hjp+quagga at wsr
May 18, 2012, 2:35 AM
Post #9 of 9
On 2012-05-16 13:32:51 -0400, Lennart Sorensen wrote:
> On Wed, May 16, 2012 at 01:14:23PM -0400, Steve Clark wrote:
> > On 05/16/2012 12:59 PM, Thomas York wrote:
> > >I know that if the interface speeds are the same and you don't have
> > >a manually defined cost (IE, Quagga derives the same cost on
> > >multiple interface), Quagga will add the route on both interfaces,
> > >but with identical weights and metrics. The kernel should do some
> > >kind of load balancing (not perfectly) in this scenario (as far as
> > >I know, someone feel free to correct me)
> > >
> > >192.168.1.0/30 dev eth0 proto kernel scope link src 192.168.1.1
> > >
> > >192.168.1.4/30 dev eth1 proto kernel scope link src 192.168.1.5
> > >
> > >10.1.17.0/24 proto zebra metric 20
> > >
> > > nexthop via 192.168.1.2 dev eth0 weight 1
> > >
> > > nexthop via 192.168.1.6 dev eth1 weight 1
> > Thanks Lennart and Thomas for taking the time to reply.
> Make sure to turn off rp_filter when using multipath. Otherwise the
> kernel will drop incoming traffic in many cases.
And in another mail you mentioned that the routes are chosen per flow.
Is this per-flow routing symmetric?
Consider this scenario:
R is router, F1 and F2 are firewalls (routers with packet filters), S is
a server. The links between R, F1, F2 and S all have the same cost.
A Client C tries to establish a TCP connection to S. The SYN packet
arrives at R which chooses to forward the packet to F1 which in turn
forwards it S. S then replies with a SYN-ACK packet. How does it
determine where the send this packet?
1) By receiving the SYN packet from F1 a flow was established. The
SYN-ACK packet belongs to the same flow, so it is sent to F1.
2) This is the first packet to be sent in this connection, so no flow
has been established yet. The packet will be sent to either F1 or F2
with equal probability.
If F1 and F2 are stateful packet filters, only case 1 would work (unless
F1 and F2 have a way to share state).
PS: Does it matter whether S can reach F1 through the same or different
_ | Peter J. Holzer | Auf jedem Computer sollte der Satz Ludwigs II
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