ITPEC IP October 2018 Question 15
There are three devices, A, B, and C, with availabilities of 0.90, 0.95, and 0.95 respectively. Among the systems combining these devices shown below, which of the following has the highest availability? Here, at least one of the parallel-connected devices must be available, and all of the serially-connected devices must be available.
Explanation
B and C in parallel — at 0.9975 it is the only arrangement with no device in series.
Multiply availabilities in series; in parallel, take 1 minus the probability that all devices are down.
- •(a) A and B in series: 0.90 × 0.95 = 0.855
- •(b) A, B and C in series: 0.855 × 0.95 = 0.81225
- •(c) B and C in parallel: 1 − (1 − 0.95) × (1 − 0.95) = 1 − 0.0025 = 0.9975; then A in series: 0.90 × 0.9975 = 0.89775
- •(d) B and C in parallel alone: 0.9975
Answer (d)
Why not others:
- •(a) A–B — 0.855; every device added in series lowers the availability below that of the weakest one
- •(b) A–B–C — 0.81225, the lowest: three devices that must all be up
- •(c) A with B and C in parallel — the parallel pair reaches 0.9975, but A in series caps the whole at 0.89775, below A's own 0.90
Key rule: A series system is less available than its weakest device, a parallel system more available than its best one; so adding a device in series always lowers availability and adding one in parallel raises it.
System Component · AI-written, may contain errors
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