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subnetting

 
 
Sorahl
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      11-25-2003
Hey all,

The subject of TCP/IP subnetting befuddles me a bit.

I have given myself the example scenario of a small company with a C-class
network ID of 212.78.160.0.
They need 10 subnets.

If I correctly understand this, encoding 11 (10+1) into binary needs 4 bits.
Therefore, I use 4 bits more for the network ID and get a subnetmaks of
255.255.255.240.

The lowest bit in 11110000 is worth 16 so I get a multiplier of 16.

Subnets are then defined as follows:
212.78.160.16
212.78.160.32... right on till
212.78.160.224 as the 14th subnet.

Each subnet allows for 14 hosts.

Sounds okay up till now?

My question is what the ranges of valid IP host addresses look like.

Greets, Sorahl




 
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-= Hawk =-
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Posts: n/a
 
      11-25-2003
On Tue, 25 Nov 2003 11:22:00 +0100, "Sorahl" <(E-Mail Removed)>
scribbled:

>Hey all,
>
>The subject of TCP/IP subnetting befuddles me a bit.
>
>I have given myself the example scenario of a small company with a C-class
>network ID of 212.78.160.0.
>They need 10 subnets.
>
>If I correctly understand this, encoding 11 (10+1) into binary needs 4 bits.
>Therefore, I use 4 bits more for the network ID and get a subnetmaks of
>255.255.255.240.
>
>The lowest bit in 11110000 is worth 16 so I get a multiplier of 16.
>
>Subnets are then defined as follows:
>212.78.160.16
>212.78.160.32... right on till
>212.78.160.224 as the 14th subnet.
>
>Each subnet allows for 14 hosts.
>
>Sounds okay up till now?
>
>My question is what the ranges of valid IP host addresses look like.


# ID Range Broadcast
1 212.78.160.16 212.78.160.17 - 212.78.160.30 212.78.160.31
2 212.78.160.32 212.78.160.33 - 212.78.160.46 212.78.160.47
3 212.78.160.48 212.78.160.49 - 212.78.160.62 212.78.160.63
4 212.78.160.64 212.78.160.65 - 212.78.160.78 212.78.160.79
5 212.78.160.80 212.78.160.81 - 212.78.160.94 212.78.160.95
6 212.78.160.96 212.78.160.97 - 212.78.160.110 212.78.160.111
7 212.78.160.112 212.78.160.113 - 212.78.160.126 212.78.160.127
8 212.78.160.128 212.78.160.129 - 212.78.160.142 212.78.160.143
9 212.78.160.144 212.78.160.145 - 212.78.160.158 212.78.160.159
10 212.78.160.160 212.78.160.161 - 212.78.160.174 212.78.160.175
11 212.78.160.176 212.78.160.177 - 212.78.160.190 212.78.160.191
12 212.78.160.192 212.78.160.193 - 212.78.160.206 212.78.160.207
13 212.78.160.208 212.78.160.209 - 212.78.160.222 212.78.160.223
14 212.78.160.224 212.78.160.225 - 212.78.160.238 212.78.160.239

--
'What Profiteth It A Kingdom If The Oxen Be Deflated?'
Riddles II, v3
- T. Pratchett
 
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why?
Guest
Posts: n/a
 
      11-25-2003

On Tue, 25 Nov 2003 11:22:00 +0100, Sorahl wrote:

>Hey all,
>
>The subject of TCP/IP subnetting befuddles me a bit.


That depends on how difficult you make it.

www.google.com look for
subnet tutorial
subnet calculator

http://www.subnetonline.com/
http://www.certguide.com/tcpip.asp
http://www.firewall.cx/ip-subnetting-mask-effect.php
http://www.microsoft.com/windows2000...b_tcp_pnmz.asp

Searched English pages for
subnet calculator.

Results 1 - 10 of about 22,400. Search took 0.36 seconds.


>I have given myself the example scenario of a small company with a C-class
>network ID of 212.78.160.0.


Try and use examples that don't belong to someone
inetnum: 212.78.160.0 - 212.78.160.255
netname: NL-COLT
descr: COLT Internet NL
country: NL


>They need 10 subnets.


It's easier to use, I think the most obvious division of 16 subnets.
Mind you it depends on how many hosts per subnet there are.
Too few meand everything needs routed and too many means larger
collission / broadcast domains.


Use http://www.faqs.org/rfcs/rfc1918.html
RFC 1918 - Address Allocation for Private Internets

If it's that small they don't need subnets, since a default Class C is
254 hosts maybe move to default B 64k hosts.

If it's your company you want to connect to the Internet, get a NAT /
Router, use the priviate address range mapping to the ISP IP.

<snip>

>Sounds okay up till now?


Yes.

>My question is what the ranges of valid IP host addresses look like.


You are fairly close already,

>Subnets are then defined as follows:
>212.78.160.16
>212.78.160.32... right on till
>212.78.160.224 as the 14th subnet.



Hint(s) -
You can't use use the 1st IP, it's the subnet address, as you said just
above.

The next IP is the 1st host.

The last but 1 IP is the last host.

The last IP for each subnet is the broadcast address.

try figuring it out or use one of the calculators, like
http://www.subnetonline.com/subcalc/subnet2.html



Me
 
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Peter H
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Posts: n/a
 
      11-25-2003

"Sorahl" <(E-Mail Removed)> wrote in message
news:3fc32d2c$0$1497$(E-Mail Removed)4all.nl...
> Hey all,
>
> The subject of TCP/IP subnetting befuddles me a bit.
>
> I have given myself the example scenario of a small company with a C-class
> network ID of 212.78.160.0.
> They need 10 subnets.
>
> If I correctly understand this, encoding 11 (10+1) into binary needs 4

bits.
> Therefore, I use 4 bits more for the network ID and get a subnetmaks of
> 255.255.255.240.
>
> The lowest bit in 11110000 is worth 16 so I get a multiplier of 16.
>
> Subnets are then defined as follows:
> 212.78.160.16
> 212.78.160.32... right on till
> 212.78.160.224 as the 14th subnet.
>
> Each subnet allows for 14 hosts.
>
> Sounds okay up till now?
>
> My question is what the ranges of valid IP host addresses look like.
>
> Greets, Sorahl
>
>
>


I remember going through this myself. I've forgotten it all now but remember
where I found the answers I needed. Try

alt.certification.network-plus

Peter H


 
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