Question: IBFT 2.0 Fault Tolerance Discrepancy between Documentation and Paper

[HaoXuan40404]

Description
I noticed a potential discrepancy regarding IBFT 2.0's fault tolerance capability between the Besu documentation and the IBFT 2.0 paper.

Current Behavior
According to the Besu documentation, a 6-node IBFT 2.0 network can tolerate 2 faulty nodes.

Expected Behavior
Based on the IBFT 2.0 paper, Section 2 (Failure Model), a 6-node network should only be able to tolerate 1 faulty node according to the traditional Byzantine Fault Tolerance formula: n = 3f + 1, where:

n is the total number of nodes
f is the number of faulty nodes

Question
Could you explain why there is a difference between the documentation and the theoretical model? Is there a specific implementation in Besu that enables higher fault tolerance than the traditional BFT model suggests?

Environment

• Consensus Protocol: IBFT 2.0

[HaoXuan40404]

PBFT consensus with the following scenario:

Total nodes (n) = 6
Byzantine nodes (f) = 1
Disconnected nodes = 1
Normal functioning nodes = 4

It satisfies the basic PBFT requirement where n ≥ 3f + 1:

With f = 1, we need n ≥ 3(1) + 1 = 4
Our n = 6 satisfies this condition (6 > 4)
For consensus, PBFT requires 2f + 1 nodes to participate:

With f = 1, we need 2(1) + 1 = 3 nodes
We have 4 normal functioning nodes available (after subtracting 1 Byzantine and 1 disconnected node)
4 available nodes > 3 required nodes

Could you please confirm if this analysis is correct? Would the system be able to reach consensus in this scenario?
Thank you for your help in clarifying this.