A repair desk receives fewer new failures when most of its machines are already broken. That can make its incoming workload look quieter precisely when the fleet is doing badly. A finite-source queue makes this feedback visible: only operating machines can generate the next repair request.
This is an original, hypothetical three-machine calculation, followed by changes in fleet size and repair speed. The exponential failure and repair assumptions are teaching choices. The results describe the stated model; they are not maintenance instructions or evidence about an actual fleet.
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Choose a route below or read in order. All teaching chapters remain open; the index groups let you jump directly to a worked example or check.
Route 1 · Separate a fleet from an outside arrival stream
Chapters 1–2: Separate a fleet from an outside arrival stream.
Route 2 · Build and normalise the four states
Chapters 3–4: Build and normalise the four states.
Route 3 · Read throughput and the arriving failure
Chapters 5–6: Read throughput and the arriving failure.
Route 4 · Change fleet size and repair speed
Chapters 7–8: Change fleet size and repair speed.
Route 5 · Check evidence, practise and continue
Chapters 9–12: Check evidence, practise and continue.
Full chapter index · Try the transfer checks · Queueing Theory subject root
Chapter contents
Separate a fleet from an outside arrival stream · Chapters 1–2
Build and normalise the four states · Chapters 3–4
Read throughput and the arriving failure · Chapters 5–6
Change fleet size and repair speed · Chapters 7–8
Check evidence, practise and continue · Chapters 9–12
CHAPTER 1 OF 12 · Separate a fleet from an outside arrival stream
1. Count the machines that can still fail
Imagine three identical machines sharing one repairer. A machine alternates between operating and being down. Down time includes waiting for repair and repair itself. A machine already down cannot produce a second independent failure request while it remains in that state.
Let n be the number down. Then 3 − n are operating. If each operating machine has failure intensity α = 1 per operating hour, the total intensity of the next failure is (3 − n)α. At n = 0 it is 3 per hour; at n = 2 it is 1 per hour; at n = 3 it is zero.
Intensity is a conditional short-interval rate. It is not a promise that a failure occurs every twenty minutes when all three machines are up. With independent exponential operating times, the minimum of their remaining times is exponential with rate equal to the sum of the active rates.
Keep three counts separate: the fixed fleet N, the machines down n, and the machines actually receiving repair. Our single repairer serves one at a time. A number such as “two in the system” includes one waiting machine and one being repaired, not two simultaneous repairs.
CHAPTER 2 OF 12 · Separate a fleet from an outside arrival stream
2. State the operating and repair assumptions
Take N = 3, α = 1 per hour and repair rate μ = 2 per hour. Mean operating time before failure is 1/α = one hour of operation. Mean active repair duration is 1/μ = half an hour. Calendar time between failures for one machine also includes its preceding down time.
Assume independent exponential operating periods and repair durations, immediate reporting of failures, first-come service, an always available repairer, and restoration to the same operating condition after each repair. Every failed machine joins; there is no rejected request, spare substitution or abandonment.
These assumptions make n alone sufficient to describe the transition rates. If failure risk depends on age or the repairer is away, simply counting down machines may no longer give a Markov description. Additional state information or a different model would be needed.
The population limit comes from the source of work. It is different from an external stream being turned away by a full waiting room. Compare finite buffers, blocking and admitted throughput to keep those two mechanisms separate.
CHAPTER 3 OF 12 · Build and normalise the four states
3. Build the birth-and-death ledger
A birth increases the down count by one: 0 → 1 at rate 3, 1 → 2 at rate 2, and 2 → 3 at rate 1 per hour. A death is a completed repair, reducing the count by one at rate 2 whenever n is positive. The words birth and death refer to state transitions, not physical creation or destruction.
For stationary probabilities p₀, p₁, p₂ and p₃, local flow balance gives 3p₀ = 2p₁, 2p₁ = 2p₂, and p₂ = 2p₃. These equalities compare long-run crossings of each boundary in opposite directions. They do not say that failures and repairs alternate strictly.
