# RE8 | anthropic/claude-opus-5 | run 2

## Prompt

Minimize total time to finish these tasks, one at a time, respecting dependencies: A (2h, no deps); B (3h, needs A); C (1h, no deps); D (2h, needs B and C). Give a valid order and the total time.

## Response

Because only one task runs at a time and no task must wait on an unfinished dependency, the total time is simply the sum of all durations: **8 hours**. Any topological order achieves this optimum.

## Recommended Order

| Step | Task | Duration | Starts | Finishes |
|---|---|---|---|---|
| 1 | A | 2h | 0:00 | 2:00 |
| 2 | B (needs A) | 3h | 2:00 | 5:00 |
| 3 | C | 1h | 5:00 | 6:00 |
| 4 | D (needs B, C) | 2h | 6:00 | 8:00 |

**Total time: 8 hours**

## Why 8h Is Optimal

- A single worker can never finish faster than the sum of work: \( 2 + 3 + 1 + 2 = 8 \) hours.
- Dependencies here never force idle time, so no schedule is worse than 8h either — every valid order ties.
- Other valid orders: C → A → B → D, and A → C → B → D. All finish at 8:00.

*Note:* If tasks could run in parallel with unlimited workers, the answer would differ — the critical path A → B → D takes 7 hours, with C fitting anywhere before D.

Would you like me to work out the parallel-execution schedule with a limited number of workers?
