2009/6/24 Rémi Villé <[email protected]> > 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"... >
I think I have understand my mistake, ETX is an estimation of the number of transmissions/retransmissions to deliver a packet. So if we have 1/2 chance a packet to be lost between a mote A and B, ETX should be 2 because we expected that we'll need two transmissions to deliver the message. And through a path (A,B,C) (with 1/2 chance between B and C too) we repeat this reasoning twice, so here, using addition has sense. I'll continue to read the paper.
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