問50
頻出度★★★
120 k ステップのソフトウェアを開発した。開発の各工程における生産性の実績が表のとおりであるとき，開発全体の工数は何人月か。ここで，生産性は1人月当たりの k ステップとする。 単位 k ステップ／人月
| 工程 | 生産性 |
|---|---|
| 設計 | 6.0 |
| 製造 | 4.0 |
令和4年度 ITパスポート試験 問50
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 50
Incorrectア
10 is simply the two productivities 6.0 and 4.0 added together, which is not an effort calculation. Adding productivities (rates) is meaningless; you must convert to effort by dividing scale by productivity.
Incorrectイ
12 is the error of treating 10 (the sum of the productivities) as the overall productivity and computing 120÷(6.0+4.0)=12. The two processes each handle the full 120 k steps, so you cannot simply add the productivities into a single rate.
Incorrectウ
24 is the error of using the average productivity (6.0+4.0)÷2=5.0 and computing 120÷5.0=24. You should compute the effort for each process and sum them; a single calculation with the average productivity comes out too small.
- エ
正解 The effort of each process is found by dividing the development scale by that process's productivity. Both processes work on the whole 120 k steps, so the design effort = 120÷6.0 = 20 person-months and the manufacturing effort = 120÷4.0 = 30 person-months. The total effort for the whole development is their sum, 20+30 = 50 person-months. Therefore 50 in choice “エ” is correct.
ここがポイント
- Effort (person-months) = development scale ÷ productivity. When productivity is given as “k steps per person-month”, dividing the scale by the productivity gives the effort of each process.
- For multiple processes, compute the effort of each against the whole scale, then sum at the end. You must not add or average the productivities themselves and calculate in one shot.
用語