Choose the question you need to answer
Choose a route or read from the start. The lessons below remain open; the five expandable index groups simply make the article easier to navigate.
Why time and arrival averages differ
Identify the two denominators: minutes in a state and incoming customers who see it.
Build a reproducible queue
Use independent Poisson arrivals, one exponential server and a stated steady-state utilisation.
Check PASTA numerically
Derive empty, busy and long-line probabilities and connect them to waiting.
Challenge the sampling rule
Select only delayed customers, change to timed appointments and inspect uncertainty.
Use the right evidence
Work a changed-rate calculation, answer misconceptions and follow reliable sources.
A service counter is busy for 40 percent of the time. Does that mean 40 percent of customers arrive to find it busy? Under a suitable Poisson-arrival model the answer is yes, but the equality is a theorem about how arrivals sample a changing system. It is not a promise about every timetable or filtered data set.
We will calculate an M/M/1 example, distinguish the state just before arrival from the state after joining, check a second queue-length event, and find the point where selected or scheduled observations stop representing the same time population. All rates, assumptions and conclusions are explicit.
Chapter index
Start with the question · Chapters 1–2
1. A quiet counter and an apparently different customer experience
2. Define the two fractions before comparing them
Build and check the model · Chapters 3–5
3. Build the simplest stable one-server model
4. Work out the queue-length distribution
5. Why an arrival can see what a clock sees
Work through the result · Chapters 6–7
6. Turn the state probabilities into practical waits
7. A second check using Little’s Law
Change the conditions · Chapters 8–10
8. Change only the selection rule and the answer changes
9. A changed arrival model needs new reasoning
10. What PASTA does not establish
Practise and continue · Chapters 11–12
11. Practise with different arrival and service rates
12. Read the theorem and build a trustworthy report
CHAPTER 1 OF 12 · Start with the question
1. A quiet counter and an apparently different customer experience
Imagine an observer making a note every minute: empty counter, occupied counter, perhaps a waiting line. After many hours, the counter is found busy in 40 percent of the observations. A manager then asks a different question: what fraction of actual customers arrive when the server is already occupied? The minute-by-minute sample and the arrivals are not the same list of observations.
A customer may be more likely to arrive during a busy period if people follow a bell, if a bus unloads a crowd at a particular time, or if a report deliberately keeps only late requests. The incoming stream could instead be Poisson, so that its future arrival moments have no information about the current state. The famous PASTA principle concerns this latter situation under the appropriate lack-of-anticipation condition.
A distinction is essential before doing arithmetic. The arriving customer examines the queue immediately before they join; afterwards the number present rises by one. Mixing those two instants can make the measured arrival-state distribution look inconsistent even when the probability model is sound.
CHAPTER 2 OF 12 · Start with the question
2. Define the two fractions before comparing them
Let N(t−) denote the number of customers in the system just before time t, including the person being served if there is one. The time-average fraction in state n is the long-run amount of time with N(t)=n divided by the long-run observation time. An arrival-average fraction is the number of arriving customers who see N(T−)=n divided by the total number of arrivals.
Both quantities can be valid descriptions of the same operating system. The time average weights durations: a long occupied interval counts more than a short one. The arrival average weights arriving events: a crowd of arrivals may contribute many observations in a short interval. Equality is therefore a substantive property to establish, not an identity obtained by cancelling fractions.
In a stable system supplied by a homogeneous external Poisson arrival process, an arrival cannot predict the next jump of that Poisson process. Under the usual stationarity and nonanticipation conditions, Poisson arrivals see time averages: for each state, the long-run arrival share equals the long-run time share. This is PASTA, associated with Ronald Wolff’s 1982 result, not an assertion that all random arrivals behave this way.
CHAPTER 3 OF 12 · Build and check the model
3. Build the simplest stable one-server model
Consider an M/M/1 service desk. The first M says arrivals are Poisson with constant rate λ = 2 customers per hour. The second M says each service time is independent and exponentially distributed with mean 1/μ = 12 minutes, so μ = 5 customers per hour. The final 1 means there is one server.
Assume customers wait if necessary, service is first-come first-served, there is no loss or abandonment, the operation does not close during the observation and the system has reached its stationary regime. Also assume the incoming Poisson process does not anticipate the queue state. Our utilisation is ρ = λ/μ = 2/5 = 0.4. Since ρ is less than one, the familiar stationary M/M/1 probability formula applies.
The exercise does not describe a real queue merely because its arithmetic seems plausible. Actual arrivals can depend on school dismissal, public transport, booking slots or staff announcements; service may be non-exponential. We use the model to isolate what PASTA adds to a clearly stated set of assumptions.
