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I am reading Raft-extended paper and above statement was there. Also I found a statement in the web saying failures of f servers can be tolerated if there were 2*f+1 servers. It's obvious to have another two servers where f=1. Is there an inductive way to prove it?

Amila Senadheera
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It is intuitive once you realize that for any operation to be successful, it must complete successfully on a majority of the servers. i.e. in case of a discrepancy, at least quorum number of servers must agree on the same value.

For 2 servers, both servers must be in agreement for a majority.

For 3 servers, at least 2 must be in agreement for a majority.

For 4 servers, at least 3 must be in agreement for a majority.

For 5 servers, at least 3 must be in agreement for a majority.

i.e. for n servers, n/2 +1 servers must be in agreement.

So for n servers, the number of servers that can afford to fail is n - (n/2+1).

Which mean for 2n servers, it is 2n - (2n/2 + 1).

Therefore for 2n +1 servers, it is 2n - (2n/2 + 1) + 1, which simplifies to n servers.

itisravi
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  • Thanks @itisravi, It makes sense to me. There should to at least (half +1) servers up and running for failed servers to agree on the replicated values are correct. – Amila Senadheera Aug 14 '19 at 23:50