined for unrecognized subnets to the best supernet route possible. Didn't see this before on the ccna books I was following.

    See http://www.cisco-gu.com/univercd/cc/...m#xtocid212805
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  25. Member
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    #24
    edited sorry carveone i was getting the ip subnet zero mixed up with the ip classless command thanks for pointing that out sorry if it caused any confusion monkey


    But as carveone correctly pointed out you should be able to do this in your head in about 30 – 40 seconds
    This look at this from a different angle
    192.168.1.0 Network default mask is 255.255.255.0
    And I want 8 networks with at least 20 hosts on each subnet
    Two to the power of what gives me 8 networks =( 2 to the power of 3= and since im using the ip subnet zero command on my router I don’t need to take away 2 (network of all 0s and all 1s)
    Add these 3 bits borrowed to my default mask = /27 or 255.255.255.224 (my new mask)
    Ok now to find out my block sizes (were my networks are)
    = take the 224 from 256 = block size of 32 my networks now are going up increments of 32
    Networks 192.168.1.0 192.192.1.32 192.168.1.64 192.168.1.96 192.168.1.128 and so on
    Now to find how many hosts on each subnet
    2 to the power of bits left = 2 to the power of 5 =32 hosts – 2 (1 for network address of each subnet and one for broadcast address of each subnet) so 30 valid hosts in each subnet. Now lets see how that works in/
    Network . 0 .32 .64 .96 .128 .160 and so on

    First add .1 .33 .65 .97 .129 .161 do this first

    Last add . 28 .62 .94 .126 .158 .192

    Broadcast add .31 .63 .95 .127 .159 .191 do this
    second


    You only take away 2 in the networks if your router does not support the ip subnet zero command so in the above example network 192.168.1.0 and network 192.168.1.224 would not be allowed ( networks of all 0s and all 1s ) so we would then have only 6 networks (2 to the power of 3 -2 = 6)
    Hope that is a bit easier to take in and remember after subnetting it is now classless as it has lost its classful mask of 255.255.255.0

    Or what network is the 192.168.5.56/26 pc on this should take about 30 second at max
    = /26 is 255.255.255.192 so 192 from 256 = block sizes of 64 and we can see the pc is on network 192.168.5.0
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  26. Member
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    #25
    I'm sure I understand now - im going to do some examples and get back to you.
    I appreciate your help.

    I really need to get subnetting mastered by today, is supernetting part of the ccna criteria?
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