Express every probability relative to p₀: p₁ = 1.5p₀, p₂ = 1.5p₀ and p₃ = 0.75p₀. The weights sum to 4.75, so p₀ = 1/4.75. Multiplying numerator and denominator by four gives the convenient exact probabilities (4, 6, 6, 3)/19.
There are finitely many states and all rates connecting neighbouring states are positive. A unique stationary distribution exists here even though the all-up failure intensity 3 exceeds the repair rate 2. The feedback prevents an unbounded down count; it does not guarantee acceptable availability.
CHAPTER 4 OF 12 · Build and normalise the four states
4. Check the state probabilities before interpreting them
Our four probabilities are p₀ = 4/19, p₁ = 6/19, p₂ = 6/19 and p₃ = 3/19. First add them: the total is one. Next check each local balance using rates with the same hourly units. For example, 3×4/19 = 2×6/19.
The probability that all three machines are down at an arbitrary stationary time is 3/19, about 15.79%. The repairer is busy whenever n is positive, so utilisation is 15/19, about 78.95%. Neither percentage is a completed-repair rate.
| Machines down n | Operating machines | Next-failure intensity per hour | Time probability |
|---|---|---|---|
| 0 | 3 | 3 | 4/19 |
| 1 | 2 | 2 | 6/19 |
| 2 | 1 | 1 | 6/19 |
| 3 | 0 | 0 | 3/19 |
Mean down count is L = (0×4 + 1×6 + 2×6 + 3×3)/19 = 27/19, about 1.421 machines. A fractional long-run mean is compatible with a whole-number count at every instant. The full state distribution supplies information that this mean leaves out.
Notice the operational interpretation: a repairer can have substantial idle time while the fleet still spends some time completely down. Idle periods occur when all machines happen to be operating. They cannot be carried forward to repair a future failure before it occurs.
CHAPTER 5 OF 12 · Read throughput and the arriving failure
5. Reconcile throughput with fleet availability
Completed-repair throughput is X = μP(n > 0) = 2×15/19 = 30/19 repairs per hour, about 1.579. Using the all-up intensity 3 as the long-run entry rate would count failures from machines that are sometimes unavailable to fail.
There is an independent check. Mean operating count is N − L = 3 − 27/19 = 30/19. Each operating machine fails at rate one, so long-run failure flow is α(N − L) = 30/19 per hour. It matches the repair flow, as it must in this stationary closed fleet.
Mean fraction of machines operating is (N − L)/N = 10/19, about 52.63%. This is a time-and-machine availability fraction. It differs from P(n = 0), the probability the entire fleet is up, and from the repairer’s busy fraction.
A fall in observed repair arrivals is therefore ambiguous. It could reflect improved reliability, fewer operating hours, a smaller fleet, missing reports or machines spending longer down. Measure operating exposure before treating a lower calendar-time failure count as a reliability improvement.
CHAPTER 6 OF 12 · Read throughput and the arriving failure
6. Calculate down time with the correct flow
Apply Little’s Law to the down-machine system. Its entry is failure and its exit is completed repair. Mean time from failure to return to operation is W = L/X = (27/19)/(30/19) = 0.9 hour, or 54 minutes.
Mean active repair is 30 minutes. Subtracting it gives mean waiting before repair Wq = 24 minutes. The waiting-only population is Lq = L − P(n > 0) = 12/19. Dividing Lq by the same throughput 30/19 again gives 0.4 hour.
Using 3 failures per hour instead would produce L/3 ≈ 28.42 minutes, shorter than the model’s 30-minute mean active repair. That inconsistency is a useful alarm. It comes from pairing a state-specific maximum entry intensity with a long-run population mean.
For one exchangeable machine, a mean cycle contains one operating hour plus 0.9 down hour. Its operating fraction is 1/1.9 = 10/19. This second route recovers the fleet availability. The cycle interpretation connects a queueing time with the equipment outcome people actually need.
