Calculating overheads

Linux Advanced Routing and Traffic Control

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Hi,

I am looking to figure out the most fool proof way to calculate stab overheads for ADSL/VDSL connections.

ppp0      Link encap:Point-to-Point Protocol
          inet addr:81.149.38.69  P-t-P:81.139.160.1 Mask:255.255.255.255
          UP POINTOPOINT RUNNING NOARP MULTICAST  MTU:1492  Metric:1
          RX packets:17368223 errors:0 dropped:0 overruns:0 frame:0
          TX packets:12040295 errors:0 dropped:0 overruns:0 carrier:0
          collisions:0 txqueuelen:100
          RX bytes:17420109286 (16.2 GiB)  TX bytes:3611007028 (3.3 GiB)

I am setting a longer txqueuelen as I am not currently using any fair queuing (buffer bloat issues with sfq)

The connection is a BT Infinity FTTC VDSL connection synced at 80mbit/20mbit. The modem is connected directly to the ethernet port on a server running a slightly tweaked HFSC setup that you folks helped me set up in July - back when I was on ADSL. I am still running pppoe I believe from my server.

The largest ping packet that I can fit out onto the wire is 1464 bytes:

# ping -c 2 -s 1464 -M do google.com
PING google.com (31.55.166.216) 1464(1492) bytes of data.
1472 bytes from 31.55.166.216: icmp_seq=1 ttl=58 time=11.7 ms
1472 bytes from 31.55.166.216: icmp_seq=2 ttl=58 time=11.9 ms

# ping -c 2 -s 1465 -M do google.com
PING google.com (31.55.166.212) 1465(1493) bytes of data.
From host81-149-38-69.in-addr.btopenworld.com (81.149.38.69) icmp_seq=1 Frag needed and DF set (mtu = 1492) From host81-149-38-69.in-addr.btopenworld.com (81.149.38.69) icmp_seq=1 Frag needed and DF set (mtu = 1492)

Based on this I believe overhead should be set to 28, however with 28 set as my overhead and hfsc ls m2 20000kbit ul m2 20000kbit I seem to be loosing about 1.5mbit of upload...

No traffic manager enabled: http://www.thinkbroadband.com/speedtest/results.html?id=141116089424883990118 HFSC traffic manager: http://www.thinkbroadband.com/speedtest/results.html?id=141116216621093133034

Am I calculating overhead incorrectly?

Alan
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