ITPEC FE Morning October 2022 Question 16

System Availability (2-of-3 and 1-of-2 redundancy) — calculate availability for a mixed-redundancy system.

The system has two parts in series:
- Left: Devices 1–3 (0.9, 0.8, 0.9) — needs 2 or more operating
- Right: Devices 4–5 (0.9, 0.9) — needs 1 or more operating

Left part (2-of-3):

Find P(failure) = P(0 working) + P(exactly 1 working):
- P(0 working) = 0.1 × 0.2 × 0.1 = 0.002
- P(only D1) = 0.9 × 0.2 × 0.1 = 0.018
- P(only D2) = 0.1 × 0.8 × 0.1 = 0.008
- P(only D3) = 0.1 × 0.2 × 0.9 = 0.018
- P(failure) = 0.002 + 0.044 = 0.046
- A_left = 1 − 0.046 = 0.954

Right part (1-of-2 parallel):
- A_right = 1 − (1−0.9)(1−0.9) = 1 − 0.01 = 0.99

Total (series):
- A_total = 0.954 × 0.99 = 0.94446 ≈ 0.94

Why not others:
- (a) 0.65 — would imply simple series of all 5 devices
- (b) 0.81 — incorrect redundancy model
- (d) 0.99 — uses 1-of-3 (standard parallel) instead of 2-of-3 for the left part

Key rule: "k-of-n" redundancy: calculate P(failure) = sum of probabilities for 0 to (k−1) devices working, then A = 1 − P(failure).

AI-generated — may contain errors