CHAPTER 4 OF 12 · Build and check the model
4. Work out the queue-length distribution
In a stationary M/M/1 queue, P(N=n)=(1−ρ)ρn for n=0,1,2,… . At ρ=0.4 the empty probability is P(N=0)=0.6. One person present has probability 0.6×0.4=0.24 and two people have probability 0.6×0.4²=0.096.
To find at least two present, sum the geometric tail rather than looking only at N=2: P(N≥2)=ρ²=0.16. Similarly P(N≥1)=ρ=0.4. These probabilities describe the state at an ordinary stationary time; we can now use PASTA to interpret what incoming Poisson customers see just before joining.
It is tempting to equate P(N=2)=0.096 with the probability that an arrival meets a line of two waiting people. That is incorrect: N counts the person in service as well as any waiting customers. With N=2 there is one in service and one waiting; two already waiting would correspond to N≥3.
| Pre-arrival N | Stationary time share | Long-run Poisson arrival share | Meaning |
|---|---|---|---|
| 0 | 0.600 | 0.600 | Server idle |
| 1 | 0.240 | 0.240 | One in service |
| 2 | 0.096 | 0.096 | One in service, one waiting |
| 3 or more | 0.064 | 0.064 | At least two waiting |
CHAPTER 5 OF 12 · Build and check the model
5. Why an arrival can see what a clock sees
An independent homogeneous Poisson stream has a probability approximately λΔt of delivering an arrival in a very short interval Δt, no matter whether the queue currently looks busy or empty. To first order, time occupied by a state therefore contributes arrival opportunities in proportion to the time spent there. The rigorous version handles dependence between earlier arrivals and the current state through nonanticipation.
A simple expectation calculation helps: in a long stationary observation lasting 100 hours, expected time empty is 60 hours and expected time busy is 40 hours. At two arrivals per hour, those two kinds of intervals contribute expected 120 and 80 arrivals respectively, provided the Poisson lack-of-anticipation conditions apply. These are illustrative expected quantities, not observed counts guaranteed to occur in a particular 100-hour sample.
Do not use the argument to claim the arrival epochs are independent of the queue. Earlier arrivals help determine N(t−). PASTA concerns how the next Poisson jump samples the pre-arrival system; it does not erase causality or make the full arrival and queue histories independent.
CHAPTER 6 OF 12 · Work through the result
6. Turn the state probabilities into practical waits
An arriving customer must wait if at least one customer is already present and receiving service. By PASTA, P(wait)=P(N≥1)=ρ=0.4 for this steady M/M/1 system. Thus four in ten arrivals wait in the long run, even though the server spends six in ten units of time idle.
Mean time in the system is W=1/(μ−λ)=1/3 hour, or 20 minutes. Mean service time is 12 minutes, leaving mean queue waiting Wq=20−12=8 minutes. Independently, the M/M/1 formula Wq=λ/[μ(μ−λ)]=2/(5×3) hour gives the same eight minutes.
These are different questions and should not be interchanged. Forty percent is the chance of any waiting, eight minutes is the unconditional mean queue wait across all arriving customers, and 20 minutes includes service. A report restricted to delayed customers will return a different conditional mean and cannot simply be labelled Wq.
CHAPTER 7 OF 12 · Work through the result
7. A second check using Little’s Law
The stationary mean number of customers in the whole M/M/1 system is L=ρ/(1−ρ)=0.4/0.6=2/3. Little’s Law independently gives L=λW=2×(1/3)=2/3 customers. For the waiting room alone, the mean number is Lq=λWq=2×(2/15)=4/15, about 0.267 customers.
Here λ is the accepted flow of two customers per hour, because no arrivals are rejected or abandoned. If you change the model to a finite capacity that loses arrivals, the arrival rate entering Little’s Law must correspond to the population and boundary you are measuring. The relevant companion reading is Finite Buffers, Blocking and Admitted Throughput.
Neither Little’s Law nor PASTA requires that every single person have the mean experience. One is a relation among consistent long-run flow, number and time measurements; the other links how Poisson arrivals sample states. Treating them as interchangeable will lead to inaccurate queue dashboards.
CHAPTER 8 OF 12 · Change the conditions
8. Change only the selection rule and the answer changes
Suppose a monitoring report records an arrival only when it finds the server occupied. Every selected arrival is then, by construction, a delayed arrival, so its observed pre-arrival busy fraction is 100 percent. The true time fraction busy remains 40 percent in our stationary example. No physical queue changed: only which event records were kept changed.
The same issue arises when a service desk analyses only cases that received a callback, only requests that generated complaints or only calls that waited long enough to be answered. These are selected populations. A statement such as “all our sampled arrivals encountered a busy counter” says nothing about unrecorded arrivals that were served immediately.
