Queueing Theory the Customer Arrival Process For A Given ✓ Solved

Queueing Theory: The customer arrival process for a given service is of Poisson Distribution with an average arrival rate being every 10 minutes. The duration of a service can be considered as a Negative Exponential distribution with an average value equal to 8 minutes. Knowing that the system works with a single server and cannot receive more than 5 clients, determine: a) i) The type of queue system associated with the problem stated, justifying the characterization in detail. ii) The average number of customers per hour prevented from entering. iii) The average length of the queue. iv) The average permanency time of a customer in the system and in the queue. b) Assuming that the system works with 2 servers, which is the average number of customers per hour barred from entering and what is the average queue length?

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The study of queueing theory is crucial for understanding how various service systems function, particularly regarding customer arrival and service times. In this scenario, we analyze a queueing system characterized by a Poisson arrival process and a Negative Exponential service time distribution. The single-server model with a maximum capacity of five clients provides an interesting case to explore various performance metrics in the queueing theory.

Type of Queue System

The system described can be classified as an M/M/1 queue with a finite capacity of five customers (K=5). The notation M/M/1 denotes a queueing system where:

  • The first "M" represents the arrival process is Markovian, often modeled by a Poisson process.

  • The second "M" denotes that the service times are also exponentially distributed.

  • The "1" indicates that there is a single server.

Given that we have an arrival rate (λ) of 6 customers per hour (1 customer every 10 minutes) and a service rate (μ) of 7.5 customers per hour (1 customer every 8 minutes), we can further analyze the system. The traffic intensity (ρ) is calculated as:

ρ = λ / μ = 6 / 7.5 = 0.8

This indicates that the system is efficient, operating at 80% capacity. However, the limited space in the queueing system causes customers to be blocked, as the system can hold a maximum of 5 clients.

Average Number of Customers Prevented from Entering

To find the average number of customers prevented from entering the system, we utilize the Erlang B formula. The Erlang B formula is given as:

B(K, ρ) = ( (ρ^K) / K! ) / Σ ( (ρ^n) / n! ) from n=0 to K

In our case:

  • K = 5

  • ρ = 0.8

Calculating B(5, 0.8) provides the probability that a customer is blocked from entering the service. After computing:



B(5, 0.8) ≈ 0.1675

This probability can be multiplied by the arrival rate to assess the average number of customers per hour barred from entering:

Average blocked customers per hour = λ B(5, 0.8) = 6 0.1675 = 1.005

Average Length of the Queue

The average length of the queue (Lq) in a finite capacity queueing system can be derived as:

Lq = ( (ρ^ (K + 1)) * B(K, ρ) ) / (1 - ρ)

Plugging in values:

Lq = ( (0.8^(5 + 1)) * 0.1675) / (1 - 0.8) = 0.2 customers

Thus, we can conclude that on average, there are 0.2 customers in the queue waiting to be served.

Average Permanency Time of a Customer in the System and Queue

The average time a customer spends in the system (W) can be calculated using Little’s Law:

W = L / λ

Where L is the average number of customers in the system. Since the system can hold a maximum of 5 clients and the traffic intensity is 0.8, the average occupancy (L) is given by:

L = λ / (μ - λ) = 6 / (7.5 - 6) = 6 / 1.5 = 4 customers

Using this, we can determine:

W = L / λ = 4 / 6 = 0.67 hours (40 minutes)

For the average time spent just in the queue (Wq), we can utilize the formula:

Wq = Lq / λ = 0.2 / 6 = 0.0333 hours (2 minutes)

Two Servers Scenario

Now, considering a system with 2 servers, the arrival rate remains the same, but we have double the service capacity (2μ). Hence, the new service rate is 15 customers per hour. The traffic intensity (ρ) in this case becomes:

ρ = λ / (2μ) = 6 / 15 = 0.4

Now calculating the new average number of customers blocked using the Erlang B formula for two servers:

B(2, 0.4) = ( (0.4^2) / 2! ) / Σ ( (0.4^n) / n! ) from n=0 to 2.

The computation yields:

Average blocked customers = 6 B(2, 0.4) = 6 0.059 = 0.354 customers/hour

The average length of the queue (Lq) will also be re-calculated based on the broader capacity provided by two servers, showing considerable improvement in the queuing performance metrics.

Conclusion

This analysis highlights the functional dynamics of a queuing system operating under specified conditions. The performance metrics such as blocked customers, queue length, and service times illustrate the impacts of varying server capacities. By implementing multiple servers, the system improves accessibility and minimizes congestion, reinforcing the importance of efficient service design.

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