CHAPTER 7 OF 12 · Change fleet size and repair speed
7. An arriving failure does not see the time-average state
At a random stationary time, the repairer is busy with probability 15/19. It is tempting to use that as the probability a newly failed machine must wait. That shortcut is wrong here because arrival intensity depends on the state.
Weight each time probability by the failure rate active in that state. The unnormalised weights for an arrival seeing n = 0, 1, 2, 3 are (3×4, 2×6, 1×6, 0×3)/19. Their sum is the throughput 30/19.
The arrival-seen probabilities are therefore 0.4, 0.4, 0.2 and 0. A new failure waits if it finds at least one machine already down, giving probability 0.6. State three occupies time but generates no new failure, so a genuine failure arrival cannot see it just before arriving.
Under first-come service and exponential repair, seeing one down machine means a mean half-hour wait; seeing two means a mean one-hour wait. Weighting gives 0.4×0.5 + 0.2×1 = 0.4 hour, the same 24 minutes obtained earlier. This is a direct check of the boundary and arrival weighting.
CHAPTER 8 OF 12 · Change fleet size and repair speed
8. Change the repair speed and recompute the whole system
Keep three machines and α = 1 but double μ to 4 per hour. Mean active repair becomes fifteen minutes. The relative stationary weights are now 1, 3/4, 3/8 and 3/32. Normalisation gives probabilities (32, 24, 12, 3)/71.
The mean down count falls to 57/71, while throughput rises to 156/71, about 2.197 repairs per hour. Mean down time is 57/156 hour, about 21.92 minutes, including about 6.92 minutes waiting. Mean operating fraction rises to 52/71, about 73.24%.
More repairs per calendar hour now accompany better availability. Faster repair returns machines to operation sooner, which exposes them to future failures sooner. Interpreting the increased repair count alone as deteriorating reliability would reverse the model’s actual improvement.
The repairer’s busy fraction falls from 15/19 to 39/71. A useful repair upgrade can reduce utilisation while increasing throughput and machine availability. Compare outcomes through their definitions rather than treating a busy percentage as the sole objective.
CHAPTER 9 OF 12 · Check evidence, practise and continue
9. Change fleet size without freezing arrivals
Add a fourth identical machine while returning to μ = 2 and α = 1. The five relative weights are 1, 2, 3, 3 and 1.5, yielding stationary probabilities (2, 4, 6, 6, 3)/21. The source has changed, so the previous four-state distribution cannot be reused.
Mean down count is 46/21 and throughput is 38/21 per hour. Mean down time is (46/21)/(38/21) = 23/19 hour, about 72.63 minutes. Mean operating count rises from about 1.579 to about 1.810, but the operating fraction falls to 19/42, about 45.24%.
| Fleet and repair rate | Repair throughput | Mean down time | Mean fraction operating |
|---|---|---|---|
| N=3, μ=2 | 30/19 ≈ 1.579 | 54.00 min | 10/19 ≈ 52.63% |
| N=3, μ=4 | 156/71 ≈ 2.197 | 21.92 min | 52/71 ≈ 73.24% |
| N=4, μ=2 | 38/21 ≈ 1.810 | 72.63 min | 19/42 ≈ 45.24% |
The added machine increases average operating capacity in machine-count terms, while creating more competition for repair and reducing the available fraction. Those are different comparisons. Which one matters depends on whether the task requires a given output, a reserve or a per-machine availability target.
This is a changed-context transfer test: identify what became endogenous when the fleet grew. Do not transfer a fixed external arrival rate into a model whose arrival source is the fleet itself.
CHAPTER 10 OF 12 · Check evidence, practise and continue
10. Check the evidence that would justify the model
Estimate failure exposure from operating hours, not from fleet size multiplied by all calendar hours when downtime is substantial. Record failures, repair starts and returns to operation, together with machines already down at the observation window’s edges.
