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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Define the service model
Chapters 1–2. Separate waiting from total time and state the arrival and service assumptions.
Build the mean-wait result
Chapters 3–5. Calculate service moments and connect residual work to queued jobs.
Compare and change demand
Chapters 6–7. Work the equal-mean cases and test another arrival rate.
Judge the model
Chapters 8–10. Separate means from tails, inspect evidence and answer common questions.
Practise and continue
Chapters 11–12. Change the mixture and choose related queueing guides.
Full chapter index · Practice and worked checks · Queueing Theory subject root
Two services each take one minute per job on average. They receive the same random arrivals and have the same long-run utilisation. Their mean waiting times can still differ by more than a factor of three.
The missing information is the spread of service durations. We will compare three invented job populations, explain why long jobs receive extra weight, and derive the single-server mean-wait relationship under explicit assumptions.
Chapter index
Define the service model · Chapters 1–2
1. Define the wait being compared
2. State the single-server model
Build the mean-wait result · Chapters 3–5
3. Construct three equal-mean service populations
4. Explain why a random arrival notices long jobs
5. Connect residual work to the mean queue
Compare and change demand · Chapters 6–7
6. Calculate the three waiting times
7. Change the arrival rate and retain the service populations
Judge the model · Chapters 8–10
8. Separate a mean result from a tail claim
9. Check whether the model fits the evidence
10. Questions readers often ask
Practise and continue · Chapters 11–12
11. Practise with a changed mixture
12. Choose a connected reading route
Waiting time Wq means time from joining the queue until service begins. Total time W includes that wait plus the job’s own service. A queue-length average Lq counts waiting jobs; it excludes the one currently being served.
These distinctions matter before any formula is chosen. A system can have a mean wait of two minutes and a mean total time of three minutes. Reporting one as the other changes the experience being described.
Our comparison concerns long-run averages in a stationary mathematical model. It does not predict the next person’s wait or establish a staffing rule for a real organisation.
Arrivals follow a homogeneous Poisson process at rate λ. Service durations S are independent and identically distributed, independent of arrivals, and have finite first and second moments. One continuously available server works whenever jobs are present.
Jobs enter an unlimited waiting area, none leave without service, and service follows first-come, first-served order without interruption. There are no setup intervals, server vacations, priorities or retries in this model.
Write m = E[S] and ρ = λm. Stability requires ρ < 1. For the first comparison, λ = 0.8 jobs/minute and m = 1 minute, so ρ = 0.8. These conditions define an M/G/1 queue: Poisson arrivals, a general service-duration distribution and one server.
CHAPTER 3 OF 12 · Build the mean-wait result
3. Construct three equal-mean service populations
Population A gives every job exactly one minute. Population B gives 90% of jobs half a minute and 10% five and a half minutes. Population C uses an exponential duration with mean one minute.
Population B’s mean is 0.9 × 0.5 + 0.1 × 5.5 = 1 minute. Its second moment is E[S²] = 0.9 × 0.5² + 0.1 × 5.5² = 3.25 min². Squaring the mean instead would give 1 min² and discard the variation.
| Population | Mean service m | E[S²] | Service variance |
|---|---|---|---|
| A: constant duration | 1 min | 1 min² | 0 min² |
| B: short–long mixture | 1 min | 3.25 min² | 2.25 min² |
| C: exponential duration | 1 min | 2 min² | 1 min² |
The means match, but the second moments do not. That is enough to change the stationary mean wait under the stated model.
CHAPTER 4 OF 12 · Build the mean-wait result
4. Explain why a random arrival notices long jobs
A long job occupies the server for longer, making it more likely to be encountered in progress. In Population B, the long jobs are 10% of the count but occupy 0.1 × 5.5/1 = 55% of busy time.
There is another weighting step. During a service lasting s minutes, its remaining duration falls from s to zero. The area under that remaining-duration graph is a triangle of area s²/2. A long job therefore contributes according to its squared duration to this time-integrated residual work.
Let R be the remaining service of the job in progress, with R = 0 when idle. In the stationary lossless system, throughput is λ. Summing those triangle areas per unit time gives E[R] = λE[S²]/2. This expectation includes idle time; it is not the mean conditional on finding the server busy.
CHAPTER 5 OF 12 · Build the mean-wait result
5. Connect residual work to the mean queue
A newly arriving job must wait for the residual service plus the service of jobs already waiting ahead. Under first-come, first-served service and the independence assumptions, the latter work has mean mLq.
Poisson arrivals see the stationary time-average state here, so the arrival’s mean residual service is the E[R] calculated above. Little’s Law gives Lq = λWq. Consequently Wq = E[R] + mλWq = E[R] + ρWq.
Rearrange to obtain Wq = λE[S²]/[2(1−ρ)]. This is the Pollaczek–Khinchin mean-wait formula for this model. MIT’s M/G/1 treatment provides the theoretical framework; our populations and comparisons are constructed teaching cases. The Northeastern M/G/1 notes provide another formal derivation.
