Triage Line Staffing: Erlang C Versus a Process Change
A regional health service runs a telephone triage line. During the evening peak, calls arrive as a Poisson process at 24 per hour and are handled by nurses whose service times are well approximated by an exponential distribution with a mean of 4 minutes. The service's standard is that no more than 20% of callers should wait longer than 10 minutes before speaking to a nurse. Currently two nurses are on shift. Each nurse costs €40 per hour including on-costs, and the service must staff the six-hour evening peak every day. Management has two proposals: add a third nurse, or buy triage-decision software that would cut the mean service time to 3 minutes while two nurses remain on shift, at a licence and training cost of €25 per hour. A board member objects that the models are unrealistic because 'real call centres are not random', and asks whether the numbers can be trusted.
1. Compute the probability that a caller waits more than 10 minutes with two nurses, and compare it with the service standard.
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Parameters. λ = 24/h = 0.4 calls/min; μ = 1/4 = 0.25 calls/min per nurse; c = 2; offered load a = λ/μ = 1.6; utilisation ρ = a/c = 0.8 < 1, so the system is stable.
Erlang C (probability of waiting at all). With a = 1.6, c = 2: C = [a^c / (c!(1 − ρ))] / [Σ_{k=0}^{c−1} a^k/k! + a^c/(c!(1 − ρ))] = [2.56 / (2 × 0.2)] / [1 + 1.6 + 6.4] = 6.4 / 9.0 = 0.711.
Waiting-time tail. For M/M/c, P(W > t) = C · exp(−(cμ − λ)t). With cμ − λ = 0.5 − 0.4 = 0.1 per minute and t = 10: P(W > 10) = 0.711 × e^{−1} = 0.711 × 0.368 ≈ 26.2%.
Against standard. The 20% target is breached: about one caller in four waits more than 10 minutes. Mean queueing delay is C/(cμ − λ) = 0.711/0.1 ≈ 7.1 minutes before service begins, on top of the 4-minute call. Two nurses do not meet the standard, even though utilisation of 80% sounds reasonable - the queueing curve is convex, and the last 20% of utilisation is where the waiting lives.
2. Evaluate the third-nurse proposal and the software proposal against the 20% standard, and recommend the cheaper option that satisfies it.
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Third nurse (c = 3). ρ = 1.6/3 = 0.533; a³/(3!(1 − ρ)) = 4.096/(6 × 0.4667) = 1.463; denominator = 1 + 1.6 + 1.28 + 1.463 = 5.343, so C = 1.463/5.343 = 0.274. cμ − λ = 0.75 − 0.4 = 0.35, so P(W > 10) = 0.274 × e^{−3.5} = 0.274 × 0.0302 ≈ 0.8%, and mean queueing delay is 0.274/0.35 ≈ 0.8 min. Comfortably inside the standard, arguably over-provisioned.
Software with two nurses (μ = 1/3). a = 1.2, ρ = 0.6, c = 2: a²/(2!(1 − ρ)) = 1.44/0.8 = 1.8; denominator = 1 + 1.2 + 1.8 = 4.0; C = 0.45. cμ − λ = 0.6667 − 0.4 = 0.2667, so P(W > 10) = 0.45 × e^{−2.667} = 0.45 × 0.0695 ≈ 3.1%, with mean queueing delay of 0.45/0.2667 ≈ 1.7 min. Also inside the standard.
Cost per peak hour. Third nurse: 3 × €40 = €120. Software: 2 × €40 + €25 = €105, a saving of €15 per hour, or €90 per day across the six-hour peak (≈ €32,850 per year).
Recommendation. Buy the software: it meets the standard with a 3.1% tail versus 0.8%, and reduces cost. The third nurse is the fallback if the 3-minute mean cannot be achieved in practice - and the 0.8% tail shows how much slack it buys, which matters if the standard is later tightened.
3. The board member says the queueing model cannot be trusted because arrivals and service times are not really random. Assess how much this undermines the recommendation, and state what you would monitor or test.
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What the objection gets right. The numbers are only as good as the M/M/c assumptions: Poisson arrivals, exponential service, a stationary arrival rate over the peak, and no abandonment or reneging. Real triage lines have a build-up at the start of the evening, calls that are not independent (a GP referral generating several calls), and service-time distributions that are typically less variable than exponential. The 4-minute mean is an estimate, and because the tail is steep in ρ, a 4.5-minute true mean changes the answer materially.
What it does not overturn. The recommendation is a comparison of two configurations, not a prediction of an exact percentage. Erlang C is known to be conservative when service times are less variable than exponential - lower variance means shorter queues - so 26.2% and 3.1% are more likely upper bounds than underestimates. The ranking (two nurses fail, software passes, third nurse passes with more margin) is robust to plausible departures from the assumptions, which is the decision-relevant property.
What to do. Validate the inputs: fit the arrival rate by 30-minute interval rather than pooling the peak, and compare the empirical service-time variance with the exponential benchmark. Test the design against a 4.5-minute mean and a 10% higher arrival rate as a sensitivity check. Then monitor live abandonment rate, the 90th-percentile wait and occupancy weekly - abandonment is the signal that the model's stability assumption has broken, and it is invisible in the Erlang C output.