[6tsch] Rank rounding error, churn and hysteresis [was Zero Objective Function discussion]

"Pascal Thubert (pthubert)" <pthubert@cisco.com> Thu, 12 September 2013 09:37 UTC

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From: "Pascal Thubert (pthubert)" <pthubert@cisco.com>
To: Qin Wang <qinwang@berkeley.edu>
Thread-Topic: Rank rounding error, churn and hysteresis [was Zero Objective Function discussion]
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Date: Thu, 12 Sep 2013 09:37:27 +0000
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Cc: Thomas Watteyne <watteyne@eecs.berkeley.edu>, 6TSCH <6tsch@ietf.org>, "xvilajosana@eecs.berkeley.edu" <xvilajosana@eecs.berkeley.edu>
Subject: [6tsch] Rank rounding error, churn and hysteresis [was Zero Objective Function discussion]
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Hello Qin:

In this particular case, a computation of DAGRank(rank) of 7 will prevent this node from attaching to other nodes that are also at 7, and will encourage more other nodes to attach to this. The end to end Rank calculation for children is only slightly affected since the 2 byte rank is used for the total rank computation.

I think that the rounding errors are part of the errors, just like the stats of transmissions may vary.
Depending on the representation of ETX in memory, you might also have system that round to 1 decimal and others to 2 decimals and the same calculation may still create a similar effect.

Is that a big deal? Well, Considering the variations on the ETX computation I do not think that this makes a lot of difference in the run time.

What is more critical is to use a good hysteresis so as to avoid jumping over the integral boundary of DAGRank(rank) all the time.

Both RFC 6552 and 6719 recommend you to do that, with RFC 6552 dereferencing RFC 6719 that recommends to use a value boundary on Rank before reparenting.
http://tools.ietf.org/html/rfc6719#section-3.2.2 :

   3.  If the smallest path cost for paths through the candidate
       neighbors is smaller than cur_min_path_cost by less than
       PARENT_SWITCH_THRESHOLD, the node MAY continue to use the current
       preferred parent.  This is the hysteresis component of MRHOF.

RFC 6552 recommends by default to follow that guidance in http://tools.ietf.org/html/rfc6552#section-4.1

                                                           it is thus
   RECOMMENDED to base the computation of the step_of_rank on dynamic
   link properties such as the expected transmission count (ETX) metric
   as introduced in [DeCouto03<http://tools.ietf.org/html/rfc6552#ref-DeCouto03>] and discussed in [RFC6551<http://tools.ietf.org/html/rfc6551>]1>].  "Minimum
   Rank Objective Function with Hysteresis" [HYSTERESIS<http://tools.ietf.org/html/rfc6552#ref-HYSTERESIS>] provides
   guidance on how link cost can be computed and on how hysteresis can
   improve Rank stability.

Additional tuning or methods exist to increase stability. We can start some work on that if we find that MRHOF's method is not sufficient in our case.

One method (Cisco IPR there) consists in hiding limited variations of the Rank to the children (around the last advertised 2 bytes value) until:
1) a delta threshold is reached in the step of rank with parent (one-hop hysteresis)
or
2) Rank has to be recomputed because of the parent (the parent changes or the Rank of the parent changes)

This method creates a laser effect by which a number of new values of Rank a recomputed and advertised as a single wave, whereby every node affected by the change re-centers its rank value to its current moving average, reducing the chances to hit condition 2) in the near future.

What do you think?

Pascal

From: Qin Wang [mailto:qinwang@berkeley.edu]
Sent: mercredi 11 septembre 2013 23:58
To: Pascal Thubert (pthubert)
Cc: xvilajosana@eecs.berkeley.edu; Thomas Watteyne; 6TSCH
Subject: Re: [6tsch] Zero Objective Function discussion

Hi Pascal,

I agree with your calculation. But, I have a small question.

