問39
頻出度★★★
プログラムの開発作業で担当者A～Dの4人の工程ごとの生産性が表のとおりのとき，4人同時に見積りステップ数が12kステップのプログラム開発を開始した場合に，最初に開発を完了するのはだれか。 単位 kステップ／月
| 担当者 | 設計 | プログラミング | テスト |
|---|---|---|---|
| A | 3 | 3 | 6 |
| B | 4 | 4 | 4 |
| C | 6 | 4 | 2 |
| D | 3 | 4 | 5 |
平成21年度 秋期 ITパスポート試験 問39
マネジメント系システム開発技術 › システム開発技術
This question's explanation and your answers so far.
解説
問 39
Incorrectア
A takes 12÷3＋12÷3＋12÷6＝4＋4＋2＝10 months. A's testing speed of 6k steps per month is the fastest of the four, but both design and programming are slow at 3k steps per month, so the total does not beat B's 9 months.
- イ
正解 The numbers in the table are productivity values, that is, how many k steps can be handled per month, so the time each process takes is 12k steps divided by the productivity for that process. All four people carry out design, then programming, then testing entirely on their own, so the time until development is finished is the sum of the months needed for the three processes. For B: design 12÷4＝3 months, programming 12÷4＝3 months, testing 12÷4＝3 months, for a total of 9 months. Of the other three, A takes 10 months, C takes 11 months and D takes 9.4 months, so the one who finishes first is B, and the correct answer is “イ”. B has no single outstandingly fast process, but has no weak point either, working at 4k steps per month in every process, and so has the shortest total time.
Incorrectウ
C takes 12÷6＋12÷4＋12÷2＝2＋3＋6＝11 months, the slowest of the four. Even though C's design is the fastest at 6k steps per month, testing is extremely slow at 2k steps per month, and that stage pushes the overall period up. Be careful: one fast process is not the same thing as finishing early overall.
Incorrectエ
D takes 12÷3＋12÷4＋12÷5＝4＋3＋2.4＝9.4 months. That is 0.4 months slower than B's 9 months, a narrow second place. D loses time with a slow design process at 3k steps per month, and testing at 5k steps per month is not enough to make up for it.
ここがポイント
- When productivity is given in k steps per month, the time required equals the amount of work divided by the productivity. When the processes are carried out one after another, the overall development time is the sum of the times for each process.
- Compare by the total time for all processes combined, not by the speed of any single process. Even if just one process is exceptionally fast, slow processes elsewhere still leave the total at a disadvantage.
用語