Hygiene Clinic Capacity and No-Show Overbooking
A dental hygiene clinic has two hygienists and operates from 8:00 a.m. to 5:00 p.m. with a one-hour lunch break, giving each hygienist 8.0 clinical hours per day. Recall appointments are scheduled in 40-minute slots, and new-patient appointments in 60-minute slots. The clinic's daily target is 16 completed recall visits and 4 completed new-patient visits. Historically, 12% of scheduled patients do not show, and no-shows are independent. The office currently schedules exactly the target number of each appointment type. The practice manager wants to know whether the schedule has enough capacity, how many recall appointments must be scheduled to expect 16 completed recalls, and whether the clinic can fit that many recalls while still scheduling 4 new patients.
1. Does the current schedule have enough capacity? Compute total chair minutes, slack, and scheduled utilization.
Read the worked answer
Daily chair capacity:
2 hygienists × 8.0 h × 60 min/h = 960 min.
Scheduled demand:
- 16 recalls × 40 min = 640 min.
- 4 new patients × 60 min = 240 min.
- Total = 880 min.
Slack = 960 − 880 = 80 min, which is 80 ÷ 960 = 8.3% of capacity.
Scheduled utilization = 880 ÷ 960 = 91.7%.
So the schedule fits on paper, but only 80 minutes of buffer remain. That buffer must absorb late starts, treatment overruns, emergency add-ons, and infection-control delays. With a 12% no-show rate, the expected completed minutes are 0.88 × 880 = 774.4 min, or 80.7% utilization, but empty slots from no-shows are not freely interchangeable with busy periods. The schedule is therefore feasible but tight, not robust.
2. How many recall appointments must be scheduled to expect 16 completed recalls? Can the clinic fit that many recalls while still scheduling 4 new patients in the existing hours?
Read the worked answer
Let S be the number of scheduled recalls. Expected completed recalls = 0.88S. Set:
0.88S ≥ 16
S ≥ 16 ÷ 0.88 = 18.18.
Since appointments are whole slots, the clinic must schedule 19 recalls to expect at least 16 completed.
Capacity required for 19 recalls plus 4 new patients:
19 × 40 min + 4 × 60 min = 760 + 240 = 1,000 min.
Available capacity is 960 min, so the schedule is 40 min over capacity. Scheduling 19 recalls and 4 new patients does not fit in the existing two-hygienist day.
With 4 new patients, the maximum number of recalls that fits is:
(960 − 240) ÷ 40 = 18 recalls.
But 18 recalls yield only 0.88 × 18 = 15.84 expected completed recalls, just under the target. Feasible adjustments include extending one hygienist by 40 minutes, opening a third chair for 40 minutes, reducing new patients to 3 that day (19 × 40 + 3 × 60 = 940 min, which fits), or shortening recall slots enough to absorb the extra visit.
3. If 4 new patients are scheduled, what is the probability that at least one no-shows? Interpret the result operationally.
Read the worked answer
The probability that a given new patient shows is 0.88. Assuming independence, the probability that all 4 show is:
0.88^4 = 0.88 × 0.88 × 0.88 × 0.88 ≈ 0.5997.
Therefore the probability that at least one of the 4 new patients no-shows is:
1 − 0.5997 = 0.4003 ≈ 40.0%.
Operationally, a 40% chance of at least one open new-patient slot is substantial. If the clinic does nothing, that time may sit idle because a 60-minute new-patient slot is hard to fill at short notice. A waitlist or a short-notice text alert can reduce idle time. The same logic explains why overbooking by one recall is tempting: with 18 scheduled recalls, the chance that all 18 show is 0.88^18 ≈ 0.10, so roughly a 90% chance exists that at least one recall slot opens. However, overbooking must respect the 960-minute capacity limit and cannot be applied blindly to both appointment types at once.