If you calculate with rank_increase=(2*xmit/ack)*256, instead of rank_increase=(512*xmit/ack), then, you will get the result as follows.

r(1)=r(0)+2.66*256 = 2.66*256               => DAGRank(rank) = 2
r(2)=r(1)+2.66*256= 5.32*256                 => DAGRank(rank) = 5
r(3)=r(2)+2.66*256=7.98*256                  => DAGRank(rank) = 7
r(4)=r(3)+2.66*256=10.64*256                => DAGRank(rank) = 10
r(5)=r(4)+2.66*256=13.30*256                 => DAGRank(rank) = 13
You can see r(3) is different from that in your calculation. Obviously, the difference comes from different implementation. In another word, following same OF, different implementation may lead to different Rank value. Then, my question is if it is Ok while the motes from different vendors work together in a DAG.
Please point out if I'm wrong.
Thanks
Qin






On Wed, Sep 11, 2013 at 3:08 AM, Pascal Thubert (pthubert) <pthubert@cisco.com<mailto:pthubert@cisco.com>> wrote:
Hello Xavi:

Great!

I'd add the following:


RFC 6550    3.5.1<http://tools.ietf.org/html/rfc6550#section-3.5.1>.1>. Rank Comparison (DAGRank())has:
"

When an Objective Function computes Rank, the Objective Function

   operates on the entire (i.e., 16-bit) Rank quantity.  When Rank is

   compared, e.g., for determination of parent relationships or loop

   detection, the integer portion of the Rank is to be used.  The

   integer portion of the Rank is computed by the DAGRank() macro as

   follows, where floor(x) is the function that evaluates to the

   greatest integer less than or equal to x:



              DAGRank(rank) = floor(rank/MinHopRankIncrease)



   For example, if a 16-bit Rank quantity is decimal 27, and the

   MinHopRankIncrease is decimal 16, then DAGRank(27) = floor(1.6875) =

   1.  The integer part of the Rank is 1 and the fractional part is

   11/16.

"
DAGRank(rank) is what we expect to place in the Flow label and the join priority.
In our case:

r(1)=r(0)+rank_increase = 0+683               => DAGRank(rank) = 2
r(2)=r(1)+683=1366                                         => DAGRank(rank) = 5
r(3)=r(2)+683=2049                                         => DAGRank(rank) = 8
r(4)=r(3)+683=2732                                         => DAGRank(rank) = 10
r(5)=r(4)+683=3415                                         => DAGRank(rank) = 13

We see that the DAGRank() always increases at least by one. This is the desired property so as to ensure that we can detect loop with just one octet in the flow label.
Note that the real 2 octets Rank does not lose the rounding info since the DIP always passes 2 octets.

Cheers,

Pascal

From: Xavier Vilajosana Guillen [mailto:xvilajosana@eecs.berkeley.edu<mailto:xvilajosana@eecs.berkeley.edu>]
Sent: mardi 10 septembre 2013 19:53
To: Pascal Thubert (pthubert)
Cc: Thomas Watteyne; 6TSCH
Subject: Re: [6tsch] Zero Objective Function discussion

Hi all,

after some study I agree that we can use the RFC6552 in the minimal 6TiSCH configuration. I would like to call for approval or more input from others before I consolidate the text in the draft.

I put here an example following Pascal suggestion:

Given

Rf = 1
Sp = 2* ETX
Sr = 0
minHopRankIncrease = 256 (default in RPL)
ETX=(xmit/ack)

r(n) = r(p) + rank_increase
rank_increase= (Rf*Sp + Sr) * minHopRankIncrease
rank_increase=(512*xmit/ack)


if we take 5 hops (95 are supported) network and r(0)=0 and xmit=100 and ack=75 for all nodes

r(1)=r(0)+rank_increase = 0+683 with f=0
r(2)=r(1)+683=1366 with f=1
r(3)=r(2)+683=2049 with f=2
r(4)=r(3)+683=2732 with f=2
r(5)=r(4)+683=3415 with f=3
...
etc...

so f is monotonically increasing and the rank function enables more than enough hops.
Please provide feedback.
cheers!
Xavi


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