When using MLPPP, the maximum size of a fragment is incorrectly
calculated with an offset of +2. When the MTU of underlying channels
minus the multilink overhead is less than the MTU of the MLPPP bundle
divided by the number of channels, the fragmentation algorithm is
supposed to send maximally sized fragments down the channels and
generate additional fragment(s) for the remainder, but due to the error
in the fragment size calculation the earlier fragments exceed the
underlying MTU and are lost. This patch reduces the maximum fragment
size by 2.
Signed-off-by: Ben McKeegan <ben@xxxxxxxxxxxxxxxx>
---
We have successfully been running this patch on production routers for
several months now using PPPoE as an underlying channel. We are
configuring a high MTU on a multilink enabled PPP device with a single
underlying PPPoE channel running over an Ethernet device with a much
lower MTU. Thus our larger PPP frames get ML-fragmented with all
fragments being sent down the same channel. Without this patch the link
does not work correctly, and tcpdump shows the kernel generating PPPoE
frames 2 bytes longer than the MTU of the Ethernet which obviously do
not make it out. With this patch, the maximum size of the PPPoE frames
correctly matches the Ethernet MTU and the link works correctly.
We believe this issue should affect other types of underlying channel in
the same way, and more usually causing problems when a multilink bundle
is running in a degraded state (i.e. with fewer than normal channels).
I cannot see why the '+2' was put in the MTU calculation in the original
code, but it looks like this feature might have always been broken.
--- linux-2.6.27.7.orig/drivers/net/ppp_generic.c 2008-11-20
23:02:37.000000000 +0000
+++ linux-2.6.27.7/drivers/net/ppp_generic.c 2008-12-03
14:43:21.000000000 +0000
@@ -1342,14 +1342,14 @@
/*
* Create a fragment for this channel of
- * min(max(mtu+2-hdrlen, 4), fragsize, len) bytes.
- * If mtu+2-hdrlen < 4, that is a ridiculously small
+ * min(max(mtu-hdrlen, 4), fragsize, len) bytes.
+ * If mtu-hdrlen < 4, that is a ridiculously small
* MTU, so we use mtu = 2 + hdrlen.
*/
if (fragsize > len)
fragsize = len;
flen = fragsize;
- mtu = pch->chan->mtu + 2 - hdrlen;
+ mtu = pch->chan->mtu - hdrlen;
if (mtu < 4)
mtu = 4;
if (flen > mtu)
--
Ben McKeegan
Netservers Limited
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