問84
頻出度★★★
図のような構成の二つのシステムがある。システムXとYの稼働率を同じにするためには，装置Cの稼働率を幾らにすればよいか。ここで，システムYは並列に接続した装置Bと装置Cのどちらか一つでも稼働していれば正常に稼働しているものとし，装置Aの稼働率を0.8，装置Bの稼働率を0.6とする。 図の構成: - システムX: 装置A（1台のみ） - システムY: 装置Bと装置Cを並列に接続
平成26年度 秋期 ITパスポート試験 問84
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 84
Incorrectア
With 0.3, the availability of the parallel part is 1 − (1 − 0.6) × (1 − 0.3) = 1 − 0.4 × 0.7 = 1 − 0.28 = 0.72, which falls short of System X's 0.8. The availability of Device C is too low, so it does not compensate for Device B enough.
Incorrectイ
With 0.4, the availability of the parallel part is only 1 − (1 − 0.6) × (1 − 0.4) = 1 − 0.4 × 0.6 = 1 − 0.24 = 0.76, which again does not match 0.8. Note also that 0.4 equals 1 − 0.6, that is, the failure rate of Device B; this wrong answer is easy to pick if you confuse failure rate with availability.
- ウ
正解 The correct answer is ウ (0.5). First, express the availability of each system as a formula. As the figure shows, System X consists of only one Device A, so its availability is simply the availability of Device A, 0.8. System Y has Device B and Device C connected in parallel, and it works normally as long as at least one of them is operating. The availability of a parallel system is found as 1 − (the probability that both have failed), so if the availability of Device C is C, then the availability of System Y = 1 − (1 − 0.6) × (1 − C) = 1 − 0.4 × (1 − C). The condition for making the two systems equal is 1 − 0.4 × (1 − C) = 0.8, and solving it gives 0.4 × (1 − C) = 0.2, 1 − C = 0.5, therefore C = 0.5. Checking the result: when C = 0.5, the availability of System Y is 1 − 0.4 × 0.5 = 1 − 0.2 = 0.8, which matches System X. Because a parallel configuration stops only when both units stop, combining devices whose individual availabilities are fairly low, such as 0.6 and 0.5, can still raise the whole system to 0.8. That is the key point of this question.
Incorrectエ
0.6 is the same availability as Device B, and in that case the parallel part becomes 1 − 0.4 × 0.4 = 1 − 0.16 = 0.84, which exceeds System X's 0.8. The requirement is to make them equal, and 0.84 is not equal, so it is not appropriate. This is also the typical mistake of simply reusing the value 0.6 given for Device B in the question.
ここがポイント
- The availability of a parallel (redundant) connection is 1 − (1 − availability 1) × (1 − availability 2). It comes from the idea that the system stops only when both units fail at the same time. For a series connection it is availability 1 × availability 2, where the system operates only when every unit is operating.
- Distinguish availability from failure rate (= 1 − availability). In parallel calculations, keep to the order of first multiplying the failure rates of the devices and then subtracting from 1. Also remember this property: making a configuration parallel always raises the overall availability above that of the individual devices, while making it series always lowers it.
用語