Question

Customers arrive at a two-server station in accordance with a Poisson process with a rate of two per hour. Arrivals finding server 1 free begin service with that server. Arrivals finding server 1 busy and server 2 free begin service with server 2. Arrivals finding both servers busy are lost. When a customer is served by server 1 , she then either enters service with server 2 if 2 is free or departs the system if 2 is busy. \(\mathrm{A}\) customer completing service at server 2 departs the system. The service times at server 1 and server 2 are exponential random variables with respective rates of four and six per hour. (a) What fraction of customers do not enter the system? (b) What is the average amount of time that an entering customer spends in the system? (c) What fraction of entering customers receives service from server \(1 ?\)

Short answer

(a) The fraction of customers who do not enter the system is \( \frac{1}{8}\) (b) The average amount of time that an entering customer spends in the system is \( \frac{49}{384} \) hours. (c) The fraction of entering customers receiving service from server 1 is \( \frac{3}{8}\).

Step-by-step solution

Model the system

First, we need to model the system using a M/M/c/C model where c = 2, since we have two servers. The arrival rate (λ) is two per hour and the service rates at server 1 (μ1) and server 2 (μ2) are four and six per hour, respectively.

Calculate the state probabilities

Next, we need to calculate the state probabilities using the steady-state balance equations: \(P_0 = \frac{1}{1 + \frac{\lambda}{\mu_1} + \frac{\lambda^2}{\mu_1 \mu_2}}\) \(P_1 = \frac{\lambda}{\mu_1} P_0\) \(P_2 = \frac{\lambda^2}{\mu_1 \mu_2} P_0\) We have: \(P_0 = \frac{1}{1 + \frac{2}{4} + \frac{2^2}{4 \cdot 6}} = \frac{1}{1 + 0.5 + \frac{1}{6}} = \frac{1}{\frac{16}{12}} = \frac{3}{4}\) \(P_1 = \frac{2}{4} P_0 = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}\) \(P_2 = \frac{2^2}{4 \cdot 6} P_0 = \frac{1}{6} \cdot \frac{3}{4} = \frac{1}{8}\)

Calculate system performance measures

(a) The fraction of customers who do not enter the system is represented by the probability of the system being in state 2 (both servers are busy). Thus, we have: Fraction of customers who do not enter the system = \(P_2 = \frac{1}{8}\) (b) To calculate the average amount of time that an entering customer spends in the system, we first need to find the average time spent in each server. For an M/M/c/C model, the expected service time for server i is given by: \(E[T_i] = \frac{1}{\mu_i}\) \(E[T_1] = \frac{1}{4} = 0.25h\) \(E[T_2] = \frac{1}{6} = \frac{1}{6}h\) Now, we can find the average amount of time that an entering customer spends in the system by considering the probabilities of going through each server: Average time in the system = \(E[T_1] P_1 + E[T_2] P_2 = 0.25 \cdot \frac{3}{8} + \frac{1}{6} \cdot \frac{1}{8} = \frac{3}{32} + \frac{1}{48} = \frac{49}{384} h\) (c) The fraction of entering customers who receive service from server 1 is represented by the probability of the system being in state 1 (only server 1 is busy): Fraction of entering customers receiving service from server 1 = \(P_1 = \frac{3}{8}\) Final answers: (a) The fraction of customers who do not enter the system is \( \frac{1}{8}\) (b) The average amount of time that an entering customer spends in the system is \( \frac{49}{384} \) hours. (c) The fraction of entering customers receiving service from server 1 is \( \frac{3}{8}\).