Queueing Theory, PERT and CPM
M/M/1 queue formulas (utilisation, queue length, waiting time, Little's law), network diagrams, critical path, floats, and PERT expected time and variance.
Queueing theory predicts waiting lines; CPM and PERT plan projects made of many activities. Both appear every year in GATE ME as short numerical questions.
The M/M/1 queue
Poisson arrivals at rate λ, exponential service at rate μ, one server, with ρ = λ/μ < 1:
| Quantity | Formula |
|---|---|
| Utilisation (probability server busy) | ρ = λ/μ |
| Probability system is empty | P₀ = 1 − ρ |
| Probability of n in system | Pₙ = (1 − ρ)ρⁿ |
| Average number in system | L = λ/(μ − λ) = ρ/(1 − ρ) |
| Average number in queue | L_q = λ²/(μ(μ − λ)) = ρ L |
| Average time in system | W = 1/(μ − λ) |
| Average time in queue | W_q = λ/(μ(μ − λ)) |
Little's law: L = λW and L_q = λW_q (valid for almost any queue).
Project networks (CPM)
Draw activities as arrows (AOA) or nodes (AON) in order of precedence. For each activity:
- Earliest start (ES) and earliest finish EF = ES + duration (forward pass, take the maximum at merges).
- Latest finish (LF) and latest start LS = LF − duration (backward pass, take the minimum at bursts).
- Total float = LS − ES = LF − EF. Free float = ES of the next activity − EF.
The critical path is the longest path through the network; its activities have zero total float, and its length is the project duration. Delaying a critical activity delays the whole project.
Crashing: shorten critical activities with the lowest cost per unit time first, until another path becomes critical.
PERT
For uncertain activity times with optimistic a, most likely m and pessimistic b:
- Expected time t_e = (a + 4m + b)/6.
- Variance σ² = ((b − a)/6)².
- Project variance = sum of the variances on the critical path. The probability of finishing by time T uses z = (T − T_e)/σ_project with the normal table.
Worked example
Customers arrive at 6 per hour; the server handles 10 per hour.
- ρ = 0.6. L = 6/(10 − 6) = 1.5 customers. W = 1/4 h = 15 minutes.
- L_q = 0.6 × 1.5 = 0.9; W_q = 0.9/6 h = 9 minutes.
Practice
An M/M/1 queue has λ = 8 per hour and μ = 12 per hour. What is the average time a customer spends in the system, in minutes?
Show answer and explanation
Answer: 15W = 1/(μ − λ) = 1/4 hour = 15 minutes.
For the same queue, what is the average number of customers waiting in the queue? (Give 3 decimal places.)
Show answer and explanation
Answer: 1.333L_q = λ²/(μ(μ − λ)) = 64/48 = 1.333.
A PERT activity has a = 2, m = 5 and b = 14 days. What is its expected time in days?
Show answer and explanation
Answer: 6t_e = (2 + 20 + 14)/6 = 36/6 = 6 days.
What is the variance of the same activity in days²?
Show answer and explanation
Answer: 4σ² = ((b − a)/6)² = (12/6)² = 4.
Activities on the critical path have:
Show answer and explanation
Answer: AThe critical path is the longest path; any delay on it delays the project, so its total float is zero.