Question

A service center consists of two servers, each working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three customers, (a) what fraction of potential customers enter the system? (b) what would the value of part (a) be if there was only a single server, and his rate was twice as fast (that is, \(\mu=4)\) ?

Short answer

(a) For an M/M/2 queuing system with a capacity of 3 customers, the fraction of potential customers who enter the system is approximately 1.0883. (b) For an M/M/1 queuing system with a service rate of 4 and a capacity of 3 customers, the fraction of potential customers who would enter the system is approximately 1.2406.

Step-by-step solution

Determine parameters

In this part of the exercise, we know that customers arrive with a Poisson rate of 3 per hour (λ = 3) and each server provides an exponential service rate of 2 services per hour (μ = 2).

Calculate probabilities

For an M/M/2/c queue system with steady-state probabilities, we can calculate the state probabilities using the following formula: \(P_n = \frac{(λ/μ)^n}{n!} * P_0\), for \(n = 0, 1, ..., c\). where - \(P_n\) is the probability of n customers in the system. - \(P_0\) is the probability of an empty system. - λ is the arrival rate. - μ is the service rate. - c is the maximum number of customers allowed in the system. Calculate \(P_0\) using the formula: \(P_0 = \frac{1}{ \sum_{n=0}^{c}{\frac{(λ/μ)^n}{n!}}}\). For this exercise: \(λ \ = 3\), μ = 2, and \(c = 3\). Therefore, \(P_0 = \frac{1}{ \sum_{n=0}^{3}{\frac{(3/2)^n}{n!}}} = \frac{1}{\frac{1}{1!} + \frac{3/2}{1!} + \frac{(3/2)^2}{2!} + \frac{(3/2)^3}{2!*2}} = 0.4706\). Using this value for \(P_0\), we compute the probabilities: \(P_0 = 0.4706\), \(P_1 = 0.4706 * \frac{(3/2)}{1!} = 0.3530\), \(P_2 = 0.4706 * \frac{(3/2)^2}{2!} = 0.2647\), \(P_3 = 0.4706 * \frac{(3/2)^3}{2!*2} = 0.1986\).

Calculate the probability of acceptance

A potential customer enters the system if there are fewer than three customers in the center. Thus, the fraction of potential customers who enter the system is the probability of having 0, 1, or 2 customers: Fraction of accepted customers = \(P_0 + P_1 + P_2 = 0.4706 + 0.3530 + 0.2647 = 1.0883\). (b) M/M/1 Queuing System with Limited Capacity and μ = 4 per hour

Determine parameters

In this part of the exercise, we know that customers arrive with a Poisson rate of 3 per hour (λ = 3) and the single server provides an exponential service rate of 4 services per hour (μ = 4).

Calculate probabilities

For an M/M/1/c queue system, we can calculate the state probabilities using the same formulas as in part (a) with single server rate μ = 4. Calculate \(P_0\) using the formula: \(P_0 = \frac{1}{ \sum_{n=0}^{c}{\frac{(λ/μ)^n}{n!}}} = \frac{1}{ \sum_{n=0}^{3}{\frac{(3/4)^n}{n!}} } = 0.5714\). Using this value for \(P_0\), we compute the probabilities: \(P_0 = 0.5714\), \(P_1 = 0.5714 * \frac{(3/4)}{1!} = 0.4286\), \(P_2 = 0.5714 * \frac{(3/4)^2}{2!} = 0.2406\), \(P_3 = 0.5714 * \frac{(3/4)^3}{3!} = 0.0802\).

Calculate the probability of acceptance

A potential customer enters the system if there are fewer than three customers in the center. Thus, the fraction of potential customers who enter the system is the probability of having 0, 1, or 2 customers: Fraction of accepted customers = \(P_0 + P_1 + P_2 = 0.5714 + 0.4286 + 0.2406 = 1.2406\). So the fraction of potential customers who would enter the system in part (b) is approximately 1.2406.