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Choose a route or begin at Chapter 1. Teaching chapters stay open; the chapter index expands by section.
Choose the sampling unit
Chapters 1–2. Define the unavailable period and the arrival question
Derive the residual
Chapters 3–5. Derive the remaining-time contribution of one interval
Separate queue mechanisms
Chapters 6–8. Average the direct residual over all arrival instants
Check records and practise
Chapters 9–10. Check long tails, partial records and arrival dependence
Questions and next routes
Chapters 11–12. Questions readers ask
Full chapter index · Worked practice · Queueing Theory subject root
A machine’s interruptions average five minutes. A customer who finds it unavailable may nevertheless face more than two and a half minutes until it returns. Longer interruptions occupy more of the clock, so they are encountered more often.
This guide calculates the remaining downtime seen by a time-sampling arrival. It distinguishes an interruption counted once from an interruption encountered in progress, and separates that residual delay from the complete queueing wait. The examples are original hypothetical schedules, with no claim about a particular service.
Chapter index
Choose the sampling unit · Chapters 1–2
1. Define the unavailable period and the arrival question
2. See why counting interruptions equally changes the answer
Derive the residual · Chapters 3–5
3. Derive the remaining-time contribution of one interval
4. Work through the two-length example
5. Separate the mean duration from its variability
Separate queue mechanisms · Chapters 6–8
6. Average the direct residual over all arrival instants
7. Distinguish a vacation already running from an arrival-triggered setup
8. Keep residual downtime separate from vacation-queue decompositions
Check records and practise · Chapters 9–10
9. Check long tails, partial records and arrival dependence
10. Practise with a different interruption mix
Questions and next routes · Chapters 11–12
11. Questions readers ask
12. Use the result to improve the measurement question
CHAPTER 1 OF 12 · Choose the sampling unit
1. Define the unavailable period and the arrival question
In queueing theory, a server vacation is a period when the server is unavailable to perform its usual service. It might represent an interruption, another task or a scheduled absence. The technical name does not describe a person’s holiday entitlement.
Let V be a vacation’s full duration, measured from its start to its return. Let R be the time remaining when an observer arrives during a vacation. These variables are measured in the same units but answer different questions.
We will sample independent vacation durations with a positive finite mean and examine a uniformly selected instant in a long record of vacation time. Equivalently, independent constant-rate Poisson arrivals sample that record without preference for a particular phase.
The conditioning is essential: the observer has found the server on vacation. We have not yet averaged over arrivals finding it available, and we have not included customers already ahead in a queue.
CHAPTER 2 OF 12 · Choose the sampling unit
2. See why counting interruptions equally changes the answer
Suppose half the interruptions last two minutes and half last eight minutes. Counting events gives E[V] = 0.5×2 + 0.5×8 = five minutes.
One two-minute and one eight-minute interruption occupy ten minutes altogether. Of those ten minutes, two belong to the short interruption and eight to the long one. A random instant within this unavailable time therefore falls in the long interruption with probability 0.8, not 0.5.
The event distribution assigns each interruption one vote. The time-sampled distribution assigns votes in proportion to duration. This is a change of sampling weights, not a change in how the server behaves.
If arrivals deliberately avoid the middle of long interruptions, or accumulate at announced reopening times, they do not sample clock time uniformly. Their observed weights must come from the actual arrival process rather than from this argument.
CHAPTER 3 OF 12 · Derive the residual
3. Derive the remaining-time contribution of one interval
Fix a vacation of length v. If the observer’s position u is uniform between its start and end, the remaining time is v − u. Its conditional mean is v/2.
The area under the remaining-time line over that interval is ∫ from 0 to v of (v − u) du = v²/2. This triangular area has units of minutes squared. Dividing by the interval’s v minutes gives v/2 minutes.
For many vacations, add these triangular areas and divide by the total vacation time. The average residual becomes [Σvj²/2]/[Σvj], not the average of the individual vj/2 values with equal event weights.
Under the independent, identically distributed model and the usual long-run averaging conditions, this ratio approaches E[V²]/(2E[V]). A finite second moment gives a finite value. The derivation also works directly on a finite completed record with its empirical sums.
