Model (M/M/1):(∞/FCFS)

Model Description

The (M/M/1):(∞/FCFS) queuing model represents a single-server system with:

Poisson arrival process (rate $ \lambda $)
Exponential service time distribution (rate $ \mu $)
Infinite system capacity
First-come-first-served (FCFS) queue discipline

The traffic intensity or server utilization is defined as:

$$ \rho = \frac{\lambda}{\mu} $$

Probability Distributions

1. Arrival Distribution (Poisson)

Probability that $ n $ customers arrive during time interval $ t $:

$$ p(n, t) = \frac{e^{-\lambda t} (\lambda t)^n}{n!} $$

2. Service Distribution (Exponential)

Probability that service time is less than or equal to $ t $:

$$ p(\bar{t}) = 1 - e^{-\mu t} $$

Performance Metrics

1. Probability of Exactly $ n $ Customers in System

$$ p_n = \rho^n (1 - \rho) $$

$ p_0 = 1 - \rho $
$ p_1 = \rho (1 - \rho) $

2. Expected Number of Customers

In the system: $$ n_s = \frac{\rho}{1 - \rho} = \frac{\lambda}{\mu - \lambda} $$
In the queue: $$ n_q = n_s - \rho = \frac{\rho^2}{1 - \rho} $$

3. Mean Waiting Time

In the queue: $$ t_q = \frac{\rho}{\mu - \lambda} $$
In the system: $$ \bar{t}_s = \frac{1}{\mu - \lambda} $$

4. Little’s Law

Little’s Law relates queue length and waiting time:

$$ n_q = \lambda t_q $$

Proof:

$$ n_q = \frac{\rho^2}{1 - \rho} = \frac{\lambda}{\mu} \cdot \frac{\rho}{1 - \lambda/\mu} = \lambda \cdot \frac{\rho}{\mu - \lambda} = \lambda t_q $$

Sample Problems – Queuing Theory

Problem 1: Server Utilization

Customers arrive at a service counter at an average rate of $ \lambda = 4 $ per hour. The service rate is $ \mu = 6 $ per hour.

Find: The traffic intensity $ \rho $.

Solution

$$ \rho = \frac{\lambda}{\mu} = \frac{4}{6} = 0.667 $$

The server is busy about 66.7% of the time.

Problem 2: Expected Number in System

For the same system ($ \lambda = 4 $, $ \mu = 6 $):

Find: Expected number of customers in the system $ n_s $.

Solution

$$ n_s = \frac{\lambda}{\mu - \lambda} = \frac{4}{6 - 4} = \frac{4}{2} = 2 $$

On average, there are 2 customers in the system.

Problem 3: Mean Waiting Time in Queue

Using the same parameters ($ \lambda = 4 $, $ \mu = 6 $):

Find: Average waiting time in the queue $ t_q $.

Solution

$$ t_q = \frac{\rho}{\mu - \lambda} = \frac{0.667}{6 - 4} = \frac{0.667}{2} = 0.333 \text{ hours} $$

On average, a customer waits about 20 minutes in the queue.

Model Description

Probability Distributions

1. Arrival Distribution (Poisson)

2. Service Distribution (Exponential)

Performance Metrics

1. Probability of Exactly \( n \) Customers in System

2. Expected Number of Customers

3. Mean Waiting Time

4. Little’s Law

Sample Problems – Queuing Theory

Problem 1: Server Utilization

Problem 2: Expected Number in System

Problem 3: Mean Waiting Time in Queue