Choose your reading route
You can begin with the puzzle, go directly to the worked calculation, or choose a check. Every chapter is already open; the five groups in the index only organise the links.
Find the distinction
Chapters 1–2. Separate arrivals of groups from arrivals of people.
State the model
Chapters 3–5. Specify the group-size distribution and the service capacity.
Calculate and check
Chapters 6–7. Calculate backlog and within-group delay with inspectable units.
Change the conditions
Chapters 8–10. Compare separate arrivals and simultaneous pairs; find the limits.
Practise and continue
Chapters 11–12. Work a check, repair common mistakes and follow sources.
Six jobs arrive every hour on average. The server can complete ten jobs an hour. Yet people still complain about the wait. Is the queue really that busy, or is the arrival pattern hiding a second problem?
This lesson holds the job throughput and service time fixed while changing only how work arrives. We calculate the consequences rather than calling every arrival a customer.
Chapter index
Find the distinction · Chapters 1–2
1. The crowd arrives in clumps
2. Define the event and the job before counting
State the model · Chapters 3–5
3. Two sizes of batch, and two different moments
4. The distribution on the page
5. Calculate load before calculating wait
Calculate and check · Chapters 6–7
6. See the workload arriving with each batch
7. Add the queue formed inside the arriving group
Change the conditions · Chapters 8–10
8. Change arrivals without changing workload
9. Which decisions transfer to real operations?
10. Repair four attractive mistakes
Practise and continue · Chapters 11–12
11. Evidence checks, a short exercise and FAQs
12. Read from the model into the system
Think of a service desk with people who arrive on their own, plus buses that unload several people at once. Six people per hour is an average workload. It does not say whether six people arrive separately or in two groups of three.
The first person in a group can be served before the second and third, even if the counter was empty a moment earlier. This extra waiting is not captured by simply multiplying the number of arrival events by a service time.
Our question is specific: for equal average job flow and equal service capacity, what does clustering do to an average job’s queueing wait?
CHAPTER 2 OF 12 · Find the distinction
2. Define the event and the job before counting
Let λb be the rate of batch-arrival events, measured in batches per hour. Let B be the random number of jobs in one batch. The individual job rate is λj = λb E[B], measured in jobs per hour.
Assume batch events follow a homogeneous Poisson process, each batch size is independent of other batches and of service, and the one server works continuously whenever jobs are waiting. Jobs are served one at a time in first-come, first-served order, with an arbitrary but unbiased order inside a simultaneous batch.
We will use deterministic six-minute service for each job, an unlimited waiting room, no abandonment and steady-state averages. None of these should be mistaken for guaranteed conditions in a real help desk.
Suppose a batch contains one job with probability ½ and three jobs with probability ½. The expected batch size is E[B] = ½(1) + ½(3) = 2 jobs. The second moment is E[B²] = ½(1²) + ½(3²) = 5 jobs squared.
That second moment tells us more than the mean because large groups carry more simultaneous work and contribute more pairs of jobs. In particular E[B(B−1)] = E[B²] − E[B] = 3.
The fraction of individual jobs that came in a three-job batch is not one half. Over many batches, three quarters of the jobs come from three-job groups: 0.5×3 divided by E[B]=2. A randomly selected job sees the larger batches more often.
The two batch sizes are equally likely when a batch is selected, but not when a job is selected. These are two legitimate sampling questions with different denominators.
Keep the labels and units through every line. An event count alone cannot reveal how many jobs walked in together.
| B: jobs per event | Event probability | Contribution to E[B] | Contribution to E[B²] |
|---|---|---|---|
| 1 | 0.5 | 0.5 | 0.5 |
| 3 | 0.5 | 1.5 | 4.5 |
| Totals | 1.0 | 2.0 | 5.0 |
Set the batch-event rate to three batches per hour. Then λj = 3×2 = 6 jobs per hour. Each job takes s = 0.1 hour, so the server can complete 10 jobs per hour when continuously busy.
Utilisation is ρ = λjs = 6×0.1 = 0.6. In our stable one-server model, the server is busy on average 60 percent of the time, leaving capacity for recovery.
The condition ρ < 1 is essential for the stationary mean formula. It does not mean every observed wait will be short: arrivals can still cluster, and a queue can grow sharply over a short window.
Represent the total work contained in a batch by Y = Bs because each service is a fixed 0.1 hour. E[Y] = 0.2 hour of service per batch and E[Y²] = s²E[B²] = 0.01×5 = 0.05 hour squared.
For a Poisson batch stream, treat each group as a workload jump. The standard stationary mean unfinished-work calculation gives V = λbE[Y²] / [2(1−ρ)]. Substituting our numbers: 3×0.05 / (2×0.4) = 0.1875 hour = 11.25 minutes.
