問52
頻出度★★★
図1のように稼働率0.9の装置Aを2台並列に接続し，稼働率0.8の装置Bをその後に直列に接続したシステムがある。このシステムを図2のように装置Aを1台にした場合，システムの稼働率は図1に比べて幾ら低下するか。ここで，図1の装置Aはどちらか一方が稼働していれば正常稼働とみなす。 なお，稼働率は小数第3位を四捨五入した値とする。 図1 装置A・装置A（2台並列）→ 装置B（直列） 図2 装置A（1台）→ 装置B（直列）
平成25年度 秋期 ITパスポート試験 問52
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 52
- ア
正解 In a parallel (redundant) configuration, "it works normally as long as either one is running," so you find its availability by subtracting from 1 the probability that both are down at the same time. The failure probability of device A is 1−0.9＝0.1, so the availability of the parallel section in Figure 1 is 1−0.1×0.1＝0.99. Multiplying this by device B (0.8), which is in series, the system availability of Figure 1 is 0.99×0.8＝0.792 → rounding off the third decimal place gives 0.79. Figure 2, on the other hand, is one device A in series with device B, so it is 0.9×0.8＝0.72. The decrease is therefore 0.79−0.72＝0.07, and the correct answer is ア (even if you calculate before rounding, 0.792−0.72＝0.072, which likewise amounts to 0.07).
Incorrectイ
0.09 is the difference of the device A part alone, 0.99−0.90＝0.09. What is being asked is the decrease in the availability of the whole system including device B, so it is necessary to multiply by device B (0.8), which is in series. 0.09×0.8＝0.072≒0.07 is the correct decrease, and 0.09 is the value obtained by forgetting to multiply by device B.
Incorrectウ
0.10 is the value obtained by estimating the effect of using two units of device A as "1.00−0.90" — that is, the difference you get if you mistakenly believe that putting them in parallel makes the availability 1.00 (never goes down at all). The actual parallel section is 0.99 and does not become 1.00. Furthermore, the step of multiplying by device B's 0.8 is also missing, so it is wrong in two ways.
Incorrectエ
0.45 is far too large, off by an order of magnitude. The upper bound of the decrease is 0.072, obtained by multiplying the difference of the device A part, 0.99−0.90＝0.09, by device B's 0.8, so it cannot even exceed 0.1 (it is the kind of value you get by mistakenly thinking that "removing the parallel redundancy halves the availability" and computing 0.9×0.5＝0.45, which has nothing to do with series/parallel availability calculation).
ここがポイント
- The availability of a series connection is the product of the availabilities of the individual devices, and the availability of a parallel (redundant) connection is 1−(1−availability A)×(1−availability B). Remember it as "normal as long as either one is running" ＝ "subtract from 1 the probability that both fail at the same time."
- For a composite system, first combine the parallel block into a single device, and then take the product for the series connection. When the question asks for a difference, as in "how much does it decrease," obtain the overall availability of each of the two configurations and then subtract (do not answer with the difference of only a part).
用語