ITPEC IP April 2020 Question 15
There is a system where two units of the same device with availability of 0.9 are connected in series. When one (1) more unit of the same device is added so that three (3) units of the device are connected in series in the system, which of the following is the appropriate description of the availability of the entire system, when the system with two units of the device in series is a reference point?
Explanation
It decreases by 10% — a third device in series multiplies the availability by another 0.9.
In series the system works only while every device works, so the availabilities multiply:
- •Two devices (the reference): 0.9 × 0.9 = 0.81
- •Three devices: 0.9 × 0.9 × 0.9 = 0.729
- •Relative to the reference: 0.729 ÷ 0.81 = 0.9, so the availability is 90% of what it was, a 10% decrease (from 0.81 down by 0.081)
Answer (c)
Why not others:
- •(a) — a device added in series can only lower the availability, because the system now also stops whenever the new device stops; only a device added in parallel raises it
- •(b) — the added device is down 10% of the time, a further way for the whole system to fail
- •(d) — as if each of the three devices took off 10%; the reference already contains two of them, and only the added one changes the result
Key rule: Every device added in series multiplies the system's availability by its own, so a series system is less available than its weakest device; in parallel it is the unavailabilities that multiply.
System Component · AI-written, may contain errors
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