In a single server, Poisson arrival, exponential service system with $\lambda=8$ and ${ }^{\mu}=10$ and a system capacity of $3, \mathrm{P}_{0}$ will be equal to
A) 0.2
B) 0.338
C) 0.8
D) 0.173

Question

Quizplus · Accepted Answer

In a single server queue with Poisson arrivals and exponential service times (an M/M/1 queue), the probability that there are 0 customers in the system ($P_0$) can be calculated using the formula for the probability of having 0 customers in a system with finite capacity $K$, where $K$ is the maximum number of customers allowed in the system (including the one in service). The formula for $P_0$ in a system with capacity $K$ is given by:$$P_0 = \left[ \sum_{n=0}^{K} \left( \frac{\lambda}{\mu} \right)^n \frac{1}{n!} \right]^{-1}$$Given $\lambda = 8$ and $\mu = 10$, and a system capacity of $3$, we substitute these values into the formula:$$P_0 = \left[ 1 + \left( \frac{8}{10} \right)^1 + \left( \frac{8}{10} \right)^2 \frac{1}{2!} + \left( \frac{8}{10} \right)^3 \frac{1}{3!} \right]^{-1}$$$$P_0 = \left[ 1 + 0.8 + 0.32 + 0.0853 \right]^{-1}$$$$P_0 = \left[ 2.2053 \right]^{-1}$$$$P_0 = 0.4533$$However, the provided options do not include this result, suggesting a possible miscalculation or misunderstanding in the application of the formula or the interpretation of the given options. Given the options, the closest match or the intended calculation might have been approached differently, possibly considering a simpler system or a misinterpretation of the system's parameters. In practice, the correct calculation should yield a result based on the formula and parameters given, but none of the provided options exactly match the correct calculation for $P_0$ based on the standard formula for an M/M/1/K queue. Therefore, a reevaluation of the calculation or the context of the question might be necessary to align with the provided options.

In a Single Server, Poisson Arrival, Exponential Service System With \lambda=8