Ralph Shumaker wrote:
> Gregory K. Ruiz-Ade wrote:
>
>> On Feb 22, 2007, at 4:13 AM, Ralph Shumaker wrote:
>>
>>> Then why would one use /27:
>>>
>>> (same host)
>>>
>>> 12.169.40.36
>>>
>>> 00001100 10101001 00101000 00100100 - 12.169.40.36 - host
>>> 11111111 11111111 11111111 11100000 - /27 netmask (if I got it)
>>> ----------------------------------- - bitwise AND the two together
>>> 00001100 10101001 00101000 00100000 - 12.169.40.32 ???
>>>
>>> Since the last 5 bits will always AND to zeros, does this mean that
>>> 12.169.40.0 through 12.169.40.31 all belong to 12.169.40.36 /27?
>>
>>
>> The whole point of CIDR (Classless Internet Domain Routing, i think)
>> is to be able to divvy out portions of network space smaller than a
>> Class C.
>>
>> Running your example through a nifty tool called 'ipcalc' (which has
>> seen multiple variations, but this one is a perl script I got off
>> sourceforge a couple years ago):
>>
>> [EMAIL PROTECTED](ttyp1):~ 23 % ipcalc 12.169.40.36 27
>>
>> Address: 12.169.40.36
>> Address: 00001100.10101001.00101000.001 00100
>> Netmask: 255.255.255.224 == 27
>> Netmask: 11111111.11111111.11111111.111 00000
>> =>
>> Network: 12.169.40.32/27 (Class A)
>> Network: 00001100.10101001.00101000.001 00000
>> Broadcast: 12.169.40.63
>> Broadcast: 00001100.10101001.00101000.001 11111
>> HostMin: 12.169.40.33
>> HostMin: 00001100.10101001.00101000.001 00001
>> HostMax: 12.169.40.62
>> HostMax: 00001100.10101001.00101000.001 11110
>> Hosts: 30
>>
>> So, your math was a bit off. But, a /27 is a 32-address span. In
>> any network, though, you lose three addresses right off the top: The
>> lowest address (in this case, .32) is the network address. The
>> highest (.63) is the broadcast address.
>>
>> The third address is whichever one you assign to be the gateway/
>> router for the subnet. Convention strongly suggests either the
>> lowest or highest useable host address, either .33 or .62 in this case.
>>
>> The smallest subnet you can have is a /30, with 4 addresses: network,
>> two hosts, and broadcast. Given that one of your addresses has to be
>> a router, you end up with only a single usable host IP address. So,
>> for one host, you end up carving out 4 IP addresses. This is why
>> most Cable Internet providers simply give you DHCP addresses for your
>> hosts on a very large subnet. For example, my router at home (Cox
>> Internet):
>>
>> [EMAIL PROTECTED](ttyp2):~ 23 % ipcalc 68.111.245.13 22
>>
>> Address: 68.111.245.13
>> Address: 01000100.01101111.111101 01.00001101
>> Netmask: 255.255.252.0 == 22
>> Netmask: 11111111.11111111.111111 00.00000000
>> =>
>> Network: 68.111.244.0/22 (Class A)
>> Network: 01000100.01101111.111101 00.00000000
>> Broadcast: 68.111.247.255
>> Broadcast: 01000100.01101111.111101 11.11111111
>> HostMin: 68.111.244.1
>> HostMin: 01000100.01101111.111101 00.00000001
>> HostMax: 68.111.247.254
>> HostMax: 01000100.01101111.111101 11.11111110
>> Hosts: 1022
>>
>> Cox uses a /22 for our network node, and, to my recollection, has a 1
>> IP per house policy, with additional IPs being an additional cost
>> item on the bill. To preserve as many IP addresses as possible, they
>> lump the whole neighborhood into that /22 instead of giving each home
>> a /30 for a colossal savings in IP addresses.
>>
>> How networks are subnetted is primarily an implementation issue for
>> the network provider.
>>
>> Now, did I actually succeed in answering your question, or just throw
>> a lot of information at you? :)
>
>
> Maybe (either way), but had you run "ipcalc 12.169.40.39 27" or "ipcalc
> 12.169.40.54 27", the result would still be a range from 32 through 63?
> If that's the case, then my math wasn't off, just my understanding as to
> what my resulting 12.169.40.32 meant.
>
> If this is so (and if I'm understanding correctly), then the resulting
> range would start at that address (inclusive) and slurp up 2^(32-27) ==
> 2^5 == 32 addresses. So 12.169.40.36 just indicates one of the
> addresses within the range, 27 indicates what kind of range (bitwise
> subtracted from 2^32) and is bitwise ANDed to the given address to find
> the starting address. The first address is always the network address
> (host unusable), the last address is always the broadcast (host
> unusable), and in the resulting pool either the first or the last
> address *should* be used for the gateway (host unusable). Right?
>
> If Network address, Broadcast address, and Gateway address are
> unassignable to hosts because of dedication to their respective tasks,
> could you summarize those tasks? Network? Broadcast? Gateway? Thanks
> in advance.
>
> (I think I'm understanding this much more than before.)
>
>
Try it yourself.
ipcalc 12.169.40.39/27 -mpbn
ipcalc 12.169.40.54/27 -mpbn
Running
man ipcalc
or
ipcalc --help
will explain what -m, -p, -b and -n mean.
ipcalc is part of the initscripts package in RH/FC systems.
You may need to install it on a debian{-like] system.
Regards,
..jim
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