Emergency Department Waiting-Time Target
A regional hospital's emergency department sees an average of 9 arrivals per hour, modeled as Poisson. Each of three physicians treats on average 4 patients per hour, with exponential service times. The waiting room has 12 seats. Administration receives complaints about long waits and sets a target that no more than 15% of patients wait more than 20 minutes. A consultant proposes either hiring a fourth physician with the same service rate or creating a fast-track nurse who would handle 30% of arrivals with a mean service time of 7.5 minutes. The medical director wants to know current performance and whether either proposal meets the target. Arrivals are not uniform: 40% of daily arrivals occur between 6 and 10 pm. The consultant used steady-state formulas for the whole day.
1. Compute current queue performance: the average wait in queue and the probability a patient waits more than 20 minutes. Does the department meet the target?
Read the worked answer
Use M/M/c with λ=9/hour, μ=4/hour, c=3, r=λ/μ=2.25, and ρ=9/(3×4)=0.75.
P0 = [1 + 2.25 + 2.25²/2! + (2.25³/3!)/(1−0.75)]⁻¹ = [1 + 2.25 + 2.53125 + 7.59375]⁻¹ = 1/13.375 = 0.0748.
Erlang C probability of waiting = (2.25³/3!)/(1−0.75) × P0 = 7.59375 × 0.0748 = 0.568.
Lq = C × ρ/(1−ρ) = 0.568 × 0.75/0.25 = 1.70 patients. Wq = Lq/λ = 1.70/9 = 0.189 hour = 11.35 minutes.
For the tail, P(Wq > 20 min) = C × exp[−(cμ−λ)t] = 0.568 × exp[−(12−9)(20/60)] = 0.568 × exp(−1) = 0.209.
About 20.9% of patients wait more than 20 minutes, which exceeds the 15% target. The average wait of 11.35 minutes looks acceptable, but the tail violates the standard; averages alone would hide the problem.
2. Evaluate the proposal to add a fourth physician. Does it meet the target, and what else should be checked before deciding?
Read the worked answer
With c=4, λ=9, μ=4, r=2.25, ρ=9/16=0.5625.
P0 = [1 + 2.25 + 2.53125 + 1.89844 + (2.25⁴/4!)/(1−0.5625)]⁻¹ = [7.67969 + 2.44085]⁻¹ = 0.0988.
C = 2.44085 × 0.0988 = 0.241. Lq = C × ρ/(1−ρ) = 0.241 × 0.5625/0.4375 = 0.310. Wq = 0.310/9 = 0.0345 hour = 2.07 minutes.
P(Wq > 20 min) = 0.241 × exp[−(16−9)(1/3)] = 0.241 × exp(−2.333) = 0.0234.
Only about 2.3% exceed 20 minutes, so the fourth physician comfortably meets the target. Before approving, check the cost of an additional physician, whether physical space and support staff exist, whether the 12-seat waiting room is adequate, and whether the pooled M/M/4 model is realistic. Also check the evening peak: 40% of 216 daily arrivals occur in four hours, about 21.6 arrivals/hour, far above the daily average of 9. The steady-state model likely understates evening waits, so the fourth physician may still be insufficient at peak unless staffing is time-varying.
3. Evaluate the fast-track nurse proposal and discuss whether the M/M/c model is appropriate given the arrival pattern.
Read the worked answer
Split the stream: 30% of 9 = 2.7/hour to the nurse, 6.3/hour to three physicians. This assumes the split is random and the nurse is not simply taking low-acuity patients, which would change service times.
Physicians: λ=6.3, μ=4, c=3, r=1.575, ρ=0.525. P0 = [1 + 1.575 + 1.575²/2 + (1.575³/6)/(1−0.525)]⁻¹ = [1 + 1.575 + 1.240 + 1.371]⁻¹ = 0.1928. C = 1.371 × 0.1928 = 0.264. P(Wq > 20 min) = 0.264 × exp[−(12−6.3)(1/3)] = 0.264 × exp(−1.9) = 0.0395.
Nurse: M/M/1 with λ=2.7, μ=8, ρ=0.3375. P(Wq > 20 min) = ρ × exp[−(μ−λ)t] = 0.3375 × exp(−5.3/3) = 0.0577.
Overall = 0.7×0.0395 + 0.3×0.0577 = 0.045, about 4.5%, so the proposal meets the target under these assumptions.
However, the model is questionable. Arrivals are time-varying; the evening peak rate is about 21.6/hour, so a steady-state daily average is misleading. Fast-track creates a priority queue, not two independent Poisson streams; if nurses see lower-acuity patients, the physicians' service mix and the nurse's utilization change. The analysis should use a time-varying or simulation model of the evening peak, include abandonment and boarding delays, and test sensitivity to the 30% split and to nurse service time.