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Re: Access to sub network unreachable.



On Fri, 2009-01-09 at 16:56 -0800, Rick Stevens wrote:

> No, the /24 covers EITHER 192.168.0.xxx or 192.168.1.xxx, but not both.
> To treat both as a single network you need a /23 netmask.
> 
> A netmask tells the system how many bits of the network address make up
> the NETWORK part.  

This is where my misunderstanding was.

> The remaining bits make up the host identifier.  An 
> IPV4 address is 32 bits.  A /24 (or 255.255.255.0) netmask says the 
> first 24 bits (the first three octets) make up the network part.  

What then, is a sub-net?

> In my
> graphic below, the netmask stuff is shown in by "x"s and the bits under
> the "x"s make up the network address:
> 
> netmask:	xxxxxxxx.xxxxxxxx.xxxxxxxx.-------- (255.255.255.0 /24)
> binary addr:	11000000.10101000.00000000.00000001 (192.168.0.1)
> binary addr:	11000000.10101000.00000001.00000001 (192.168.1.1)
> 
> So you can see that they're separate networks.  Now, with a /23 netmask:
> 
> netmask:	xxxxxxxx.xxxxxxxx.xxxxxxx-.-------- (255.255.254.0 /23)
> binary addr:	11000000.10101000.00000000.00000001 (192.168.0.1)
> binary addr:	11000000.10101000.00000001.00000001 (192.168.1.1)
> 
> You can see here that they're the same network now, and you're using 9
> bits as the host identifier.

Okay, for a small private network of up to 2 dozen boxes, is there a
standard or convention for selecting the final numbering system,  eg
192.168.0.101 to 125 vs 192.168.9.1 to 25 ?  Is there a need to
distinguish between computer,printer or router in the numbering
heirarchy?
> 
> Also note that the netmask must be consecutive 1 bits...the first 0 bit
> marks the end of the netmask. 255.255.254.0 is OK (the last octet is
> 11111110 binary), but 255.255.253.0 isn't (the last octet is 11111101)
> and would be treated the same as 255.255.255.252 (a /22 or 11111100).
> 
> Does it make sense now?

Getting there!

-- 
Hooroo,
Simon
Registered Linux User #463789. Be counted at: http://counter.li.org/



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