Unit 05 — Queuing Theory
The single-server Poisson–exponential queue: converting the story into λ and μ, the six operating characteristics, every probability question the assignment asks, stability, and the cost of waiting.
Queuing was Q5A (5 marks) in both papers, and both times it was the single-server Poisson–exponential model: given λ and μ, find the traffic intensity and all queue parameters, then say what happens if arrivals rise Final QP p. 4 Re-Exam QP p. 4. Assignment 5 adds the probability questions (P₀, Pₙ, "at least k") and one cost comparison.
- ρ, Ls, Lq, Ws, Wq — six formulas, all from λ and μ. Both papers solved →
- "What if λ increases?" — recompute and comment on the waiting time, and on stability (ρ must stay below 1).
- Assignment 5 — five full problems with every probability. Solved →
1. The Structure of a Queuing System Core syllabus concept
1Understand the Concept
A queue forms whenever customers arrive faster, at some moment, than they can be served. Queuing theory does not remove the queue — it predicts how long it will be and how long people will wait, so management can decide whether adding capacity is worth the cost.
2The parts of every queuing system
| Component | What it describes | In the M/M/1 model |
|---|---|---|
| Calling population | Where customers come from | Infinite — one more arrival does not change the arrival rate |
| Arrival process | How customers arrive | Random, Poisson-distributed, mean rate λ per unit time |
| Queue discipline | Who is served next | FCFS — first come, first served |
| Service mechanism | How long service takes, how many servers | One server, service time exponential, mean rate μ per unit time |
| Queue capacity | How many can wait | Unlimited |
In Kendall's notation this is written (M/M/1) : (∞/FCFS) — M = Markovian (Poisson arrivals / exponential service), 1 server, infinite capacity, first-come-first-served.
3The two inputs, and the conversion that catches people out
| Symbol | Meaning | How it is usually given |
|---|---|---|
| λ (lambda) | Mean arrival rate — customers per hour | "10 patients per hour" → λ = 10. "One every 10 minutes" → λ = 60/10 = 6 per hour. |
| μ (mu) | Mean service rate — customers one server can handle per hour | "Average service time 4 minutes" → μ = 60/4 = 15 per hour. |
Times in minutes are times between events, not rates. Rate = 60 ÷ time-in-minutes. Write λ and μ per hour at the top of every answer, then every W you compute comes out in hours — multiply by 60 for minutes.
2. The M/M/1 Formulas Asked in the 2025-26 papers
1Understand the Concept
Everything follows from one ratio, ρ = λ/μ, the traffic intensity or utilisation: the fraction of time the server is busy. It also equals the probability that an arriving customer has to wait.
2Memorise these
3Worked Example — Final Exam Q5A
Stock transactions arrive at a broker's desk at a mean rate of 20 per hour. Each order takes on average 2 minutes to process. (1) Traffic intensity? (2) All queue parameters. (3) If arrivals rise to 25 per hour, how are the waiting times affected?
- Step 1 — Rates per hour λ = 20 per hour Service time 2 min ⟹ μ = 60/2 = 30 per hour
- Step 2 — Traffic intensity ρ = λ/μ = 20/30 = 0.667 The broker is busy 66.7% of the time; idle P₀ = 0.333
- Step 3 — Queue parameters Ls = λ/(μ − λ) = 20/10 = 2 orders Lq = λ²/[μ(μ − λ)] = 400/(30 × 10) = 1.33 orders Ws = 1/(μ − λ) = 1/10 hr = 0.1 hr = 6 minutes Wq = λ/[μ(μ − λ)] = 20/300 hr = 0.0667 hr = 4 minutesCheck: Ls − Lq = 2 − 1.33 = 0.667 = ρ ✓ Ws − Wq = 6 − 4 = 2 min = 1/μ ✓
- Step 4 — Arrivals rise to 25 per hour
ρ = 25/30 = 0.833
Ws = 1/(30 − 25) = 0.2 hr = 12 minutes
Wq = 25/(30 × 5) = 0.167 hr = 10 minutes
A 25% rise in arrivals more than doubles the waiting time (4 → 10 minutes). Queues are non-linear: as ρ approaches 1, waiting explodes. That is the comment the examiner wants. ✓ Matches the key Key p. 5.
- Using minutes for one rate and hours for the other. λ = 20/hr and "μ = 2 minutes" cannot go in the same formula.
- Swapping Ls and Lq formulas. Lq has λ² on top; Ls has λ. Ws has 1 on top; Wq has λ.
