Subnet Questions from the IP Sub-Networking Mini-Howto

Subnet Questions from the IP Sub-Networking Mini-Howto

Post by Bill McEachr » Thu, 30 Aug 2001 22:57:12



Can anyone explain to me how the third netmask (126) is arrived at in
the example below cut and pasted below from the
mini-HOWTO "IP Sub-Networking". For the life of me I don't understand it.

Also, I believe the third network address should be "192.168.1.128" and
that the document has a typo.


-----------------------------------------------------------------------

  For the sake of this example, let us assume that you have decided to
  subnetwork you C class IP network number 192.168.1.0 into 4 subnets
  (each of 62 usable interface/host IP numbers). However, two of these
  subnets are being combined into a larger single network, giving three
  physical networks.
  These are :-  

  ______________________________________________________________________
  Network         Broadcast       Netmask                 Hosts
  192.168.1.0     192.168.1.63    255.255.255.192         62
  192.168.1.64    192.168.1.127   255.255.255.192         62
  182.168.1.128   192.168.1.255   255.255.255.126         124 (see note)
  ______________________________________________________________________

  Note: the reason the last network has only 124 usable network
  addresses (not 126 as would be expected from the network mask) is that
  it is really a 'super net' of two subnetworks. Hosts on the other two    
  networks will interpret 192.168.1.192 as the network address of the
  'non-existent' subnetwork. Similarly, they will interpret
  192.168.1.191 as the broadcast address of the 'non-existent'
  subnetwork.

  So, if you use 192.168.1.191 or 192 as host addresses on the third
  network, then machines on the two smaller networks will not be able to
  communicate with them.    
------------------------------------------------------------------------

 
 
 

Subnet Questions from the IP Sub-Networking Mini-Howto

Post by James Knot » Fri, 31 Aug 2001 04:52:55


Are you sure those numbers are correct?  that 126 doesn't sound like
part of a proper netmask, as it would represent 01111110 in binary.  
It would more likely be 128 or 10000000.  I don't know why the hosts
on the other 2 networks would get confused about 192.168.1.192, as
it's not on their network and they should let the router handle it.  
As far a host is concerned, there's it's own local network and the
rest of the world.


> Can anyone explain to me how the third netmask (126) is arrived at
> in the example below cut and pasted below from the
> mini-HOWTO "IP Sub-Networking". For the life of me I don't
> understand it.

> Also, I believe the third network address should be "192.168.1.128"
> and that the document has a typo.


> thanks

-----------------------------------------------------------------------
Quote:

>   For the sake of this example, let us assume that you have decided
>   to subnetwork you C class IP network number 192.168.1.0 into 4
>   subnets (each of 62 usable interface/host IP numbers). However,
>   two of these subnets are being combined into a larger single
>   network, giving three physical networks.
>   These are :-

______________________________________________________________________
Quote:>   Network         Broadcast       Netmask                 Hosts
>   192.168.1.0     192.168.1.63    255.255.255.192         62
>   192.168.1.64    192.168.1.127   255.255.255.192         62
>   182.168.1.128   192.168.1.255   255.255.255.126         124 (see
>   note)

______________________________________________________________________
Quote:

>   Note: the reason the last network has only 124 usable network
>   addresses (not 126 as would be expected from the network mask) is
>   that it is really a 'super net' of two subnetworks. Hosts on the
>   other two networks will interpret 192.168.1.192 as the network
>   address of the 'non-existent' subnetwork. Similarly, they will
>   interpret 192.168.1.191 as the broadcast address of the
>   'non-existent' subnetwork.

>   So, if you use 192.168.1.191 or 192 as host addresses on the third
>   network, then machines on the two smaller networks will not be
>   able to communicate with them.

------------------------------------------------------------------------

--

james.knott.

 
 
 

Subnet Questions from the IP Sub-Networking Mini-Howto

Post by Bill McEachra » Fri, 31 Aug 2001 11:23:25


Yep, it's a cut a cut and paste job from the mini-HOWTO.
It contradicts everything else I've read....making this
a _*_ to learn (actually I thought it was simple
until I came across this).

I think the answer is, as you say, 128.
I'm going to send a copy of this to the linuxdoc.org people and
hopefully they'll pull the doc or make some corrections.

Thanks for the comfirmation ... I thought I'd really
misunderstood this concept.

> Are you sure those numbers are correct?  that 126 doesn't sound like
> part of a proper netmask, as it would represent 01111110 in binary.
> It would more likely be 128 or 10000000.  I don't know why the hosts
> on the other 2 networks would get confused about 192.168.1.192, as
> it's not on their network and they should let the router handle it.
> As far a host is concerned, there's it's own local network and the
> rest of the world.


> > Can anyone explain to me how the third netmask (126) is arrived at
> > in the example below cut and pasted below from the
> > mini-HOWTO "IP Sub-Networking". For the life of me I don't
> > understand it.

> > Also, I believe the third network address should be "192.168.1.128"
> > and that the document has a typo.


