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QUANTITATIVE TECHNIQUES›ASSIGNMENTS›ASSIGNMENT 5

Assignment 5 — Queuing Theory (M/M/1)

Five single-server queues: every probability the questions ask for, the four averages, the arrival rate that justifies a second counter, and a before-and-after cost comparison.

PROBLEMS: 5 UNIT: UNIT 05 — QUEUING THEORY Original: Assignment 5
◆ The method for every problem on this page
  1. Convert to rates per hour. λ = arrivals per hour ("one every 10 min" ⟹ 6/hr). μ = 60 ÷ average service time in minutes.
  2. ρ = λ/μ — must be < 1. It is P(server busy) = P(an arrival waits); P₀ = 1 − ρ is P(idle).
  3. Probabilities: Pₙ = (1 − ρ)ρⁿ · P(n ≥ k) = ρᵏ · P(n > k) = ρᵏ⁺¹ · P(n < k) = 1 − ρᵏ.
  4. Averages: Ls = λ/(μ − λ) · Lq = λ²/[μ(μ − λ)] · Ws = 1/(μ − λ) · Wq = λ/[μ(μ − λ)] — W in hours, so ×60 for minutes.
  5. Check Ls − Lq = ρ and Ws − Wq = 1/μ before writing the answer.

All formulas and the reasoning: Unit 5 · Queuing Theory.

★ Tricks that save time in the exam
  • "Arriving person finds exactly 2 already in the system" is P₂ — the arriving person is not counted.
  • "More than 4" means 5 or more ⟹ ρ⁵. "Between 2 and 4 inclusive" = P(n ≥ 2) − P(n ≥ 5) = ρ² − ρ⁵.
  • Fractions keep it exact: with ρ = 2/3, P₂ = (1/3)(4/9) = 4/27 — convert to a decimal only at the end.
  • Cost questions use Wq (time waiting before service) unless the question says time in the system.

Q1 — Hospital pharmacy counter

Question

Patients arrive (Poisson) at 10 per hour; service is exponential with mean 4 minutes. Find: (1) P(pharmacist idle); (2) P(an arrival waits); (3) P(arrival finds exactly 2 in the system); (4) P(fewer than 3 in the system); (5) P(at least 4); (6) Lq; (7) Ls; (8) Wq; (9) Ws.

Assignment 5 p. 1

Solution
  1. Rates and ρ λ = 10 patients per hour Service time 4 min ⟹ μ = 60/4 = 15 per hour ρ = λ/μ = 10/15 = 0.6667 (< 1, so the queue is stable)
  2. (1)–(5) Probabilities (1) P₀ = 1 − ρ = 1/3 = 0.3333 (2) P(wait) = ρ = 2/3 = 0.6667 (3) P₂ = (1 − ρ)ρ² = (1/3)(4/9) = 4/27 = 0.1481 (4) P(n < 3) = 1 − ρ³ = 1 − 8/27 = 19/27 = 0.7037 (5) P(n ≥ 4) = ρ⁴ = 16/81 = 0.1975
  3. (6)–(9) Averages Ls = λ/(μ − λ) = 10/5 = 2.0000 patients Lq = λ²/[μ(μ − λ)] = 100/(15 × 5) = 1.3333 patients Ws = 1/(μ − λ) = 1/5 hr = 0.2000 hr = 12 min Wq = λ/[μ(μ − λ)] = 10/75 hr = 0.1333 hr = 8 min
  4. Interpretation

    The pharmacist is busy two-thirds of the time; a patient typically waits 8 minutes and spends 12 minutes at the counter in all, with about 1.33 patients waiting at any moment.

Q2 — Airline centre — when is a second counter justified?

Question

Arrivals are Poisson with a mean of 10 minutes between arrivals; service is exponential with mean 3 minutes. Find: (1) P(an arrival waits); (2) average time in the system; (3) average queue length; (4) by how much must the arrival rate rise before an arrival waits at least 5 minutes (the trigger for a second centre)? (5) P(fewer than 3 persons at the centre).

Assignment 5 p. 1

Solution
  1. Rates and ρ One arrival every 10 min ⟹ λ = 60/10 = 6 per hour Service time 3 min ⟹ μ = 60/3 = 20 per hour ρ = λ/μ = 6/20 = 0.3000 (< 1, so the queue is stable)
  2. (1)–(3) (1) P(wait) = ρ = 0.3000 (2) Ws = 1/(20 − 6) = 1/14 hr = 4.29 min (3) Lq = 36/(20 × 14) = 0.1286 persons
  3. (4) Arrival rate that makes Wq = 5 minutes Set Wq = λ/[μ(μ − λ)] = 5 min = 1/12 hr with μ = 20 λ / [20(20 − λ)] = 1/12 ⟹ 12λ = 400 − 20λ ⟹ 32λ = 400 λ = 12.5 per hour (one arrival every 4.8 minutes) Increase needed = 12.5 − 6 = 6.5 arrivals per hour

    Once arrivals exceed 12.5 an hour — roughly double today's flow — the expected wait passes 5 minutes and the second centre is justified.

