91Ó°ÊÓ

Using the notation M/M/s/K, we can represent the system as an M/M/2/3 model (two servers, three customers capacity). The steady-state probabilities are given by the formula: \[P(n) = \frac{(1-\rho) \rho^n}{n!(1-\rho^{s+1})}\] Here, \(P(n)\) represents the probability that the system has n customers, and the system utilization, \(\rho\), is given by: \[\rho = \frac{\lambda}{s\mu}\] _Determine the system utilization_ With \(s = 2\) servers, each with \(\mu = 2\) services per hour, and \(\lambda = 3\) customers per hour, we can find \(\rho\): \[\rho = \frac{3}{2\cdot2} = \frac{3}{4}\] _Calculate the steady-state probabilities_ Using the formula for the steady-state probabilities \(P(n)\) for \(n = 0, 1, 2\), and \(3\): \[P(0) = \frac{(1-\rho) \rho^0}{0!(1-\rho^{3})} = \frac{1-\frac{3}{4}}{1-\left(\frac{3}{4}\right)^3}\] \[P(1) = \frac{(1-\rho) \rho^1}{1!(1-\rho^{3})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^1}{1!(1-\left(\frac{3}{4}\right)^3)}\] \[P(2) = \frac{(1-\rho) \rho^2}{2!(1-\rho^{3})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^2}{2!(1-\left(\frac{3}{4}\right)^3)}\] \[P(3) = \frac{(1-\rho) \rho^3}{3!(1-\rho^{3})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^3}{3!(1-\left(\frac{3}{4}\right)^3)}\]

The losing probability is given by the probability that the system has reached capacity, which is equivalent to \(P(3)\). Calculate \(P(3)\) using the expression obtained in the previous step: \[P(3) = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^3}{3!(1-\left(\frac{3}{4}\right)^3)}\] Now, the fraction of potential customers who enter the system is given by: \[1 - P(3)\] (b) A single server, working at an exponential rate of 4 services per hour

Using M/M/s/K, we can represent the system as an M/M/1/3 model (one server, three customers capacity). Utilizing the same formula for the steady-state probabilities \(P(n)\), we first need to determine the new system utilization \(\rho\). The new service rate is \(\mu = 4\) services per hour: \[\rho = \frac{3}{1\cdot4} = \frac{3}{4}\] Now, calculate the steady-state probabilities \(P(n)\) for \(n = 0, 1, 2\), and \(3\): \[P(0) = \frac{(1-\rho) \rho^0}{0!(1-\rho^{1+1})} = \frac{1-\frac{3}{4}}{1-\left(\frac{3}{4}\right)^2}\] \[P(1) = \frac{(1-\rho) \rho^1}{1!(1-\rho^{1+1})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^1}{1!(1-\left(\frac{3}{4}\right)^2)}\] \[P(2) = \frac{(1-\rho) \rho^2}{2!(1-\rho^{1+1})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^2}{2!(1-\left(\frac{3}{4}\right)^2)}\] \[P(3) = \frac{(1-\rho) \rho^3}{3!(1-\rho^{1+1})} = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^3}{3!(1-\left(\frac{3}{4}\right)^2)}\]

The losing probability for the single server is given by \(P(3)\). Calculate \(P(3)\) using the expression obtained in the previous step: \[P(3) = \frac{\left(1-\frac{3}{4}\right)\left(\frac{3}{4}\right)^3}{3!(1-\left(\frac{3}{4}\right)^2)}\] The fraction of potential customers who enter the system for the single server case is: \[1 - P(3)\]