On Mon, 15 Apr 2002, Mark J Roberts wrote: > Kevin Atkinson: > > Fail in what way? Not being able to find the block on the > > network. There is little chance of that happening unless the data > > (as in the complete key) is about to fall of the network. > > Yeah, that's what I meant by fail. > > Say someone inserts a 1GB divx movie. That's a neat 32,768 blocks of > 32,768 bytes each. And say we want to be 99% certain that all the > blocks will be successfully downloaded. How reliable must each block > be? .99 = x^32768 - it approaches 1.
That equation only holds if the probability of failure for each block are independent from one another which is certainly not the case. giving non-independent events A,B,C P(A,B) = P(A|B)*P(B) P(A,B,C) = P(A|B,C)*P(B|C)*P(C) where A,B means A and B or A intersect B and A|B A given B For the sake of argument assume that if n blocks have already been retrieved the probability of the next block failing is: 1-(1-p)*(1/2)^n So for instance if the initial probability was 90% and 1 block has already been retrieved then the probability the next block will be available is 1-(1-.9)*(1/2) = 1-(.10)(1/2) = 0.95. Given the formula the equation then becomes: P(A,B,C) = (1-(1-p)*(1/2)^2)(1-(1-p)*(1/2)^1)(1-(1-p)*(1/2)^0) or more generally: product(1 - (1-p)*(1/2)^i, i=0..n) where n is the number of blocks to retrieve Solving: .99 = product(1 - (1-p)*(1/2)^i, i=0..n) gives a solution of .995 for just about any n, verified via maple. Using a decay factor larger than (1/2) increased p but it still seams to converge (or at least grows very slowly). For example the solution to: product(1 - (1-p)*(9/10)^i, i=0..n) is about .999. --- http://kevin.atkinson.dhs.org _______________________________________________ freenet-tech mailing list [EMAIL PROTECTED] http://lists.freenetproject.org/mailman/listinfo/tech
