ITPEC FE Morning October 2018 Question 16
ITPEC FE Morning October 2018 — Question 16 of 80
System Availability with Parallel Redundancy — calculate compound availability for series-parallel topology.
The system requires all three subsystems (series connection):
- Server: 1 unit, must work → a
- Clients: 3 units, at least 1 must work → parallel
- Printers: 2 units, at least 1 must work → parallel
- LAN: availability = 1 (ignored)
Parallel availability (at least 1 of N works):
- 3 clients: 1 − (1 − b)³
- 2 printers: 1 − (1 − c)²
Total (series = multiply):a × (1 − (1 − b)³) × (1 − (1 − c)²)
Why not others:
- (a) ab³c² — requires ALL clients and ALL printers to work (no redundancy)
- (b) a(1 − b³)(1 − c²) — incorrect parallel formula; 1 − b³ ≠ 1 − (1−b)³
- (c) a(1−b)³(1−c)² — probability all clients AND all printers are DOWN
Key rule: Parallel availability of N identical devices with availability X = 1 − (1 − X)^N
AI-generated — may contain errors
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