IPv6 day and tunnels
Owen DeLong
owen at delong.com
Tue Jun 5 06:18:16 UTC 2012
On Jun 4, 2012, at 7:47 PM, Jimmy Hess wrote:
> On 6/4/12, Owen DeLong <owen at delong.com> 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
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