問64
頻出度★★★
三つの装置 A，B，C の稼働率はそれぞれ 0.90，0.95，0.95 である。これらを組み合わせた図のシステムのうち，最も稼働率が高いものはどれか。ここで，並列に接続されている部分はどちらかの装置が稼働していればよく，直列に接続されている部分はすべての装置が稼働していなければならない。
平成22年度 秋期 ITパスポート試験 問64
テクノロジ系システム構成要素 › システムの評価指標
This question's explanation and your answers so far.
解説
問 64
Incorrectア
This is A and B in series, so the availability is 0.90×0.95=0.855. Because a series connection takes the product, the value is always smaller than the availability of the lowest device (A, at 0.90). It has no redundancy at all and it includes the low-availability device A, so it is the second lowest of the four.
Incorrectイ
This connects all three devices A, B, and C in series, giving an availability of 0.90×0.95×0.95=0.81225. In a series connection, adding devices only makes the product smaller, and if any single one of them stops, the whole system stops. It has the lowest availability of the four, which is the exact opposite of what this question, asking for the highest, requires.
Incorrectウ
The parallel block of B and C (0.9975) is the same as in エ, but A is connected in series ahead of it, so the result is 0.90×0.9975=0.89775. The gain obtained by the parallel connection is cancelled out by the series device A, so it does not reach the 0.9975 of エ. If you overlook the fact that adding even one part in series always lowers the overall availability, it is easy to pick this option.
- エ
正解 The correct answer is エ. Availability is calculated by multiplying the availabilities of the devices for a series connection, and, for a parallel connection, by subtracting the probability that all of the devices stop at the same time from 1. エ is a configuration that simply puts B and C in parallel, and it does not include A. The probability that both stop at the same time is (1−0.95)×(1−0.95)=0.05×0.05=0.0025, so the availability of the system is 1−0.0025=0.9975. The other three all include device A, whose availability of 0.90 is the lowest, in series, so the availability of the whole system inevitably falls to 0.90 or below. Working the values out gives ア=0.855, イ=0.81225, and ウ=0.89775, so エ, which does not use A at all and makes only the highly reliable B and C redundant, is the highest at 0.9975.
ここがポイント
- The basic formulas for availability. A series connection works only when every device is operating, so you multiply the availabilities of the devices (for example, 0.90×0.95=0.855); a parallel connection works as long as either device is operating, so you compute 1−(1−availability of A)×(1−availability of B) (for example, 1−0.05×0.05=0.9975).
- A series connection always lowers availability, while a parallel connection always raises it. Therefore, when a question asks which configuration has the highest availability, you only need to look for the one that has the fewest parts in series and does not include the device with low availability (here, A at 0.90).
用語