Consider a system in which the processes arrive at a Poisson rate,…
Consider a system in which the processes arrive at a Poisson rate, a=0.1/second and the service rate of the system is, b=0.01/second. The total number of processes that are present in the system after t=1 hour of processing is
Answer: 324 — Concept: When processes enter a system at a constant arrival rate and leave (are serviced/completed) at a constant service rate, the net number of processes…
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Correct answer: 324
Concept: When processes enter a system at a constant arrival rate and leave (are serviced/completed) at a constant service rate, the net number of processes accumulated in the system over an interval of time equals the difference of the two rates multiplied by the elapsed time: N(t) = (arrival rate − service rate) × t, provided the service process runs continuously over that interval.
Application:
Arrival rate a = 0.1 process/second and service rate b = 0.01 process/second.
Convert t = 1 hour to seconds: t = 1 × 3600 = 3600 seconds.
Net accumulation rate = a − b = 0.1 − 0.01 = 0.09 process/second.
Total processes present after t = 3600 s: N = (a − b) × t = 0.09 × 3600 = 324.
Cross-check: Total processes that arrived in 3600 s = a × t = 0.1 × 3600 = 360; total processes serviced (departed) in that time = b × t = 0.01 × 3600 = 36. Processes still present = 360 − 36 = 324, matching the direct computation.
Answer: 324 processes.