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Cisco Exam Question Clarification

Posted on 2010-08-12
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For the Cisco CCNA exam there are going to be questions regarding subnets.  The formula for calculating subnets is 2^n unless the router IOS shows the command "no ip classless" which makes the formula 2^n-2.

So my question is simple.  Which mode is Cisco going to use on their testing?  By default a Cisco router does enable "ip classless" and you have to specify "no ip classless" manually.  The lastest training I went through also said the Cisco test questions would specify "no ip classless", however on the latest pass4sure testing engine the question assumes "no ip classless" although that is not the router default.
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Question by:captclam
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by:RPPreacher
ID: 33423900
Ip classes has nothing to do with zero subnet.  See http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml

For ccna, always use 2^n-2

Ip classes command has to do with classful ranges, not zero subnet
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by:captclam
ID: 33424014
Sorry, my mistake, I meant "ip subnet zero".  So will the test assume the router default of ip subnet zero, or is the pass4sure exam question correct is assuming the use of  "no ip subnet zero".  I know ip subnet zero is on by default.
Thanks
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RPPreacher earned 50 total points
ID: 33424322
Assume 'no ip subnet zero'
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by:RPPreacher
ID: 33424502
Anything else before closing this question?
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by:captclam
ID: 33536355
This same question was posted on the Cisco Learning network.  Current exams, unless specified, you are to assume ip subnet-zero is being used.  This is the current default for the Cisco IOS.  So the default formula for calculating subnets should be 2^n.  The CCIE's also commented that they didn't believe the Cisco CCNA exam would include both possibilities in one multple choice question.  You should not see a question like:
How many usuable subnets are in 192.168.1.0 / 28?
A.  16  B.  14   C.  4  D.  24
If you see this type of question the answer should be A, not B.  You should also not see this type of question which clarifies the answer.
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