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Tricky subnet question

GobwiseGobwise Member Posts: 7 ■□□□□□□□□□
Hi,

Can anyone advise on this question from a Boson test 640-801?

2 routers are connected point-to-point over a serial link, both by their S0 interface. Router A's S0 interface has a subnet mask of 255.255.255.248. Router B's S0 interface has an ip of 172.20.3.3. Calculate the subnet for Router A's S0 interface. Also assume that you are not using subnet zero.

The answer given is 172.20.3.1, with a quite illogical explanation. Can any give me a reason why this answer is the case. I thought the answer would be 172.20.3.0........ icon_confused.gif:

Thanks

(Taking the exam on Thursday)[/img]

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    keenonkeenon Member Posts: 1,922 ■■■■□□□□□□
    Gobwise wrote:
    Hi,

    Can anyone advise on this question from a Boson test 640-801?

    2 routers are connected point-to-point over a serial link, both by their S0 interface. Router A's S0 interface has a subnet mask of 255.255.255.248. Router B's S0 interface has an ip of 172.20.3.3. Calculate the subnet for Router A's S0 interface. Also assume that you are not using subnet zero.

    The answer given is 172.20.3.1, with a quite illogical explanation. Can any give me a reason why this answer is the case. I thought the answer would be 172.20.3.0........ icon_confused.gif:

    Thanks

    (Taking the exam on Thursday)[/img]

    its seems to be right if your not using subnet 0.. because u still have 3 unused bits for host 4, 2, 1 if there using the host address of .3 u are using bits 2 and 1.. remember all 1 and all 0 are not usable so host addresses u have available are 6, 5, 4, 2, 1
    110=6
    101=5
    100=4
    010=2
    001=1
    Become the stainless steel sharp knife in a drawer full of rusty spoons
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    GobwiseGobwise Member Posts: 7 ■□□□□□□□□□
    I appreciate that you'll have those hosts left available, but why does the subnet have to be .1, as opposed to .2?

    Maybe I'm missing something here.
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    keenonkeenon Member Posts: 1,922 ■■■■□□□□□□
    i take it the question was mutliple choice... if so u have to look at the choices available..
    Become the stainless steel sharp knife in a drawer full of rusty spoons
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    netheadnethead Member Posts: 43 ■■□□□□□□□□
    Calculate the subnet for Router A's S0 interface

    As it specifies SUBNET then I would say you are correct to say 172.20.3.0, 172.20.3.1 is an ip address not a subnet.

    It is correct that the ip address does not have to be .1 as there are several addresses available for use on this subnet. I think this is a badly worded question.
    "The time for talking is over. Now call it extreme if you like, but I propose we hit it hard, and we hit it fast, with a major, and I mean major, leaflet campaign. "
    - Rimmer, Polymorph
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    JerzJerz Member Posts: 86 ■■□□□□□□□□
    Huh, I'm not sure but I thought for the subnets you would say 256-248=8 so the subnets would increase by 8 starting at 0:
    172.20.3.0
    172.20.3.8
    172.20.3.16
    172.20.3.24
    ...
    172.20.3.248


    I would think 172.20.3.0 would be the answer also.... Let's see 3 bits for the host would be 2^3-2=6hosts per subnet and 2^5=32 subnets

    172.20.3.0 - subnet1
    172.20.3.1 - host1
    172.20.3.2 - host2
    172.20.3.3 - host3
    172.20.3.4 - host4
    172.20.3.5 - host5
    172.20.3.6 - host6
    172.20.3.7 - broadcast address
    172.20.3.8 - subnet2

    icon_confused.gif:

    You got me...
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    NocturnalNocturnal Member Posts: 44 ■■□□□□□□□□
    First of all, it doesn't make any sense to have a point to point serial connection with a 255.255.255.248 (/29) mask; it's a waste. There are six valid host addresses, 172.20.3.1 through 172.20.3.6. Dot zero is the network address, and even though the question says not to use subnet zero, you have to if the IP address of Router B is 172.20.3.3. There's no such thing as a network address with a dot one. All network addresses are even numbers. All broadcast addresses are odd numbers.