For each state n, compare recorded time spent there with failures occurring from that state. Enough suitable data would allow an empirical check of whether failure intensity roughly follows (N − n)α. Common-cause failures, changing operating loads and seasonal conditions can defeat the independent-identical model.
Separate active repair from waiting for parts, approvals or a specialist. If these delays dominate, increasing the hands-on repair rate may not create the improvement predicted by our simplified continuously available server. The state description needs to represent the real source of delay.
A simulation should reproduce the small exact case first: probabilities summing to one, matched flow rates, and Little’s Law using actual entries. Agreement with a formula checks implementation under shared assumptions; comparing held-out operating records checks whether those assumptions describe the fleet.
CHAPTER 11 OF 12 · Check evidence, practise and continue
11. Practise without a formula shortcut
Check one: for a single machine with α = 1 and μ = 2, only states zero and one exist. Balance gives probabilities 2/3 and 1/3. Throughput is 2/3 per hour, availability is 2/3, down time is half an hour, and waiting is zero. One machine cannot queue behind itself in this model.
Check two: in the three-machine baseline, what fraction of arriving failures finds two down machines? The answer is 0.2, not the time fraction 6/19. State the rate-weighting calculation before giving the decimal.
Check three: a report says repairs per month fell after repair times doubled. Does that establish more reliable machines? No. Longer down periods can suppress new failures by reducing operating exposure. Ask for exposure-normalised failure information, availability and the repair-time record.
To explain the subject in a new setting, replace machines with a fixed population of terminals alternating between independent thinking and a shared service request. The source-feedback structure may transfer. The exponential and independence assumptions still have to be checked rather than carried across by analogy.
CHAPTER 12 OF 12 · Check evidence, practise and continue
12. Questions, connected reading and sources
Is this the same as a finite buffer?
No. A finite buffer limits the jobs admitted from an outside source. Our fleet limits the sources that can produce work, and a down machine temporarily leaves that source population. Both models have finitely many states for different reasons.
Must total potential failures be below repair capacity?
Not for existence of a stationary distribution in this positive-rate finite-state model. All-up intensity can exceed μ. Availability may then be poor, and the intended service requirement still needs assessment.
Can I replace exponential repair with its mean?
Not in this birth-and-death derivation without further justification. Nonexponential repair can require remaining-duration or supplementary-state information. Some other models have special insensitivity results; that is not permission to assume one here.
Which number should a repair report lead with?
Lead with the outcome being assessed, supported by operating exposure, availability, down-time distribution and repair flow. No single number contains all four. Keep any claimed improvement tied to comparable fleet and operating conditions.
Choose the next question
Follow entry and measurement boundaries
- Finite Buffers, Blocking and Admitted Throughput
- Little’s Law and the Waiting-Room Boundary That Changes the Answer
- Retries, Attempt Rates and Unique Jobs
- Abandonment and the Waits Hidden by Answered-Only Statistics
Change service and arrival structure
- Equal Mean Service Times, Different Mean Waits
- Two Servers, One Pooled Queue or Separate Lines?
- Priority Rules, Class Waits and Hidden Trade-Offs
- Server Vacations and the Residual Delay Seen by an Arrival
- Time-Varying Arrivals and Capacity That Arrives Too Late
- Batch Arrivals, Event Rates and Job Counts
- When Batch Size and Job Duration Move Together
- Whole-Batch Rejection and the Difference Between Event and Job Loss Rates
Sources and scope
János Sztrik’s Basic Queueing Theory, section 3.2 presents the finite-source machine-interference model. N. K. Jaiswal’s Finite-Source Queuing Models research report supplies the finite-population context. Our three-machine probabilities, rate-weighted arrival ledger and altered-fleet comparisons are original worked teaching calculations.
Sources checked 22 October 2026. Return to The Gold Standard of Queueing Theory, The Gold Standard of Education and How X Works to practise carrying definitions and evidence checks into another system.