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Choose a route or begin at Chapter 1. Teaching chapters stay open; the chapter index expands by section.
Define the waiting population
Chapters 1–2. Name the two ways to leave the waiting area
Derive the competing events
Chapters 3–5. Define an intentionally simple competing-times model
Read the performance trade-off
Chapters 6–8. Compare faster reported waits with worse access
Check records and transfer
Chapters 9–10. Handle unfinished records and repeated contacts
Questions and further models
Chapters 11–12. Questions readers ask
Full chapter index · Worked practice · Queueing Theory subject root
A service reports that answered callers now wait one minute instead of two. That sounds encouraging. But what if the share reaching service has fallen from one half to one quarter?
An answered-only average describes a selected group. This guide separates time until service, time until abandonment and time spent in the waiting area. It builds an exact competing-times model, then changes the assumptions to show its limits. All numbers are original hypothetical teaching cases, not performance measurements.
Chapter index
Define the waiting population · Chapters 1–2
1. Name the two ways to leave the waiting area
2. Keep both outcome groups in the ledger
Derive the competing events · Chapters 3–5
3. Define an intentionally simple competing-times model
4. Derive the observed waiting distribution
5. Calculate the answered-only mean
Read the performance trade-off · Chapters 6–8
6. Compare faster reported waits with worse access
7. Use a service target with the correct denominator
8. Connect waiting-area occupancy to every exit
Check records and transfer · Chapters 9–10
9. Handle unfinished records and repeated contacts
10. Practise with a changed rate pair
Questions and further models · Chapters 11–12
11. Questions readers ask
12. Use the evidence to choose the next improvement
CHAPTER 1 OF 12 · Define the waiting population
1. Name the two ways to leave the waiting area
Start the clock when an admitted caller enters the waiting area. Stop it when service begins or the caller abandons. Call that elapsed time Q. Time spent receiving service is outside this particular boundary.
Let S mean eventual service and A mean abandonment. For now every admitted caller eventually has exactly one of those outcomes. Immediate rejection, callbacks and unresolved records will need separate treatment later.
An answered-only mean is E[Q | S]: average waiting among the callers who reach service. The all-entrant mean is E[Q]. They refer to different populations even when both are expressed in minutes. Neither is the mean time to problem resolution.
Before comparing two reports, write the entry event, exit event and denominator beside each number. This small step prevents a low waiting statistic from silently becoming a claim about everyone’s access.
CHAPTER 2 OF 12 · Define the waiting population
2. Keep both outcome groups in the ledger
Partition a completed cohort into those served and those abandoning. If p is the served fraction, the average across the whole cohort is p times the served-group average plus (1 − p) times the abandonment-group average.
The corresponding expectation identity is E[Q] = P(S)E[Q | S] + P(A)E[Q | A], when the means exist. It follows by adding the waiting minutes in the two disjoint groups, then dividing by the total number of entrants.
For 100 fictional entrants, suppose 80 are served after an average two minutes and 20 abandon after an average twelve minutes. Total waiting is 80×2 + 20×12 = 400 caller-minutes, giving four minutes per entrant.
Reporting only the two-minute answered average omits 240 of those 400 waiting minutes. The calculation is valid for its stated subgroup; the problem arises when that subgroup is presented as the whole service experience.
CHAPTER 3 OF 12 · Derive the competing events
3. Define an intentionally simple competing-times model
Give each entrant a potential service-start time T and a patience time P, both measured from admission. If T is smaller, service starts. If P is smaller, abandonment occurs. Thus Q = min(T,P).
Assume T has an exponential distribution with rate ν per minute and P has an independent exponential distribution with rate θ per minute. Both rates are positive. Ties have probability zero in this continuous model.
Here T is a supplied service opportunity, not a waiting time derived from a staffed queue. We deliberately keep its distribution fixed when changing patience. This is an exact selection model, not an Erlang staffing calculation.
That distinction matters. In a real queue, one caller leaving may let another begin earlier, so potential start times and patience can interact through the service system. Our model isolates the reporting effect before taking on that feedback.
