The Ten-Hour Roster
Sergeant Okafor has four officers for a single 10-hour shift, 07:00 to 17:00, all four working the whole shift. Minimum staffing is two officers on foot patrol at all times. Each officer takes one 30-minute meal break, one at a time. From 12:00 to 13:00 one officer must sit on the station phone; that officer is on duty but not on patrol, so patrol numbers still apply. A community event means three officers must be on patrol from 14:00 to 16:00. Two officers are booked for a 45-minute training session at 10:00, both at once. Okafor must publish a timetable that satisfies every constraint.
1. Is the timetable feasible? Identify the tightest windows and state how many officers can be off patrol at 10:15 and at 12:30.
Read the worked answer
Total demand against supply:
| Commitment | Officer-hours |
|---|---|
| Patrol 07:00–14:00, two officers | 14.0 |
| Patrol 14:00–16:00, three officers | 6.0 |
| Patrol 16:00–17:00, two officers | 2.0 |
| Phone duty 12:00–13:00 | 1.0 |
| Meals, 4 x 0.5 h | 2.0 |
| Training, 2 x 0.75 h | 1.5 |
| Total | 26.5 of 40 available |
So it is feasible in aggregate - the difficulty is timing, not hours.
Tightest windows:
- 10:00–10:45. Two officers in training leaves exactly two, which is exactly the patrol minimum. Spare capacity: zero. No meal, no phone duty, no appointment can sit here.
- 12:00–13:00. Two on patrol plus one on the phone needs three of the four, so at most one officer can be off - that is, at most one meal at a time.
- 14:00–16:00. Three required on patrol, so at most one officer off at a time.
Answer: at 10:15 zero officers can be off patrol; at 12:30 one officer can be off (eating a meal).
2. Build a timetable that satisfies every constraint, and justify the two placements that would break it if you moved them.
Read the worked answer
| Time | Off-patrol activity | Officers on patrol |
|---|---|---|
| 07:00–07:30 | O1 meal | 3 |
| 07:30–08:00 | O2 meal | 3 |
| 08:00–08:30 | O3 meal | 3 |
| 10:00–10:45 | O3 + O4 training | 2 (minimum, no spare) |
| 12:00–13:00 | O1 phone duty | 2 (minimum, no meal) |
| 14:00–15:00 | O4 meal 15:00–15:30 | 3 (minimum) |
| All other periods | none | 3 or 4 |
Placements that carry the plan:
- Training at 10:00 is immovable only because two officers go at once. Any meal overlapping 10:00–10:45 would drop patrol to one and breach the minimum. Meals must therefore be finished before 10:00 or taken after 10:45.
- Phone duty at 12:00 consumes the fourth officer, so the officer on the phone cannot also be the officer on a meal. Putting O1 on the phone 12:00–13:00 and finishing his meal at 07:30 keeps the two roles separate.
Check: patrol never drops below 2; the 14:00–16:00 requirement of 3 is met because only one officer (O4) is off and only for 30 minutes; all four officers get exactly one 30-minute meal; the 45-minute training block is intact.
3. One officer calls in sick. What now fails, and what are your two realistic options?
Read the worked answer
What fails: the 10:00 training block. With three officers, two in training leaves one on patrol against a minimum of two - an immediate breach. The rest survives: 12:00–13:00 needs exactly three (two patrol, one phone) and 14:00–16:00 needs exactly three on patrol, so both windows are met only if nobody takes a meal inside them. Meals must be squeezed into 07:00–10:00, 10:45–12:00 and 13:00–14:00, one at a time.
Option 1 - split the training. Run it as two separate 45-minute sessions, one officer at a time (e.g. 10:00 and 15:00), so only one officer is ever off and two remain on patrol. Cost: the trainer attends twice or delivers it twice.
Option 2 - backfill. Call in overtime or move the sergeant onto the foot patrol for 10:00–10:45, keeping the training block intact. Cost: budget and a supervisor off their own duties.
What not to do: run the training with two officers and leave one on patrol, or quietly drop the patrol minimum for 45 minutes. Both are defensible only on paper and indefensible after an incident at 10:20.