On Jun 4, 2012, at 7:47 PM, Jimmy Hess wrote: > On 6/4/12, Owen DeLong <[email protected]> wrote: > [snip] >> Probing as you have proposed requires you to essentially do a binary search >> to arrive >> at some number n where 1280≤n≤9000, so, you end up doing something like >> this: > [snip] >> So, you waited for 13 timeouts before you actually passed useful >> traffic? Or, perhaps you putter along at the lowest possible MTU until you > [snip] > Instead of waiting for 13 timeouts, start with 4 initial probes in > parallel, and react rapidly to the responses you receive; say > 9000,2200, 1500, 830. >
What's the point of an 830 probe when the minimum valid MTU is 1280? > Don't wait until any timeouts until the possible MTUs are narrowed. > > > FindLocalMTU(B,T) > Let B := Minimum_MTU > Let T := Maximum_MTU > Let D := Max(1, Floor( ( (T - 1) - (B+1) ) / 4 )) > Let R := T > Let Attempted_Probes := [] > > While ( ( (B + D) < T ) or Attempted_Probes is not Empty ) do > If R is not a member of Attempted_Probes or Retries < 1 then > AsynchronouslySendProbeOfSize (R) > Append (R,Tries) to list of Attempted_Probes if not exists > or if (R,Tries) already in list then increment Retries. Did I miss the definition of Tries and/or Retries somewhere? ;-) > else > T := R - 1 > Delete from Attempted_Probes (R) > end > > if ( (B + D) < T ) AsynchronouslySendProbeOfSize (B+ D) > if ( (B + 2*D) < T ) AsynchronouslySendProbeOfSize (B+ 2*D) > if ( (B + 3*D) < T ) AsynchronouslySendProbeOfSize (B+ 3*D) > if ( (B + 4*D) < T ) AsynchronouslySendProbeOfSize (B+ 4*D) > Shouldn't all of those be <= T? > Wait_For_Next_Probe_Response_To_Arrive() > Wait_For_Additional_Probe_Response_Or_Short_Subsecond_Delay() > Add_Probe_Responses_To_Queue(Q) Not really a Queue, more of a list. In fact, no real need to maintain a list at all, you could simply keep a variable Q and let Q=max(Q,Probe_response) > R := Get_Largest_Received_Probe_Size(Q) Which would allow you to eliminate this line altogether and replace R below with Q. > If ( R > T ) then > T := R > end > > If ( R > B ) then > B := R > D := Max(1, Floor( ( (R - 1) - (B+1) ) / 4 )) > end > done > > Result := B > > > # > > If you receive the response at n=830 first, then wait 1ms and send the > next 4 probes 997 1164 1331 1498, and resend the n=1500 probe > If 1280 is what the probe needs to detect. You'll receive a > response for 1164 , so wait 1ms then retry n=1498 > next 4 probes are 1247 1330 1413 1496 > if 1280 is what the probe needs to detect, You'll receive a > response for 1247, so wait 1ms resend n=1496 > next 4 probes are 1267 1307 1327 1347 > if 1280 is what you neet to detect, you'll receive > response for 1267, so > retry n=1347 wait 1ms > next 4 probes are: 1276 1285 1294 1303 > next 4 probes are: 1277 1278 1279 1280 > next 2 parallel probes are: 1281 1282 > > You hit after 22 probes, but you only needed to wait for n=1281 n=1282 > and their retry to time out. > But that's a whole lot more packets than working PMTU-D to get there and you're also waiting for all those round trips, not just the 4 timeouts. The round trips add up if you're dealing with a 100ms+ RTT. 22 RTTs at 100ms is 2.2 seconds. That's a long time to go without first data packet passed, Owen
