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24 posts

Geek


# 18353 7-Jan-2008 20:18
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        Hello, guys how are you doing?


    i have a calss "C" network and i want to creat 15 subnets with at least 5 hosts per subnet for different offices.
    i don't know how to create subnets. so, could you please guys help me how to do this?


                 thanks

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51 posts

Master Geek


  # 103459 7-Jan-2008 21:52
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Due to the fact that subnetting works by changing the 8 bits that "mask" each octet of the IP address, you have 2 options:

Netmask 255.255.255.240 (/28) will give 16 networks of 14 hosts.
Netmask 255.255.255.248 (/29) will give 32 networks of 6 hosts.

Every time you slice up the network, you lose more available host addresses to the network and broadcast components.

Assuming you go with the first option, and your class C network is 192.168.1.0, the subnets break down like this:

Subnet 1 = 192.168.1.0
Hosts for Subnet 1 = 192.168.1.1 to 192.168.1.14
Broadcast address for Subnet 1 = 192.168.1.15

Subnet 2 = 192.168.1.16
Hosts for Subnet 2 = 192.168.1.17 to 192.168.1.30
Broadcast address for Subnet 2 = 192.168.1.31

---

Subnet 16 = 192.168.1.240
Hosts for Subnet 16 = 192.168.1.241 to 192.168.1.254
Broadcast address for Subnet 16 = 192.168.1.255

I'll leave the intermediate math to you as a learning experience. (You should also review the "Subnetwork" page on wikipedia for a more detailed explanation).

So if you create a host with IP address 192.168.1.5 and netmask 255.255.255.240 it can directly connect to hosts 1-14 but will expect a route to see anything above 15.

To be honest, it actually makes more sense once you start doing the calculations in binary! You can then see the "bits in action" and how the "masking" actually works.

HTH and have fun!
Andy.

268 posts

Ultimate Geek

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  # 103462 7-Jan-2008 21:58
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A /29 will give you 6 usable addresses, one of these would need to be used by the router (assuming you require all the subnets to be able to talk to one an other.

so

192.168.0.1 - 192.168.0.6
192.168.0.9 - 192.168.0.14
192.168.0.17 - 192.168.0.22
192.168.0.25 - 192.168.0.30
192.168.0.33 - 192.168.0.38
192.168.0.41 - 192.168.0.46
192.168.0.49 - 192.168.0.54
192.168.0.57 - 192.168.0.62
192.168.0.65 - 192.168.0.70
192.168.0.73 - 192.168.0.78
192.168.0.81 - 192.168.0.86
192.168.0.89 - 192.168.0.94
192.168.0.97 - 192.168.0.102
192.168.0.105 - 192.168.0.110
192.168.0.113 - 192.168.0.118

They would all need to have a subnet of 255.255.255.248. 

If you require more that 6 usable addresses, the next step is to have 14 - they are basically two of the above blocks...  so

192.168.0.1 - 192.168.0.14
192.168.0.17 - 192.168.0.30
etc

You will notice for each subnet you are loosing two addresses, the subnet ID (the first address) and the broadcast address (the last address).  Technically a /29 is 8 IP Addresses and a /28 is 16 IP Address ie...

192.168.0.0 - 192.168.0.7 is the first address range in the /29 list above, however 192.168.0.0 becomes the Subnet ID and 192.168.0.7 becomes the Broadcast Address, for the /28 192.168.0.0 is also the Subnet ID but the broadcast address becomes 192.168.0.15

With subneting you will need something to route the required traffic between the subnets.

Good luck
David

 
 
 
 




24 posts

Geek


# 103488 8-Jan-2008 03:07
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                        Thank you very much guys, i really Appreciated

                                             God bless you

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