A replicated store has 7 nodes with quorum reads and writes. It must survive 2 simultaneous node failures and still read the latest write. State the smallest read quorum that works, the write quorum it forces, and why moving to 8 nodes does not help.

A replicated store has 7 nodes with quorum reads and writes. It must survive 2 simultaneous node failures and still read the latest write. State the smallest read quorum that works, the write quorum it forces, and why moving to 8 nodes does not help.

Approach: Write the overlap condition that makes a read see the latest write, add the availability constraint that two failures must block neither operation, then minimise the read side.

3. Overlap requires R + W > N, so with N = 7 the pair must satisfy R + W >= 8, which forces the read quorum and the write quorum to share at least one node holding the newest write. Surviving 2 failures means both must be reachable on 5 live nodes, so R <= 5 and W <= 5. Minimising the read side gives R >= 8 - W >= 8 - 5 = 3, so R = 3 with W = 5. That configuration reads 3 nodes, writes 5, keeps fault tolerance at 2 for both operations, and still observes every committed write. Moving to N = 8 does not help: the condition becomes R + W >= 9 with W <= 6, so R >= 3 again, the same read quorum, while every write now waits on a sixth node and the cluster still tolerates only 2 failures. An even N buys a replica's storage and write cost without buying fault tolerance, which is why quorum systems are sized at odd numbers.

Follow-up: How does the answer change if reads must also survive a partition isolating 3 of the 7 nodes?

Key concepts: quorum overlap, read quorum, write quorum, fault tolerance.