For the two- and eight-minute durations, E[V²] = 0.5×4 + 0.5×64 = 34 minutes squared. Thus E[R | vacation] = 34/(2×5) = 3.4 minutes.
The same result follows from the encounter weights. With probability 0.2 the observer encounters a two-minute interruption, whose average remainder is one minute. With probability 0.8 the observer encounters an eight-minute interruption, whose average remainder is four minutes.
The weighted result is 0.2×1 + 0.8×4 = 3.4 minutes. Taking half the event-average duration instead gives 2.5 minutes, which understates this particular residual by 0.9 minutes.
The encountered full interval has mean 0.2×2 + 0.8×8 = 6.8 minutes. Half of that is 3.4. Halving can work after the correct time-sampled distribution has been identified; it fails when applied prematurely to the event distribution.
| Quantity | Fixed 5 min | 2 or 8 min, equally frequent |
|---|---|---|
| Event mean | 5 min | 5 min |
| Second moment | 25 min² | 34 min² |
| Conditional mean residual | 2.5 min | 3.4 min |
| Mean encountered full interval | 5 min | 6.8 min |
CHAPTER 5 OF 12 · Derive the residual
5. Separate the mean duration from its variability
Write E[V²] = Var(V) + E[V]². The residual formula becomes E[R | vacation] = E[V]/2 + Var(V)/(2E[V]).
At a fixed mean, variability raises the residual time seen by the observer. For the example, Var(V) = 34 − 25 = nine minutes squared. Its extra contribution is 9/10 = 0.9 minutes.
Compare a deterministic five-minute interruption. It has the same event-average length but zero variance, so its mean remainder is 2.5 minutes. The varied schedule’s residual is 36% larger: 0.9/2.5 = 0.36.
This comparison holds the mean duration fixed. A real redesign might also change interruption frequency, the proportion of time unavailable, setup work or service speed. Those changes belong in a wider comparison before an operational decision is made.
CHAPTER 6 OF 12 · Separate queue mechanisms
6. Average the direct residual over all arrival instants
Now let q be the long-run fraction of clock time spent on vacation. Define D as the current vacation’s remainder if the server is on vacation, and zero if it is available. Under time-sampling arrivals, E[D] = qE[R | vacation].
For illustration, suppose an independent alternating cycle has average vacation time five minutes and average available time twenty minutes. Its unavailable fraction is 5/(5+20) = 0.2. The direct residual contribution is then 0.2×3.4 = 0.68 minutes per arrival.
This is the average remainder of an interruption already in progress. It is not a complete customer waiting time. A customer may face earlier jobs, service interruptions later in its wait or a rule that makes the server leave again.
An arrival finding an available but busy server has D = 0 while still waiting. That one example shows why D cannot automatically replace queueing delay in a performance report.
CHAPTER 7 OF 12 · Separate queue mechanisms
7. Distinguish a vacation already running from an arrival-triggered setup
Consider a different mechanism: the server sleeps when empty, and the first new job triggers a fixed four-minute startup. That triggering job faces the whole four minutes. It has not arrived at a uniform position within a startup already under way.
A later arrival during the startup can face a remainder. But the initiating arrival and later arrivals belong to different conditional groups. Their weights depend on arrivals, startup policy and the busy periods that follow.
Assigning every job two minutes because half of four is two ignores the first job’s role in starting the clock. Assigning every job four minutes ignores arrivals that encounter an already active server.
Write the triggering rule before choosing the formula. Fixed scheduled downtime, repeated vacations until work appears, and startup only after a new arrival are distinct models even when a maintenance log gives them the same mean duration.
CHAPTER 8 OF 12 · Separate queue mechanisms
8. Keep residual downtime separate from vacation-queue decompositions
Some classical vacation queues admit decomposition results connecting an ordinary queue to an additional vacation component. Such results require a specified arrival process, service discipline, vacation policy and independence conditions.
The calculation in this guide needs no assertion that every queue admits that decomposition. It measures a particular clock-time residual. A full waiting-time formula can include the continuing effects of accumulated work after the server returns.
