同じ装置が複数接続されているシステム構成 a~c について,稼働率が高い順に並べたものはどれか。ここで,—□— は装置を表し,並列に接続されている場合はいずれか一つの装置が動作していればよく,直列に接続されている場合は全ての装置が動作していなければならない。
平成28年度 春期 ITパスポート試験 問60
This question's explanation and your answers so far.
問 60
Incorrectア
“a, b, c” lists them in increasing order of availability (a<b<c). The question asks for decreasing order, so this order is reversed and wrong.
Incorrectイ
“b, a, c” puts b first, but the highest availability belongs to the fully parallel c (e.g., 0.999). Since b (0.891) is lower than c, it cannot come first, so this is wrong.
Incorrectウ
“c, a, b” correctly puts the highest c first, but places a (two in series = the lowest) second. In reality b (0.891) is higher than a (0.81), so the order of a and b is reversed, making it wrong.
- エ
正解 Let the availability of a single device be k. In a series connection the whole system operates only when every device is working, so availability is the product of the values and drops; in a parallel connection it is enough that any one device works, so availability rises as 1−(probability that all stop). Configuration a (two in series) is k², configuration b (one in series plus two in parallel) is k×{1−(1−k)²}, and configuration c (three in parallel) is 1−(1−k)³. For example, with k=0.9 we get a=0.81, b=0.9×0.99=0.891, c=1−0.001=0.999, so the availability order is c>b>a. Therefore エ, which lists them from highest to lowest as “c, b, a”, is correct. This ordering always holds regardless of the value of k (more parallel means higher, more series means lower, so the fully parallel c is highest and the fully series a is lowest).
- The availability of a series connection is the product of each device's availability (the system works only when all devices work), which lowers it; the availability of a parallel connection is 1−(1−k)ⁿ (any one device working is enough), which raises it.
- For the same number of devices, a more parallel configuration has higher availability and a more series configuration has lower. Thus the fully parallel c is highest, the fully series a is lowest, and b is in between, giving c>b>a.