Final answer:
When utilization is 60% and there is no variability in processing or inter-arrival times, the average time in the queue would be zero hours as per the deterministic nature of the M/D/1 queueing model.
Step-by-step explanation:
In the context of queueing theory, if the theoretical utilization of a system is 60%, with no variability in inter-arrival or processing times, the system can be modeled using the M/D/1 queue (where M stands for Markovian arrivals, D for deterministic service times, and 1 for a single server). When there's no variability in inter-arrival and service times, we essentially have deterministic processing, meaning customers are served at a constant rate and arrive at a constant rate. However, to calculate the average time in the queue, we would typically use the formula Lq = λ²σ2 / (2(1-ρ)), where Lq is the average number of customers in the queue, λ is the arrival rate, μ is the service rate, and ρ (ρ = λ/μ) is the utilization rate. Given that the utilization is 60% (ρ=0.6) and there is no variability, the service rate is constant, so the time in the queue would theoretically be zero because arrivals match the service rate perfectly. Thus, the deterministic nature of the arrival and service processes leads to an average time in the queue of zero hours.