V is the work already in the system seen just before a batch arrives, under these assumptions. It does not yet count the earlier jobs arriving in the same batch as the particular job whose wait we want.
CHAPTER 7 OF 12 · Calculate and check
7. Add the queue formed inside the arriving group
A random job in a batch of size B has, on average, (B−1)/2 same-batch predecessors. Because jobs in larger batches are more likely to be selected, the mean number of predecessors of a random job is E[B(B−1)] / [2E[B]] = 3/4 = 0.75.
Every predecessor needs six minutes, so the extra mean wait inside the arriving group is 0.75×6 = 4.5 minutes. Add this to the 11.25-minute pre-existing workload: E[Wq] = 15.75 minutes.
For this teaching case, average sojourn time from batch arrival to service completion is 15.75 + 6 = 21.75 minutes. Waiting, service and total time are three different measurements.
| Quantity | Calculation | Result |
|---|---|---|
| Batch events | λb | 3 batches/hour |
| Individual flow | 3 × E[B] | 6 jobs/hour |
| Utilisation | 6 × 0.1 | 0.60 |
| Mean prior workload | 3 × 0.05 / 0.8 | 11.25 min |
| Within-batch predecessors | (5−2)/(2×2) × 6 | 4.5 min |
| Mean job wait | 11.25 + 4.5 | 15.75 min |
CHAPTER 8 OF 12 · Change the conditions
8. Change arrivals without changing workload
Now let six jobs per hour arrive as separate Poisson individual events with exactly the same six-minute service requirement. This is an M/D/1 queue, not the same arrival model as the batch queue. Its mean queueing wait is λjs²/[2(1−ρ)] = 6×0.01/0.8 = 0.075 hour = 4.5 minutes.
Or let exactly two jobs arrive simultaneously in every batch, three batches per hour. Then E[B²] = 4; the mean existing workload is (3×0.04)/0.8 = 9 minutes, and the within-pair mean predecessor time is 3 minutes. Total mean wait becomes 12 minutes.
The server remains 60 percent utilised in all three cases. The mean waiting times are 4.5, 12 and 15.75 minutes. Equal rates and equal capacity do not imply equal queueing experience.
CHAPTER 9 OF 12 · Change the conditions
9. Which decisions transfer to real operations?
The same difference appears at clinic check-in, customer support, vehicles arriving at a loading bay, server messages released in batches or students submitting work at a deadline. Record the batch size and its time stamp rather than silently converting each group into an imaginary stream of independent jobs.
Possible responses include appointment smoothing, staging release times, flexing staff when a group arrives, splitting tasks into separate streams, or setting service-level expectations that allow for bursts. These are design hypotheses requiring observation, not an instruction to use one policy everywhere.
If there are several servers, time-varying arrivals, customer abandonment, priority classes or interruptions, use a corresponding model or a simulation; do not transplant this one-server stationary result unchanged.
Mistake one: “Three arrival events per hour means three customers per hour.” It means three groups per hour here; jobs per hour require E[B].
Mistake two: “The average group has two, so every group has two.” Our distribution includes only one or three, and the second moment differs from a constant-size pair.
Mistake three: “Sixty percent utilisation guarantees no delays.” It does not describe the variance or clustering of incoming work. Mistake four: “An arriving batch and a randomly selected job sample the same batch sizes.” They do not; the job-weighted distribution favours larger batches.
CHAPTER 11 OF 12 · Practise and continue
11. Evidence checks, a short exercise and FAQs
For an actual service station, collect an arrival-event timestamp, number of jobs per event, actual service start and end times, queue discipline, interruptions, service variability and time-of-day. Recalculate from the observed event and job denominators. Check whether the arrival process resembles independent Poisson batch events.
Practice: use constant two-job batches at three events per hour and six-minute deterministic service. Compute E[B]=2, E[B²]=4, ρ=0.6 and a 12-minute mean wait under this model. Compare with the answer in Chapter 8.
FAQ: Is the mean wait a maximum? No, it is an average. Are jobs in the same group “Poisson arrivals”? No, they arrive simultaneously. Can a small arrival rate still cause a long queue? Yes, when service-time or batch-size variation and available capacity are unfavourable.
The Queueing Theory subject root supplies the broader vocabulary of arrival rates, utilisation, Little’s Law and service design. The companion equal service means, different mean waits isolates service-time variability rather than group-size variability.
For related thinking, visit Operations Research and Probability. Read the Hong Kong Baptist University queueing lecture notes for M/G/1 workload theory and a section explicitly introducing random-size batch arrivals.
The practical transfer question is simple: do the data describe the same unit of arrival that the formula describes? If not, change the model before changing the staffing.
Continue with Queueing Theory — the subject root, The Gold Standard of Education and How X Works.
Sources and reference links checked 22 October 2026. Numerical examples are deliberately simplified learning models, not predictions for an unverified real facility or material.