問67
頻出度★★★
あるコンピュータシステムを 1,200 時間稼働させたとき，正常稼働と故障修理の状況は表のとおりであった。このシステムの平均修復時間は何時間か。
| 経過時間 | 状態 |
|---|---|
| 0 ～ 250 | 正常稼働 |
| 250 ～ 265 | 故障修理 |
| 265 ～ 580 | 正常稼働 |
| 580 ～ 600 | 故障修理 |
| 600 ～ 990 | 正常稼働 |
| 990 ～ 1,000 | 故障修理 |
| 1,000 ～ 1,200 | 正常稼働 |
平成23年度 秋期 ITパスポート試験 問67
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 67
Incorrectア
10 hours is merely the repair time of the single repair from 990 to 1,000, that is, the shortest of the three repair times. It is the mistake made when you look at only one failure repair in the table and answer with it, or when you mistake the minimum value for the representative value. MTTR is the value obtained by averaging all three repair times, so you must not give one of the individual repair times (10, 15, 20) as the answer.
- イ
正解 The correct answer is イ. MTTR (Mean Time To Repair) is the average of the time required from the moment a failure occurs until the repair is completed and the system is restored, and it is obtained as MTTR = total repair time ÷ number of repairs. First, take from the table only the rows whose state is “failure repair”: 250 to 265 is 15 hours, 580 to 600 is 20 hours, and 990 to 1,000 is 10 hours, which makes 3 repairs. The total repair time is 15+20+10=45 hours and the number of repairs is 3, so MTTR = 45 ÷ 3 = 15 hours. There are two key points here. First, the value in each row is an elapsed time (a point in time), not a length, so you must subtract the start time from the end time to convert it into the length of the interval. Second, do not stop at the total of 45 hours; you must divide by the number of failures, 3, that is, take the average. Incidentally, MTBF (mean time between failures) can also be obtained from the same table. The normal operation times are 250, 315, 390 and 200, totaling 1,155 hours (which matches the whole 1,200 hours minus the 45 hours of repair), and dividing by the 3 failures gives MTBF = 385 hours. The availability is MTBF ÷ (MTBF+MTTR) = 385 ÷ 400 = 0.9625, that is 96.25%, the same as 1,155 ÷ 1,200. Therefore the mean time to repair is 15 hours, and イ is correct.
Incorrectウ
20 hours is the repair time of the single repair from 580 to 600, and it is the longest of the three repair times. The worst-case value (the longest recovery time) is a separate metric that has its own meaning when you examine availability, but what this question asks for is the mean time to repair. You must answer with the average obtained by dividing the total by the number of repairs, not with the maximum value, so this is wrong.
Incorrectエ
45 hours is the total repair time 15+20+10, not the average. It is the typical wrong answer chosen when you forget the step of dividing by the number of repairs, 3. The total of 45 hours is the value representing the time during which the system was stopped (the downtime) out of the whole 1,200 hours, and it is used in the form (1,200−45) ÷ 1,200 when calculating availability, but as the mean time to repair it must be divided further by the number of failures.
ここがポイント
- How to find MTTR (mean time to repair) — MTTR = total repair time ÷ number of failures (repairs). In this question it is (15+20+10) ÷ 3 = 15 hours. Note that what the table lists are elapsed times (points in time), so convert each interval into its length with end time − start time before adding them up.
- The relationship among MTBF, MTTR and availability — MTBF (mean time between failures) = total normal operation time ÷ number of failures = 1,155 ÷ 3 = 385 hours. Availability = MTBF ÷ (MTBF + MTTR) = 385 ÷ 400 = 0.9625 (96.25%). The longer the MTBF, the higher the reliability; the shorter the MTTR, the better the maintainability.
用語