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Operations Management and Strategic Management · Scheduling and Queuing Models

M/M/1 Single-Channel Queuing Model: Formulas and Numericals

Updated 10 October 2026 · Fact-checked

The M/M/1 model describes one server with random arrivals (rate λ) and random service times (rate μ), served first come, first served. Check that λ < μ, find utilisation ρ = λ ÷ μ, then apply Ls = λ ÷ (μ − λ), Lq = λ² ÷ [μ(μ − λ)], Ws = 1 ÷ (μ − λ), Wq = λ ÷ [μ(μ − λ)], and P(n) = (1 − ρ)ρⁿ.

Understand Single-Channel Queuing Model (M/M/1)

A queue forms when customers arrive faster than they can be served for a while. Examples are cars at a toll booth, patients at one counter, or jobs waiting for one machine. Queuing theory gives numbers for how long people wait and how busy the server is, so a manager can decide whether to add capacity.

The label M/M/1 has three parts. The first M means arrivals follow a Poisson process (random, independent arrivals at an average rate λ per unit time). The second M means service times are exponential (random, with an average service rate μ customers per unit time). The 1 means a single server, or single channel.

The usual assumptions are: an infinite population of customers, an infinite queue length allowed, first come first served discipline, no balking or reneging, and a steady state where the system has settled down. Steady state needs λ < μ. If λ ≥ μ, the queue grows without limit and the formulas do not apply.

The key ratio is utilisation ρ = λ ÷ μ. It is the fraction of time the server is busy. Everything else is built from λ and μ. Keep the time unit the same for both, such as per hour. Convert minutes to hours if needed.

Key rules to remember

Utilisation (traffic intensity)
ρ = λ ÷ μ
Probability the server is busy. Must be less than 1. Idle probability = 1 − ρ = P(0).
Average number in the system
Ls = λ ÷ (μ − λ) = ρ ÷ (1 − ρ)
Includes the customer being served.
Average number in the queue
Lq = λ² ÷ [μ(μ − λ)] = ρ² ÷ (1 − ρ)
Only those waiting, not the one in service. Also Lq = Ls − ρ.
Average time in the system
Ws = 1 ÷ (μ − λ)
Waiting plus service. Also Ws = Ls ÷ λ.
Average waiting time in the queue
Wq = λ ÷ [μ(μ − λ)] = ρ ÷ (μ − λ)
Also Wq = Lq ÷ λ and Wq = Ws − 1/μ.
Probability of n customers in the system
P(n) = (1 − ρ) × ρⁿ
For n = 0, P(0) = 1 − ρ. Probability of more than k customers = ρ^(k+1).
Average waiting time of those who must wait
W(wait | wait > 0) = 1 ÷ (μ − λ)
Average time in the queue for customers who actually have to wait. Use only when asked.

How to solve Single-Channel Queuing Model (M/M/1) questions

Use this order for any M/M/1 question. It keeps units consistent and avoids formula mix-ups.

  1. 1Identify λ (arrival rate) and μ (service rate). If the question gives an average service time, take μ = 1 ÷ that time. If it gives average time between arrivals, take λ = 1 ÷ that time.
  2. 2Convert both to the same time unit, usually per hour.
  3. 3Check λ < μ. State it in one line, as it shows the steady state condition.
  4. 4Compute ρ = λ ÷ μ and, if asked, idle probability 1 − ρ.
  5. 5Compute Ls and Lq, then Ws and Wq using the formulas, or use Little's relations (L = λ × W) as a cross-check.
  6. 6For probabilities use P(n) = (1 − ρ)ρⁿ. For 'more than k', use ρ^(k+1).
  7. 7Convert times back to minutes if the question asks for it, and write the answer with units and a one-line interpretation.

Quickest way: Compute ρ first, then derive everything from it

When to use it: Use when several measures are asked in one question and time is short.

  1. Write λ, μ and ρ = λ ÷ μ.
  2. Get Ls = ρ ÷ (1 − ρ).
  3. Get Lq = Ls − ρ.
  4. Get Ws = Ls ÷ λ and Wq = Lq ÷ λ.
  5. Check Ws − Wq = 1 ÷ μ. If it fails, you made an arithmetic slip.

Common mistakes in Single-Channel Queuing Model (M/M/1)

  • Using different time units for λ and μ, for example λ per hour and service time in minutes.

    The question mixes units, and students plug numbers straight in.

    Fix: Convert both to per hour first. Write the units next to λ and μ.

  • Taking μ as the service time instead of the service rate.

    The question says 'service time 5 minutes' and students write μ = 5.

    Fix: μ = 1 ÷ average service time. For 5 minutes, μ = 12 per hour.