- Leaving W in hours. 0.0667 hr means nothing to a manager — write 4 minutes.
- Applying the formulas when λ ≥ μ. They give negative or infinite answers — see concept 4.
3. Probability Questions — Pₙ, "at least", "fewer than" Core syllabus concept
1The four formulas
- "finds the counter idle" → P₀ = 1 − ρ
- "has to wait" → n ≥ 1 → ρ
- "at least 4" → n ≥ 4 → ρ⁴ · "more than 4" → n ≥ 5 → ρ⁵
- "fewer than 3" → n = 0, 1, 2 → 1 − ρ³
- "finds exactly 2 already in the system" → P₂ — the arriving person is not counted
2Worked Example
Customers arrive at 12 per hour; one clerk serves 16 per hour. Find P(idle), P(exactly 2 in system), P(at least 3), P(more than 3) and P(fewer than 3).
- Step 1 — ρ ρ = 12/16 = 0.75
- Step 2 — Each probability P₀ = 1 − 0.75 = 0.25 P₂ = (1 − 0.75)(0.75)² = 0.25 × 0.5625 = 0.1406 P(n ≥ 3) = 0.75³ = 0.4219 P(n > 3) = P(n ≥ 4) = 0.75⁴ = 0.3164 P(n < 3) = 1 − 0.75³ = 0.5781
- Step 3 — Sanity check
P(n < 3) + P(n ≥ 3) = 0.5781 + 0.4219 = 1 ✓ — complementary events must add to one.
4. Stability — Why ρ Must Be Below 1 Core syllabus concept
1Understand the Concept
If customers arrive as fast as, or faster than, the server can work (λ ≥ μ), the queue never clears: it grows without limit, and the formulas break down (μ − λ ≤ 0). A queue is stable only when ρ = λ/μ < 1.
The Re-Exam asked exactly this: "Analyse how an increase in arrival rate during peak hours (say 20 customers per hour) would affect system stability" for a mall counter with μ = 60 per hour Re-Exam QP p. 4.
2How to answer a stability question
- Step 1 — Compute ρ before and after Before: λ = 10, μ = 60 ⟹ ρ = 0.167 After: λ = 20, μ = 60 ⟹ ρ = 0.333
- Step 2 — Judge ρ = 0.333 < 1 ⟹ the system stays STABLE
- Step 3 — Quantify and comment
Ws: 1/(60 − 10) = 1.2 min → 1/(60 − 20) = 1.5 min
Wq: 0.2 min → 0.5 min
Doubling arrivals keeps the counter stable but multiplies the queue wait by 2.5. Add the general rule: if peak λ ever reached 60 per hour (ρ = 1), the queue would grow indefinitely and a second counter would be essential.
If arrivals and service are both constant (one customer exactly every 5 minutes, service exactly 4 minutes), there is no randomness: as long as service time ≤ inter-arrival time nobody ever waits, and the moment service time exceeds it the queue grows by a fixed amount every cycle. Real queues form even when ρ < 1 only because arrivals and service vary — which is what the Poisson–exponential model captures. The session plan lists deterministic models; one paragraph like this is the expected depth.
5. Cost Analysis — Is a Faster Server Worth It? Core syllabus concept
1Understand the Concept
Faster service costs money; waiting customers cost goodwill. The decision rule is simply: install the faster system if the waiting cost it saves is greater than the extra cost of running it. Compare both per day (or per hour) — never mix the two.
2The method
3Worked Example
Customers arrive at 6 per hour; the clerk serves 8 per hour. A new system raises the service rate to 12 per hour at an extra ₹40 per day. Goodwill lost while a customer waits is ₹0.10 per minute. The office works 8 hours a day. Should the new system be installed?
- Step 1 — Wq under each system Old: Wq = 6/[8(8 − 6)] = 0.375 hr = 22.5 min New: Wq = 6/[12(12 − 6)] = 0.0833 hr = 5 min
- Step 2 — Customers per day N = 6 × 8 = 48 customers
- Step 3 — Waiting cost per day Old: 48 × 22.5 × 0.10 = ₹108 New: 48 × 5 × 0.10 = ₹24 Saving = ₹84 per day
- Step 4 — Decide
Saving ₹84 > extra cost ₹40 ⟹ install; net gain ₹44 per day
Note it is Wq (time waiting before service) that carries the goodwill cost, unless the question says time in the system.