CHAPTER 4 OF 12 · Derive the competing events
4. Derive the observed waiting distribution
For an entrant still to be waiting after t minutes, both potential events must lie beyond t. Independence gives P(Q > t) = P(T > t)P(P > t) = exp(−νt)exp(−θt).
Therefore Q is exponential with rate ν + θ, and E[Q] = 1/(ν + θ). The rates add because either event can end the wait. Adding the mean times would answer a different question.
The probability density of service occurring at time t is ν exp(−(ν + θ)t). Integrating from zero to infinity gives P(S) = ν/(ν + θ). Similarly P(A) = θ/(ν + θ), and the two probabilities sum to one.
The rate ν is not an arrival rate and θ is not a fraction. Both have inverse-time units. Checking these units catches the mistaken calculation ν/(1 + θ), whose denominator combines incompatible quantities.
Divide the service-event density by the probability of eventual service. The conditional density of Q among served entrants becomes (ν + θ)exp(−(ν + θ)t).
Hence E[Q | S] = 1/(ν + θ). The abandonment group has the same conditional mean in this special independent exponential model. Substituting them into the cohort identity recovers E[Q].
The equality of the two conditional means is a consequence of these distributions. It is not a general rule about callers. A fixed patience deadline, dependence between patience and required service, or a changing service opportunity can break it.
An answered caller has beaten the patience clock. Increasing the abandonment rate removes more entrants before their later service opportunities. The observed answered sample can therefore become faster even though the underlying T distribution has not improved.
CHAPTER 6 OF 12 · Read the performance trade-off
6. Compare faster reported waits with worse access
Set ν = 0.25 per minute, so the model’s potential service opportunity has mean four minutes. First let θ = 0.25. Half the entrants reach service, and both the answered-only and all-entrant waiting means are two minutes.
Now increase θ to 0.75, leaving ν fixed. The served fraction falls to one quarter and the observed mean wait falls to one minute. The potential service opportunity still has mean four minutes.
The shorter observed wait reflects earlier exits, including more abandonment. It cannot establish improved access. These scenarios change the patience distribution rather than adding service capacity.
Do not interpret the four-minute potential mean as something directly measured for abandoners. Once a caller leaves, the ordinary record does not show when that caller would have been served had they stayed. The counterfactual T is known here because we constructed the model.
| Model quantity | θ = 0.25/min | θ = 0.75/min |
|---|---|---|
| Potential opportunity mean | 4 min | 4 min |
| Served fraction | 50% | 25% |
| Abandoned fraction | 50% | 75% |
| Answered-only mean wait | 2 min | 1 min |
| All-entrant mean time waiting | 2 min | 1 min |
CHAPTER 7 OF 12 · Read the performance trade-off
7. Use a service target with the correct denominator
Suppose the target is service within two minutes of admission. Among all entrants, the model probability is P(S and Q ≤ 2) = [ν/(ν + θ)]×[1 − exp(−2(ν + θ))].
In the first scenario this is 0.5×(1 − exp(−1)), approximately 31.6%. In the second it is 0.25×(1 − exp(−2)), approximately 21.6%.
Among answered entrants alone, the corresponding percentages are approximately 63.2% and 86.5%. The answered-only target rate rises while the all-entrant service-within-target rate falls.
State which result the service intends to measure. Excluding abandoners may be a declared reporting convention, but it changes the question. A transparent report can show the answered-only measure beside admission counts, abandonment and the all-entrant target rate.
| Target statistic | First scenario | Second scenario |
|---|---|---|
| Within two minutes among served | 63.2% | 86.5% |
| Served within two minutes among all admitted | 31.6% | 21.6% |
CHAPTER 8 OF 12 · Read the performance trade-off
8. Connect waiting-area occupancy to every exit
In a suitable long-run regime, Little’s Law for the waiting area uses all admitted entrants and their times until either exit. Let λ be that entry rate, in callers per hour. Then average waiting occupancy is Lq = λE[Q], converting minutes to hours.
