ppp_generic: fix multilink fragment MTU calculation (again)

When using MLPPP, the maximum size of a fragment is incorrectly
calculated with an offset of -2.
This patch reverses the changes in the patch found here:
http://marc.info/?l=linux-netdev&m=123541324010539&w=2

The value of hdrlen includes the size of both the 2-byte PPP protocol
field and the 2- or 4-byte multilink header (2+4=6 for long sequence
numbers, 2+2=4 for short sequence numbers). Section 2 of RFC1661 says
that the MRU that is negotiated (i.e., the MTU of the sending system)
includes only the PPP payload but not the protocol field, thus the
correct MTU should be the link's MTU minus the multilink header (mtu -
(hdrlen-2)).

The incorrect calculation causes Linux to fragment packets to a size two
bytes smaller than the allowed MTU. While not technically illegal, this
behaviour confounds MRU-tuning to avoid PPP-layer fragmentation.

Signed-off-by: Henry Wong <henry@stuffedcow.net>
Signed-off-by: David S. Miller <davem@davemloft.net>

1 files changed, 6 insertions, 1 deletions

diff --git a/drivers/net/ppp_generic.c b/drivers/net/ppp_generic.c
index 10e5d985afa3..edfa15d2e795 100644
--- a/drivers/net/ppp_generic.c
+++ b/drivers/net/ppp_generic.c
@@ -1465,7 +1465,12 @@ static int ppp_mp_explode(struct ppp *ppp, struct sk_buff *skb)
                         continue;
                 }
 
-                mtu = pch->chan->mtu - hdrlen;
+                /*
+                 * hdrlen includes the 2-byte PPP protocol field, but the
+                 * MTU counts only the payload excluding the protocol field.
+                 * (RFC1661 Section 2)
+                 */
+                mtu = pch->chan->mtu - (hdrlen - 2);
                 if (mtu < 4)
                         mtu = 4;
                 if (flen > mtu)