  • Confusing Ls with Lq, or Ws with Wq.

    The formulas look alike and the 's' and 'q' labels are forgotten.

    Fix: s means in the system (includes service), q means in the queue only. Remember Ws = Wq + 1/μ and Ls = Lq + ρ.

  • Applying the formulas when λ ≥ μ.

    Students skip the stability check.

    Fix: Always check λ < μ first. If it fails, say the queue grows without limit and steady-state measures do not exist.

  • Finding P(more than k) by adding many P(n) values and making errors.

    Students do not know the shortcut.

    Fix: For M/M/1, P(n > k) = ρ^(k+1). For 'at least k', use ρᵏ.

  • Leaving the time answer in hours when the question wants minutes.

    Formulas give results in the same unit as the rates.

    Fix: Multiply hours by 60 at the end and write the unit.

Worked examples

Example 1

Customers arrive at a single-counter service desk at an average of 6 per hour (Poisson). Service time is exponential with an average of 6 minutes per customer. Find: (a) utilisation, (b) average number in the system, (c) average number in the queue, (d) average time in the system, (e) average waiting time in the queue.

Show the solution
  1. λ = 6 per hour. Average service time = 6 minutes, so μ = 60 ÷ 6 = 10 per hour.
  2. Check: λ = 6 < μ = 10, so steady state exists.
  3. (a) ρ = λ ÷ μ = 6 ÷ 10 = 0.6.
  4. (b) Ls = λ ÷ (μ − λ) = 6 ÷ 4 = 1.5 customers.
  5. (c) Lq = λ² ÷ [μ(μ − λ)] = 36 ÷ (10 × 4) = 0.9 customers. Check: Ls − ρ = 1.5 − 0.6 = 0.9.
  6. (d) Ws = 1 ÷ (μ − λ) = 1 ÷ 4 = 0.25 hour = 15 minutes.
  7. (e) Wq = λ ÷ [μ(μ − λ)] = 6 ÷ 40 = 0.15 hour = 9 minutes. Check: Ws − 1/μ = 15 − 6 = 9 minutes.

Answer: Utilisation 60%; Ls = 1.5; Lq = 0.9; Ws = 15 minutes; Wq = 9 minutes.

Example 2

A bank has one teller. Customers arrive at an average of 8 per hour and the teller serves 10 per hour, both following the M/M/1 assumptions. Find (a) the probability that the teller is idle, (b) the probability that exactly 2 customers are in the system, (c) the probability that more than 3 customers are in the system, and (d) the average waiting time in the queue in minutes.

Show the solution
  1. λ = 8 per hour, μ = 10 per hour. λ < μ, so the system is stable.
  2. ρ = 8 ÷ 10 = 0.8.
  3. (a) Idle probability = P(0) = 1 − ρ = 0.2.
  4. (b) P(2) = (1 − ρ) × ρ² = 0.2 × 0.64 = 0.128.
  5. (c) P(n > 3) = ρ⁴ = 0.8⁴ = 0.4096.
  6. (d) Wq = λ ÷ [μ(μ − λ)] = 8 ÷ (10 × 2) = 0.4 hour = 24 minutes.

Answer: (a) 0.2; (b) 0.128; (c) 0.4096; (d) 24 minutes.

Exam tips

  • In the MCQ section, the question usually needs one formula. Convert units, get ρ, and apply the right formula. Check the options for the minutes versus hours trap.
  • In written answers, list the given data, the stability check, each formula with substitution, and the final answer with units. Step marks are earned for the formula and substitution.
  • Use Little's relations (L = λ × W) to cross-check your answers. A quick check can save a mark lost to arithmetic.
  • Read the wording carefully: 'in the system' includes the customer being served, 'in the queue' does not.
  • If the question asks for a decision, such as adding a second counter, compare waiting times or cost, and state your recommendation in one line.

Practice questions from Scheduling and Queuing Models

Single-Channel Queuing Model (M/M/1): frequently asked questions

What do M/M/1 and the letters in it mean?

The first M is Markovian (Poisson) arrivals, the second M is Markovian (exponential) service times, and 1 is the number of servers. It is the simplest queuing model with one channel.

How do I calculate average waiting time in the queue in M/M/1?

Use Wq = λ ÷ [μ(μ − λ)]. Keep λ and μ in the same unit, and multiply by 60 if you want minutes. You can also use Wq = Lq ÷ λ.

What is the difference between Ws and Wq?

Wq is the average time spent waiting in the queue before service begins. Ws is the average total time in the system, which is Wq plus the average service time 1 ÷ μ.

What if the arrival rate is higher than the service rate?

Then ρ is 1 or more and the queue grows without limit. The steady-state formulas do not apply, so you must increase the service rate or add servers.