Peter Humphrey wrote:
On Monday 20 December 2010 09:37:48 Dale wrote:
I set it up like this. The modem uses DHCP to get the IP from AT&T.
My local IP from the modem is 192.168.1.2. Then the router has the
IP 192.168.2.1 for my connection to the puter. The IP of my puter
is 192.168.2.5. The next puter will be 192.168.2.6 or something
different anyway.
The one thing you didn't mention there is the outer address of your
router. It needs to be 192.168.1.x where x is anything other than 2. It
needs to be on the same network segment as the inner side of your modem.
Yep, front end is 192.168.1.4 or 5 I think. Something close to that.
The last number may be different. It may be .2 or something.
I need to read up on the netmask thing some more. It's still murky
for sure.
(What follows has grown rather long. I hope it doesn't come over too
much as a lecture.)
It's fairly straightforward once you get the hang of it. The address of
a device is a 64-bit number, expressed as four 16-bit numbers joined
with dots. It's just easier to read when split into chunks, but it is
really a 64-bit number. As in decimal arithmetic, the right-hand digit
is the least significant.
An interface address consists of two parts: the leftmost part defines a
group of addresses (the network part) and the rightmost part specifies
the number of the interface in that group (the host part). The function
of the network mask is to specify where the boundary is between the
network part and the host part.
Two conventions are used for expressing where that boundary is: the
older method is to write, say, 255.255.255.0, which indicates that the
first 24 bits (three eight-bit numbers - 255 is all-ones in eight bits)
belong to the network and anything to the right of those can be
allocated to interfaces in that network. That convention dates from the
era of plenty of IP addresses in the world and goes along with Class A,
B, C or D. A class A network has a mask of 255.0.0.0, class B has
255.255.0.0, class C has 255.255.255.0 and a class D (never used in the
wild as far as I know) would have 255.255.255.255.
Since the meteoric growth of the Internet this class scheme has become a
handicap, and a finer division of network scope has become necessary, to
allow use of, say, 255.255.255.248 as a net mask. Rather than specifying
a plethora of new classes (we'd need anything up to 60), a shorthand
notation has been invented in which we just append a number to an
address to specify the number of bits that identify the network, with
the rest identifying the host on it (strictly speaking, a host's
interface on the network, as a host may have more than one interface -
sometimes even on the same network). This scheme is known as CIDR
notation. Thus your modem's inner address is, I assume, 192.168.1.2/24,
which is the same as writing 192.168.1.2 with a mask of 255.255.255.0.
That mask 255.255.255.248 I mentioned specifies 29 bits for the network
address and three for the hosts on it; that's enough for six computers
once the ..0 and ..7 addresses are reserved for network address and
broadcast address. A lot of ISPs use such a scheme for allocating
address ranges to their customers.
How's it look? Think it will work for a while?
Once you've set your router's outer address correctly, yes.
Sorry I was asleep overnight and had to leave you to the tender mercies
of your compatriots. :-)
Again, apologies if I've seemed to want to teach my grandmother to suck
eggs.
Taking a nap is fine. I do that sometimes myself. If worse came to
worse, I could set my puter to DHCP and hooked straight to the modem.
That always works.
I seem to have the stuff set up correctly now. I may try to hook up the
second rig at least for testing anyway. I should have set the IP and
set it to start ssh before the last shutdown. I didn't tho. Oh well.
I'm starting to grasp the netmask thingy. It just has to soak in a
little bit. lol
Thanks for the help.
Dale
:-) :-)
