Calling All Networkers!
Discussion
Hi all. This really is a last resort, as I really am struggling here!
Can't seem to get my head around classless IP addresses, masks & blocks and all the rest of it. I have composed myself a network LAN diagram, but need to figure out a suitable classless IP address for the network that I have made. Now really, where do I start!?
Can't seem to get my head around classless IP addresses, masks & blocks and all the rest of it. I have composed myself a network LAN diagram, but need to figure out a suitable classless IP address for the network that I have made. Now really, where do I start!?
This is about the best tutorial I've ever seen on the topic : http://www.3com.com/other/pdfs/infra/corpinfo/en_U...
marshalla said:
This is about the best tutorial I've ever seen on the topic : http://www.3com.com/other/pdfs/infra/corpinfo/en_U...
Cheers, I'll have a look and see what I can make of it! 
Mini_Lund said:
marshalla said:
This is about the best tutorial I've ever seen on the topic : http://www.3com.com/other/pdfs/infra/corpinfo/en_U...
Cheers, I'll have a look and see what I can make of it! marshalla said:
Ok - how many subnetworks do you have, and how many hosts on each ?
I have a LAN of 5 subnets, with 50 computers in each network. The computers are then all linked through a switch per subnet, then file server and all to a main Central switch, linked to a web server & email server.OK - so for 50 hosts you need to allocate enough space for 64 (closest larger power of 2) - which means your host space will be at least the last 6 bits of the address.
5 subnets - closest larger power of 2 is 8 so you'll be using 3 bits for your own subnet address.
So - given that IPv4 addressing is 32 bit, you need your ISP to allocate you 32-3-6 = 32-9 = 23 bits of address space
(i.e. a /23) OR at least 5 /26s (23+3) which you can use for each of your subnets.
Do you need your 5 subnets to look like a single network to the outside world ? If so - you need the /23 really. If not - a batch of /26s will do.
All this /23 and /26 stuff, btw - refers to the number of bits, counting from the left hand side, which make up the network address. The rest of the bits are the host number. Understanding it in binary really DOES make it easier because you can start to see the split between network and host.
e.g. 10.1.19.12 = 00001010.00000001.00010011.00001100 (binary)
Taken as a /23 - the network address is 00001010.00000001.0001001 0.00000000 (note how the first 23 bits match the ones in the full address ? ) = 10.1.18.0
and the host address within it is 00000000.00000000.0000000 1.00001100 = 1.12 = 278 (of 512 available as the host portion is 9 bits covering 0 to 511)
Thing is - once you've been allocated a network you can subdivide it by adding bits to the network section and reducing the number of hosts per subnet as a consequence - so if you were to divide the /23 into a /26, the boundary between network and host address just moves 3 places to the right. To do this properly you really need to be able to configure your router(s) to handle the supernet(s) properly (supernet is the shortest network address which can be used to identify just your networks as a group - if you are allocated a /23, the the /23 is the supernet for everything you create from it).
That's pretty much it apart from avoiding using addresses which are all 1s and 0s in the host portion because they are used as broadcast & network addresses (which means you always need to allow for at least 2 more addresses than you have hosts in each subnet).
(Someone please check the binary - I'm doing it mentally here).
If we were to subnet the /23 above into /26s we'd get the following networks :
10.1.18.0
10.1.18.64
10.1.18.128
10.1.18.192
10.1.19.0
10.1.19.64
10.1.19.128
10.1.19.192
Each would have up to 64 hosts (numbered 0 to 63), but two should not be used.
And I'll let you explain why by looking at the bit patterns.
[Edited to correct typos & incorrect numbering in the new subnets]
5 subnets - closest larger power of 2 is 8 so you'll be using 3 bits for your own subnet address.
So - given that IPv4 addressing is 32 bit, you need your ISP to allocate you 32-3-6 = 32-9 = 23 bits of address space
(i.e. a /23) OR at least 5 /26s (23+3) which you can use for each of your subnets.
Do you need your 5 subnets to look like a single network to the outside world ? If so - you need the /23 really. If not - a batch of /26s will do.
All this /23 and /26 stuff, btw - refers to the number of bits, counting from the left hand side, which make up the network address. The rest of the bits are the host number. Understanding it in binary really DOES make it easier because you can start to see the split between network and host.
e.g. 10.1.19.12 = 00001010.00000001.00010011.00001100 (binary)
Taken as a /23 - the network address is 00001010.00000001.0001001 0.00000000 (note how the first 23 bits match the ones in the full address ? ) = 10.1.18.0
and the host address within it is 00000000.00000000.0000000 1.00001100 = 1.12 = 278 (of 512 available as the host portion is 9 bits covering 0 to 511)
Thing is - once you've been allocated a network you can subdivide it by adding bits to the network section and reducing the number of hosts per subnet as a consequence - so if you were to divide the /23 into a /26, the boundary between network and host address just moves 3 places to the right. To do this properly you really need to be able to configure your router(s) to handle the supernet(s) properly (supernet is the shortest network address which can be used to identify just your networks as a group - if you are allocated a /23, the the /23 is the supernet for everything you create from it).
That's pretty much it apart from avoiding using addresses which are all 1s and 0s in the host portion because they are used as broadcast & network addresses (which means you always need to allow for at least 2 more addresses than you have hosts in each subnet).
(Someone please check the binary - I'm doing it mentally here).
If we were to subnet the /23 above into /26s we'd get the following networks :
10.1.18.0
10.1.18.64
10.1.18.128
10.1.18.192
10.1.19.0
10.1.19.64
10.1.19.128
10.1.19.192
Each would have up to 64 hosts (numbered 0 to 63), but two should not be used.
And I'll let you explain why by looking at the bit patterns.
[Edited to correct typos & incorrect numbering in the new subnets]
Edited by marshalla on Wednesday 21st April 20:05
marshalla said:
BliarOut said:
Why bother, just use the private class C addresses, one per subnet. No point making life more complicated than necessary.
Depends whether you need external access or not. NATing that many machines is not fun either.
There's lots of ifs buts and maybes, depends if we're talking real world or theory. I'd only start playing with masks if it was absolutely necessary.BliarOut said:
marshalla said:
BliarOut said:
Why bother, just use the private class C addresses, one per subnet. No point making life more complicated than necessary.
Depends whether you need external access or not. NATing that many machines is not fun either. There's lots of ifs buts and maybes, depends if we're talking real world or theory. I'd only start playing with masks if it was absolutely necessary.
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