The denominator describes the remaining capacity margin. The numerator retains the squared-duration effect. Replacing a general service distribution by an exponential one merely because its mean is known changes the numerator without evidence.
With λ = 0.8 and ρ = 0.8, the multiplier λ/[2(1−ρ)] is 2 per minute. Population A therefore gives Wq = 2 minutes, B gives 6.5 minutes, and C gives 4 minutes.
Adding the same one-minute mean service gives mean total times of 3, 7.5 and 5 minutes respectively. Population B’s mean wait is 3.25 times A’s, even though their average service durations and utilisations match.
For B, Lq = λWq = 0.8 × 6.5 = 5.2 waiting jobs on average. This is a time average, so it need not be a whole number. It does not mean that each arrival finds exactly 5.2 jobs.
CHAPTER 7 OF 12 · Compare and change demand
7. Change the arrival rate and retain the service populations
Keep m = 1 minute but vary λ. The two-point mixture always has 3.25 times the constant-duration second moment, so its mean wait is 3.25 times as large at each common stable arrival rate.
| Arrival rate λ | Utilisation | Constant-duration Wq | Mixture Wq |
|---|---|---|---|
| 0.2 jobs/min | 0.2 | 0.125 min | 0.40625 min |
| 0.5 jobs/min | 0.5 | 0.5 min | 1.625 min |
| 0.8 jobs/min | 0.8 | 2 min | 6.5 min |
| 0.9 jobs/min | 0.9 | 4.5 min | 14.625 min |
The absolute difference grows near full utilisation. At λ = 1 the stationary finite-mean formula is not applicable. Substituting beyond that boundary and interpreting a negative result as a waiting time would be a model error.
The formula uses only two service moments, but it gives a mean wait. It does not say that two distributions with those same moments have identical waiting-time distributions or matching 95th percentiles.
Nor does a smaller average justify excluding long or complicated work from a service. That changes the admitted population and the access outcome. A useful comparison keeps service quality, completion and fairness alongside delay.
Service variability can sometimes be reduced by removing avoidable interruptions or missing information. Whether a proposed change achieves that requires measurement. The equation explains a conditional mechanism; it does not identify which operational change will work.
Record arrival, service-start and service-end events using consistent boundaries. Check time variation in demand, server availability and unfinished jobs at the observation boundary. A completion-only sample can miss long jobs still in progress.
Inspect the service distribution rather than reporting its mean alone. A few very long observations can strongly affect the squared-duration average. Conversely, a short sample that contains no rare long job can understate the second moment.
The arrival and scheduling assumptions also need attention. Appointments, bursts, dependence, multiple servers, priorities and abandonment can require another model. Poisson-arrival sampling is doing specific work in the derivation; an average arrival rate alone does not establish it.
Does 80% utilisation fix the mean wait?
No. Even within M/G/1, mean service scale and its second moment matter. Utilisation alone leaves out information needed for the calculation.
Is constant service always best?
At fixed mean service and arrival rate in this first-come model, E[S²] is minimised by a constant duration. That minimises this particular mean wait. It is not a universal judgement about service quality or every scheduling objective.
Is the average remaining service half the average service?
Conditional on being busy, the mean residual is E[S²]/(2m), not generally m/2. They agree for constant service. For the mixture it is 1.625 minutes, although m is only one minute.
Suppose half the jobs take 0.5 minute and half take 1.5 minutes. The mean remains one minute, and E[S²] = (0.25 + 2.25)/2 = 1.25 min². At λ = 0.8, Wq = 2.5 minutes. It lies between the constant and exponential cases in the table.
Now keep Population B but reduce arrivals to 0.5 per minute. Calculate E[R] first: 0.5 × 3.25/2 = 0.8125 minute. Divide by 1−ρ = 0.5 to obtain Wq = 1.625 minutes. Explain why the residual mean includes idle intervals while the conditional busy residual stays 1.625 minutes.
Finally, change the policy to interrupt a long job for a short one. Identify the assumption that has changed before attempting another calculation. The first-come derivation cannot simply be reused unchanged.
For the model map, return to Queueing Theory. The random-arrivals and inspection-paradox guide develops the related reason that a randomly encountered interval can be longer than a randomly selected interval. Here that residual-work idea is combined with a queue of earlier jobs.
For arrival assumptions and the broader queue model, use Poisson processes and arrival rates and waiting times. The call-centre and Erlang C guide explains why parallel servers require a different model.
For predicting a particular arrival rather than a stationary mean, compare waiting-time estimation before joining. The queues and service-decisions guide connects the model with boundaries, simulation and outcome measures.
For the mathematical source, use MIT’s M/G/1 chapter. Keep its stationary model assumptions attached when transferring the result to another service.
Continue through How X Works and The Gold Standard of Education. The habit worth carrying is to inspect the distribution behind an average before treating two systems as equivalent.
Sources checked 22 October 2026. Worked examples are educational models with the assumptions stated above.