System Design Fundamentals
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Trade-offs in CAP Theorem

We now know what the three properties mean. This lesson answers the question underneath them: why can a system not simply have all three?

Walking Through the Impossible Case

Take a system with two replicas, R1 and R2, holding the same value. The link between them goes down. Neither can reach the other, and neither can tell whether the other has crashed or is merely unreachable.

A client now sends a write to R1.

R1 has exactly two options.

Option one: accept the write. The client is happy. But R2 knows nothing about it, and a client reading from R2 gets the old value

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manoranjan das

manoranjan das

· 3 months ago

When we claim CA is possible only when there is systems with out any partition meaning no nodes or systems talking over n/w then how it can be A. As we need redundant systems for A as per definition. So in that case how does it work ? Seems A and P always together otherwise there is no A.

This part: "No matter which node you contact, if it hasn’t crashed, it will respond within a reasonable time. An available system never resorts to saying “sorry, I can’t serve your request right now” due to a node or network failure.

In practical terms, this often means there are multiple redundant nodes, so if one goes down or can’t be reached, another can serve the request."