ITPEC FE Morning October 2022 Question 16
Which of the following is the approximate availability of the entire system in the figure below? The availability is calculated by rounding off to two decimal places. The numeric value within parentheses is the availability of each device, and the system is available when two or more of the devices connected in parallel are operating in the left part (device 1 to 3) and one or more of the devices connected in parallel is operating in the right part (device 4 to 5).
Explanation
0.94 — the left part needs any two of three devices, the right part any one of two, and the two parts are in series.
Left part, at least two of devices 1 to 3 (0.9, 0.8, 0.9) working:
- •All three: 0.9 × 0.8 × 0.9 = 0.648
- •Exactly two: 0.9 × 0.8 × 0.1 + 0.9 × 0.2 × 0.9 + 0.1 × 0.8 × 0.9 = 0.072 + 0.162 + 0.072 = 0.306
- •Left availability: 0.648 + 0.306 = 0.954
Right part, at least one of devices 4 and 5: 1 − 0.1 × 0.1 = 0.99
Whole system: 0.954 × 0.99 = 0.94446, which rounds to 0.94.
Answer (c)
Why not others:
- •(d) 0.99 — treats the left part as an ordinary parallel group needing only one device, 1 − 0.1 × 0.2 × 0.1 = 0.998, and 0.998 × 0.99 ≈ 0.99
- •(a) 0.65 — close to 0.648, the chance that all three left devices work, which the system does not require
- •(b) 0.81 — lower than it can be: the left part alone is 0.954 and the right part 0.99, and their product is 0.94
Key rule: For "k of n" redundancy, add the probabilities of every case with at least k devices working, then multiply the parts that are in series.
System Component · AI-written, may contain errors
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