Failed again 809 --> 834 --> 3rd time a charm? maybe..

mobri09mobri09 Users Awaiting Email Confirmation Posts: 723
Failed 640-801 again on my second attempt - Was a little bummed out, but I have to remind myself it's only a test. I know my stuff, but when it comes to exams and time limits I have horrible anxiety.

Attempt 1 (809)

Planning and Designing 83%
Implementation and Operation 75%
Troubleshooting 72%
Technologoy 61%

Attempt 2 (834)
Planning and Designing 75%
Implementation and Operation 83%
Troubleshooting 66%
Technologoy 84%

My gray areas are subnetting larger network problems...Actually finding out what the problem is on a network with subnetting involved. I can subnet, but I can't subnet large networks fast.
Either way pass or fail I have learned so much
The CCNA is a toughy icon_lol.gif


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    miller811miller811 Member Posts: 897
    I struggled with subnetting. I understood the concept using the boolean AND method, but it was time consuming. After finding these, it all clicked and I can almost subnet any problem in my head.
    Hope it helps


    Typical questions you may see are those asking what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

    What subnet does belong to?

    You may wonder where to begin. Well to start with let's find the next boundary of this address.

    Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size.
    We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

    Our address is so it must sit on the subnet.

    What subnet does sit on?

    Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.
    We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

    Our address is so it must sit on the subnet. Easy eh?

    What subnet does sit on?

    Our mask is 12. Our next bounday is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.
    We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

    Our address is which must sit on the subnet.

    We will now change the type of question so that we have to give a particular host range of a subnet.

    What is the valid host range of of the 4th subnet of

    Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

    Therefore the 4th subnet is and the host range must be to, remembering that the subnet and broadcast address cannot be used.

    What is the valid host range of the 1st subnet of

    /17 tells us that the block size is 24-17 = 7 and 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

    The first subnet is and the valid host range is to You must remember not to include the subnet address ( and the broadcast address (

    What is the valid host range of the 7th subnet of address

    The block size is 4, from 16 - 14 = 2 then 2^2 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

    The seventh subnet is Our valid host range must be to again remebering not to include our subnet ( and the broadcast address (

    I don't claim to be an expert, but I sure would like to become one someday.

    Quest for 11K pages read in 2011
    Page Count total to date - 1283
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    borumasborumas Member Posts: 244 ■■■□□□□□□□
    Sorry to hear that, keep trying and you will get it. For subnet practice hit this site: www.subnettingquestions.com/default_int.asp after doing many questions it really helped me to nail subnetting down. I've been low on funds but hope to schedule my test tomorrow evening for either November 5th or 6th.
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    impelseimpelse Member Posts: 1,237 ■■■■□□□□□□
    I studied subnetting with the Bryant Advantage, it was very good and the exercises helped me a lot to nail it.

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    PlazmaPlazma Member Posts: 503
    That was an issue of mine long ago

    You subnet the bigger networks the exact same way you do the class C ones

    Remember your block sizes

    Class C = 256 128 64 32 16 8 4 2 1
    Class B = 131072 65536 32768 16384 8192 4096 2048 1024 512
    Class A = you get it

    Keep in mind that 256 and 131072 are right on the line but are still included
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