The linked Boxma and coauthors paper, for example, studies vacations after a specified number of completed services. It illustrates why the rule determining departure matters. Its model should not be substituted for an arrival-triggered setup without checking the assumptions.
For an actual service, record vacation starts and ends, arrival timestamps, service starts, job order and cancellations. Those observations distinguish immediate residual downtime from the backlog and later events that complete the customer’s experience.
CHAPTER 9 OF 12 · Check records and practise
9. Check long tails, partial records and arrival dependence
A duration distribution can have a finite mean but an infinite second moment. In that case this stationary conditional mean residual is infinite, even though an event-counted average duration is finite. This is a mathematical possibility, not a diagnosis based on a short log.
In finite records, one long interruption can dominate Σvj². Report sensitivity to extreme events and check whether they are valid observations. Deleting them merely because they affect the answer changes the population being described.
An interruption still open at the observation cutoff has no known completed duration. Treating elapsed time as its full length truncates both the duration and its square. A completed-interval analysis also needs to acknowledge selection caused by the cutoff.
Finally, arrivals responding to operating hours or outage messages need arrival-weighted analysis. The length-biased clock result cannot establish what such arrivals actually experienced.
CHAPTER 10 OF 12 · Check records and practise
10. Practise with a different interruption mix
Take event probabilities 0.75 for a one-minute interruption and 0.25 for a nine-minute interruption. The event mean is 0.75+2.25 = three minutes. The second moment is 0.75+20.25 = 21 minutes squared.
The conditional mean remainder is 21/(2×3) = 3.5 minutes. The encountered full-duration probabilities are 0.75×1/3 = 0.25 and 0.25×9/3 = 0.75.
Check independently: the two conditional remainders are 0.5 and 4.5 minutes. Their encounter-weighted mean is 0.25×0.5 + 0.75×4.5 = 3.5 minutes. The mean remainder can exceed the event-average full duration because they describe different sampling populations.
If the unavailable time fraction is 0.1, E[D] is 0.35 minutes per time-sampling arrival. State both the conditioning and the units when reporting the answer. Neither number includes an assumed backlog.
| Event duration | Event probability | Encounter probability | Mean remainder |
|---|---|---|---|
| 1 min | 0.75 | 0.25 | 0.5 min |
| 9 min | 0.25 | 0.75 | 4.5 min |
| Weighted total | 1 | 1 | 3.5 min |
Is the average remainder always half the average full interruption? It is half the mean full interruption under the encounter distribution. That is generally different from the event-counted distribution.
Does a smaller mean interruption guarantee a smaller residual? No. A sufficiently larger second moment can outweigh the reduction in the mean. Compare both moments under the same sampling rule.
Does an idle server always count as on vacation? No. An idle server may be ready to serve immediately. A vacation makes it unavailable under the stated model.
Can a customer’s full wait be calculated from vacation lengths alone? Usually not. Service times, arrivals, existing workload and vacation rules are also needed. The residual calculation supplies one clearly defined component rather than a universal replacement for a queue model.
CHAPTER 12 OF 12 · Questions and next routes
12. Use the result to improve the measurement question
Begin by deciding whether the question concerns interruptions as events, the server’s time allocation or arrivals’ experience. Keep each distribution visible instead of using one average for all three.
The examples above were derived from explicit two-point duration ledgers. Columbia’s renewal-process notes give further background on equilibrium residual lifetimes. The research reference provides a route into a complete queue with a particular vacation policy.
For learning, ask a student to derive the triangle area, replace event weights by encounter weights, then reconcile the two calculations. If those agree, change the arrival rule and ask which step stops being valid.
For transfer, examine any process that is sampled while in progress: a repair, a delivery gap or a download. Long intervals appear more often in clock-time samples. The useful habit is to identify the sampling unit before interpreting the average.
References checked 22 October 2026
Columbia University, Ward Whitt: equilibrium residual lifetimes and the inspection paradox
Boxma and coauthors: a queueing system with vacations after N services
Continue through Queueing Theory
Equal mean service times, different mean waits
Little’s Law and the waiting-room boundary
Abandonment and answered-only waiting statistics
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