On Jun 18, 2009, at 4:41 AM, Rémi Villé 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...
No -- the cost assumes retransmissions. So the expectation is that you'll have 5 transmissions on b,c and 2 on a,b, for a total of 7. 10 assumes no retransmissions, which, as your math shows, is a bad policy, as it leads to multiplicative rather than additive costs. Phil _______________________________________________ Tinyos-help mailing list [email protected] https://www.millennium.berkeley.edu/cgi-bin/mailman/listinfo/tinyos-help
