From: Rick Burts (burts@xxxxxxxxxxxxxx)
Date: Wed Apr 25 2001 - 03:46:45 GMT-3
This has been a very interesting thread.
My original response was based on my understanding of inverse masks and
my memory of what I thought worked. The message indicating it was an
error made me re-evaluate and re-test. When I test with routers running
12.1 code I also get this error but when I test with routers running
12.0(9) the command is accepted without error.
I think there is both an immediate lesson and a bigger picture message in
this. The immediate lesson is that the code supporting OSPF masks has
changed so sometimes it works with discontiguous masks and sometimes not.
The bigger picture message may be that in preparing for the lab to be
very sensitive to release dependent behaviours.
Rick
Rick Burts, CCSI CCIE 4615 burts@mentortech.com
Mentor Technologies 240-568-6500 ext 6652
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On Fri, 20 Apr 2001, Shane Miles wrote:
> When I try 0.0.6.255 I get the follwing:
>
> Router(config-router)#network 192.168.2.0 0.0.6.255 area 0
> OSPF: Invalid address/mask combination (discontiguous mask)
> Router(config-router)#
>
> What am I missing?
>
> -----Original Message-----
> From: Rick Burts [mailto:burts@mentortech.com]
> Sent: Thursday, April 19, 2001 2:03 PM
> To: louie kouncar
> Cc: 'David Bader'; 'CCIE Lab'; 'Chuck Larrieu'
> Subject: RE: wildcard mask problem
>
>
> This is not correct.
> There is a requirement that the ones be contiguous in a subnet mask but
> not a requirement in an inverse mask such as this. The 0.0.6.255 is a
> somewhat more sophisticated and more efficient solution, and it would
> work. The original question asked how to combine network statements
> for network 2 and network 4 in an OSPF network statement. Any solution
> that combines into fewer statements will include some excess networks.
> The .6 mask will include 2 excell networks (0 and 6) while the 7 mask
> will include 6 excess networks (0,1,3,5,6,7). Both solutions are valid.
> One is slightly optimized.
>
> Rick
>
>
> On Thu, 19 Apr 2001, louie kouncar wrote:
>
> > Dave,
> >
> > You are right, This should be 192.168.0.0 0.0.7.255, 6 in the third octet
> > will break the subnetting rules and will not work, the reason is if you do
> > the 255-248 you will get 7 and not 6. I am not aware of a 255.255.249.0
> > mask, which will be the case if this was 0.0.6.0
> >
> > I hope I am right.....
> >
> >
> > Thanks
> >
> >
> > Louie J. Kouncar (CCIE)/Written
> > TCO3 Senior Data Center Engineer
> > UUNET
> > W-703-343-6645
> > C-703-304-2460
> >
> >
> >
> > -----Original Message-----
> > From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
> > David Bader
> > Sent: Thursday, April 19, 2001 12:08 PM
> > To: CCIE Lab; Chuck Larrieu
> > Subject: AW: wildcard mask problem
> >
> >
> > This wouldn't work, because it it not a contignuous wildcard:
> > 00000000.00000000.00000110.11111111
> >
> > dave
> >
> > -----Urspr|ngliche Nachricht-----
> > Von: Chuck Larrieu [mailto:chuck@cl.cncdsl.com]
> > Gesendet: Donnerstag, 19. April 2001 08:40
> > An: David Bader
> > Betreff: RE: wildcard mask problem
> >
> >
> > Try 192.168.0.0 0.0.6.255
> >
> > Tell me what that covers.
> >
> > to quote a sage - think different.
> >
> > Chuck
> >
> > -----Original Message-----
> > From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf
Of
> > David Bader
> > Sent: Wednesday, April 18, 2001 11:27 PM
> > To: "louie kouncar" <lkouncar@UU.NET>
> > Subject: RE: wildcard mask problem
> >
> > This is correct, but you can only summarise 192.168.0.0 - 192.168.7.255.
> so
> > you either summarise all of them, or you need two statements.
> >
> > Dave
> >
> >
> > > Bruce,
> > >
> > > you are totally correct, my mistake, I was more concerned with the third
> > > octet of the wildcard mask since that was where the question was
> directed
> > > at.
> > >
> > >
> > > it should be 192.168.0.0 0.0.7.255 area 0
> > >
> > >
> > >
> > > Thanks
> > >
> > >
> > > Louie J. Kouncar (CCIE)/Written
> > > TCO3 Senior Data Center Engineer
> > > UUNET
> > > W-703-343-6645
> > > C-703-304-2460
> > >
> > >
> > >
> > > -----Original Message-----
> > > From: Bruce Williams [mailto:bruce@williamsnetworking.com]
> > > Sent: Wednesday, April 18, 2001 9:37 PM
> > > To: louie kouncar; 'Bob Dixon'; ccielab@groupstudy.com
> > > Subject: Re: wildcard mask problem
> > >
> > >
> > > Louie,
> > >
> > > Please correct me if I am wrong. Should not the summarized network
> address
> > > be 192.168.0.0 because 192.168.2.0 does not summarize 192.168.4.0.
> > >
> > > The third octet for both networks.
> > > 192.168.2.0 is 00000010
> > > 182.168.4.0 is 00000100
> > >
> > > If that is not correct, could you please explain.
> > >
> > > Bruce
> > > bruce@williamsnetworking.com
> > >
> > >
> > > ----- Original Message -----
> > > From: "louie kouncar" <lkouncar@UU.NET>
> > > To: "'Bob Dixon'" <bobdixon@mediaone.net>; <ccielab@groupstudy.com>
> > > Sent: Wednesday, April 18, 2001 9:24 PM
> > > Subject: RE: wildcard mask problem
> > >
> > >
> > > > Ops,
> > > >
> > > > That should be 0.0.7.255
> > > >
> > > > sorry for the typo
> > > >
> > > > Louie J. Bouncer (CCIE)/Written
> > > > TCO3 Senior Data Center Engineer
> > > > UUNET
> > > > W-703-343-6645
> > > > C-703-304-2460
> > > >
> > > >
> > > >
> > > > -----Original Message-----
> > > > From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
> > > > Bob Dixon
> > > > Sent: Wednesday, April 18, 2001 8:38 PM
> > > > To: ccielab@groupstudy.com
> > > > Subject: wildcard mask problem
> > > >
> > > >
> > > > Folks,
> > > >
> > > > I should know how to do this, but I can't seem to get it.
> > > >
> > > > Question:
> > > > take the following:
> > > > network 192.168.2.0 0.0.0.255 area 0
> > > > network 192.168.4.0 0.0.0.255 area 0
> > > >
> > > > and create one network statement that summarizes the two.
> > > >
> > > > Thanks,
> > > > Bob
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