Re: [6tsch] Zero Objective Function discussion

["Pascal Thubert (pthubert)" <[e-mail]>]

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:[e-mail]]
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