2009/6/23 Omprakash Gnawali <[email protected]> > 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 >
Thanks a lot, it's exactly the paper I was looking for. I have a maybe strange question, I would like to know the exact sense of "expected" in ETX, because I don't speak english fluently. In an wordReference dictionary it's translated by "wait for" or "hope". I think it's more "estimated"...
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