Research — Illustrative (draft, unvalidated)
K-of-N availability — auto
Two components at 0.995 availability must both be up (N), then a third is added so any two will do (N+1), then a whole second side is added so either side will do (2N). Watch the nines gauge: needing both is worse than a single component, and the spare is what pays off.
Redundancy topology
N (both needed)first step better than one component2N (two sides)
Reading —
Units, required slots and nines gauge (fixed illustrative drawing)
System availability status
System availability
Topology
Expected downtime (derived)
How the check read it
Client-side port:the logic and table in this page are a TypeScript port of the harness function, checked against values exported from the Python tests. Not a live API call. Domain maturity, quoted from the research page: “Draft; no reliability or controls engineer has reviewed it yet. The availability math is cross-checked against closed-form series, parallel and 2-of-3 cases, and an AI-assisted pre-review found and fixed three issues. IEEE 493 and ISA-75.19.01 were not read directly.” Only the system availability comes from the harness function; nines and downtime per year are plain arithmetic from it (365-day, 8760 h year), labeled as derived. The 0.995 component availability is a value chosen for the walk, matching the harness’s own worked example, not catalog data.Real function this maps to:
calculate_redundant_system_availability.Not modeled: common-cause failures (shared feeds, shared cooling loops, shared fire zones), which make every figure here optimistic; repair times and failover timing; real component reliability data; and Uptime Institute tier mapping. The tick marks the first step whose availability beats a single 0.995 component; it is a comparison, not a pass/fail threshold.See also:K-of-N availability, interactive · Failover race, auto