# Average busy time with Poisson arrival

We have a factory that can process jobs. Each job takes an hour to complete. Jobs arrive according to a Poisson arrival process, with a mean of $\lambda$ jobs per hour. If the factory is free when a job arrives, it accepts the job with probability $p$, independently of other jobs. Over the long run, what is the average proportion of time that the factory is busy?

I'm not sure how to set up the calculation for this. I think we have to calculate the amount of time that, starting from any point where the factory is free, we need to wait until the next job is accepted. If a job is always accepted when the machine is free ($p=1$), then the expected waiting time should be $1/\lambda$. But here matters are complicated because we have a probability $p\leq 1$.

You can use Poisson thinning to divide arriving jobs (if you want, a priori) into two separate streams: one containing "acceptable" jobs (a Poisson process with $\lambda p$ events per hour), and one containing "unacceptable" jobs (a Poisson process with $\lambda (1-p)$ events per hour). The factory only processes acceptable jobs.
• Renewal theory deals with renewal processes, which are a generalization of Poisson process in which the times between "arrivals" can be i.i.d. from any distribution (rather than just exponential distributions). Suppose $X_1, X_2, \ldots$ represent the times between successive arrivals, and the $R_i$ represents a (possibly random) reward that's collected during the $i$\textsuperscript{th} epoch. The renewal reward theorem is a statement about the long-run reward collected per unit time: $$\lim_{n \to \infty} \frac{\sum_{i=1}^n R_i}{\sum_{i=1}^n X_i} = \frac{E[R_1]}{E[X_1]}.$$ – Kenneth Chong Feb 12 '17 at 21:59
• So in this case, you could let $R_i = 1$ for each $i$ (to reflect the time spent processing each job), and $X_i$ be the duration of an epoch, as described above. – Kenneth Chong Feb 12 '17 at 22:07