A second way to break the required sampling premise is to schedule inspections at moments when workers deliberately pause. Such inspections may see a busier system than a uniform time sample. Do not assign the PASTA label to those observations without showing the process that generated the sampling times.
| Observation protocol | Busy share seen | What it estimates |
|---|---|---|
| Long-run random-time observation | 40% | Fraction of time busy |
| All eligible Poisson customer arrivals | 40% | Unfiltered pre-arrival fraction by PASTA |
| Arrivals kept only when server busy | 100% | Selected delayed-arrival subgroup |
| Inspection scheduled at known peaks | Not determined from ρ | Peak-conditioned observation |
CHAPTER 9 OF 12 · Change the conditions
9. A changed arrival model needs new reasoning
Imagine a classroom booking desk receiving a batch of six enquiries every Monday morning rather than a Poisson stream spread over the week. The number of arrivals per week might match a Poisson system, but the sampling of queue states will be concentrated around the release period. There is no basis for asserting that the week-long fraction of busy minutes equals the fraction of those Monday enquiries encountering a wait.
Similarly, when arrivals have a time-dependent Poisson intensity, an arrival is more likely to fall in a high-intensity hour. A raw arrival average is weighted toward high-arrival periods; a clock-time average is not. Analysts must define the relevant time regime or compute intensity-weighted quantities before importing a homogeneous stationary PASTA conclusion.
This does not mean Poisson arrivals always fail in a system that has peaks. It means the exact version of the process and the conditioning matter. The right check is to preserve arrival timestamps, operational state immediately before arrival, service starts, staffing changes and the definition of the observation period.
PASTA is not a guarantee that an observed sample of 25 customers will match stationary proportions exactly. Sampling variation remains, as do possible data gaps, incorrect clocks and slow convergence to an operating regime. The equality concerns suitable limiting distributions or long-run fractions under stated conditions.
Nor does seeing a busy server prove causation for a long waiting time. If the person ahead needs little service, the delay can be short. In a complex service environment, busy probability is only one dimension of experience; service-time tails, multiple staff, routing, priorities and disruptions can matter more to the right operational decision.
Finally, a queue seen immediately after an arrival includes the newcomer. A new arrival to an empty server sees N(T−)=0 but leaves N(T)=1. If a dataset mixes before and after states, it can manufacture a discrepancy with PASTA that no amount of statistical averaging will repair.
CHAPTER 11 OF 12 · Practise and continue
11. Practise with different arrival and service rates
Exercise: keep μ=5 customers per hour but let λ=3 customers per hour. Under the same independent Poisson and steady M/M/1 assumptions, ρ=3/5=0.6. Calculate P(N=0)=0.4, P(N≥1)=0.6 and P(N≥2)=0.36. PASTA says those are also the respective long-run pre-arrival shares of eligible Poisson arrivals.
Check the time calculations without copying the earlier example. W=1/(5−3)=0.5 hour=30 minutes; mean service remains 12 minutes, so mean queue waiting is 18 minutes. The increase is not a contradiction: moving utilisation from 40 to 60 percent produces disproportionately longer waits in this stochastic model.
FAQ: Does the server’s 60 percent busy time imply an arriving customer waits exactly 60 percent of the time? In this changed model, yes in the long run under PASTA, not automatically in an arbitrary model. Do deliberate appointment slots obey PASTA? Not generally. Can the observed share be different for small samples? Of course; compare uncertainty intervals as well as point estimates.
CHAPTER 12 OF 12 · Practise and continue
12. Read the theorem and build a trustworthy report
For the source of the principle, read Ronald W. Wolff’s “Poisson Arrivals See Time Averages” (1982). An accessible worked explanation appears in Applied Probability Notes on PASTA. The condition that arrivals cannot anticipate future Poisson jumps matters more than the slogan.
For linked educational applications use the Queueing Theory root, Little’s Law and a boundary that changes the count, and batch arrivals and event rates. They use different measurement and arrival assumptions, so compare them rather than mixing their formulas.
Before publishing an arrival-state chart, name the denominator: all arrivals, admitted arrivals, selected complaint cases, sampled minutes or scheduled inspections. State when the count is taken, whether the operation was stationary and how the arrivals were generated. A believable percentage is not yet an adequately defined measurement.
Under a nonanticipating stationary Poisson arrival stream, the share of arrivals seeing each pre-arrival state agrees with the time share. Everything else—sample filtering, time-of-day peaks, different queue boundaries and service performance—requires its own evidence. Continue through How X Works and The Gold Standard of Education for further worked learning routes. Source definitions checked 23 October 2026.