Linux - NetworkingThis forum is for any issue related to networks or networking.
Routing, network cards, OSI, etc. Anything is fair game.
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I am kinda new to the whole subnetting thing. I am learning and as such came across this comment:
The binary arithmetic involved in using CIDR block addressing is painful. It is much easier to start with another private network address space, say 172.16.0.0, and use the third byte to represent internal subnets: 172.16.1.0, 172.16.2.0, 172.16.3.0, and so on. Now you use the default ( or "classful") subnet mask of 255.255.255.0 (or /24) and each subnet includes 254 host addresses. See how much easier it is to start with a Class B (/16) network, and split it into 254 Class Cs (/24)? Perhaps the designers of TCP/IP weren't so dumb after all when they chose to use byte boundaries in the initial network addressing scheme
I am not sure I see how it is easier using the numbers with the example above.
It isn't really any easier, just depends on which system you find easier. If you work with IP a lot its best to get used to translating between both systems as different kit uses different notation in different circumstances.
It isn't really any easier, just depends on which system you find easier. If you work with IP a lot its best to get used to translating between both systems as different kit uses different notation in different circumstances.
Thanks I went to this site to start on my subnetting education; did the binary stuff etc etc but got stuck on the above paragraph as to why it was easier for the OP. Link is here http://www.watchguard.com/infocenter...ial/135191.asp The pargraph is located right after the 192.x.x.x table on the site.
If you for example start with the private network 172.16.0.0, and use the third octet for subnetting: E.G.
172.16.10.0/255.255.255.0, 172.16.11.0/255.255.255.0, 172.16.12.0/255.255.255.0; you don't need to do any math.
Now compare if you use the network 172.16.144.0/36 and you subnet it into 12 subnets. The first 4 bits of the 3rd octet are added to the network address part.
You need to add the network part of the third octet (144) to the host part of the third octet to obtain the final IP address for a host. The third is a combination of the subnet address and the upper bits of the host address. No math involved to express it in dotted decimal notation. Just look at the 3rd octet to see which subnet it is.
If you use byte boundaries, then you only need to change the third octet to designate a different subnet. Otherwise, to look at an IP address, you need to use the netmask to see which subnet it belongs to. Not difficult, but you don't see it at a glance.
If you don't need more than 254 hosts per subnet, using the entire third octet for subnetting makes it extremely easy to enter routes, etc. without having to do any math.
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