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| Mickael Dickinson 2002-07-28, 9:25 pm |
| iv almost go this cracked now.
if you have a 140-node network, then you divide the network into 10 subnets
and each subnet must be able to acommodate up to 14 nodes. How should you
configure the Ip addressing structure to allow for this if the companys IP
address is 194.194.194.0.
is it 194.194.194.0/26?
other options are /25 /27 and /28
______________________________
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______ Mike
ICQ#: 48866977 Current ICQ status: + More ways to contact me i See more
about me: ______________________________
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| |
| _Mike_ 2002-07-29, 4:25 am |
|
"Mickael Dickinson" <michael.dickinson@ntlworld.com> wrote in message
news:IZ119.26479$vN6.1324316@newsfep2-win.server.ntli.net...
> iv almost go this cracked now.
>
> if you have a 140-node network, then you divide the network into 10
subnets
> and each subnet must be able to acommodate up to 14 nodes. How should you
> configure the Ip addressing structure to allow for this if the companys IP
> address is 194.194.194.0.
>
> is it 194.194.194.0/26?
>
> other options are /25 /27 and /28
Michael,
You need to go back to drawing board :-)
I strongly recommend that in your studying process, you first understand
the binary involved and how subnetting works. Once you have mastered
this, I have a simple table that is easy to draw up from memory that makes
subnet calculations easy. I just attached it on Friday for some one else,
but hopefully I will annoy too many people by attaching it again, it is only
17Kb.
See a previous thread on Friday named
Re: Is the calculator applet available...
for details on using this table.
Simply put, in the columns on the left, use the third column to
find the closest match to 14 (you want 14 nodes) you want
the closest number to 14 that is 14 or greater, in this case in
the spreadsheet Row 9, the first colum of this shows the
number 4. As this is a Class C IP address, this means you
will use 4 bits of the last (4th) octet for nodes.
In the rows on top of the spreadsheet, count 4 accross
and this gets to the base2 column of 5 which has a
running total of 240, this means you will need a subnet
mask of 255.255.255.240 or alternativley you will
use the first thee Octects plus 4 bits of the 4th octec
for the host , which is 8 + 8 + 8 + 4 = 28 so the
correct answer in your question is 194.194.194.0/28
Some good links for learning to subnet are (links may wrap)
http://infocenter.cramsession.com/T...542&GetDes=&Cat
ID=309
http://infocenter.cramsession.com/T...526&GetDes=&Cat
ID=309
http://www.howtosubnet.com/
To test your answers to practice questions, use this calculator
http://www.wildpackets.com/products/ipsubnetcalculator
Cheers
Mike
begin 666 Subnet Mask.xls
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end
| |
| Consultant® 2002-07-29, 9:25 am |
| www.learntosubnet.com
"Mickael Dickinson" <michael.dickinson@ntlworld.com> wrote in message
news:IZ119.26479$vN6.1324316@newsfep2-win.server.ntli.net...
> iv almost go this cracked now.
>
> if you have a 140-node network, then you divide the network into 10
subnets
> and each subnet must be able to acommodate up to 14 nodes. How should you
> configure the Ip addressing structure to allow for this if the companys IP
> address is 194.194.194.0.
>
> is it 194.194.194.0/26?
>
> other options are /25 /27 and /28
> ______________________________
______________________________
______ Mike
> ICQ#: 48866977 Current ICQ status: + More ways to contact me i See more
> about me:
______________________________
______________________________
______
>
>
| |
| Doug Allen 2002-07-29, 7:25 pm |
| Hi Michael,
There is a good tutorial at http://digitalrm.com/tips/tcpip.html that
explains binary and subnetting in easy to read/undertand terms.
Hope this helps,
Doug
"Mickael Dickinson" <michael.dickinson@ntlworld.com> wrote in message
news:IZ119.26479$vN6.1324316@newsfep2-win.server.ntli.net...
> iv almost go this cracked now.
>
> if you have a 140-node network, then you divide the network into 10
subnets
> and each subnet must be able to acommodate up to 14 nodes. How should you
> configure the Ip addressing structure to allow for this if the companys IP
> address is 194.194.194.0.
>
> is it 194.194.194.0/26?
>
> other options are /25 /27 and /28
> ______________________________
______________________________
______ Mike
> ICQ#: 48866977 Current ICQ status: + More ways to contact me i See more
> about me:
______________________________
______________________________
______
>
>
|
|
|
|
|