Re: Two NICs in one machine and same LAN?
From: InBan (InBan_at_discussions.microsoft.com)
Date: 07/10/04
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Date: Fri, 9 Jul 2004 18:23:01 -0700
If redundancy is what your looking for maybe you should investigate clustering the SQL servers.
Ian Bagnald
MCSE:Security Windows 2000
MCSA:Security Windows 2000
COMPTIA A+
"RQ" wrote:
> Thanks Kevin, it helps.
>
>
> "Kevin D. Goodknecht Sr. [MVP]" <admin@nospam.WFTX.US> wrote in message
> news:OGmv6$5YEHA.2432@tk2msftngp13.phx.gbl...
> > In news:uGiowr5YEHA.2480@tk2msftngp13.phx.gbl,
> > RQ <r.qian@inetmail.att.net> posted a question
> > Then Kevin replied below:
> > > Would you please provide a answer why it's not a good
> > > idea?
> > Because they are different networks, and should be addressed acordingly.
> > Lets say that there is a machine with IP 10.1.1.3, which network provides
> > the way to that machine?
> > If one is 10.1.1.x and the other is also 10.1.1.x the machine won't know
> > which way to go to get to a machine that is on one or the other networks.
> > One should be 10.1.1.x and the other could be 10.1.0.x.
> > If both NICs are on the same subnet, they should be teamed and would only
> > have one IP address for both NICs. If you have two NICs on the same
> machine
> > with different IP addresses looking at each other (not teamed) only one
> will
> > get used at a time. Plus, if you have NetBIOS enabled, one of the NICs if
> > not both will be disabled due to a NetBIOS name conflict. It is like
> having
> > to machines on the same Network with the same NetBIOS name.
> >
> > Suppose a query is sent to DNS for this machine and DNS gives the 10.1.1.1
> > address but that address is not on the subnet that has that address?
> > Ther is no way for DNS to know which address to give, it will just use
> round
> > robin. If they are on different subnets, DNS will use Netmask ordering to
> > give the correct record. (based on 10.1.1.x or 10.1.0.x)
> >
> >
> >
> > --
> > Best regards,
> > Kevin D4 Dad Goodknecht Sr. [MVP]
> > Hope This Helps
> > ============================
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