On Thu, Jun 18, 2009 at 4:41 AM, Rémi Villé<[email protected]> wrote: > Hi, > > I would like to discuss about how the path cost is calculated from the cost > of its arcs. > Currently this path is accumulated with an addition. I think it's bizarre > because we try to create the best path in term of ratio (msg received/msf > sent). So the cost of a path should be proportional to the chance of a > packet to reach the sink through this path, i.e. 1/q(a,b)*q(b,c) for the > path (a,b,c), and not 1/q(a,b) + 1/q(b,c). (q = PRR(a,b)*PRR(b,a)). > > I tried to find incoherent cases with this accumulation and I found this one > : > 3 motes : a, b and c (a is the sink). > q(a,b) = 0.5 > q(b,c) = 0.2 > > The cost of (a,b,c) should be 1/(0.5*0.2) = 10, but the effective cost is > 1/0.5 + 1/0.2 = 7. (q(a,b,c) = 0.5*0.2 = 0.1). > If we have cost(a,c) = 8, then q(a,c) = 1/8 = 0.125. > We have q(a,c) best then q(a,b,c) but cost(a,c) > cost(a,b,c), and c choose > a bad best parent/path (the path (a,b,c) instead of (a,c)). > > If I take ETX instead of EETX (10/q) it doesn't change this reasoning. > > I would like to know if I miss something, maybe there's a good reson to use > the addition instead of multiplication to accumulate the cost path, I would > like to understand...
This paper goes into the details of additive path cost: http://pdos.csail.mit.edu/papers/grid:mobicom03/paper.pdf ETX gives you the expected number of transmissions for a link. So, the path cost is the sum of these per-link costs. - om_p _______________________________________________ Tinyos-help mailing list [email protected] https://www.millennium.berkeley.edu/cgi-bin/mailman/listinfo/tinyos-help