At sixty entrants per hour, the first model gives Lq = 60×2/60 = 2 callers. The second gives Lq = 1 caller. A smaller waiting area population can accompany worse access because more callers leave.
The total can also be partitioned by eventual outcome. In the first scenario, served and abandoning callers each contribute one caller to the time-average occupancy. In the second, their contributions are 0.25 and 0.75, respectively.
Using the served rate alone with the answered-only waiting mean counts only the occupancy attributable to eventual service. It omits the time spent by eventual abandoners. A consistent system boundary must retain those waiting minutes.
CHAPTER 9 OF 12 · Check records and transfer
9. Handle unfinished records and repeated contacts
A caller still waiting when a report closes has an unresolved outcome. Dropping that record from a completed-outcomes average can remove the longest waits. Alternatively, treating elapsed time so far as the final wait understates that caller’s eventual duration.
Follow an entry cohort to completion when practical, or use an appropriate method for incomplete observations and state its assumptions. Patience among served callers is also incompletely observed: they reached service before their full patience was revealed.
Repeated attempts create another denominator. One person may abandon and call again. Attempt-level access, unique-person resolution and service workload are related but different quantities. The retry guide explains why deduplicating people does not erase work already consumed.
The present arithmetic does not identify a caller’s motive for leaving. A disconnect can reflect impatience, a technical fault or another event. Outcome labels should follow the actual record and evidence.
CHAPTER 10 OF 12 · Check records and transfer
10. Practise with a changed rate pair
Use ν = 0.4 and θ = 0.1 per minute. The combined rate is 0.5, so the all-entrant mean wait is two minutes. The served probability is 0.4/0.5 = 0.8, and the abandoned probability is 0.2.
For a three-minute target, the served-within-target fraction among all entrants is 0.8×(1 − exp(−1.5)), approximately 62.1%. Among served entrants only, it is approximately 77.7%.
At thirty admitted entrants per hour, mean waiting occupancy is one caller. The eventual-service portion contributes 0.8 and the abandonment portion 0.2. Reconcile both the probability ledger and occupancy ledger.
Now replace exponential patience with a fixed two-minute deadline. The derivation no longer applies unchanged: the abandonment time has a point mass, and the conditional waiting distributions differ. Explain the failed assumption before attempting a new calculation.
Does a low answered-only mean prove the service is poor? No. It is a useful measure of one group’s experience. It needs the access and outcome measures required by the actual question.
Does abandonment always lower the mean wait? The worked model has that direction when θ increases and ν stays fixed. Actual systems can change capacity, arrival mix, routing and patience together; the direction must be investigated.
Can we infer how long abandoners would have waited from the observed abandonment mean? Not generally. Their time to departure is observed; their later potential service time is not.
Is this an Erlang A model? No. Erlang A models a queue with parallel servers and exponential patience. This guide supplies independent competing times to isolate selection and denominator effects.
CHAPTER 12 OF 12 · Questions and further models
12. Use the evidence to choose the next improvement
Begin with admitted attempts, completed outcomes, unresolved records, waiting time by outcome and service within a declared target. Check whether counts reconcile before fitting a richer queue model.
The competing-times calculations and numerical comparisons here are original derivations under stated assumptions. Ward Whitt’s linked research provides a route into queue models that explicitly include abandonment and distinguish conditional performance measures.
Use the waiting-room boundary guide to define the population, the retry guide to separate attempts from jobs, and the priority guide when the next service opportunity depends on class. Each owns a different mechanism.
For transfer, examine any average reported only among completers. Ask who left before completion, how their experience was recorded and whether the outcome being celebrated still matches the original purpose. A better measure makes the next operational question visible.
References checked 22 October 2026
Ward Whitt, Columbia: queues with abandonment and conditional waiting measures
Continue through Queueing Theory
Choose the waiting-room boundary for Little’s Law
Separate retry attempts from unique jobs
Compare priority rules and class waits
Call centres and the no-abandonment Erlang C baseline
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