How do you calculate the availability of components in series and in parallel?

Series

A = A1 x A2 x A3. Three components at 99.9 percent: 0.999 x 0.999 x 0.999 = 0.997, about 99.7 percent or roughly 26 hours a year. Every box in a single path lowers the total.

Parallel

A = 1 - (1 - A1)(1 - A2). Two links at 99.5 percent: 1 - (0.005)(0.005) = 1 - 0.000025 = 99.9975 percent, about 13 minutes a year instead of about 44 hours for one link.

Combine them

Those two links terminating on one router at 99.95 percent: 0.999975 x 0.9995 = about 99.9475 percent, roughly 4.6 hours a year. The single router now dominates. That is the point of the exercise: find the series element that limits you.

Single component

A = MTBF / (MTBF + MTTR). MTBF 50,000 hours with a 4 hour repair gives 50,000 / 50,004, about 99.992 percent. With a 24 hour repair it drops to about 99.95 percent. MTTR is the lever you control: spares on site, support contracts, out-of-band access, runbooks.

Where the math lies

• It assumes independent failures. Two circuits in one duct, two routers on one UPS or one software bug on both boxes break the parallel formula.
• It assumes failover is instant and always works. A 40 second reconvergence counts as downtime.
• It ignores human error and changes, which cause a large share of real outages.