    The way to set up a point to point connection is with a 255.255.255.252, ( /30) mask. There are only two valid host addresses with this mask.

    That question sounds messed up. If I'm wrong, I hope someone corrects me.

    I agree with you that the network address, given the information you presented, is 172.20.3.0.
    "...a long habit of not thinking a thing wrong, gives it a superficial appearance of being right,..."
    --Tom Paine
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    GobwiseGobwise Member Posts: 7 ■□□□□□□□□□
    Thanks everyone, I'm glad to hear I'm not going mad. The question was actually a text box one, not multiple choice. But I agree that the subnet should be .0

    As a side note, I've been very impressed with the Boson tests I bought and the CCNP netsim as well. The tests have all the simulation, drag and drop and multiple choice stuff - I'd highly recommend them.

    Cheers
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    tunerXtunerX Member Posts: 447 ■■■□□□□□□□
    Nocturnal wrote:
    First of all, it doesn't make any sense to have a point to point serial connection with a 255.255.255.248 (/29) mask; it's a waste.

    If the other side of the connection was multipoint then it wouldn't be a waste.

    If both sides were point-to-point then you would be better off with ip unnumbered pointing to a loopback interface.
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    rossonieri#1rossonieri#1 Member Posts: 799 ■■■□□□□□□□
    Gobwise wrote:
    Also assume that you are not using subnet zero.

    The answer given is 172.20.3.1, with a quite illogical explanation. Can any give me a reason why this answer is the case. I thought the answer would be 172.20.3.0........

    172.20.3.1 is valid ip add -> from network 172.20.3.0
    172.20.3.2 is also a valid ip add since it is the member of the same network.

    but for explanation,
    subnet consisting a network id, some member ip add, and a broadcast id - before it go to the next network id. and so on.
    the More I know, that is more and More I dont know.
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    sunny_evandersunny_evander Member Posts: 126
    Gobwise wrote:
    Hi,

    Can anyone advise on this question from a Boson test 640-801?

    2 routers are connected point-to-point over a serial link, both by their S0 interface. Router A's S0 interface has a subnet mask of 255.255.255.248. Router B's S0 interface has an ip of 172.20.3.3. Calculate the subnet for Router A's S0 interface. Also assume that you are not using subnet zero.

    The answer given is 172.20.3.1, with a quite illogical explanation. Can any give me a reason why this answer is the case. I thought the answer would be 172.20.3.0........ icon_confused.gif:

    Thanks


    It's given in the question icon_confused.gif Also assume that you are not using subnet zero.




    (Taking the exam on Thursday)[/img]
    :santa:
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    tunerXtunerX Member Posts: 447 ■■■□□□□□□□
    Sunny, you have it all wrong. Even if you cannot use subnet zero 172.20.3.0 is still a valid subnet.

    Everyone else was right in saying that 172.20.3.0 is the correct answer and the question was wrong.
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    GobwiseGobwise Member Posts: 7 ■□□□□□□□□□
    I think the confusion over subnet zero is quite common ( I wasn't sure for a while). The point of subnet zero is that the subnet is the zero network with the classful mask, i.e. 10.0.0.0 172.16.0.0 192.168.0.0

    I sent a fair few corrections back to the authors of the boson tests, but on reflection it's actually the wrong ones which you end up debating like this and learn most from. Just a form of reinforcement I suppose - an evaluable part of teaching.

    Not sure who said it, but failure is so much more interesting than success.
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    GobwiseGobwise Member Posts: 7 ■□□□□□□□□□
    Just found out it was a guy called Max Beerbohm BTW...

    evening all...
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    rossonieri#1rossonieri#1 Member Posts: 799 ■■■□□□□□□□
    about ip subnet-zero : /9 - /17 - /25
    if you are working on win2k , the TCP/IP properties let you do an IP subnet-zero ex : 192.168.0.1 mask 255.255.255.128.
    and afaik, the newer IOS release ( 12.0 so on ) has ip subnet zero enabled by default.
    the More I know, that is more and More I dont know.
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