A server with an average service time of 0.2 seconds is treated as an M/M/1 queue. To keep the average response time (waiting time plus processing time) at or below 1.0 second, which of the following is closest to the maximum arrival rate that can be accepted?

Answer: A. 4.0 per second

The average response time is W = Ts/(1 − ρ), so 0.2/(1 − ρ) ≤ 1.0 gives 1 − ρ ≥ 0.2, that is, ρ ≤ 0.8. Since ρ = λ × Ts, λ ≤ 0.8 ÷ 0.2 = 4.0 per second. At an arrival rate of 5.0 per second, the utilization would be 1 and the queue would grow without limit.

• A. Correct. ρ ≤ 0.8, so λ ≤ 0.8 ÷ 0.2 = 4.0 per second.
• B. This is the value when only the waiting time Wq is kept at or below 1.0 second (ρ ≤ 5/6, λ ≤ about 4.17 per second), which leaves out the 0.2-second processing time.
• C. 1 ÷ 0.2 = 5.0 is the arrival rate at which the utilization reaches 1, where the waiting time grows without limit.