  4. (5) P(n < 3) = 1 − ρ³ = 1 − 0.027 = 0.9730

Q3 — Reservation clerk — should the computer be installed?

Question

Customers arrive at 8 per hour; the clerk can serve 12 per hour. (1) Average wait before service? (2) Average number waiting? (3) Other queue parameters. (4) A computer system would cut service time from 5 to 3 minutes for an extra ₹50 per day. Goodwill lost is 12 paise per minute a customer waits before being served; the day is 8 hours. Install it?

Assignment 5 p. 1

Solution
  1. Rates and ρ λ = 8 customers per hour μ = 12 per hour (service time 5 min) ρ = λ/μ = 8/12 = 0.6667 (< 1, so the queue is stable)
  2. (1)–(3) Present system Ls = λ/(μ − λ) = 8/4 = 2.0000 customers Lq = λ²/[μ(μ − λ)] = 64/(12 × 4) = 1.3333 customers Ws = 1/(μ − λ) = 1/4 hr = 0.2500 hr = 15 min Wq = λ/[μ(μ − λ)] = 8/48 hr = 0.1667 hr = 10 minP₀ = 0.3333 — the clerk is idle one-third of the time
  3. (4) With the computer: μ = 60/3 = 20 per hour Wq = 8/[20(20 − 8)] = 8/240 hr = 2 min
  4. (4) Cost comparison per day Customers per day = 8 × 8 = 64 Goodwill now: 64 × 10 min × ₹0.12 = ₹76.8 Goodwill with it: 64 × 2 min × ₹0.12 = ₹15.36 Saving = ₹61.44 per day > extra cost ₹50

    Install the computer system — it saves ₹61.44 of goodwill for ₹50, a net gain of ₹11.44 per day.

Q4 — Restaurant pickup counter — before and after a faster handover

Question

Delivery partners arrive at 15 per hour; handover takes 3 minutes on average. Find P(idle), P(wait), P(exactly 1), P(fewer than 3), P(at least 3), P(more than 4), P(between 2 and 4 inclusive), Lq, Wq, Ls, Ws. Then with handover cut to 2.5 minutes: new P(wait), P(more than 4), Lq, Wq — and comment.

Assignment 5 p. 2

Solution
  1. Rates and ρ λ = 15 per hour Service time 3 min ⟹ μ = 20 per hour ρ = λ/μ = 15/20 = 0.7500 (< 1, so the queue is stable)
  2. Probabilities (ρ = 0.75) P₀ = 0.25 P(wait) = ρ = 0.75 P₁ = 0.25 × 0.75 = 0.1875 P(n < 3) = 1 − 0.75³ = 0.5781 P(n ≥ 3) = 0.75³ = 0.4219 P(n > 4) = 0.75⁵ = 0.2373 P(2 ≤ n ≤ 4) = 0.75² − 0.75⁵ = 0.5625 − 0.2373 = 0.3252
  3. Averages Ls = λ/(μ − λ) = 15/5 = 3.0000 partners Lq = λ²/[μ(μ − λ)] = 225/(20 × 5) = 2.2500 partners Ws = 1/(μ − λ) = 1/5 hr = 0.2000 hr = 12 min Wq = λ/[μ(μ − λ)] = 15/100 hr = 0.1500 hr = 9 min
  4. Faster handover: 2.5 min ⟹ μ = 60/2.5 = 24 per hour ρ = 15/24 = 0.6250 P(wait) = 0.6250 P(n > 4) = 0.625⁵ = 0.0954 Lq = 225/(24 × 9) = 1.0417 Wq = 15/(24 × 9) hr = 4.17 min
  5. Comment
    Measure3 min2.5 minChange
    P(wait)0.750.6250−17%
    P(more than 4)0.23730.0954−60%
    Lq2.251.0417−54%
    Wq9 min4.17 min−54%

    A 17% faster handover cuts the queue and the wait by more than half — the non-linearity of queues working in the restaurant's favour.

Q5 — College examination section

Question

Students arrive at 8 per hour; the clerk serves 12 per hour. Find P(idle), P(wait), P(arrival finds at least 2), P(arrival finds fewer than 4), P(exactly 3), P(more than 5), Lq, Ls, Wq, Ws.

Assignment 5 p. 2

Solution
  1. Rates and ρ λ = 8 students per hour μ = 12 per hour ρ = λ/μ = 8/12 = 0.6667 (< 1, so the queue is stable)
  2. Probabilities (ρ = 2/3) P₀ = 1/3 = 0.3333 P(wait) = 2/3 = 0.6667 P(n ≥ 2) = ρ² = 4/9 = 0.4444 P(n < 4) = 1 − ρ⁴ = 65/81 = 0.8025 P₃ = (1/3)(8/27) = 8/81 = 0.0988 P(n > 5) = ρ⁶ = 64/729 = 0.0878
  3. Averages Ls = λ/(μ − λ) = 8/4 = 2.0000 students Lq = λ²/[μ(μ − λ)] = 64/(12 × 4) = 1.3333 students Ws = 1/(μ − λ) = 1/4 hr = 0.2500 hr = 15 min Wq = λ/[μ(μ − λ)] = 8/48 hr = 0.1667 hr = 10 min