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

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

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Comments

miller811

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

http://www.****.com/easy_way_to_subnet.html

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 192.168.12.78/29 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:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

What subnet does 172.16.116.4/19 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:-

172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc

Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

What subnet does 10.34.67.234/12 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:-

10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc

Our address is 10.34.67.234 which must sit on the 10.32.0.0 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 192.168.10.0/28?

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.

192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc

Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

What is the valid host range of the 1st subnet of 172.16.0.0/17?

/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:-

172.16.0.0
172.16.128.0

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

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 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remebering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

HTH

borumas

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.

impelse

I studied subnetting with the Bryant Advantage, it was very good and the exercises helped me a lot to nail it.

Raul

Plazma

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