> > thanks

> -----------------------------------------------------------------------

> >   For the sake of this example, let us assume that you have decided
> >   to subnetwork you C class IP network number 192.168.1.0 into 4
> >   subnets (each of 62 usable interface/host IP numbers). However,
> >   two of these subnets are being combined into a larger single
> >   network, giving three physical networks.
> >   These are :-

> ______________________________________________________________________
> >   Network         Broadcast       Netmask                 Hosts
> >   192.168.1.0     192.168.1.63    255.255.255.192         62
> >   192.168.1.64    192.168.1.127   255.255.255.192         62
> >   182.168.1.128   192.168.1.255   255.255.255.126         124 (see
> >   note)

> ______________________________________________________________________

> >   Note: the reason the last network has only 124 usable network
> >   addresses (not 126 as would be expected from the network mask) is
> >   that it is really a 'super net' of two subnetworks. Hosts on the
> >   other two networks will interpret 192.168.1.192 as the network
> >   address of the 'non-existent' subnetwork. Similarly, they will
> >   interpret 192.168.1.191 as the broadcast address of the
> >   'non-existent' subnetwork.

> >   So, if you use 192.168.1.191 or 192 as host addresses on the third
> >   network, then machines on the two smaller networks will not be
> >   able to communicate with them.

> ------------------------------------------------------------------------

> --

> james.knott.

 
 
 

Subnet Questions from the IP Sub-Networking Mini-Howto

Post by James Knot » Fri, 31 Aug 2001 20:56:48


One thing I've learned over many years of reading technical material,
is that you can't always believe everything you read.  Just the other
day I was browsing through a new book on mobile communications and
found several errors about basic radio communications.  It makes you
wonder about how good the rest of the book is.  I've occasionally
come across an article, where the author was clearly wrong.



> Yep, it's a cut a cut and paste job from the mini-HOWTO.
> It contradicts everything else I've read....making this
> a _*_ to learn (actually I thought it was simple
> until I came across this).

> I think the answer is, as you say, 128.
> I'm going to send a copy of this to the linuxdoc.org people and
> hopefully they'll pull the doc or make some corrections.

> Thanks for the comfirmation ... I thought I'd really
> misunderstood this concept.

>> Are you sure those numbers are correct?  that 126 doesn't sound
>> like part of a proper netmask, as it would represent 01111110 in
>> binary.
>> It would more likely be 128 or 10000000.  I don't know why the
>> hosts on the other 2 networks would get confused about
>> 192.168.1.192, as it's not on their network and they should let the
>> router handle it. As far a host is concerned, there's it's own
>> local network and the rest of the world.


>> > Can anyone explain to me how the third netmask (126) is arrived
>> > at in the example below cut and pasted below from the
>> > mini-HOWTO "IP Sub-Networking". For the life of me I don't
>> > understand it.

>> > Also, I believe the third network address should be
>> > "192.168.1.128" and that the document has a typo.


>> > thanks

-----------------------------------------------------------------------
Quote:

>> >   For the sake of this example, let us assume that you have
>> >   decided to subnetwork you C class IP network number 192.168.1.0
>> >   into 4 subnets (each of 62 usable interface/host IP numbers).
>> >   However, two of these subnets are being combined into a larger
>> >   single network, giving three physical networks.
>> >   These are :-

______________________________________________________________________
Quote:>> >   Network         Broadcast       Netmask                 Hosts
>> >   192.168.1.0     192.168.1.63    255.255.255.192         62
>> >   192.168.1.64    192.168.1.127   255.255.255.192         62
>> >   182.168.1.128   192.168.1.255   255.255.255.126         124
>> >   (see note)

______________________________________________________________________
Quote:

>> >   Note: the reason the last network has only 124 usable network
>> >   addresses (not 126 as would be expected from the network mask)
>> >   is that it is really a 'super net' of two subnetworks. Hosts on
>> >   the other two networks will interpret 192.168.1.192 as the
>> >   network address of the 'non-existent' subnetwork. Similarly,
>> >   they will interpret 192.168.1.191 as the broadcast address of
>> >   the 'non-existent' subnetwork.

>> >   So, if you use 192.168.1.191 or 192 as host addresses on the
>> >   third network, then machines on the two smaller networks will
>> >   not be able to communicate with them.

------------------------------------------------------------------------

>> --

>> with james.knott.

--

james.knott.
 
 
 

Subnet Questions from the IP Sub-Networking Mini-Howto

Post by Bill McEachra » Mon, 03 Sep 2001 01:44:47



> One thing I've learned over many years of reading technical material,
> is that you can't always believe everything you read.  Just the other
> day I was browsing through a new book on mobile communications and
> found several errors about basic radio communications.  It makes you
> wonder about how good the rest of the book is.  I've occasionally
> come across an article, where the author was clearly wrong.

or sloppy ... as I think was the case here. I did contact
the Linux Documention Project about the errors and they
promptly responded indicating that they've made the needed corrections.

For the masses of documentation they deal with I think they do a fairly
good job ... they just need us to tell
them when we find an error. I think that's fair enough.
Their on line books (such as "Securing and Optimizing RH Linux") are
excellent.
--
Bill McEachran

 
 
 

1. IP Sub-Networking Mini-HOWTO & /proc/ksyms on 2.2.14-5.0 kernel?

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Frank

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