Overtime Split on a Two-Shift Detail
A unit has 7 officers available for a detail that runs 10 hours. Department policy pays the first 8 hours at $32/hour and any hours beyond 8 at time-and-a-half ($48/hour). The detail needs exactly 2 officers present at all times for the full 10 hours, and no officer may work more than 8 hours total. Officer A and Officer B each agree to work the full 10 hours, but the supervisor points out that would violate the 8-hour cap. Instead, the supervisor must build a rotation using the 7 officers so that coverage is continuous and no one exceeds 8 hours, while keeping total labor cost as low as possible. All 7 officers are equally available and interchangeable for cost purposes.
1. What is the minimum total labor cost to staff the detail, and how should hours be allocated?
Read the worked answer
Coverage requirement: 2 officers × 10 hours = 20 officer-hours must be worked.
Cost logic: Straight time ($32) is cheaper than overtime ($48), so we want to avoid any hours beyond 8 per officer. Since the cap is 8 hours and every officer can work up to 8, we can cover all 20 hours with straight time alone.
Allocation: e.g., Officer A works hours 0–8 (8 hrs), Officer B works hours 2–10 (8 hrs) to cover the overlap, and a third officer fills the remaining 4 hours. More cleanly: two officers cover the first 8 hours (16 officer-hours) and a rotation covers the last 2 hours with 2 officers × 2 hours = 4 officer-hours, none exceeding 8.
Minimum cost: 20 officer-hours × $32 = $640, with zero overtime. Any plan using overtime would cost more ($48 > $32), so $640 is the floor.
2. Suppose only 3 officers were available instead of 7. Could the detail still be staffed without anyone exceeding 8 hours? Show the arithmetic.
Read the worked answer
Total capacity check: 3 officers × 8 hours max = 24 officer-hours of capacity. The detail needs 20 officer-hours. Since 24 ≥ 20, it is theoretically possible by raw capacity.
But check the per-instant constraint: At any moment 2 officers must be present. With only 3 officers, at most 1 can be off-duty at a time. Over 10 hours each officer can work at most 8, so each must rest at least 2 hours. Total rest needed = 3 × 2 = 6 officer-hours of rest, but the schedule only allows 1 officer resting at a time for 10 hours = 10 rest-slots - that part is fine.
Construct: A works 0–8, B works 0–8, C works 2–10. At hour 0–2: A,B (ok). Hours 2–8: A,B,C all present but we only need 2 - we can instead stagger: A 0–8, B 0–8, C 8–10 covers only 2 hours. The gap is hours 8–10 needing 2 officers, but A and B are capped at 8. So C alone can cover only 1 slot. Not feasible - the last 2 hours need 2 fresh officers but only C remains under cap. Conclusion: 3 officers cannot do it; at least 4 are needed (two to cover the tail).
3. If the detail were extended to 12 hours with the same 7 officers and same rules, what is the minimum cost?
Read the worked answer
Officer-hours needed: 2 × 12 = 24 officer-hours.
Capacity without overtime: 7 officers × 8 hours = 56 hours of straight-time capacity, far more than 24. So we can still avoid overtime entirely by rotating fresh officers.
Cost: 24 officer-hours × $32 = $768, zero overtime.
The key insight: overtime only becomes forced when required officer-hours exceed total straight-time capacity (56 here) or when the per-instant staffing can't be met by fresh officers - neither applies. So the answer scales linearly at straight time.