ITPEC FE Subject A October 2025 Question 12
ITPEC FE Subject A October 2025 — Question 12 of 60
System Availability with Parallel Redundancy — calculate availability when "at least 1 of N" devices must work.
The system has components connected in series (all required):
- 1 Server (availability a)
- 3 Clients — at least 1 must work
- 2 Printers — at least 1 must work
- LAN (availability = 1)
Parallel availability (at least 1 of N identical devices):
- 1 - (1 - x)^n
- 3 clients: 1 - (1 - b)^3
- 2 printers: 1 - (1 - c)^2
Series availability (all must work): multiply all components.
Result: a × (1 - (1-b)³) × (1 - (1-c)²) → (d)
Why not others:
- (a) ab³c² — assumes ALL clients and ALL printers must work (no redundancy)
- (b) a(1-b³)(1-c²) — wrong formula: 1 - b³ ≠ 1 - (1-b)³
- (c) a(1-b)³(1-c)² — probability that ALL clients AND ALL printers are DOWN
Key rule: "At least 1 of N" → 1 - (1-x)^n, then multiply in series.
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.