my probability is not so great, but i think that at 2:27 there might be an error. The speaker says that if you add another server (S-2) then the probability of failure is divided by 2. But based on my (non-expert) calculations, I think the probability of failure is reduced by more than half. Please let me know if my reasoning is flawed.. here it is: Let's say the cluster has one node, and the the probability of that node going down is .25. Now we add another node, and for the sake of argument lets say that the failure of one node happens completely independent of the other (admittedly, not a perfect assumption since they may share power supplies, since both will be wiped out if a mega-asteroid hits earth, etc.) Now with two nodes (failing independently), the probability of cluster failure is the probability of these 2 independent events occurring at once (within some time window). Thus P(cluster failure) = P(node1 fails) * P(node 2 fails) = .25 * .25 = .0625. .0625 is significantly less than .25/2. (.125) Right ? If wrong, apologies in advance ! -chris
You are a great teacher. Explained extremely clearly. Thank You.
awesome lecture, a lot of information condensed in just 30 odd mintues... truly grateful to the author
Great lecture. Thank you. The witness and round concept makes things much easier to understand.
good lecture, clear and detailed.
my probability is not so great, but i think that at 2:27 there might be an error. The speaker says that if you add another server (S-2) then the probability of failure is divided by 2. But based on my (non-expert) calculations, I think the probability of failure is reduced by more than half. Please let me know if my reasoning is flawed.. here it is: Let's say the cluster has one node, and the the probability of that node going down is .25. Now we add another node, and for the sake of argument lets say that the failure of one node happens completely independent of the other (admittedly, not a perfect assumption since they may share power supplies, since both will be wiped out if a mega-asteroid hits earth, etc.) Now with two nodes (failing independently), the probability of cluster failure is the probability of these 2 independent events occurring at once (within some time window). Thus P(cluster failure) = P(node1 fails) * P(node 2 fails) = .25 * .25 = .0625. .0625 is significantly less than .25/2. (.125) Right ? If wrong, apologies in advance ! -chris
agreed, the probability of both servers failing is the product of each individual server failure probability
One lecture to rule all the others, thanks.
Really excellent and clear explanation. Take my like and my subscription
you have put up a nice foundation. I shall take it up forward and understand more. Thanks
Thank you very much for the informative lecture.
One of the better explanation of paxos.
same as Raft which is used by etcd and Kafka 3
Very clear explanation!
Excellent explanation! Thanks!
We can us Paxos BF to resolve the problem of intrusion detection (IDS) !?
En quoi un système distribué de type paxos montre-t-est elle une incohérence du système financier mondial?
thanks you, 30 minutes explains the same as 3 hours of reading with the same pratical information.
Excellent!
This is very similar to the Chandra-Toueg consensus algorithm. en.wikipedia.org/wiki/Chandra%E2%80%93Toueg_consensus_algorithm
Horrible.