Final answer:
Using queuing theory principles, the mean time in the system is 1/3 of a day per machine, the mean number in the system is 1 machine, mean waiting time in the queue is 1/2 of a day per machine, the probability of finding two machines in the system is a calculation using the Poisson formula, and the expected number in the queue is 0.5 machines.
Step-by-step explanation:
To address the question regarding the maintenance service facility with Poisson arrival rates and negative exponential service times, we need to apply queuing theory formulas. Here's how we approach each part of the question:
- Meantime in the system (T): This is given by T = 1 / (μ - λ), where μ is the service rate and λ is the arrival rate. With μ = 6 machines per day and λ = 3 machines per day, T would be 1 / (6 - 3) = 1/3 of a day per machine.
- Mean number in the system (L): L = λ / (μ - λ). Substituting the given values, L = 3 / (6 - 3) = 1 machine, on average, in the system.
- Mean waiting time in the queue (W): W = λ / μ * (μ - λ), giving W = 3 / 6 * (6 - 3) = 1/2 of a day per machine.
- Probability of finding two machines in the system (Pn): For Poisson processes, Pn = (λn * e-λ) / n! where n is the number of machines. Calculating for n=2, P2 would be (32 * e-3) / 2!.
- Expected number in the queue (Lq): Lq = (λ2) / μ * (μ - λ), which simplifies to Lq = (32) / 6 * (6 - 3) = 3 / 6 = 0.5 machines.
It's important to note that these calculations assume steady-state conditions and that arrivals and services are truly Poisson and exponentially distributed respectively.