Hello Juan,

On Tue, 9 Jul 2002, Juan Luis Baptiste wrote:

> Hi Grigory,
>
> First of all thanks for your answer, it helped me very  much to understand the
> seq number renumbering problem.
>
> But I have a cuestion about your  example, I think it only works when all
> packets arrive ordered, not when one or  more get delayed and the packets
> start  arriving in desorder. Please see this modified  version of your
> example, please look at packet 5.
>
>
>
> 1) (OUT) IPv4    <-  IPv6     Dout = 0, Din = 0
>
> L4 = 20          <- L6 = 50 D'out = -30, D'in = 0
>
> SN'out = SNout + Dout = 100 <-  SNout = 100
> AN'out = ANout - Din = 500  <-  ANout = 500
>
> Result: Dout += D'out = -30, Din += D'in = 0
>
>
>
> 2) (IN) IPv4    -> IPv6      Dout = -30, Din = 0
>
> L4 = 40         -> L6 = 60  D'out = 0, D'in = 20
>
> SNin = 500     -> SN'in = SNin + Din = 500
> ANin = 120     -> AN'in = ANin - Dout = 150
>
> Result: Dout += D'out = -30, Din += D'in = 20
>
>
>
> 3) (OUT) IPv4    <- IPv6    Dout = -30, Din = 20
>
> L4 = 40        <- L6 = 40 D'out = 0, D'in = 0
>
> SN'out = SNout + Dout = 120  <- SNout = 150
> AN'out = ANout - Din = 540   <- ANout = 560
>
> Result: Dout += D'out = -30, Din += D'in = 20
>
>
>
> 4) (IN) IPv4    -> IPv6    Dout = -30, Din = 20
>
> L4 = 40         -> L6 = 60 D'out = 0, D'in = 20
>
> SNin = 540     -> SN'in = SNin + Din = 560
> ANin = 160     -> AN'in = ANin - Dout = 190
>
> Result: Dout += D'out = -30, Din += D'in = 40
>
>
>
> 5) (IN) IPv4    -> IPv6    Dout = -30, Din = 40
>
> L4 = 40         -> L6 = 60 D'out = 0, D'in = 20
>
> SNin = 580     -> SN'in = SNin + Din = 620
> ANin = 160     -> AN'in = ANin - Dout = 190
>
> Result: Dout += D'out = -30, Din += D'in = 60
>
>
>
> 6) (OUT) IPv4   <- IPv6   Dout = -30, Din = 60
>
> L4 = 30       <- L6 = 40 D'out = -10, D'in = 0
>
> SN'out = SNout + Dout = 160  <- SNout = 190
> AN'out = ANout - Din = 560   <- ANout = 620
>
> Result: Dout += D'out = -40, Din += D'in = 60
>
> As you can see, packets 4 and 5 comes from the same direction, both modify
> Din and when packet 6 arrives in the other direction, it takes the last known
> Din to calculate AN'out. As the packet 6 is the answer
> for packet 4, it should have AN'out=580 and not 560.
>
> What do you think of this? I hope I made my self clear.
>

You are right! I didn't take this variant into account.
The solution of such problem will be more complex. I think we need to keep
a track of each changes of Din or Dout and save corresponding real sequence
numbers. In this case we can correct SN and AN based on the valid Din and Dout.
But we need to have a mechanism to control the size of such list (Din, Dout,
SN, AN) and to remove older records from the head, because such list may be
very big. I haven't time to do any decision, it's only my first opinion.
If you have any thinks, please, tell me.

P.S. Sorry for my delay... I reinstalled my computer (it take much time :)...

Best regards,
Grigory Klyuchnikov, System Engineer,
Institute for System Programming
Russian Academy of Sciences



---------------------------------------------------------------------
The IPv6 Users Mailing List
Unsubscribe by sending "unsubscribe users" to [EMAIL PROTECTED]

Reply via email to