This question's explanation and your answers so far.
自分のデスクにある PC と共有スペースにあるプリンタの起動を1人で行う。PC とプリンタの起動は図の条件で行い，それぞれの作業・処理は逐次実行する必要がある。自動処理の間は，移動やもう片方の作業を並行して行うことができる。自分のデスクにいる状態で PC の起動を開始し，移動してプリンタを起動した上で自分のデスクに戻り，PC の起動を終了するまでに必要な時間は，最短で何秒か。 〔条件〕 移動：片道60秒 自分のデスクにある PC ⇔ 共有スペースにあるプリンタ PCの起動の流れ
| 作業・処理内容 | 所要時間 | 処理種別 |
|---|---|---|
| （起動開始） | ||
| A 電源を入れる | 3秒 | 手作業 |
| B ログイン画面起動処理 | 150秒 | 自動処理 |
| C ログイン操作 | 10秒 | 手作業 |
| D ログイン後のアプリケーション起動処理 | 60秒 | 自動処理 |
| （起動終了） |
プリンタの起動の流れ
| 作業・処理内容 | 所要時間 | 処理種別 |
|---|---|---|
| （起動開始） | ||
| E 電源を入れる | 3秒 | 手作業 |
| F 起動処理 | 60秒 | 自動処理 |
| （起動終了） |
正解 The PC startup must execute A→B→C→D sequentially, and this chain determines the overall minimum time. A Turn on the power (3 sec) + B Login screen startup process (150 sec) + C Login operation (10 sec) + D Application startup process (60 sec) = 223 sec. Concretely: at the desk, do A (0→3 sec), which starts automatic process B (3→153 sec), during which you are free to move around. While B is running, move to the printer (60 sec) → do E Turn on the power (3 sec); since F is an automatic process, you return to the desk (60 sec) without waiting. The outbound 60 + E 3 + return 60 = 123 sec fits inside B's 150-second window, so you can get back to the desk before the 153-second mark (at second 126). After B finishes (153 sec), do C (153→163 sec), then automatic process D (163→223 sec) ends the startup. All of the printer-side work is hidden inside B's automatic-processing time and is not added to the total, so the minimum is 223 sec = ア.
If you wait for the printer's automatic process F (60 sec) to finish before returning to the desk, your return arrival slips past the completion of B (153 sec) to 186 sec, so C (186→196 sec) → D (196→256 sec) are all pushed back, giving 256 sec. Since F is an automatic process that needs no human presence, you can return right after doing E — this value overlooks that point.
This does not run the outbound movement to the printer (60 sec) and E Turn on the power (3 sec) in parallel with B's automatic process; instead it adds 63 sec on top of the correct 223 sec (223+63=286). It fails to use the condition that “during an automatic process you can move or carry out the other task in parallel”, so it never hides the printer work in B's idle time.
This runs nothing in parallel — neither the movement nor the printer work — and simply sums every step sequentially. PC side 223 sec + printer side (move 60 + E 3 + F 60 + move 60 = 183 sec) = 406 sec. It is the most naive mistake, taking no account of parallel execution during automatic processes.