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Network Management: Question about calculating network,host and broadcast a

I would like to say i am new to this forum and I do not know if this is the right place to post this but i am currently in a cisco academy for CCNA and if anyone else is in this class i am stuck @ 6.2.2 in the online cisco material.....

given the following address/prefix of 141.124.88.174/30

enter the last octet in binary of the network,

I am totally lost @ how to find the network octet i know /30 means the first 30bits are network and the last 2 are host i also know the last octet is 10101100 or 172

my question how do you get .174 to .172??? if you are wondering how i know the answer --it is in the book i just dont know how to do it on my own :)

thanks much

2 ACCEPTED SOLUTIONS

Accepted Solutions
Hall of Fame Super Silver

Re: Network Management: Question about calculating network,host

Eric

My sense is that you may have skipped a step (or maybe I skipped in understanding your post). The best way to explain this is to refer to binary - which you have started to do.

For address ...174 the last octet is 101011 10 (in which 101011 represent the network bits and 10 represent the host bits). To find the network bits make the host bits all binary 0 - which means that the subnet address is 101011 00 or 172.

So the answer to your question is that you get from 174 (host address ) to 172 (network address) by changing all host bits to binary zero.

HTH

Rick

Hall of Fame Super Silver

Re: Network Management: Question about calculating network,host

Eric

To figure out the broadcast address just take the host bits of the address (in this case the last 2 bits of the address) and put binary 1s in those bits. So the address of 101011 11 or 175 is the broadcast address.

Thanks for using the rating system to indicate that your question was resolved (and thanks for the rating). It makes the forum more useful when people can read a question and can read responses which successfully answered the question.

I encourage you to continue your participation in the forum.

HTH

Rick

4 REPLIES
Hall of Fame Super Silver

Re: Network Management: Question about calculating network,host

Eric

My sense is that you may have skipped a step (or maybe I skipped in understanding your post). The best way to explain this is to refer to binary - which you have started to do.

For address ...174 the last octet is 101011 10 (in which 101011 represent the network bits and 10 represent the host bits). To find the network bits make the host bits all binary 0 - which means that the subnet address is 101011 00 or 172.

So the answer to your question is that you get from 174 (host address ) to 172 (network address) by changing all host bits to binary zero.

HTH

Rick

New Member

Re: Network Management: Question about calculating network,host

i knew it was some simple math thing

thanks much

now how do u figure out the broadcast address

thanks

Hall of Fame Super Silver

Re: Network Management: Question about calculating network,host

Eric

To figure out the broadcast address just take the host bits of the address (in this case the last 2 bits of the address) and put binary 1s in those bits. So the address of 101011 11 or 175 is the broadcast address.

Thanks for using the rating system to indicate that your question was resolved (and thanks for the rating). It makes the forum more useful when people can read a question and can read responses which successfully answered the question.

I encourage you to continue your participation in the forum.

HTH

Rick

Re: Network Management: Question about calculating network,host

Rick makes an imprtant point here - if in doubt go to the binary. That is fundamental.

After a while, and after doing a bit of work with subnetting, it will become a lot easier, and by an large you will take a look and "just know" what the numbers are.

I'll throw in one of my little things here. Subtract the mask from 256 and you get the increment of the subnets - in this case a /30 the mask is 252. Take that from 256 and you get 4. The subnets will be 0,4,8,12...168,172,176,180 etc.

Paul.

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