cs-133 — The Late Bus Ledger
| Field | Value |
|---|---|
| id | cs-133 |
| difficulty | Foundational |
| topics | scheduling, communication, assumptions |
| lecture | lecture-15 |
| estimated time | 5-minute attempt + 10-minute discussion |
| prerequisites | lecture-05 (remainders and cycles), lecture-06 (traces) |
| objective | check a public claim ("a bus every ten minutes") against schedule arithmetic and report it honestly |
| solution | held by the instructor; discussed after the attempt |
Problem
The university shuttle poster says: "a bus every ten minutes." The operational facts:
- Departures from the depot are scheduled at :00, :10, :20, :30, :40, :50.
- One loop — depot → stop A → stop B → stop C → depot — takes exactly 25 minutes (legs: 4, 6, 7, 8).
- The fleet is two buses. A bus departs only when it is back at the depot; a bus that returns late departs late.
Your deliverables:
- The actual depot departure times for the first six scheduled slots.
- The gap between consecutive departures for those slots.
- The verdict: does "a bus every ten minutes" hold, and if not, exactly which gaps break it?
- The two-sentence notice to students that is honest — no cheerleading, no excuses.
If the problem is ambiguous anywhere, state the assumption you are making and carry on — a stated assumption is part of a good answer.
Examples
- Input: the schedule, loop legs, fleet size
- Output:
departures 0, 10, 25, 35, 50, 60— gaps10, 15, 10, 15, 10— verdict + notice
Solve for the worked instance first. Then ask what would change if one number, name, or condition were replaced by something stranger.
What to notice
- Departure slots are a wish; the fleet is the constraint. The actual departure is
max(scheduled, back-at-depot time). - Once one departure slips, every later gap inherits the slip — lateness propagates through the schedule like a carried error.
- The fix is arithmetic, not driving faster: a loop of 25 minutes with headway 10 needs a fleet of at least ⌈25/10⌉ = 3.
Discussion starter
After the solution is revealed, the class spends five minutes on this prompt: "Operations proposes shortening the loop by three minutes instead of buying a third bus. Does the poster claim become true? Show the departure ledger that decides it."
Constraints and assumptions
- Difficulty: Foundational (multi-step schedule reasoning).
- Topics: scheduling, communication, assumptions.
- Legs are exact (no traffic variance) — this is a timetable claim, checked arithmetically.
- Classroom-scale inputs unless the problem says otherwise.
- Time-box: about five minutes for a first attempt; partial answers with clear reasoning are worth more than silent struggling.
- Optional stretch (not required): describe an alternative approach and say when it beats the obvious one.
Hints and extension
- First nudge: write the return time of the bus that left at :00 before worrying about any other slot.
- If stuck after three minutes, write the input/output line you expect and trace the smallest example by hand.
- After the reveal, attempt the extension below before the next lecture.
- Extension challenge: with a three-bus fleet, find the loop length at which the ten-minute promise breaks even at the depot.